mathematics workbook - frank edu · 2019. 9. 30. · 8 smmms workbook – 7 time taken to ascend 50...

(Student's Support Materials)

MATHEMATICS WORKBOOK

7

3SM-MATHEMATICS WORKBOOK – 7

Chapter 1

Exercise 1.1 1. (a) –16, –20, –24 (b) –15, –18, –21 (c) –6, –10, –14 (d) 19, 25, 31 2. (a) Srinagar = –20°C, Mandi = –10°C, Ooty = –5°C, Shimla = 5°C, Pune = 20°C (b) Difference = 5°C – (–10°C) = 5°C + 10°C = 15°C (c) Hottest is Pune = 20°C Coldest is Srinagar = – 20°C Difference = 20°C – (–20°C) = 20°C + 20°C = 40°C (d) Temperature of Shimla = 5°C Temperature of Ooty = –5°C Temperature together = (5 + (–5))°C = 0°C Temperature of Mandi = –10°C Yes Temperature of Shimla and Ooty together is more than Temperature of Mandi. 3. –250, –205, –200, –165, –73. 4. Score of Team A = –10 + (–5) + 20 + 25 + (–7) = 23

Score of Team B = –15 + (–10) + 30 + (–20) + 17 = 2 Team A performed better. 5. Temperature on Monday = 45°C

Temperature drop on Monday night = 5°C ∴ Temperature on Monday night = 45°C – 5°C = 40°C Temperature on Tuesday night = 40°C – 7°C = 33°C

6. Distance of Submarine = – 1500 m Distance of aeroplane above the sea level = 2650 m. Vertical distance between Submarine and aeroplane = [2650 – (–1500)] m = 4150 m 7. Melting point = 39°C

Freezing point = – 20°C Difference = 39°C – (–20°C) = 59°C

8. Distance towards south is represented negative. Her position from the original position = –30 + 40 = 10 km ∴ She is 10 km towards north from her original position. 9. Position of Submarine = 1500 m above sea level.

Descend in 1 minute = 450 m Descend in 2 minutes = 450 m × 2 = 900 mPosition of submarine after 2 minutes = (1500 – 900) m = 600 m above sea level

10. Netprofitorlossinthreedays=47+(–15)+(–13)=47–28=19\ Netprofit=` 19

11. Vertical distance between the two points = 8848 – (–10911) m = 8848 + 10911 = 19759 m

SolutionS

4 SM-MATHEMATICS WORKBOOK – 7

12. (a) True, (b) False the sum of two negative integers is always negative (c) True (d) False, Zero is neither positive nor negative (e) False, Every negative integer is smaller than positive integers. 13. Refer answers given in book

d = 4 14. (a) a + b = a – (–b)

25 + (–18) = 25 – (– (– 18)) = 25 – (18) 25 –18 = 25 – 18 7 = –7(b) 215 + (–350) = 215 – (– (– 350)) 215 –350 = 215 – 350 –135 = –135(c) 75 + 85 = 75 – (– 85) 75 + 85 = 75 + 85 160 = 160(d) 28 + 15 = 28 – (– 15) 28 + 15 = 28 + 15 43 = 43

15. (a) –18 + (–15) –25 + 4 (b) –36 + (–5) 40 – (–2) – 33 < –21 – 41 < +42 (c) 24 – 40 + 11 –40 – 24 + 11 (d) 48 + (–25) – 200 –200 + 48 – 25

–16 + 11 –64 + 11 23 – 200 –152 – 25 –5 > –53 –177 = –177 (e) –231 + 75 – 51 –450 + 17 – 25

–282 + 75 –475 + 17

–207 > –458 16. (a) We take jumping down + jumping back in total. We take + for down and – for up 3 – 2 + 3 – 2 + 3 – 2 + 3 – 2 + 3 – 2 + 3 – 2 + 3 – 2 + 3 – 2 + 3 – 2 + 3 = 12 \ Total jumps = 19 (b) (–4 + 2) + (–4 + 2) + (–4 + 2) + (–4 + 2) + –4 = – 12 Hence he will take a total of 9 jumps. 17. (a) (–16, –8) as – 16 – (–8) = –8 (b) (–20, 5) as –20 + 5 = –15 (c) (–5, 5) as (–5) – (5) = –10 (d) (–2, 25) = –50 (e) (–12 ÷ 2) = –6 Answer may vary. 18. Refer answers given in book 19. (a) (–4, –2) as (–4) + (2) = – 2 (b) (–4, 4) as = (– 4) + 4 = – 4 + 4 = 0 (c) (–2, –5) as (–2) + (–5) = – 7 and –7 < – 2, –7 < –5 (d) (–2, 5) as (–2) + 5 = 3 and –2 < 3 < 5 (e) (2, 5) as 2 + 5 = 7 There exists other pair also. 20. (a) (–8, 4) –8 – 4 = –12 (b) (–2, +3) (–2 – 3) = –5 and –5 < –2, –5 < –3


(c) (+3, –2) = 3 – (–2) = 5 and 5 > 3, 5 > –2 (d) (6, 2) and 6 – 2 = 4 and 2 < 4 < 6 (e) (5, 5) as 5 – 5 = 0 Answer may vary.

Exercise 1.2 1. Refer answers given in the book 2. (a) – 24 × 10 × –9 = –240 × – 9 = +2160 (b) –40 × 2 × (–5) × 5 = –80 × –25 = 2000 (c) –2 × –4 × –6 × –8 = 8 × 48 = 384 (d) – 810 ÷ 9 = –90 (e) – 1331 ÷ 11 = –121 (f) 0 ÷ 13567 = 0 (g) –400 ÷ 50 = –8 (h) (–39) ÷ [(–39) ÷ 13] = –39 ÷ [–3] = 13 (i) (–48 ÷ 8) ÷ 3 = –6 ÷ 3 = –2 (j) (–18 + 7) ÷ (–156 + 155) = – 11 ÷ –1 = 11 (k) (–63 ÷ 7) ÷ 9 = (–9) ÷ 9 = –1 3. Refer answers given in the book 4. (a) 15 × – 3 = –3 × 15 –45 = –45 (b) (15 + (–3) + 4 = 15 + ((–3) + 4) 12 + 4 = 15 + 1 16 = 16 (c) 15 + 4 = 4 + 15 19 = 19 (d) 15 × (–3 + 4) = 15 × (–3) + 15 × 4 15 × 1 = –45 + 60 15 = 15 5. (a) (–4, –5) as (–4) + (–5) = 9 (b) (–14, –6) as –14 – (–6) = – 8 (c) (–4, 9) as –4 + 9 = 5 (d) (8, –2) as 8 × (–2) = –16 (e) (–15, 3) as – 15 ÷ 3 = – 5 No, there exist more than one pair. 6. (a × b) × c = a × (b × c) (– 5 × 3) × (–7) = – 5 × (3 × –7) – 15 × –7 = – 5 × (–21) 105 = 105 So, the given values satisfy the equation. 7. (a ÷ b) ÷ c = a ÷ (b ÷ c) LHS = (125 ÷ 25) ÷ – 5 = 5 ÷ – 5 = – 1

RHS = 125 ÷ (25 ÷ – 5) = 125 ÷ (–5) = – 25 RHS≠LHS So, the given values does not satisfy the equation.


8. a × (b + c) = a × b + a × cLHS = 25 × (–17 + 2) = 25 × (– 15) = –375RHS = 25 × (–17) + 25 × 2 = – 425 + 50 = – 375

9. a × (b – c) = a × b – a × c LHS 40 × (25 – (–5)) = 40 × (25 + 5) RHS = 40 × 25 – (40 × –5))

= 40 × 30 = 1200) = 1000 – (– 200) = 1000 + 200 = 1200

10. (a) 52 × (–76) + 52 × 26 = 52 × (– 76 + 26) (Distributive property of multi) = 52 × (– 50) = – 2600

(b) –8 × 75 × – 25 = (–8 × –25) × 75 (Associative c of multilication) = 200 × 75 = 15000(c) –48 × 103 = –48 × (100 + 3) (Distributive property of multiplication) = –48 × 100 + (–48) × 3 = – 4800 + (– 144) = – 4944

(d) –625 × –45 + 625 × –55 = 625 × 45 + 625 × – 55 = 625 × (45 + (– 55)) (Distributive property) = 625 × (–10) = – 6250.

(e) –189 × 999 = – 189 × (1000 – 1) (Distributive property) = – 189000 + 189 = – 188811

(f) –157 × (–19) + 157 = – 157 × (–19) + 157 × 1 = 157 × 19 + 157 × 1 = 157 × (19 + 1) (Distributive property) = 157 × 20 = 3140

(g) 1693 × 99 – (– 1693) = 1693 × 99 + 1693 × 1 = 1693 × (99 + 1) (Distributive property) = 1693 × 100 = 169300

(h) –85 × 29 = – 85 × (30 – 1) (Distributive property) = – 85 × 30 – 85 × – 1 = – 85 × 30 + 85 = – 2550 + 85 = –2465

11. Initial temperature = 45°C Decrease in temperature in 1 hr = 3°C. (a) Decrease in temperature 5 hrs = 3°C × 5 = 15°C Temperature after 5 hrs = 45°C –15°C = 30°C (b) Decrease in temperature in 15 hrs = 3°C × 15° = 45°C Temperature after 15 hrs = 45°C – 45°C = 0°C (c) Decrease in temperature in 20 hrs = 3°C × 20° = 60°C Temperature after 20 hrs = 45°C – 60°C = –15°C 12. (a) Marks obtained by Saurabh = 43 × 5 + 7 × (–2)

= 215 – 14 = 201(b) Marks obtained by Sunita = 30× 5 + 15 × (–2) = 150 – 30 = 120


(c) Anita = 15 × 5 + (15) (–2) = 75 + (–30) = 75 – 30 = 45

13. (a) Score of Team A = Marks for correct + Marks of in correct + Marks of not attempted = 10 × 2 + 5 × (–2) + 5 × (–1) = 20 + (–10) + (– 5) = 20 – 15 = 5

(b) Score of Team B = 12 × 2 + 7 × (–2) + 1 × (–1) = 24 + (–14) + (–1) = 24 – 14 – 1 = 24 – 15 = +9

(c) Score of Team C = 15 × 2 + 5 × (–2) = 30 + (–10) = 30 – 10 = 20Team C won the Quiz

14. (a) Temperature at midnight = – 14°C Increase in temperature in 1 hr = + 3°C Temperature at 7 am i.e after 7 hrs

= –14°C + 3°C × 7 = – 14°C + 21°C = + 7°C.(b) Temperature at 12 noon = –14°C + 3°C × 12 = –14 + 36°C = 22°C

15. Initial temperature of coldroom = 15°C Decrease in temperature in 1 hr = 5°C Decrease in temperature from 15°C to –10°C = (15 – (– 10))°C = 25°C

Temperature required to reduce 25°C = 25 ÷ 5 = 5 h.\ In 5 hrs the temperature of the coldroom will reach –10°C

16. (a)ProfitonTelevision=` 800. Loss on refrigerator = ` 500 Profiton5000piecesofTV= 5000 × 800 = 4000000 Loss on 2000 refrigerators = 2000 × 500 = 1000000 Profit>Loss Overallprofit=4,00,0000–1000000=` 3000000 (b)Profiton4000pieceofcolourTV=4000×800=` 32,00,000 Let x be the number of refrigerators. sold. Loss on x refrigerators = x × 500 Fornoprofitnoloss Profit=loss 3200000 = x × 500

3200000

500 = x

6400 = x ∴ 6400refrigeratorsshouldbesoldtohavenoprofitnoloss. 17. Position of elevator = 250 m below the ground. Total distance to cover to reach 500 m above the ground

= 500 – (– 250) = 750 m.


Time taken to ascend 50 m = 1 minute

Time taken to ascend 1 m = 150

minutes

Time taken to ascend 750 m = 150

75015

× = 15 minutes

18. (a) Temperature at 8 pm = 15°C. Decrease in temperature in 1 hr = 2°C. Final temperature = – 5°C. Total decrease in temperature = 15°C – (–5°C) = 20°C Time taken to decrease 2°C = 1 hr.

Time taken to decrease 1°C = 12

hr.


× 20 = 10 hrs

\ Temperature will be – 5°C at 6 am

(b) Temperature at 8 pm = 15°C Final temperature = 0°C. Decrease in temperature = 15°C. Time taken to decrease 2°C = 1 hr.


hr.


× 15 = 7.5 hrs

i.e., after 7:30 hr i.e., at 3.30 am temperature will be 0°C. (c) Duration of time from 8 pm to 4 am = 8 hrs. Decrease in temperature in 1 hr = 2°C Decrease in temperature in 8 hr = 2°C × 8 = 16°C Final temperature at 4 am = 15°C – 16°C = –1°C 19. Marks given for correct answer = +4 Marks given for incorrect answer = –2 (a) Marks scored by Sanjana = 42 Marks for correct answer = 15 × 4 = 60 Let questions answer incorrect be x Marks for incorrect answer = –2 × x

Score = 60 + (–2x) According to questions

60 + (–2x) = 42 –2x = 42 – 60 –2x = –18 x = 9.

\ She answered 9 questions wrong.


(b) Score of Monika = – 8 Marks for correct answer = 8 × 4 = 32 Let the no. of incorrect answers = x. Marks for incorrect answers = –2 × x.

Her score = 32 + (–2x) According to question

32 + (–2x) = –8 –2x = –8 –32 –2x = –40 x = 20

Monika answered 20 questions in correctly. (c) If equal no of questions are answered correctly and incorrectly, the score of Anita

will be positive because the score of correct answers is more than score incorrect answers.

20. Initial position of elevator = +40m Final position of elevator = –760 m. Total distance to cover = 40 – (– 760)m = 40 + 760 m = 800 m

Time taken to descend 16 m = 1 minute

Time taken to descend 1 m = 1

16 minute

Time taken to descend 800 m = 1

16 × 800 minute = 50 min

\ Elevator will take 50 minutes 21. (a) Loss = ` 100 Number of pens sold = 200 Profitonselling1pen=` 2 Loss on selling 1 pencil = 50 paise Profit=2×200=` 400 Let x no of pencil sold on that particular week Loss = + 0.5 × x Loss>Profit ∴ Loss–Profit=` 100 0.5x – 400 = 100 0.5x = 500

x = 5000 5.

= 1000

∴ Number of pencils sold = 1000 (b) ProfitorGain = ` 100 Number of pencils sold = 500 Loss on pencils = ` 0.5 × 500 = ` 250 Let number on pen sold = x


Profitonpens=2x Profit>loss ∴ Profit–loss=` 100 2x – 250 = 100 2x = 350 x = 175 So he sold 175 pens. (c) No. of pens sold = 50 Profitonpens=` 2 × 50 = ` 100 Let the number of pencil sold = x Loss on pencils = ` 0.5 × x 0.5x = 100

x = 1000 5.

= 200

∴ 200 pencils were sold.

22. Ground level

–25 ft –20 ft

–2 ft

–3 ft

In 1st year In 2nd year Next year

Water level in start = – 25 ft Drop in water level in 2 years = (–2)ft + (–3)ft = –5 ft So the water level after 2 years = – 25 ft + (–5)ft = –30 ft Next year, the level = –20 ft \ Rise in water level that year = (–20)ft – (–30)ft = + 10 ft.

Worksheet 1 1. Refer to answer given in the book. 2. (a) LHS = – 50 + 47 = –3 and RHS = – 47 + 50 = 3 – 3 < 3 \ – 50 + 47 < – 47 + 50 (b) LHS = (60) + (–70) + (–15) = 60 – 85 = –25 and RHS = (–80) + 100 + (–95) = 100 – 175 = –75 –25 > –75 \ (80) + (–70) + (–15) > (–80) + 100 + (–95) (c) LHS = –4 × –5 × 7 = +20 × 7 = +140 and RHS = –7 × –4 × –5 = –7 × (+20) = –140 +140 > –140 \ –4 × –5 × 7 > –7 × –4 × –5


(d) LHS = 360 ÷ (60 ÷ 3) = 300 ÷ 20 = 18 and RHS = (480 ÷ 4) ÷ 12 = 120 ÷ 12 = 10 18 > 10 \ 360 ÷ (60 ÷ 3) > (480 ÷ 4) ÷ 12 (e) LHS = 50 – (–75) = 50 + 75 = 125 and RHS = 75 – (–50) = 75 + 50 = 125 125 = 125 \ 50 – (–75) = 75 – (–50) 3. Maxi temperature = 45°C Mini temperature = –5°C Difference = 45°C – (–5°C) = (45 + 5)°C = 50°C 4. Score of A Team = 15 + (–20) + (–25) + 75 = –5 + 50 = 45. Score of B Team = (–45) + (–70) + 10 + 105 = (–115) + 115 = 0 Team A scored more by (45 – 0) = 45 5. (a) 25 × (–15 × 30) = (25 × 15) × 30

25 × (–450) = (–375) × 30 –11250 = –11250

(b) –15 × (25 + 30) = – 15 × 25 + (–15) × 3 – 15 × 55 = –375 – 450 –825 = –825

(c) 25 + (30 + (–15)) = –15 + (30 + 25) 25 + (15) = –15 + 55 40 = 40

(d) a × b = b × a 25 × – 15 = – 15 × 25 –375 = –375

6. (a) –40 × 54 × 125 = (–40 × 125) × 54 (Ass. prop of multi) = –5000 × 54 = –270000

(b) 136 × (–54) + 136 × (–46) = 136 × (–54 + (–46)) (Distribute prop.) = 136 × (–100) = –13600

(c) 833 × 99 + 833 = 833 × 99 + 833 × 1 = 833 × (99 + 1) = 833 × 100 = 83300

(d) 956 × 101 – 956 × 1 = 956 × (101 – 1) (Distribute prop.) = 956 × 100 = 95600

(e) 75 × 10 × – 256 + 25 × 3 × 10 × –244 = 75 × 10 × (–256) + 75 × 10 × (–244)

= 750 × (–256 + (–244) = 750 × (–500) = –375000 7. Temperature of water of certain time = 25°C Temperature cools in 1 hr = 5°C Temperature cool in 8 hr = 5 × 8 = 40°C Temperature after 8 hrs = 25°C – 40°C = –15°C 8. Marks for correct ans = 3 Marks for incorrect ans = –1 (a) Score of Rohit = 8 × 3 + 9 × (–1) = 24 – 9 = 15 (b) Marks of Sangeeta = 9 × 3 + 5 × (–1) = 27 + (–5) = 22 (c) Score of Abdul = 30 Marks for correct ans = 15 × 3 Let the number of correct questions = x


∴ Marks for incorrect questions = x × –1 ∴ 45 – x = 30 –x = 30 – 45 = –15 ⇒ x = 15

So he attempted 15 questions incorrectly (d) Score of Sultan = 0 Marks for wrong ans = 30 × –1 = –30 Let the number of correct ans = x Marks for correct ans = 3x

∴ 3x – 30 = 0 3x = 30 x = 10

∴ He answered 10 questions correctly. 9. Position of elevator = 250 m below the ground = –250 m.

Ascend in 1 minutes = 5 m (a) Ascent in 60 minutes = 5 × 60 = 300 m Position after an hr = –250 + 300 = 50 m. (b) Initial position = –250 m Final position = 30 m Total Ascend = 30 – (–250) = 280 m. Time for 5 m ascend = 1 minutes

Time for 1 m ascend = 15

minutes

Time for 280 m = 15

× 280 = 56 min.

(c) Final position = Ground level = 0 Initial = –250 m Total distance = (–250) = 250 m Total for 5 m descend = 1 m

Total for 5 m descend = 15

Time for 250 m descend = 15

× 250 = 50 minutes.

10. (a) 48°C – (23°C) = 25°C (b) 5°C – (–10°C) = 5°C + 10°C = 15°C (c) 11°C – (–18°C) = 11°C + 18°C = 29°C (d) 10°C – (–10°C) = 10°C + 10°C = 20°C

Chapter 2Exercise 2.1

1. (a) 4 – 125

= 4 – 75

= 20 75− =

135

= 235

(b) 6 + 238

= 6 + 198

= 48 19

8+

= 678

= 838

(c) 45

+ 37

= 28 1535+ = 43

35 = 1

835

(d) 4

15 –

112

= 16 5

60−

= 1160


(e) 9

10 +

25

+ 312

= 9

10 +

25

+ 72

= 9 4 35

10+ +

= 4810

= 48

10 = 4

45

(f) 315

+ 423

= 165

+ 143

= 48 70

15+

= 11815

= 71315

(g) 923

– 358

= 293

– 298

= 29 8 29 3

24× − ×

= 29 8 3

24× −( )

= 29 5

24×

= 14524

= 6124

(h) 6 – 234

+ 15

= 6 – 114

+ 15

= 6 20 11 5 1 4

20× − × + ×

= 120 55 4

20− +

= 6920

= 3920

(i) 16 + 215

= 16 + 115

= 80 11

5+

= 915

= 1815

2. (a) 23

, 45

, 79

, 12

L.C.M. (3, 5, 9, 2) = 90

2 303

, 4 185 18

, 7 109 10

, 1 452 45

××

⇒×

×

×

×

×

×306090

7290

7090

4590

, , ,

∴ 4590

6090

7090

7290

< < < i.e. 12

23

79

45

< < <

(b) 25

, 57

, 3

10,

27

L.C.M. (5, 7, 10) = 70

2 145 14

5 107 10

3 710 7

2 107 10

2870

5070

2170

2070

××

××

××

××

⇒, , , , , ,

2070

2170

2870

5070

< < < i.e., 27

310

25

57

< < <

3. (a) 52

, 43

, 75

, 34

L.C.M. (2, 3, 5, 4) = 60

⇒ 5 302 30

, 4 203 20

, 7 125 12

, 3 154 15

×

×

×

×

×

×

×

× =

15060

, 8060

, 8460

, 4560

= 15060

> 8460

8060

> 4560

> i.e., 52

75

43

34

> > >

(b) 43

, 29

, 15

, 52

L.C.M.(3, 9, 5, 2) = 90.

4 303 30

, 2 109 10

, 1 185 18

, 5 452 45

×

×

×

×

×

×

×

× ⇒

12090

, 2090

, 1890

, 22590

22590

12090

> 2090

> 1890

> i.e. 52

43

29

15

> > >


4. Length = 1512

m = 312

m

Breadth = 1015

m = 515

m

Distance covered in 1 complete round = Perimetre of rectangle

= 2 (l + b) = 2 × 312

515

+

= 2 × 31 5 51 2

10× + ×

= 2 × 155 102

10+

= 2 ×

25710

= 257

5 = 51

25

m

5. (a) Perimetre of triangle ABC = AB + BC + CA

= 65

m + 118

m + 214

m = 65

98

94

+ +

m

= 6 8 9 5 9 10

40× + × + ×

m (LCM (5, 8, 4) = 40)

= 48 45 90

40+ +

m = 18340

= 42340

m

(b) Perimetre of hexagon ACDEFGA = AC + CD + DE + EF + FG

= 214

+35

+32

+43

+95

+65

= 94

+35

+32

+43

+95

+65

= 9 15 + 3 12 + 3 30 + 4 20 + 9 12 + 6 12

60× × × × × ×

= 135 + 36 + 90 + 80 + 108 + 7260

= 52160

= 84160

.

6. Length of book = 1715

cm

Breadth of book = 1512

cm

Perimeter = 2 (l + b) = 2 (1715

+ 1512

)

= 2 865

+ 312

= 2 172 + 155

10

= 2

32710

= 6525

cm×

7. Length of frame wood = 1523 m

Frame wood required = 1325 m

Length of frame wood left = 1523

– 1325

= 473

675

−

= 47 5 67 3

15× − ×

= 235 20115− = 34

15 = 2

415

m

24

15 m of frame wood is left.


8. Total length cloth = 100 m. Cloth required for 1 dress = 1

34

m

Cloth required for 50 dress = 50 × 134

m

= 5074

25

2

× = 25 7

2×

= 175

2 = 87

12

m.

Cloth left = 100 – 175

2 =

200 1752−

= 252

= 1212

m.

9. Sohan’s shelf = 715

Rohan’s shelf = 34

Let us compare 7

15 and

34

.

LCM (15, 4) = 60

Now 7

157 4

15 42860

= ××

= and 34

3 154 15

4560

= ××

= . We know, 2860

< 4560

So, Rohan’s shelf is more full

So Difference in the two shelves = Rohan self – Sohan shelf

= 34

– 7

15 =

45 2860−

= 1760

\ Rohan shelf is 1760

more full than Sohan’s.

10. (a) 34

12

+ 14

32

+

3 2

4+

1 3 2

4+ ×

54

<74

(b) 615

213

+ 215

613

+

315

73

+ 115

193

+

31 3 7 5

15× + ×

11 3 19 5

15× + ×

93 35

15+

33 95

15+

12815

= 12815


11. Side of square park = 4015 m

Perimeter of square park = 4 × side = 4 × 2015

= 804

5m.

Length of wire required for making three rounds = 3 × 804

5 =

24125

m = 48225

m

Cost of 1 m wire = ` 50

Cost of 2412

5 m wire = ` 50 ×

24125

= ` 24,120

12. (a) 27

367

× = (b) 98

6274

6344

3× = = (c) 45

(d) 23

49

827

× =

13. (a) (b)

(c)

14. (a) (b) (c) (d)

15. (a) (i) 15

of 56

= 15

561

1

× = 16

(ii) 15

1013

× = 15

1013

× = 2

13

(iii) 15

of 5

11 = 1

5 ×

511

= 1

11 (iv)

15

of 157

= 15

1571

3

× = 37

(b) (i) 38

of 16 = 38

× 16 = 3 × 2 = 6 (ii) 38

of 56

= 38

562

× = 5

16

(iii) 38

of 135

= 38

of 85

= 38

85

× = 35

(iv) 38

of 516

= 38

of 316

= 38

3162

× = 3116

= 11516

16. Abhinav solved 45

of maths exercise.

Aditi solved 78

of the same.

Let us compare 45

and 78

.


LCM of (5, 8) = 40

45

= 4 85 8

××

= 3240

78

= 7 58 5

××

= 3540

3540

> 3240

i.e., 78

> 45

Thus Aditi solved greater part.

Now 78

– 45

= 3540

– 3240

= 3

40

∴ Aditi, 3

40 parts greater than Abhinav.

17. Duration of each period = 34

of an hour = 34

of 60 minutes = 34

6015

× = 45 minutes

\ Duration of 5 periods = 5 × 45 minutes = 225 minutes or 334

hours 18. Capacity of tank = 520 litres

Water used during the day = 34

of 520 = 34

520130

× = 3 × 130 = 390 litres

\ Water left in the tank = 520 – 390 litres = 130 litres 19. Total sapling planted = 15

Distance between 2 saplings = 35

m.

Distance between 15 saplings = 35

14425

× = m = 825

m

20. In one day Lavanya reads = 234 hrs

In 10 days she will read = 10 × 234

hrs = 10 × 114

hrs = 5 112

× hrs

= 552

hrs = 2712

hrs

\ Lavanya took 2712

hours to read the whole book.

21. (a) 25

× 347

= 25

257

21

57

107

137

5

× = × = = (b) 37

145

= 3 2

5 =

65

= 115

2

× ×

(c) 49

67

= 4 23 7

= 8213

2

××

× (d)

1

1

3

25151122

= 32

= 112

×

(e) 325

57

=175

57

= 177

= 237

× × (f) 235

12 = 135

12 = 156

5 = 31× ×

15

(g) 345

=195

= 95

= 1× ×9

199

1945

(h) 4157

157

93 3

15

= 215

= × ×

(i) 3215

23 215

45

327

= 7

= 695

= 13× ×


22. (a) 25

314

>15

73

× ×

= 25

134

715

×2

= 13 7

1510

= 13 × 15 7 × 10 = 195 > 70

(b) 245

112

312

215

− −

= 145

32 2

115

− − 7

= 14 2 3 5

10 10× − × × − ×

7 5 11 2

= 28 15

10 10− −

35 22

= 1310

=1310

(c) 34

1657

215

43

× × = 12 3>

(d) 312

+ 415

412

+ 315

= 72

+ 215

92

+ 165

= 7 5 + 21 210

9 5 + 16 2

10× × × ×

= 35 + 42

10

45 + 3210

= 7710

7710

=

(e) 12

of 87

23

of 37

= 12

87

23

4

× × 37

= 47

27

>

(f) 34 of 1

13

45 of 1

34

= 34

43

45

74

× × = 1 75

<

(g) 515

312

+ 115

−

= 25 1

5

72

+ 65

−

= 245

7 5 + 6 2

10× ×

= 245

35 + 12

10 = 4

45

4710

= 48

410

710

>

23. (a) 53

210

= 1030

× and the simplest form of 210

is 15

(b) 37

125

= 3635

×

Simplest form of number is 125

and mixed form is 225

.

Exercise 2.2

1. (a) 18 ÷ 95

= 21859

× = 10 (b) 14 ÷ 73 = 2 14

37

× = 6

(c) 40 ÷ 135

= 40 ÷ 85

= 5 4058

× = 5 × 5 = 25

(d) 15 ÷ 347

= 15 ÷ 257

= 3

5

157

25 =

215

= 415

×

(e) 5 ÷ 313

= 5 ÷ 103

= 53

10 =

32

= 1122

×

(f) 75

÷ 2 = 75

12

= 7

10× (g)

2

5

615

÷ 7 = 25

17

= 2

35×


(h) 435

÷ 5 = 235

÷ 5 = 235

15

=2325

×

(i) 312

215

=72

115

=72

511

=3522

÷ ÷ × = 11322

(j) 315

145

=165

95

=165

59

=169

= 179

÷ ÷ ×

(k) 412

83

=92

83

=92

38

=2716

= 11116

÷ ÷ ×

2. Sugar bought for ` 4012

i.e. `812

= 1 kg.

Sugar bought for ` 1 = 1 ÷ 812

= 12

81=

281

× kg

Sugar bought for ` 810 = 2

81810× = 20 kg.

3. Weight of 16 boxes = 1245

kg. = 645

kg.

Weight of 1 box = 645

÷ 16 = 4 64

51

16 =

45

× kg.

4. For ` 515

Subham Buy = 1 pencil.

For ` 1 Subham Buy = 1 ÷ 515

= 1 ÷ 265

= 1 × 5

26 =

526

For ` 52 Subham Buy = 526

52×2 = 10 pencils.

5. Product of two numbers = 2881

One number = 1427

Other number = product ÷ One number

= 2881

1427

÷ = 2

3

2881

2714

= 23

×

\ Other numbers is 23

.

6. Cost of 623

m of iron wire is ` 1623

.

Cost of 1 m wire = 1623

÷ 623

= 503

÷ 203

= 503

320

52

× = = ` 212


7. Length of rope = 21 m.

Length of 1 piece = 312

m

Number of pieces made = 21 ÷ 312

m

= 21 ÷ 72

= 21 × 27

= 3 × 2 = 6.

8. Sum of 6512

+83

=65 + 32

12 =

9712

Now 9712

512

=9712

125

=975

= 1925

÷ ×

9. Reciprocal of 53

is 35

.

∴ 315

÷ 35

= 165

35

=165

53

=163

= 513

÷ ×

10. Oil in 1 tin = 514

litres

Oil in 25 tins = 514

× 25 = 214

× 25 = 5254

= 13114

litres

11. Total students = 3500

No. of boys = 57

× 3500500

= 5 × 500 = 2500

No. of girls = Total students – No. of boys = 3500 – 2500 = 1000.

12. Time taken to cover 313

km i.e., 103

km = 1 hr

Time taken to cover 1 km = 1013

= 1 ÷ 103

= 1 × 3

10 = 3

10

Time taken to cover 10 km = 3

10 × 10 = 3 hours

Exercise 2.3 1. Refer answers given in the book.

2. (a)-(d). Refer answers given in the book

(e) 412

= 4 +12

= 4 + 5

10 = 4.5 (f)

6 25 2

=1210

= 1.2××

3. (a) 50 paise = ` 50

100 = ` 0.50 (b) 40 cm =

40100

m = 0.4 m

(c) 5 m = 5

1000 km = 0.005 km (d) 800 gm =

8001000

kg = 0.8 kg


(e) 75m as km = 75

1000 = 0.075 km (f) 9 cm as m =

9100

m = 0.09 m

(g) 3500 mL = 35001000

litres = 3.5 litres (h) 175 paise = ` 175100

= ` 1.75

(i) 9 mg = 9

1000 g = 0.009 g

4. (a) 284.57 = 7 hundredths (b) 10.975 = 7 hundredths

(c) 30.756 = 7 tenths (d) 7.935 = 7 ones

(e) 75.4 = 7 tens

5. (a) Greatest = 75.001

Ascending order: 3.009, 7.75, 10.75, 15.75, 18.09, 75.001 (b) Greatest 19.831 Ascending order is: 11.398, 11.839, 11.938, 13.948, 18.931, 19.831

6. (a) 3 . 7 5× 2 . 51 8 7 57 5 0 ×

9 . 3 7 5

(b) 5 7 . 5 6× 0 . 0 1

0 . 5 7 5 6

(c) 4 . 1 5× 0 . 3

1 . 2 4 5

(d)

112 × 55 = 616011.2 × 5.5 = 61.60

1 1 . 2× 5 . 5

5 6 05 6 0 ×6 1 6 0

(e) 1 0 1 0 14

4 0 4 0 4∴ 101.01 × 0.004

= 0.40404

(f) 1 7 5× 5

8 7 5

1.75 × 0.005 = 0.00875

7. (a) 85.7 ÷ 100 = 0.857 (b) 97.563 ÷ 10 = 9.7563

(c) 0.082 ÷ 1000 = 0.000082 (d) 5.72 ÷ 0.2 = 5 720 2..

= 57 2

2.

= 28.6

(e) 0.090 ÷ 1.5 = 0 0901 5

0 9015

..

.= = 0.06

(f) 0.1705 ÷ 0.5 = ××

0.1705 100.5 10

= 1.705

5 = 0.341


(g) 16.2 ÷ 50 = 162 ÷ 500 = 0.324 (h) 30.94 ÷ 0.07 = 3094 ÷ 7 = 442

(i) 0.505 ÷ 0.01 = 50.5

1 = 50.5 (j)

3.4680.24

= 346 8

24.

= 14.45

(k) 96.48 ÷ 0.24 = 9648

24 = 402 (l) 3.28 ÷ 0.08 = 328 ÷ 8 = 41

8. (a) Area of Square = Side × Side = 5.25 × 5.25 = 27.5625 cm2

(b) Area of rectangle = l × b = 12.5 × 7.5 = 93.75 cm2

9. Distance covered in 1 litre petrol = 16.25 km. Distance covered in 12.5 litres of petrol = 16.25 × 12.5 = 203.125 km 10. Cost of 1 m cloth = ` 165.75 Cost of 7.5 m cloth = `165.75 × 7.5 = ` 1243.125 11. Cast of 25 pens = `675.50 Cost of 1 pen = ` 675.50 ÷ 25 = ` 27.02

Cost of 5 such pens = ` 27.02 × 5 = `135.10 12. In 3.25 m, number of gifts packed = 1

In 1 m, number of gifts packed = 1

3 25.

In 48.75 m number of gifts packed = 1

3 25. × 48.75 = 15

∴ 15 gifts can be packed. 13. Products of two decimals = 5.922

One of the number = 1.2 Other number = 5.922 ÷ 1.2 = 4.935

14. (a) Initial fraction of girl = 300

10003

10=

(b) New fraction of girls = 300 200

1000 300500

13005

13++

= =

(c) Importance of women literacy.

Worksheet

1. (a) 134

67

74

67

32

1122

3

× = × = = (b) 215

10115

10115

22

× = × = × = 22

(c) 7

151011

14333

2

× = (d) 123

38

53

38

58

× = × =

(e) 1006.2 × 100 = 100620 (f) 1.88 × 1.2 = 2.256 (g) 0.09 × 1.4 = 0.126 (h) 1.32 × 2.58 = 3.4056

2. (a) 37

1237

112

17 4

1284

÷ = × =×

= (b) 1518

59

1518

95

32

112

3

2

÷ = × = =

(c) 214

2134

94

2134

94

3421

5114

39

14

3

2

17

7

÷ = ÷ = × = =


(d) 425

1115

225

1511

2 3 62 3

÷ = × = × = (e) 0.72 ÷ 10 = 0.072

(f) 16.2 ÷ 4 = 16 2

416240

. = = 4.05 (g) 0.956 ÷ 0.4 = ××

=0 956 100 4 10

9 564

..

. = 2.39

(h) 0.8432 ÷ 0.8 = 0 8432 10

0 8 108 432

8.

..×

×= = 1.054

(i) 846 ÷ 0.009 = 846 10000 009 1000

8460009

××

=.

= 94000

(j) 98.372 ÷ 0.4 = 98 372 10

0 4 10..

××

= 983 72

4.

= 245.93

3. Refer answers given in the book.

4. Cost of 1 kg onion = ` 3012

Cost of 825

kg onion = ` 3012

825

× = `612

425

21

× = ` 1281

5 = ` 256

15

5. Distance covered in one day = 512

km.

\ Distance covered in Oct. i.e., 31 days = 512

× 31 km = 112

31× km = 17012

km

6. Weight of Sugar in one bag = 90.250 kg \ Weight of Sugar in 80 bags = 80 × 90.250 = 7220 kg 7. Length of 1 curtain = 3.25 m Length of 5 curtains = 3.25 × 5 m = 16.25 m Length of the roll = 50 m Length of cloth left = 50 – 16.25 = 33.75 m

8. Perimeteroffigure=12 1012

615

512

615

1012

+ + + + +

= 12212

315

112

315

212

+ + + + +

= 120 105 62 55 62 10510

+ + + + + = 50910

m = 509

10m

9. Total time for studies = 10 hrs

Total time for Maths and Physics = 414

312

+ = 174

72

174

7 22 2

+ = + ××

= 17 144

314

734

+ = =

Time for other subjects = 10 – 734

= 40 31

494

214

− = = h


10. Weight of onion = 30 kg 500 g. Weight of tomatoes = 30 kg 600 g Weight of potatoes = 45 kg 450 sugar Total weight of vegetables = 106.55 kg In order to carry load. she should avoid = 106.55 – 100 = 6.55 kg wt.

11. Sum of two sides of D = 1215

1012

+ = 615

212

+ = 122 105

1022710

+ = = 22.7 m

Perimeter of triangle = 25

Third side = (25 – 22.7) m = 2.3 m.

12. Total students = 400

Number of students who like to play cricket = 15

400of = 80

Number of students who like to play football = 25

400of = 160

Number of students who like to play hockey = 400 – (160 + 80) = 400 – 240 = 160.

Chapter 3Exercise 3.1 1. (a) Maximum marks = 17, Minimum marks = 05 (b) Range = Maximum marks – minimum marks = 17 – 5 = 12

(c) Arithmetic mean =

13 14 15 10 12 13 9 7 9 5 1517 11 14 16

15

+ + + + + + + + + ++ + + +

= 18015

= 12.

(d) Number of students who secured marks more than arithmetic mean = 8 2. Firstfivemultiplesof4are4,8,12,16,20.

Mean = 4 8 12 16 205

605

+ + + + = = 12.

3. First six prime numbers are 2, 3, 5, 7, 11, 13 = 416

= 6.83

4. Firstfivecompositenumbersare4,6,8,9,10

Mean = 375

= 7.4.

5. (a) Maximum score = 100 Minimum score = 15 Range = 100 – 15 = 85

(b) Arithmetic mean = 60 75 80 45 65 15 100 80

8520

8+ + + + + + + = = 65


6. (a) Mean score of A = 60 87 90 954

3324

+ + + = = 83

Mean score of B = 82 0 95

3177

3+ + = = 59

Mean score of C = 68 75 80 904

3134

+ + + = = 78.25

(b)TofindthemeanofplayerBhisscorewillbedividedby3asheplayedonly3games. (c) Player A performed best. 7. (a) Highest score = 96, Lowest score = 45 (b) Range = Highest score – Lowest score = 96 – 45 = 51

(c) Arithmetic mean = 45 75 50 60 80 92 96 86 76 8010

+ + + + + + + + +

= 74010

= 74

8. Mean of students of different sections = Sum total of students in all sections

7

= 50 45 43 48 60 50 54

7+ + + + + +

= 50

9. (a) Maximum weight = 75.2 kg (b) Minimum weight = 40.5 kg (c) Range = Maximum weight – Minimum weight = 75.2 – 40.5 = 34.7 kg

(d) Arithmetic mean = Sum total of weights

Total number of students =

877 815

. kg = 58.52 kg

(e) Number of students whose weight is less than the arithmetic weight = 8 10. Arranging the data in ascending order. 19, 20, 20, 20, 25, 28, 30, 30, 31, 31, 32, 38, 38, 39, 40, 41, 41, 46, 46, 48

Median = 21 12+

th

observation = 11th observation

∴ Median = 32. 11. Mode = 44 size

Average = 50 45 36 60 48 406

2796

+ + + + + = = 46.5

12. (a) Mean = 25 26 32 30 28 30 42 27 309

+ + + + + + + = 2709

= 30

Tofindmedian;arrangethedatainascendingorder 25, 26, 27, 28, 30, 30, 30, 32, 42

Median = 9 12+

th

= 5th observation = 30

Mode = 30

(b) Mean = 40 60 80 90 70 50 40 30 809

5409

+ + + + + + + + = = 60

Arranging the data in ascending order, we get 30, 40, 40, 50, 60, 70, 80, 80, 90


Median = 9 1

2+

th

observation = 5th observation = 60.

Mode = 40 and 80

13. Mean = 38 40 41 40 48 45 42

72947

+ + + + + + = = 42°C

Median: Arranging the data in descending order we get 48, 45, 42, 41, 40, 40, 38

Median = 7 12+

th

observation = 4th observation = 41°C

Mode = 40°C 14–15. Refer answer given in the book.

16. (a) Average time spent on studies = 2 1

12

2 212

2 212

6

+ + + + +

= 2

32

252

252

6

6132

6

12 1326

+ + + + +=

+=

+

= 2512

21

12= h

(b) Time Management

(c) (i) Total time = 1 112

1 112

112

332

12

32

372

+ + + + = + + + = + = 132

612

= h

(ii) Total time spent = 12

12

34

12

12

12

52

34

314

+ + + + + = + = h

(d) Yes, because she spend more time with family, then with friends.

Exercise 3.2 1. (a) The bar graph gives the information about the production of wheat (in tonnes) from

the year 2008 to 2014. (b) 550 tonnes (c) Maximum = 600 tonnes in 2014, Minimum = 100 tonnes in 2013 (d) Range = (600 – 100) = 500 tonnes 2–6. Refer answers given in the book.

7. (a)


(b) On Thursday

(c) Mean = 150 100 200 300 250 200

6+ + + + +

= 1200

6 = 200

8–9. Refer answers given in the book.

10. (a) Possible outcomes are 8, 12

∴ P (a multiple of 4) = 28

= 14

(b) Two digit numbers are 10, 11, 12, 13, 14, 15

∴ P (two digit number) = 68

= 34

.

(c) Single digit numbers are = 8, 9

P (single digit number = 28

14

=

11. (a) Even number on a dice are 2, 4, 6

P (even number) = 36

12

=

(b) Odd numbers are 1, 3, 5

P(odd number) = 36

12

=

(c) Composite number are 4, 6

P (Composite number) = 26

13

=

(d) Prime number are 2, 3, 5

P (Prime number) = 36

12

=

12. (a) Vowels in the word EXPERIMENT are E, E, and I i.e., 4.

∴ P (a vowel) = 410

25

=

(b) P (getting E) = 310

(c) Consonants are X, P, R, M, N, T. P (a consonant) = 6

1035

=

13. WhenacoinisflippedtherearetwooutcomesHeadortail.

∴ Probability of team A starting the game = 12


14. Total balls = 80 + 60 + 10 = 150

(a) P (a blue ball) = 80

1508

15= (b) P (a red ball) = 60

1506

1525

= =

(c) P (a green ball) = 10

1501

15= (d) P (not blue) =

70150

715

=

(e) P (not red) = 90150

35

= (f) P (not green) = 140150

1415

=

15. Number of times the coin is tossed up = 10 Number of times the head up = 7

(a) P (head up) = 7

10 (b) P (tail up) = 3

10 16. Possible outcomes when two coins are tossed up are HH, TH, HT, TT

(a) P (getting two heads) = 14

(b) P (getting one head) = 24

12

=

(c) P (getting no head) = 14

Worksheet

1. (a) Mean = Sum of the weights of a 7 students

7 =

2877

kg = 41 kg

(b) Arranging data in ascending order we have 36 kg, 38 kg, 39 kg, 40 kg, 41 kg, 45 kg, 48 kg

Median = 7 1

2+

th

observation = 4th observation = 40 kg.

There does not exits a mode 2. Mode = 14 years as it occurs maximum (two) number of times

3. Name of game Tally marks Frequency

Volleyball (V) 10

Cricket (C) 12

Football (F) 7

Hockey (H) 11

Mode of the data is cricket as it is liked by maximum number of students.

4. Mean = Sum total of marksNumber of students

= 92 5 94 89 2 95 1 84 5 87 1 94 5 96 2 97 8 88 7

10. . . . . . . . .+ + + + + + + + +

= 919 6

10.

= 91.96


Median: Arranging data in descending order, we get, 97.8, 96.2, 95.1, 94.5, 94, 92.5, 89.2, 88.7, 87.1, 84.5

Median = Average of two middle observations = 94 92 52

186 52

+ =. . = 93.25 Mode: Mode does not exist. 5. (a) Height of tallest boy = 154 cm (b) Height of shortest boy = 128 cm (c) Range of data = (154 – 128) cm = 26 cm (d) For median, arranging data in ascending order we get 128, 129, 136, 138, 140, 140, 142, 142, 143, 144, 144, 146, 150, 152, 154

Median = 15 12+

th

observation = 162

th

= 8th observation = 142 cm

6. (a) The bar graph represents the information about the students having pet animals. (b) 5 students have dog as pet (c) Rabbit is most popular as pet (d) 5 + 9 + 12 + 11 = 37 (e) Dog (f) Range = 12 – 5 = 7 7. Refer answers given in the book.

8.


(b) 210 (c) Electronics (d) Yoga (e) Dramatic (f) Fine Arts

10. There are 11 total balls

(a) Even numbers are 10, 12, 14, 16, 18, 20, i.e., 6

P (even number) = 6

11 (b) Odd numbers are 11, 13, 15, 17, 19

P (odd number) = 5

11


(c) Prime number are 11, 13, 17, 19.

P (prime number) = 4

11 (d) Composite numbers = 10, 12, 14, 15, 16, 18, 20

P (composite number) = 7

11 11.

40

20

10

30

50

1 2 3 4 5 6

= 10 units

(a) P (getting 4) = 50250

= 15

(b) P (getting 6) = 50250

15

=

(c) P (getting a number less than 6) = 200250

45

=


Chapter 4

Exercise 4.1 1–2. Refer answers given in the book. 3. (a)L.H.S.=2×6×3=15=R.H.S.;Yes (b)L.H.S.=7×5+15=35+15=50≠40=R.H.S;No (c)L.H.S.=4×4–3=16–3=13≠12=R.H.S.;No.

(d) L.H.S. = 4 6

3×

+ 2 = 8 + 2 = 10 = R.H.S.;Yes

(e)L.H.S.=7×4–3=28–3=25=R.H.S.;Yes

(f) L.H.S.=5×2+3=13+18=R.H.S.;No (g)L.H.S.=8×5–5=35=R.H.S.;Yes (h)L.H.S.=60–10=50=R.H.S.;Yes

(i) L.H.S. = 142 + 2 = 7 + 2 = 9 ≠8=R.H.S.;No

(j) L.H.S.=5×11–10=55–10=45≠55=R.H.S.;No.

(k) L.H.S. = 2 5×

5 + 4 = 2 + 4 = 6≠5=R.H.S.;No


4. (a) x LHS(2x + 5)

RHS LHS = R.H.S?

1 2 × 1 + 5 = 7 15 No2 2 × 2 + 5 = 9 15 No3 2 × 3 + 5 = 11 15 No4 2 × 4 × 5 = 13 15 No5 2 × 5 + 5 = 15 15 Yes6 2 × 6 + 5 = 17 15 No

\ x = 5 is the solution.

(b) x LHS(7x + 5)

RHS LHS = R.H.S?

2 7 × 2 – 5 = 11 30 No3 7 × 3 – 5 = 16 30 No4 7 × 4 – 5 = 23 30 No5 7 × 5 – 5 = 30 30 Yes6 7 × 6 – 5 = 37 30 No


(c) x LHS(4x – 3)

RHS LHS = R.H.S?

2 4 × 2 – 3 = 5 13 No3 4 × 3 – 3 = 8 13 No4 4 × 4 – 5 = 13 13 Yes5 4 × 5 – 5 = 17 13 No


(d) m LHS(4x – 3)

RHS LHS = R.H.S?

333

× 1 = 2 10 No

993

× 1 = 4 10 No

15153

× 1 = 6 10 No

21213

× 1 = 8 10 No

27273

× 1 = 10 10 Yes



(e) x LHS(4x – 3)

RHS LHS = R.H.S?

52 5

5×

+ 2 = 4 6 No

102 10

5×

+ 2 = 6 6 Yes

152 15

5×

+ 2 = 8 6 No

\ x = 10 is the solution.(f) p LHS

(4p – 5)RHS LHS = R.H.S?

2 4 × 2 + –5 = 3 23 No4 4 × 4 – 5 = 11 23 No6 4 × 6 – 5 = 19 23 No7 4 × 7 – 5 = 23 23 Yes8 4 × 8 – 5 = 27 23 No

\ p = 7 is the solution.

(g) z LHS

(z5

– 7)

RHS LHS = R.H.S?

555

+ 7 = 3 9 No

10105

+ 7 = 9 9 Yes

15155

+ 7 = 10 9 No

\ z = 10 is the solution.

(h) m LHS3

5m

RHS LHS = R.H.S?

53 5

5×

= 3 6 No

103 10

5×

= 6 6 Yes

15 3 155

× = 9 6 No

\ z = 10 is the solution.


5. (a) Let Iqbal has x story books. No of books Sheena has = 8 + 5x \ 63 = 5x + 8 (b) Let age of Shyam = x years

Father’s age = 3x – 4 ∴ Also we have 71 = 3x – 4

(c) Lowest marks obtained in class = x Highest marks = 3x – 1 \ we have 98 = 3x – 1. (d) Let the base angle = x° Vertex Angle = 4x° According to angle sum property of

triangles We have x + x + 4x = 180

6x = 180 (e) Let the no. of Girls = x

Boys = 3 + 2x \ 3 + 2x = 23

(f) Let Kavita’s age = x years Kavita’s Mother age = 2 + 5x \ 2 + 5x = 87

(g) Let the breadth = b m. Length = 5 + 4b \ 5 + 4b = 49

(h) Let Mohan’s age = x year Rohan = 2x – 10 50 = 2x – 10

(i) Let Ravi’s score = x Shri’s score = 4x ∴ 4x + x = 150 ⇒ 5x = 150

6. (a) Add 4 to both sides x + 4 + 4 = 7 + 4 ⇒ x = 11

(b) Add 3 to both sides ∴ 2x – 3 + 3 = 8 + 3

⇒ 2x = 11

⇒ x = 112

(c) Divide both sides by 8

∴ 8 8

=368

y ⇒ y = 92

(d) 20p – 5 = 55 Add 5 to both sides

20p = 55 + 5⇒ 20p = 60

⇒ p = 6020

= 3

(e) 4m + 2 = 10 Subtract 2 from both sides

4m + 2 – 2 = 10 – 2⇒ 4m = 8⇒ m = 2

(f) x5

=920

Multiply both sides by 5

5 5

= 920

54

× ×x

x = 94

(g) 3x – 2 = 46 Adding 2 both sides

3x – 2 +2 = 46 + 2 3x = 48

x = 483

= 16

(i) p LHS RHS LHS = R.H.S?

3 6 × 3 – 3 = 15 33 No

4 6 × 4 – 3 = 21 33 No

5 6 × 5 – 3 = 27 33 No

6 6 × 6 – 3 = 33 33 Yes

7 6 × 7 – 3 = 39 33 No



(h) 34p

= 6


34

4p × = 6 × 4

3p = 24 Divide both seats by 3

p = 8. (i) zv – 6 = 8 Add 6 to both sides. 2q – 6 + 6 = 8 – 6 ⇒ 2q = z ⇒ q = 1 7. (a) 3x = 48 Dividing both sides by 3 x = 16 (b) 5m – 7 = 23 Adding 7 to both sides 5m = 30 Dividing both sides by 5 m = 30 ÷ 5 = 6

(c) x5

= 4


x5

× 5 = 4 × 5

⇒ x = 20(d) 2y – 5 = 7

⇒ 2y = 7 + 5⇒ 2y = 12⇒ y = 12 ÷ 2 = 6

(e) 17 + 4y = 45 Subtracting 17 from both sides

4y = 28 Dividing both sides by 4

44

=284

y

⇒y = 7 (f) 3t – 22 = – 1 Adding 22 to both sides 3t – 22 + 22 = t + 22

⇒ 3t = 21

Dividing both sides by 3 t = 7

8. (a) 3x + 5 = 17 Subtracting 5 from both sides 3x = 17 – 5 ⇒ 3x = 12 Dividing both sides by 3

x = 123

= 4

(b) 23b – 5 = –7

Adding 5 to both sides

23b = 12.

Dividing both sides by 23

23b

÷ 23

= 12 ÷ 23

b = 12 × 32

b = 18

(c) 34x

= 9

Dividing both sides by 34

We get 34x

÷ 34

= 9 ÷ 34

x = 9 × 43

= 12

x = 12 (d) 6y – 10 = 14 Adding 10 to both sides

6y = 14 + 10 6y = 24

Dividing both sides by 6 we get y = 24 ÷ 6 = 4. (e) 9x + 5 = 14 Subtracting 5 from both sides

9x = 14 – 5 9x = 9

Dividing both sides by 9 x = 1


(f) 45z

= 20


4z = 20 × 5 Dividing both sides by 4

z = 20 54×

⇒ z = 25

(g) q5

– 3 = 5

Adding 3 to both sides

q5

= 5 + 3

⇒ q5

= 8


q5

5× = 8 × 5

⇒ q = 40

(h) x3

+ 4 = 15 subtracting 4 from both

sides

x3

= 15 – 4

x3

= 11

Multiply both sides by 3 we get x = 11 × 3 = 33

9. (a) 3t = 42 ⇒ t = 42 ÷ 3 ⇒ t = 14(b) z – 5 = 6 ⇒ z = 6 + 5 = 11(c) 3x – 2 = 13 ⇒ 3x = 13 + 2 ⇒ 3x = 15 ⇒ x = 15 ÷ 3 = 5(d) m – 2 = 8 ⇒ m = 8 + 2 ⇒ m = 10(e) 4p + 3 = 19

⇒ 4p = 19 – 3 ⇒ 4p = 16 ⇒ p = 16 ÷ 4 = 4(f) 9q – 7 = 56 ⇒ 9q = 56 + 7 ⇒ 9q = 63 ⇒ q = 63 ÷ 9 = 7(g) 4p – 3 = 21 ⇒ 4p = 21 + 3 ⇒ 4p = 24 ⇒ p = 24 ÷ 4 = 6(h) 6p – 5 = 31 ⇒ 6p = 31 + 5 ⇒ 6p = 36 ⇒ p = 36 ÷ 6 = 6

(i) 45p = 8

⇒ 4p = 8 × 5

⇒ p = 2 8 5

4×

= 10

Exercise 4.2 1. (a) 2y = 37

454

−

= 37 54

324

8− =

⇒ 2y = 8 ⇒ y = 8 ÷ 2 = 4 ⇒ y = 4(b) 5t – 28 = 12 ⇒ 5t = 12 + 28 ⇒ 5t = 40 ⇒ t = 40 ÷ 5 = 8 ⇒ t = 8

(c) a5

+ 4 = 9

⇒ a5

= 9 – 4

⇒ a5

= 5

⇒ a = 5 × 5 ⇒ a = 25


(d) 6y + 2 = 14 ⇒ 6y = 14 – 2 ⇒ 6y = 12 ⇒ y = 12 ÷ 6 = 2 ⇒ y = 2

(e) 34

53

l =

⇒ 3l = 53

× 4

⇒ l = 5 43 3

209

229

××

= =

⇒ l = 229

(f) 23b – 5 = 7

⇒ 23b = 12

⇒ 2b = 12 × 3

⇒ b = 12 3

2×

⇒ b = 6 × 3 ⇒ b = 18

(g) 7z = 13 + 92

⇒ 7z = 26 92

352

+ =

⇒ z = 352 7

52

212

5

×= =

z = 212

(h) 53

152

y =

⇒ y = 152

÷ 53

= 3 15

235

×

⇒ = 92

412

=

⇒ y = 412

(i) 452

72

p − =

⇒ 472

52

p = +

⇒ 4122

6p = =

⇒ p = =64

32

2. (a) 3 (n – 5) = 9 ⇒ (n – 5) = 9 ÷ 3 ⇒ n – 5 = 3 ⇒ n = 8(b) 2(p – 5) = 12 ⇒ p – 5 = 6 ⇒ p = 6 + 5 ⇒ p = 11(c) –4(2 + y) = 12

⇒ 2 + y = 124−

= – 3

⇒ y = – 3 – 2 = –5 ⇒ y = –5

(d) 3(n – 5) = 42 ⇒ n – 5 = 42 ÷ 3 ⇒ n – 5 = 14 ⇒ n = 14 + 5 ⇒ n = 19(e) 3(x – 1) = 2x – 11 ⇒ 3x – 3 = 2x – 11 ⇒ 3x – 2x = – 11 + 3 ⇒ x = – 8

(f) 4(m – 6) + 24 = 0 ⇒ 4(m – 6) = –24 ⇒ m – 6 = –24 ÷ 4 ⇒ m = –6 + 6 = 0(g) 16 = 4 + 2 (t + 4) 16 – 4 = 2 (t + 4) ⇒ 12 = 2 (t + 4) ⇒ 6 = t + 4 ⇒ 6 – 4 = t ⇒ 2 = t


(h) 3(x – 1) = 2x + 1 ⇒ 3x – 3 = 2x + 1 ⇒ 3x – 2x = 1 + 3 ⇒ x = 4(i) 4 + 5 (p ) = 31 ⇒ 5 (p + 1) = 27

⇒ p + 1 = 275

– 1

⇒ p = 225

3. (a) 4(5x + 2) = 28 ⇒ 5x + 2 = 28 ÷ 4 ⇒ 5x + 2 = 7 ⇒ 5x = 7 – 2 ⇒ x = 5 ÷ 5 ⇒ x = 1(b) 3(x – 5) = 27 ⇒ x – 5 = 27 ÷ 3 ⇒ x – 5 = 9 ⇒ x = 9 + 5 ⇒ x = 14 (c) 28 – 4 = 3 (t – 5) ⇒ 24 = 3 (t – 5) ⇒ 24 ÷ 3 = t – 5 ⇒ 8 = t – 5 ⇒ 13 = t (d) 5(2x – 3) = 3 (x – 7) ⇒ 10x – 15 = 3x – 21 ⇒ 10x – 3x = – 21 + 15 ⇒ 7x = –6

⇒ x = −67

(e) 8x – 3 = 9 + 4x ⇒ 8x – 4x = 9 + 3 ⇒ 4x = 12 ⇒ x = 12 ÷ 4 ⇒ x = 3

(f) 23

53

x + = 4

⇒ 23x

= 4 – 53

⇒ 23x

= 12 5

3 =

73

−

⇒ x = 73

÷23

= 73

= × 32

72

⇒ x = 312

(g) 34 – 5 (p + 1) = 4⇒ –5 (p + 1) = 4 – 34⇒ + 5 (p + 1) = +30⇒ p + 1 = 30 ÷ 5 ⇒ p + 1 = 6⇒ p = 6 – 1⇒ p = 5

(h) 4(2x + 3) – 5 (3x + 4) = 2⇒ 8x + 12 – 15x – 20 = 2⇒ 8x + 15x = 2 – 12 + 20⇒ – 7x = 10

⇒ x = −107

= −137

(i) 3(x + 1) – 2 (x + 1) = 5 ⇒ 3x + 3 – 2x – 2 = 5 ⇒ 3x – 2x + 3 – 2 = 5 ⇒ x = 5 – 1 ⇒ x = 4.

4. (a) x = 4 Multiply by 2 both sides

2x = 8 Adding 1 to both sides

⇒ 2x + 1 = 8 + 1⇒ 2x + 1 = 9 (I equation)

Again we have x = 4 Subtracting 2 from both sides, we

get⇒ x – 2 = 4 – 2⇒ x – 2 = 2Multiply both sides by 3, we get⇒ 3 (x – 2) = 2 × 3⇒ 3 (x – 2) = 6 (II equation)

Again considers x = 4

Adding 12

from both sides

⇒ x + 12

= 4 + 12

⇒ x – 12

= 92



3 (x – 12

) = 92

× 3

3 (x – 12

) = 272

(III equation)

(b) and (c). Similar work to be done. 5. (a) Let the number be x.

x + 3x = 16⇒ 4x = 16

⇒ x = 4 16

4 = 4

x = 4(b) Let the number = x

According to question 7 + 5x =47 5x = 47 – 7 5x = 40 x = 40 ÷ 5 x = 8(c) Let the number = x

15

x – 5 = 45

15

x = 45 + 5

x = 50 × 5 x = 250(d) Let the number be x

According to question 4x – 6 = 42 4x = 42 + 6 4x = 48 x = 48 ÷ 4 = 12 x = 12(e) Let the number be x

According to question 6x = 42 ⇒ x = 42 ÷ 6 x = 7(f) Let the number be x

According to question 4x = 36 + x 4x – x = 36 3x = 36 x = 12

(g) Let the no. of pencils Abinav has = x. According to question

100 – 3x = 19⇒ –3x = 19 – 100⇒ – 3x = –81⇒ x = 27

(h) Let the no. Ibennal thinks = x According to Question

(x – 9) ÷ 4 = 5⇒ x – 9 = 5 × 4⇒ x = 20 + 9 x = 29

(i) Let the no. Anwar thinks = x According to question

32

x – 4 = 14

⇒ 32

x = 18

⇒ x = 18 23×

⇒ x = 12(j) Let the no. = x

According to question.⇒ 2x – 15 = 13⇒ 2x = 28⇒ x = 14

6. (a) Let the no = x According to question

50 + 7x = 300 – 40 7x = 260 – 50 7x = 210 x = 30

(b) Let the no = x

14

x = 4 + 11

14

x = 15

x = 60(c) Let the no = x 5x + 15 = 30 5x = 15 x = 3


(d) Let the lowest marks = x According to the question

Highest marks = 7 + 3 lowest marks 97 = 7 × 3 × x 97 – 7 = 3x 90 = 3x 30 = x(e) Let the vertex Angle = x° Base angle = 2x°

Sum of three angle of ∆ = 180° x° + 2x° + 2x° = 180° 5x = 180° x = 36°

So the angles are 72°, 72°, 36° (f) Let the runs scored by Rahul = x Sachin’s score = 2x + 5 According to question Rahul score + Sachin score = 300 – 13

x + 2x + 5 = 287 3x = 287 – 5 3x = 282 x = 282 ÷ 3 = 94 Sachin’s score = 2 × 94 + 5 = 188 + 5 = 193

(g) Let Parmeet has = x marbles Irfan has = 7x – 5 Irfan = 51∴ 7x – 5 = 51 7x = 51 + 5 7x = 56 x = 56 ÷ 7 = 8

Permeet has 8 marbles. (h) Let the no of Non-fruit trees = x then no. of fruit trees = 3x – 5 Total trees planted = 403 i.e., x + 3x – 5 = 403

4x – 5 = 403 4x = 408 x = 102.

No od non-fruit trees = 3 × 102 – 5 = 306 – 5 = 301

(i) Let Raja’s age = x years Father’s age = 4x – 5

According to question 4x – 5 = 43 4x= 43 + 5 4x= 48 x = 12

7. (a) Number of dogs = 36 and other animals = 11

(b) Kindness and sincerity 8. (a) Number of boys = 46 and no. of girls

= 14 (b) For good health and sound mind

Worksheet 1. (a) x + 9 = 15

x = 1, 2 , 3 ... We get 6 + 9 = 15 ∴ x = 6 (b) x – 1 = 6 Putting x = 1, 2,... We get 7 – 1 = 6

6 = 6 x = 7

(c) 12y = 36 Putting x = 1, 2, 3... We get 12 × 3 = 36

x = 3

(d) x − 1

3 = 4

Put x = 1, 2, ... We get 13 1

313 1

3123

4− −

= = =

x = 13(e) x + 7 = 13

Put x = 1, 2, 3,... We get 6 + 7 = 13

∴ x = 6

(f) m + 1

4 = 5

Put x = 1, 2...

We get 19 14

204

+= = 5

∴ m = 19


(g) Put m = 12

, 22

, 32

...

We get 72

– 12

= 62

= 3

∴ x = 72

(h) 3x + 2 = 11 Put x = 1, 2, 3, ... We get

3 × 3 + 2 = 11 9 + 2 = 11 ∴ x = 3(i) 4x – 5 = 23 Put x = 1, 2, 3...

We get 4 × 7 – 5 = 28 – 5 = 23 x = 7

2. (a) x + 3 = 5 3+3≠6 ∴ x = 3 is not a solution. (b) LHS – 4x + 2 –4, (–2) + 2 = +8 + 2 = 10

Yes x = –2 is a solution. (c) LHS 2x – 3 =2×2–3=4–3=1≠3.

∴ x = 2 is not a solution.

(d) 4 63× = 4 × 2 = 8

Yes x = 6 is a solution.

(e) LHS 2x – 7 = 2 × 11 – 7 = 22 – 7 = 15 Yes x = 11 is a solution.

3. (a) Let the no. be = x. According the question

5 + 5x = 65 5x = 60 x = 60 ÷ 5 x = 12(b) Let the no be = x.

According the question

1 + x7

= 8

x7

= 7

x = 7 × 7 = 49

(c) Let the no. be = x According the question

3x – 20 = 40 3x = 40 + 20 3x = 60 x = 20(d) Let the number be = x.

According the question

5 + x2

= 12

x2

= 12 – 5

x2

= 7⇒ x = 7 × 2 = 14

4. (a) 2 added to 5 times a number t gives 15.

(b) 7 added to a number x gives 15. (c) 2 added to one third of m gives 6. (d) Half less than 2 times m is 3. (e) Adding half to 3 times a number

gives half of 9. (f) Subtracting 5 from 9 times x

gives 40. 5. (a) 3x + 2 = 17

3x = 17 – 2 3x = 15 x = 15 ÷ 3 = 5(b) 9m – 6 = 48 9m = 48 + 6 m = 54 ÷ 9 = 6

(c) 7x + 12

= 4

7x = 4 – 12

7x = 8 12−

7x = 72

x = 72

÷ 7 = 12

(d) x3

– 5 = 7

x3

= 7 + 5


x = 12 × 3 x = 36(e) 3x + 2 = 11 3x = 9

x = 93

= 3

(f) 12y = 7y + 15 12y – 7y = 15 5y = 15 y = 15 ÷ 5 y = 3(g) 3(x + 6) = 2x + 6 3x + 18 = 2x + 6 3x – 2x = 6 – 18 x = –12(h) 2(x – 1) = 4 (x + 2) 2x – 2 = 4x + 8 2x – 4x = 8 + 2 –2x = 10 x = –5

6. (a) Let the no. of girls = x

has no of boys = 35

x

According to question = x + 35

x = 48

5 35

x x+ = 48

85x = 48

x = 6 48 5

8×

= 30

(b) Let the no. = x According to question

12 + 3x = 48 3x = 48 – 12 3x = 36 x = 36 ÷ 3 = 12(c) Let the no. be = x

According to question 3x – 7 = 11 3x = 11 + 7 3x = 18 3x = 18 ÷ 3 = 6

(d) Let the no be = x According to question

2x – 60 = 4 2x = 64 x = 64 ÷ 2 = 32

(e) Let the third side = x Two equal side = 3x Perimeter of triangle = x + 3x+ 3x According the question

28 = 7x 4 = x

Sides of triangles are 4, 12, 12 (f ) Let age of Aman = x year Mother’s age = 7x According to question

x = 7x = 48 8x = 48 x = 48 ÷ 4 x = 6

(g) Let the runs scored by Yuvraj = x. Gautam’s score = 3x According to question

3x + x = 300 – 16 4x = 284 x = 71 Yuvraj Scored = 71

7. (a) –3 (4 + x) = 2x + 5 –12 – 3x = 2x + 5 –12 – 5 = 2x + 3x –17 = 5x

−175

= x

(b) 4x – 2 = 5 + 3x 4x – 3x = 5 + 2 x = 7

(c) 4(5x – 4) + 3 (2x – 1) = – 7 20x – 16 + 6x – 3 = – 7 26x – 19 = 7 26x = 7 + 19 26x = 26 x = 1

(d) 52 3

12

− y =

− −y3

12

52

=


− −y3

1 52

=

+y = +42

× 3

y = 6(e) 34 – 5(n –1) = 2n

34 – 5n + 5 = 2n

34 + 5 = (2 + 5)n

397

= n

547

= n

(f) 3p – 2 (2p – 5) = 2 (p + 3) – 8 3p – 4p + 10 = 2p + 6 – 8 –p + 10 = 2p – 2 –p – 2p = –2 – 10 – 3p = –12 p = 4

Chapter 5

Exercise 5.1

1. (a) 45° (b) 30° (c) 32° (d) 54° (e) 7° (f) 65° 2. (a) 78° (b) 91° (c) 35° (d) 115° (e) 59° (f) 85°

3. (a) Let an angle = x° Complement = 90 – x° According to given condition

x° = 2(90° – x)° x° = 180° – 2x° 3x° = 180° x° = 60°(b) Let the angle = x° Supp = 180° – x° x° = 3(180° – x)° x° = 3 × 180° – 3x x° + 3x = 3 × 180° 4x = 540°

x = 5404

= 135°

(c) Let the angle = x Complement = 90° – x

According to given condition x = 90° – x° 2x = 90° x = 45°(d) Let the angle = x Supp = 180° – xAccording to given condition x = 180° – x 2x = 180° x = 90°

4. (a) 120° + 60° = 180° Supplementary (b) 40° + 50° = 90° Complimentary (c) 115° + 65° = 180° Supplementary (d) 70° + 20° = 90° Complimentary (e) 36° + 54° = 90° Complimentary (f) 112° + 68° = 180° Supplementary 5. (a) x + 62° = 180° (Linear pair)

x = 180° – 62° x = 118°(b) 2x + x = 180° (Linear pair) 3x = 180° x = 60°


(c) 2x + x – 30°= 180° (Linear pair) 3x = 210° x = 70°

(d) 2x + 10 + x + 20°= 180° 3x + 30 = 180° 3x = 150° x = 50°

6. (a) Yes (b) No (c) No (d) Yes 7. (a) No (b) Yes (c) Yes (d) Yes 8. (a) ∠AOD, ∠BOC;∠AOE, ∠BOF;∠EOC, ∠DOF;∠AOC, ∠BOD;∠EOD, ∠COF;∠EOB,

∠AOF (b) ∠AOE, ∠EOC;∠DOF, ∠FOB (many case are there) (c) ∠AOD, ∠AOC;(Answermayvary) (d) ∠EOD and ∠EOB (Answer may vary) 9. (a) No, as they do not have common vertex (b) No, No common vertex (c) No, No common arm (d) No, L1 and L2 do not have non overlapping interiors. 10. (a) x + 40° = 180° (Linear pair)

x = 180° – 40° x = 140° y = 40°(V.O.A) z = x = 140° (VOA)(b) 2x + 60° + x = 180° (Linear angles) 3x + 60° = 180° 3x = 120° x = 40°

(c) 50° + x + 40° = 180° (Linear angles) x = 90° y = 50° z = 130°

(d) 30 +y +40° = 180° y = 180° – 70° = 110° z = 40°(VOA) x = 30°(VOA)

11. 90° + x + y = 180° (As PQ ia a line) x : y = 3 : 2∴ x = 3a y = 2a 90° + 3a + 2a = 180° 90° + 5a = 180° 5a = 180° – 90° 5a = 90°


a = 905

= 18°

x = 54° y = 36° z = 90° + x (VOA) z = 90° + 54° z = 144°

12. (a) 40° and 50° (b) Let the angle = x° Supplementary angle = 180° – x° According to Question x° = 180° – x° + 50° 2x = 230° x = 115° \ angles are 115° and 65°

Exercise 5.2 1. (a) No (b) Yes (c) No (d) Yes 2. (a) l, m, n and p are all lines. (b) (l, m, n) lines, p is a transversal. (l, m, p) lines, n is a transversal. (p, m, n) lines, l is a transversal. (p, l, n) lines and n is a transversal. (c) (p, l) (p, m) (p, n) (l, m) (l, n) (m, n) are all intersecting lines. 3. (a) (∠1,∠2) are Corresponding angles (b) ∠1and ∠2are Vertically Opposite Angles (c) ∠1and ∠2are adjacent angles (d) ∠1,∠2are interior angles on the same side of transversal (e) ∠1and ∠2 are alternate interior angles (f) ∠1and ∠2 are alternate exterior angles 4. (a) x = 75° (as all corresponding angles are equal) (b) 70 + x = 180° x = 110° (As interior angles on the same side are supplementary if lines are parallel) (c) x = 60° as l ||m and hence alternative exterior angles are equal.

(d) a = 60° (Corresponding angles) x + a = 180° (linear pair) x + 60° = 180° x = 120°(e) x + 3x = 180° 4x = 180°

x = 1804

= 45°

60°

x

a

(Interior angles the some side of trans one supplementary)


5. (a) All pairs of corresponding angles are. (a, e) (b, f), (c, g), (d, h) (b) All pairs of alternate interior angles are (d, f) and (c, e) (c) (d, e) and (c, f) are interior angles are on the same side of transversal. (d) (a, b) (a, d) (b, c) (d, c) (e, f) (f, g) (g, h) (h, e) are all pairs as linear angles. 6. (a) ∠3 = 65° (as V.O.A) ∠3 = 7 (Corresponding angles) ∠7 = 65° ∠7 = ∠5 (V.O.A) ∠5 = 65°

Also ∠4 + 65° = 180° ∠4 + 180° – 65° ∠4 = 115°

∴ ∠2 = ∠4 = 115° (V.O.A) ∠2 = ∠8 (Corresponding Angles) ∠8 = 115° ∠8 = ∠6 (V.O.A) ∴ ∠6 = 115° (b) x + 3x = 180°

4x = 180° x = 45° ∴ ∠1 = 45° ∠2 = 135° ∠1 = ∠3 (V.O.A) ∠5 = x = 45° (Corresponding Angles) ∠7 = ∠3 = 45° ∴ ∠1 = ∠3 = ∠5 = ∠7 = 45°

Similarly ∠2 = ∠6 (Corresponding Angles) ∠6 = ∠8 (V.O.A) ∠2 = ∠4 (V.O.A) ∴ ∠2 = ∠4 = ∠6 = ∠8 = 135°

7. (a) If two parallel lines are intersected by a transversal then, alternate interior angles are equal.

(b) If sum of interior angles on the same side of transversal is 180° then the lines are parallel.

(c) If two parallel lines are intersected by a transversal then corresponding angles are equal.

(d) If two parallel lines are intersected by a transversal then consecutive interior angles are supplementary or interior angles on the same side of transversal are supplementary.

8. (a) a = 120° (Corresponding angles) a + x = 180° (Linear angles) 120° + x = 180° x = 180° – 120° = 60°


(b) 100° = p (Corresponding angles as a || b) p + x = 180° (Linear pair) x = 180° – 100° = 80° x + y = 180° (Consecutive interiors on the

same side of transversal are supplementary as l || m) 80° + y = 180° y = 1000

9. (a) a = 65° (V.O.A) a+65°=65°+65°=130°≠180°

As the sum of interior angles on the same side of transversal is not supplementary hence the line l is not parallel to m.

(b) a = 62° (Vertically Opposite Angels) a+116°=62°+116°=178°≠180°

Since interior angles on the same side of transversal are not supplementary hence the lines l and m are not parallel.

(c) a = 81° (Vertically Opposite Angels) a + 99° = 81° + 99° = 180°

∴ line l and m are parallel as the sum of consecutive interior angles on the same side of transversal is supplementary.

99°l

8Pm

t

a

62°l

a116°

m

t

a


10. x + 115° = 180° (Linear pairs) x = 65° a = x (Alternate interior angles) a = 65° ∠a = ∠d (As Alternate exterior angles are

equal)⇒ d = 65° ∠d = ∠e (Corresponding angles as p ||q) ∠e = 65° ∠c + e = 180° (Consecutive interior angles)∴ ∠c + 65° = 180° ∠c = 180° – 65° ∠c = 115°

∠c = ∠b (V.O.A) ∠b = 115° ∠a = ∠d = ∠e = 65° ∠b = ∠c = 115°

11. (a) ∠DGC = ∠DEF (Corresponding angles as BC || EF) ∴ ∠DGC = 68°

∠ABC = ∠DGC (Corresponding angles are equal AB|| DE)

= 68° ∠BGE = ∠DGC = 68° Also ∠BGE = ∠DGC = 68° (V.O.A) ∠DGC + ∠DGB = 180° (Linear pair) ∠DGB + 68° = 180° ∠DGB = 180° – 68° = 112°

(b) ∠DGB = ∠DEF (Corresponding angles) ∠DGB = 68° ∠ABC + ∠DGB = 180° (Consecutive interior

angles) ∠ABC + 68° = 180° ∠ABC = 180° – 68° = 112° ∠BGE + ∠DGB = 180° (Linear pair) ∠BGE + 68° = 180° ∠BGE = 112°

Worksheet 1 1. (a) 180° (b) Complimentary (c) adjacent (d) angle (e)adjacent;supplementary (f) transversal (g) supplementary (h) parallel 2. 2x + x = 180° (Linear pair)

3x = 180° x = 60°

A

B C

D

E F

G

68°

4x + x = 180° 5x = 180°

x = 180

5°

= 36°

115°l

m

q

x

p

a

cd e

b


3. Angle Complement Supplement45° 45° 135°80° 10° 100°70° 20° 110°48° 42° 132°0 90° 180°

4. (a) V.O.A of ∠COB is ∠DOA. V.O.A of ∠DOF is ∠FOC. (b) Adjacent complimentary are (∠AOE and ∠COE) and (∠DOF, ∠BOF) (c) Equal supplementary angle are (∠AOC, ∠BOC), (∠BOC, ∠BOD), (∠AOC, ∠AOD),

(∠AOD and ∠BOD) (d) Acute V.O. Angles are ( ∠AOE, ∠BOF), (∠COE ∠FOD) 5. (a) Let the angle = x° its supplementary is 180° – x° According to Q

x – (180° – x) = 30° x – 180° + x = 30° 2x = 30 + 180° x = 210° ÷ 2 x = 105°

∴ 105° and 75° are two supplementary angles whose difference is 30°

(b) Let the angle = x angle Supp. angle = 180 – x comp. angle = 90° – x 2(180 – x) = 7(90 – x) 360° – 2x = 630° – 7x

7x – 2x = 630° – 360° 5x = 270°

x = 270

5°

= 54°

(c) Let the angle = x° compliment = 90° – x x – (90° – x) = 50° x – 90° + x = 50° 2x = 50° + 90°

x = 140

2°

= 70°

6. x = 45° (V.O. Angle) y = a (V.O. A) 45° + 30° + a = 180° a = 105° y = 105° z = 30° (V.O.A)

7. l || m ∠6 = 135° (V.O.A) ∠6 = ∠4 (A.I. Angles) ∠4 = 135° ∠4 = ∠2 (V.O.A) ∠2 = 135°


70°≠60° hence alternate interior angles are not equal so l is not parallel to m.

l

ma

60°

70°

110°

(d) 125°+65°=190°≠180° Since consecutive

interior angles are not supplementary so l and m are not parallel.

∠135° + ∠7 = 180° (Linear pair) ∠7 = 180° – 135° = 45° ∠7 = ∠3 = 135° (Corresponding Angles) ∠3 = 135° ∠3 = ∠1 (V.O.A) ∠1 = 135° ∠5 = ∠1 (Corresponding Angles) ∠5 = 135°

8. (a) ∠a (b) ∠f (c) ∠h (d) ∠c and ∠a (e) ∠d 9. 125° = ∠e (V.O.A)

∠e + ∠c = 180° (Interior angles on the same sides of transversal are supplementary)

∠c = 180° – ∠e = 180° – 125° = 55° ∠c + ∠d = 180° ∠d = 125° 85° + ∠a = 180° ∠a = 95° ∠a = ∠f (V.O.A) ∠95° = ∠f ∠a = ∠b (Corresponding angles)∴ ∠b = 95°

10. ∠95° + a = 180° ∠a = 180° – 95° = 85° ∠a = ∠2 = 85° (Corresponding Angle) ∠2 + ∠3 = 180° (Linear pair) ∠3 = 180° – 85° = 95° ∠4 = 120° (V.O.A) ∠4 = ∠5 = 120° (Corresponding Angle) ∠5 + ∠6 = 180° (Linear pair) 120° + ∠6 = 180° ∠6 = 180° – 120° = 60°

11. (a) a = 67° (V.O.A) 67° + 113° = 180°

∴ l || m (b) a = 102°

102°≠98° ∴ Corresponding

angles are not equal ∴ l is not parallel to m.

(c) a + 110° = 180° (Linear pair) a = 70°


Worksheet 2 1. (a) False (b) False (c) True (d) True (e) True (f) False (g) True (h) True (i) False (j) False (k) True (l) False (m) True 2. ∠QPS = 35° ∠PQU = 35° (A.I. Angles are equal as l || m) ∠QRT = 55° ∠RQU = 55° (A.I. Angles are equal as m || n) ∴ ∠PQR = ∠PQU + ∠RQU = 35° + 55° = 90° 3. (a) Adjacent angles are (∠SOR, ∠ROQ) (∠SOR, ∠ROP) (∠ROQ, ∠QOP) (∠SOQ, ∠QOP) (b) (∠ROQ, ∠POQ) 4. No, because sum of two acute angles cannot be equal to 180° 5. Compliment of an angle = 62° So the angle = 90 – 62 = 28° ∴ its supplement = 180° – 28° = 152°. 6. AB || CD AF || ED ∠AFC = 68° ∠FED = 42° ∠EFA = ∠FED = 42° (A.I. Angles are equal as AF || ED) No ∠AFC + ∠AFE + ∠EFD = 180° (Linear angles) 68° + 42° + ∠EFD = 180° ∠EFD = 180° – 110° = 70° 7. ∠BOC = 49° ∠AOC = 90° – 49° ∠AOC = 41° ∠AOC + ∠AOD = 180° ∠AOD = 180° – 41° = 139° 8. A, O, B are collinear ∠AOD + ∠DOC + ∠COB = 180° (Linear angles) x – 10° + 3x – 25° + (x – 5°) = 180° 5x° – 40° = 180° 5x = 180° + 40° = 220° x = 44° ∴ ∠BOC = x – 5° = 44 – 5° = 39°. 9. a = 132° (Corresponding angels are equal as l || m) b = a (Alternat exterior angle are equal as l || m) = 132° ∴ 2a + b = 2 × 132 + 132 = 396° 10. EF|| GH as 123° + 57° = 180° i.e., consecutive interior angles are supplementary GH is

not parallel to KP because corresponding angle are not equal also AB is not parallel to CD because consecutive interior angles are not supplementary.


Chapter 6

Exercise 6.1 1. Three altitudes and three median can be drawn in a triangle. 2. Yes median of a triangle always lies inside of a triangle. 3. No, altitude of a triangle does not always lie inside the triangle. 4. (a) Acute angle triangle and Right angle triangle (c) Obtuse angle triangle. 5. Exterior angle = 100° One of the interior opposite = 60° Let the other interior opposite angle = x

∴ x + 60° = 100°(Exterior angle prop) x = 40°

6. (a) x = 40° + 70° = 110° (Exterior angle prop) (b) 150° = 90° + x ⇒ x = 150° – 90° = 60° (Exterior angle prop) (c) 135° = 90° + x° (Exterior angle prop)

x = 135° – 90° = 45° (Exterior angle prop) (d) 120° = 60 + x (Exterior angle prop)

120° – 60° = x ⇒ 60° = x(e) 110° = x + 30° (Exterior angle prop.) 110° – 30° = x ⇒ 80° = x(f) 85° = 45° + x (Exterior angle prop.) 85° – 45° = x ⇒ 40° = x

Adjacent interior angle = x + 100° = 180° x = 180° – 100° = 80°

7. (a) Both will be acute (b) One obtuse and other acute (c) Both will be acute 8. Exterior angle = 120° Interior opposite angles are x, 3x

∴ x + 3x = 120° 4x = 120° x = 30° 3x = 90°

So the angles are 30°, 90° and (180 – (30° + 90°)) = 60°

9. (a) ∠PRS = ∠RPQ + ∠PQR (b) ∠TPQ = ∠UQR + ∠QRP (c) ∠UQR = ∠QPR + ∠QRP. 10. (a) x + 40° + 50° = 180° (ASP)

x = 180° – 90° x = 90°

(b) x + 90° + 30° = 180° (ASP) x + 120° = 180° x = 180° – 120° = 60°

(c) x + 40° + 40° = 180° x + 80° = 180° (ASP) x = 100°

(d) x + 90° + 50° = 180° (ASP) x + 140° = 180° x = 180° – 140° = 40°(e) x + x + x = 180° (ASP) 3x = 180°

x = 1803

= 60°

(f) 90 + 3x + x = 180° 90 + 4x = 180° 4x = 180° – 90° x = 90° ÷ 4 x = 22.5°

11. (a) y + 130° = 180° (Linear pair) y = 180° – 130° = 50° x + 60° + y = 180° (ASP) x + 60° + 50° = 180° x + 110° = 180° x = 70°


(b) y = 50° (V.O.A) y + 70° + x = 180° (ASP) 50° + 70° + x = 180° x = 180° – 120° x = 60°(c) 60° + 70° + x = 180° (ASP) 130° + x = 180° x = 50° x y = 180° (Linear pair) 50° + y = 180° y = 180° – 50° = 130°(d) y = 50° (V.0.A) x + 30° + y = 180° (ASP) x + 30° + 50° = 180° x + 30° + 50° = 180° x = 180° – 80° x = 100°(e) y = 75° (V.O.A) x + x + y = 180° (ASP) 2x + 75° = 180° 2x = 105 x = 52.5°(f) x = y ∴ x + x + x = 180° 3x = 180°

x = 180

3 = 60°

x = y = 60° 12. Let the three angles of a triangle be x,

y, and z. According to question x + y = z Also x + y + z = 180° (ASP) z + z = 180°

⇒ 2z = 180° z = 90° Yes

13. The three angles of a triangle are 80°, x° and 3x° 80° + x° + 3x° = 180° (ASP) 80 + 4x = 180° 4x = 100° x = 25°∴ The angles are 80°, 25° and 75°.

14. Let the three angles of a triangle are be x°, 2x and 3x

∴ x + 2x + 3x = 180° x = 180° x = 30°

∴ the angles are 30°, 60°, and 90° 15. Let the third angle be x° then the other

two angles are 2x°, 2x°.∴ x° + 2x° + 2x° = 180° (ASP) 5x = 180° x = 180 ÷ 5 = 36°

∴ the angles are 36°, 72°, 72° 16. In ∆ABC

∠A + ∠B + ∠C = 180° (ASP) 40° + x + 60° = 180° x = 180° – 100° = 800°∴ x = y⇒ y = 80°

(Corresponding Angles) z = 60°

(Corresponding Angles)

Exercise 6.2 1. (a) 70° + x + x = 180° (ASP)

2x = 180° – 70°

2x = 110°

x = 55°

(b) 70° + 70° + x = 180° (ASP) 140° + x = 180° x = 180° – 140° = 40° x + y = 180° (Linear pair) 40° + y = 180°


y = 180° – 40° = 140°

70°

x

y

(c) y = 50° (Angles opposite equal side are equal)

50 + y = x (Ex. A prop) 50 + 50° = x = 100° = x(d) y = 72° (Angles opposite

equal sides) x = 72° + y (Ex. Angles prop) y = 72° + 72° x = 144°

72°

y

x

(e) y + 110° + 180° (Linear pair) y = 180° – 110° = 70° x + y = 110° x + 70° = 110° x = 40°

(f) 40° + x + 40° = 180° (ASP) x = 180° – 80° = 100°

x

40°

(g) a = 70° (VOA) a+ x+ x = 180° 70° + x + x = 180°

70°

a

x xy

2x = 180° – 70° 2x = 110° x = 55° x + y = 180° (Linear pair) 55°+ y = 180° y = 180° – 55° = 125°(h) y + 130° = 180° (Linear pair) y = 50° x + y + y = 180° x + 50° + 50° = 180° x = 180° – 100° = 80°

130°y

x

y

(i) x + x + 90° = 180° (Linear pair) 2x = 180° – 90° 2x = 90° x = 45° y = 90° + x = 90° + 45° = 135°

2. (a) ∆ABC AB = AC ∠ACB = ∠ABC (b) ∆PQR QR = PR ∠QPR = ∠PQR (c) ∆XYZ YZ = XZ ∠X = ∠Y (d) ∆UVW UV = VW ∠W = ∠U 3. (a) Two smaller sides are 3 and 5 ∴ 3 + 5 = 8 > 5 (third side) So it is possible to have a triangle with given sides


(b) 5 cm 6 cm 7 cm Two smaller sides are 5 cm and 6 cm Their sum = (5 + 6) cm = 11 cm Third side = 7 cm We have 11 cm > 7 cm hence the sum of two sides is greater than third side, so it is possible (c) 6 cm 3 cm 2 cm Two smaller sides are 3 cm and 2 cm Their sum = 3 cm + 2 cm = 5 cm Also 5 cm < 6 cm So it is not possible to have triangle with the given sides (d) 3 cm 7 cm 4 cm Two smaller sides are 3 cm and 4 cm Their sum = 3 cm + 4 cm = 7 cm Third side = 7 cm ∴ Sum of two smaller sides = third side ∴ It is not possible to have a triangle with the given sides. Note:Tocheckifthegivensidescanmakeatriangle,itissufficienttocheckifthe

sum of two smaller sides is greater than the third side or not. If it is greater than the third side then only it is possible to have a triangle otherwise it is not possible.

4. Ifthelengthoftwosidesofatriangleareknownthenwecanfindtwonumbersbetweenwhich the third side may lie. Here given length are 11 cm and 14 cm. So the third side C will lie between 14 – 11 and 14 + 11 cm i.e. 3 cm and 25 cm.

5. (a) In DOAB we have OA + OB > AB (b) In DOBC we have OB + OC > BC (c) In DAOC we have OA + OC > AC Because sum of any two sides is greater then third side.

6. In ∆PQM, PQ + OM > PM. (Sum of two sides greater than third)...(i)

In DPRM We have PM + MR > PM ...(ii) Adding both sides of (i) and (ii) we get PO + QM + PM + MR > 2 PM PO + OM + MR + PM > 2 PM PQ + QR + PM > 2 PM Yes. 7. (a) In ∆PQR PQ + QR > PR ...(i) In ∆PSR PS + SR > PR ...(ii) Similarly In ∆RSQ SR + QR > SQ ...(iii) In ∆PSQ PS + PQ > SQ ...(iv) Adding both the sides of (i), (ii), (iii) and (iv) we get 2 (PQ + QR + PS + SR) > 2PR + 2SQ Or PQ + QR + PS + SR > PR + SQ.

A

B C

O

P

Q M R

S R

P Q


(b) In ∆PSO PO + SO > PS ...(i) In ∆POQ PO + OQ > PQ ...(ii) In ∆ROQ RO + OQ > RQ ...(iii) In ∆SOR SO + OR > SR ...(iv) Adding both the sides of (i), (ii), (iii), and (iv) we get PO + SO + PO + OQ + RO + OQ + SO + OR > PQ + PS + RQ + RS Or 2(PO + OR) + 2(SO + OQ) > PQ + PS + RQ + RS ⇒ 2PR + 2SQ > PQ + PS + RQ + PS or PQ + PS + RQ + RS. 8. OA + OB > AB OA + OC > AC OB + OC > BC ∴ 2 (OA + OB + OC) > AB + AC + BC

OA + OB + OC > 12

(AB + AC + BC)

9. (a) Given triangle is right angle triangle ∴ x2 = (42 + 32)cm2 = (16 + 9)cm2

x2 = 25 cm2

x = ± 5 cm ∴ Since it is length of side of a ∆ hence it is positive ∴ x = 5 cm(b) 82 + 62 = x2 (Right angle triangle prop or Pythagoras property) (64 + 36)cm2 = x2

±10 cm = x 10 cm = x (Length is always positive)(c) 52 + 122 = x2 (Right angle triangle prop. or Pythagorean property) ∴ 25 + 144 = x2

169 = x2

±13 = x 13 = x (Since it is length so it can not be negative)

S R

P Q

O

A

B C

O

10. PQ2 +PR2 = RQ2

x2 + 102 = 262 (Pythagorean) x2 = 262 –102

= (26 – 10) (26 + 10) x2 = 16 × 36 x = 4 × 6

x = 24

10 cmP Q

R

26

0

11. AB2 + AC2 = BC2

242 + 72 = BC2

576 + 49 = BC2

625 = BC2

25 × 25 = BC2

25 = BC

7 cmA B

C

24 cm


12. AC2 = AB2 + BC2

172 = 152 = x2

172 – 152 = x2

289 – 225 = x2

64 = x2

8 = x

17

A

B C

15

x

13. Height of Tree = AB + BC ABC is a right angle ∆

∴ AB2 + AC2 = BC2

102 = 242 = BC2

100 + 576 = BC2

676 = BC2

26 = BC

A C

x

B

24 m

10 m

∴ The original height of tree

= AB + BC

= 10 m + 26 m = 36 m.

14. Let A be the position of window and AB is the ladder.

A

B C

x17 m

8 m

∴ AB2 = AC2 + BC2

172 = x2 + 82

172 – 82 = x2

289 – 64 = x2

225 = x2

15 = x∴ Height of the window is 15 m above the ground.

15. Let A be the position of the window AB be the ladder then. ABC is a right angle triangle

∴ AB2 = AC2 + BC2

= 122 + 352

AB2 = 1369 37 = AB

∴ Length of the Ladder is 37 m.A

B C

35 m

12 m

16. In ∆ADB BD2 = AD2 + AB2

172 = AD2 + 152

172 – 152 = AD2

289 – 225 = AD2

64 = AD2

8 m = ADA B

CD

17 m

15 m

∴ Perimeter of rectangle ABCD = 2(AD + AB) = 2 (15 + 8) = 2 × 23 = 46 m.

17. ABCD is a rhombus. AC and BD are its diagonals AC and BD are perpendicular and

bisect each other.


A

B

C

DO

AO = 12

AC = 12

× 24 = 12 cm.

BO = 12

BD = 12

10 cm = 5 cm.

∆AOD ∠AOD = 90°∴ (AO)2 + (OD)2 = AD2

(12)2 + (5)2 = AD2

144 + 25 = AD2

169 = AD2 = 13 cm = AD 18. ∠P = 40° ∠Q = 50° ∠R = ?

∠P + ∠Q + ∠R = 180°∴ 40° + 50° + ∠R = 180° ∠R = 180° – 90° ∠R = 90°∴ PQ2 = PR2 + RQ2

C is correct.

19. Rohini walks from A to B and then from B to C. Her distance from initial position is AC.∴ AB2 + BC2 = AC2

92 + 122 = AC2

81 + 144 = AC2

225 = AC2

15 m = AC2

∴ She is 15 m away from her initial position.

20. AD = 23 m BC = 28 m

A E

B

CD

28 m23 m

12 m

28 – 23 = 5 m

ABE is a right angle ∆ ∴ AB2 = AE2 + BE2

AB2 = 122 + 52

AB2 = 144 + 25 AB2 = 169 AB = 13 cm

9 mA B

C

12 m

Worksheet 1. (a) Sum of two smaller sides = 12 + 5 = 17 cm < 18 cm. So these can not be the sides of

a triangle. (b) Sum ...... = 10 + 10 = 20 > 17 cm Yes these can be the sides of a triangle. (c) Sum...... = 12 + 13 = 25 cm = 25 ∴ These can not be the sides of a triangle. 2. (a) 40° + 60° + 80° = 180°. Yes these can be the three angles of a triangle. (b)80°+100°+10°=190°≠180°.Sothesecannotbetheanglesofatriangle. (c)100°+40°+60°=200°≠180°.Sothesecannotbetheanglesofatriangle. (d) 50° + 60° + 70° = 180°. Yes these can be the three angles of a triangle. 3. (a) False (b) False (c) True (d) False (e) False (f) False (g) True


4. (a) 102 = 100 62 + 82 = 36 + 64 = 100 ∴ 102 = 62 + 82

Yes these are sides of a right angle triangle.

(b) 7, 8, 15. 152 = 225 72 = 49 82 = 64 72 + 82 = 49 + 64 = 103. 152 = 225

∴ 152≠72 + 82

∴ 7, 8, 15 can not be the sides of a right angle triangle.

(c) 40, 41, 9 412 = 1681 402 = 1600 92 = 81

∴ 402 + 92 = 412

Yes these can be the sides of a triangle.

5. (a) x + 80° = 135° (Ex. Angle prop) x = 135° – 80° x = 55°

135°80°

x

y

y + 135° = 180° (Linear angles) y = 180° – 135° = 45°(b) 5x + 3x = 160° 8x = 160° x = 20°Also 160° + y = 180° y = 180° –160° = 20°(c) 70° + x = 180° (Linear pair) x = 180° – 70° x = 110° x + 30° + 70° = 180° (ASP)

x + 100° = 180° x = 180° – 100° x = 80°

x70°y

30°

(d) 100° + 30° + x = 180° (ASP) x = 180° – 130° = 50° x = 50°(e) x = 70° (Angles opposite

equal sides are equal) x = 70° (Angles opposite

equal sides are equal.) ∴ x + 70° + y = 180° (ASP) 70° +70° + y = 180° y = 180° –140° y = 40°

70°

x y

6. In a right angle triangle = 90° One of the two angles is 60°. Let the

other triangle is x. x + 60° + 90° = 180° x = 180° – 150° = 30°

7. The lengths of the third side will lie between (18 – 12) cm and (18 + 12) cm ie 6 cm and 30 cm.

8. Let AB be the length of the hypotenuse∴ 172 = BC2 + AC2

172 = (8)2 + AC2

172 – 82 = AC2

289 – 64 = AC2

225 = AC2

15 = AC


A

B C

17 m

8 m

9. ABC is an isosceles triangle AB = AC ∠C = ∠B⇒ 55° = x∴ x + y + 55° = 180° (ASP) x + 55° + 55° = 180° x = 180° –110° = 70°

A

B C55° x

y

10. (a) ∠ABC + ∠EBC = 180° (Linear pair)

∠ABC + 120° = 180° ∠ABC = 60° ∠ACD = ∠ABC + ∠BAC

(Ex. Angles prop) 136° = 60° + ∠BAC 136° – 60° = ∠BAC 76° = ∠BAC

(b) 120° + ∠ABC = 180° ∠ABC = 180° –120° = 60° ∠ACB + 135° = 180° ∠ACB = 180° – 135° = 45° y + ∠ABC + ∠ACB = 180° (ASP) y + 45° + 60° = 180° y = 180° – 105° y = 75°

11. In ∆ABM AB + BM > AM ...(1)

A

B C

In AMC AC + MC > AM ...(2) Adding (1) and (2) then we get AB + BM + AC + MC > 2AM AB + AC + (BM + MC) > 2AM AB + AC + BC > 2AM 12. The lengths of two sides of a triangle

are given than the third side will lie between the sum and difference of the two given sides, i.e., third side will lie between 18 – 12 and 18 + 12 i.e. 6 and 30 cm.


1. Cong. ∆ Corr. Cong. Angle Corr. Cong. Angle(a) ∆ABC ≅∆PQR ∠A = ∠P

∠B = ∠Q∠C = ∠R

AB = PQBC = QRCA = RP

(b) ∆BAC ≅∆RQP ∠A = ∠Q∠B = ∠R∠C = ∠P

BA = RQAC = QPBC = RP


(c) ∆CAB ≅∆QPR ∠A = ∠P∠B = ∠R∠C = ∠Q

AB = PRBC = RQCA = QP

2. (a) PQ (b) XY 3. (a) ∠C (b) ∠K (c) ∠E (d) ∠E

4. Symbolic form Condition of congruence (a) ∆ABC ≅ ∆PQR SSS (b) ∆MNP ≅ ∆RST SAS (c) ∆PQR ≅ ∆XZY ASA (d) ∆RST ≅ ∆XYZ RHS (e)Notcongruentasgivenconditionsarenotsufficient. 5. If ∆ABC ≅∆RPQ by RHS cong. conditions and we are given ∠B = ∠R = 90° and AB = RP

then their hypotenuses i.e. AC and RQ should also be equal i.e, AC = RQ 6. If ∆PQR = ∆FED by SAS and PQ = FE and RP = DF then angle included between the

equal sides should be equal. ∴ ∠P = ∠F. 7. If ∠X = ∠R, ∠Y, ∠P and ∠XYZ ≅ ∆RPS by ASA, then side included i.e. XY = RP. 8. (a) ∆ABC ≅∆QPR SSS (b) ∆XYZ ≅∆ABC SAS (c) ∆XYZ ≅∆RQP ASA (d) ∆BOY ≅∆GRL RHS 9. Symbolic form Condition (a) ∆ABC ≅ ∆ZXY SSS congruence (b) ∆XYZ ≅ ∆QPR SAS congruence (c) ∆KML ≅ ∆STR ASA (use ASP of triangels) (d) ∆UVW ≅ ∆SUM RHS (e) ∆LMK ≅ ∆XZY ASA 10. (a) Yes,

AB = AC (Given) AD = AD (Common) BD = CD (Given)

∴ by SSS Congruence condition. (b) Yes by cpct (i.e., all corresponding parts of congruent triangle are equal.)

Exercise 7.2 1. (a) QP = QR (Given) PM = RM (M is the mid point of PR) QM = QM (Common) (b) Yes (c) Yes, CPCT 2. (a) Yes (b) AB = DC (Opposite side of rect. are equal) BC = DA (Opposite side of rect. are equal) ∠ABC = ∠CDA (Each angle of a rect. is right angle) (c) Yes, by CPCT


3. (a) Yes (b) XO = YO (XY is bisected by PQ at O.) PO = QO (As PQ is bisected by XY at O.) ∠XOP = ∠YOQ (V.O Angles are equal) (c) Yes, PX = QY (by CPCT) 4. (a) XU = XU (Common) ∠YXU = ∠ZXU (XU bisects ∠YXZ) YU = ZU (Given) (b) Yes, SAS congruence condition as three matching parts are given in (a). 5. (a) ∠QPR = ∠SRP PR = PR (Common) PQ = RS (Opp. sides of a || gm are equal) (b) Yes (SAS done in a) and by SSS congruence condition we have PR = PR (Common) PQ = RS (Opp. of a || gm given are equal) PS = RQ (Opp. of a || gm given are equal) 6. (a) ∠QPR = ∠SPR ( PR bisects ∠SPQ) ∠QRP = ∠SRP ( PR bisects ∠QRS) PR = PR (Common) (b) ∠PQR = ∠PSR (by CPCT) (c) PQ = PS (by CPCT) (d) QR = SR (by CPCT) 7. (a) ∆ABC ≅∆ADC, Yes (b) ∠ABC = ∠ADC = 90° (Given) AB = AD (Given) AC = AC (Common) ∴ RHS congruence condition (c) Yes, BC = DC (By CPCT) 8. (a) BD = CE (Given) ∠BDC = ∠CEB = 90° (Given) CB = CB (Common) (b) Yes, ∆CBD ≅∆BCE ∴ by RHS congruence condition (c) ∠DCB = ∠EBC (By CPCT) 9. (a) PQ = PR (Given) ∠PMQ = ∠PMR (90°) PM = PM (Common) (b) Yes, ∆PQM = ∆PRM (by RHS congruence condition) (c) Yes, ∠PQR = ∠PRQ (By CPCT) (d) Yes, QM = RM (By CPCT)


10. (a) PM = OM (Reason CPCT) (b) ∠PMA = ∠QMA (Given) (c) AM = AM (Common) (d) ∆AMP ≅∆AMQ (By SAS congruence condition) (e) ∠PAM = ∠QAM (By CPCT)

Worksheet I 1. (a) 110° (b) 70° (c) SAS (d) ASA (e)BC;∠A 2. (a) Yes (b) ∠ADC = ∠CBA = 90° (Given) CD = AB (Given) AC = CA (Common) (c) AB = CD (by CPCT) 3. (a) Yes (b) AO = BO as CD bisects AB. CO = DO as AB bisects CD. ∠AOC = ∠BOD (Vert. opp. angles) ∴ By SAS congruence condition is applicable to prove (a). (c) Yes, AC = BD (by CPCT) 4. (a) Yes, ∆QRS ≅∆RQT PQ = PR Given ...(i) ∴ ∠PQR = ∠PRQ (Angles opposite equal sides are equal) ...(ii) PS = PT (Given) ...(iii) Subtracting (iii) from (i) we get PQ – PS = PR –PT QS = TR ...(iv) and QR = RQ (Common)...(v) ∴ by (ii), (iv) and (v) Using SAS we can say that ∆QRS ≅ ∆RQT (b) Yes QT = RS (by CPCT) 5. (a)Yes;∆RPO ≅ ∆SQO by AAS (b) PR = QS (Given) ∠POR = ∠QOS (V.O.A.) ∠PRO = ∠QSO = 90° (c) Yes ∠RPO = ∠SQO by CPCT and RO = SO (by CPCT) 6. (a) PQ = DR (b) Yes by CPCT 7. (a) Yes by ASA cong. condition


(b) ∠BAD = ∠CAD (AD bisects ∠BAC) AD = AD (Common) ∠ADB = ∠ADC = 90° (c) Yes as AB = AC (by CPCT) 8. Yes ∆ADB ≅ ∆ADC In ∆ABD ∠BAD + 50° + 90° = 180° ∠BAD = 40° ∴ ∠BAD = ∠CAD = 40° ∠ADB = ∠ADC = 90° AD = AD (Common) ∴ by ASA ∆ADB ≅ ∆ADC. 9. AB = AD (Given) ∠BAC = ∠DAC (Given) AC = AC (Common) ∴ ∆ABC ≅ ∆ADC (by SAS cong. condition) 10. In a Paralelogram ABCD we join AC AB = CD (Opposite sides of a ||gm are equal) AD = CB (Opposite sides of a ||gm are equal) AC = AC (Common) ∴ By SSS ∆ABC ≅ ∆ADC.


1. (a) 15075

21

= = 2 : 1

(b) 36 362 60

36120

620

310

min minmin5 hr

=×

= = = = 3 : 10

(c) 400 m2 km

=×

= =4 002 10 00

210

15

= 1 : 5

(d) 300 mL1 Litre

= =3001000

310

= 3 : 10 (e) 10 cm4 m

10 cm400 cm

= = 140

= 1 : 40

(f) 750 gm2.50 kg

75 0250 0

= =3

10

310

= 3 : 10 (g) 3 years 7 months

= × =3 127

367

= 36 : 7

2. (a) Riya marks

Shiya marks= =65

951319

= 13 : 19

(b) (i) No. of girls No. of boys

= =6040

32

= 3 : 2


(ii) No. of girls Total Students

= =60100

35

= 3 : 5

(iii) No. of boys No. of girls

= =4060

23

= 2 : 3

(c) (i) Income Saving

= =180 00065 000

3613

, = 36 : 13

(ii) Expenditure

Saving= =115000

1800002336

= 23 : 36

(iii) 65000 115000

= 1323

= 13 : 23

3. BooksStudents

= =124

31

Number of the books is 3 times of no. of students. ∴ For 400 students books required = 400 × 3 = 1200.

Also Books

Students= =12

4 400x

Comparing we get x = 1200. 4. 2 cm = 100 km 12 cm = x km.

2100

= 12x

⇒ 2 × x = 12 × 100

x =6 12 100

2× = 600 km

∴ x = 600 km.

5. Speed of Cheetah = Distance

time = 93.9 m

3 sec= 31.3 m/s

Speed of Horse = 42.6 m2 sec

= 21.3 m/sec.

Speed of Cat = 43.6 m4 sec

= 10.9 m/s

Fastest = Cheetah

Minimum Speed = Cat

6. (a) Population of City A = 940 Lakh

Area of City A = 400 km2

Population of City A per km2 = 235000 940,00,0 00

4 00 = 2,35,000 people/km2


Population of City B per km2 = 180000 1080,00,0 00

6 00 = 1,80,000 people/km2

(b) City A is thickly populated.

7. (a) 1 : 2 = 12

= 24

3 : 4 = 34

24≠

34

No

(b) 25

512

LCM of (5, 12) = 60

25

2460

= , 512

2560

=

2460

2560

¹ ∴ 25

512

¹ or2:5≠5:12

(c) 3 : 4 and 12 : 16.

34

, 1216

.

Lowest form of 1216

is 34

∴ 3 : 4 and 12 : 16 are equal. (d) 4 : 5 and 3 : 7 both are in lowest form and are not equal.

So 4 : 5 is not equal to 3 : 7. 8. Cost of 5 registers = ` 300 Cost of 1 register = ` 300 ÷ 5 = 60

Cost of 12 registers = ` 60 × 12 = 720 9. Number of words in 27 sec = 15 words.

Number of words in 1 sec = 1527

Number of words in 3 minutes (3 × 60 Sec) = 1527

3 605

9

20

3

× × = 100 words.

10. Sarita’s earning for 10 days = ` 8000

Sarita’s earning for day = 800010

= ` 800

Sarita’s earning for Oct (31 day) = ` 800 × 31 = ` 24800

11. Ratio of Rohit and Mohits 5 : 6 Rohit’s age = 25 years


Mohit’s age = x years

56

25=x

5 56 5

2530

××

=

∴ Mohit’s age = 30 years.

12. lb

x= ××

=5 2 33 2 3 6 9

.

. . (

6 93.

= 2.3)

= 11 56 9

..

So length = 11.5 m. 13. Distance covered in 5 hrs = 3000 km Distance covered in 1 hr = 3000 ÷ 5 km = 600 km Distance covered in 8 hrs = 600 × 8 km = 4800 14. In 2015 ratio of wins: Loss = 10 : 3 In 2016 ratio of wins = 6 : 2

Let us compare 103

and 62

i.e., 10 23 2

××

and 6 32 3

××

So 206

and 186

i.e., 206

> 186

∴ Performance in 2015 was better.

Exercise 8.2 1. Fraction Percentage

(a) 10100

110

= 10100

× 100 = 10%

(b) 25100

14

= 25100

× 100 = 25%

(c) 12100

325

= 12100

× 100 = 12%

(d) 17100

17100

= 17100

× 100 = 17%

(e) 22100

1150

= 22100

× 100 = 22%

(f) 14100

750

= 14100

×100 = 14%


2. Fraction Percentage Decimal

20100

15

= 20100

× 100 = 20% = 0.2

15

1003

20= 15

100× 100 = 15% = 0.15

20100

15

= 20100

× 100 = 20% = 0.20

12100

325

= 12

100100× = 12% = 0.12

18100

950

= 18100

100× = 18% = 0.18

15

1003

20= 15% = 0.15

3. Fraction Fraction with denominator 100 Percentage

(a) 1550

310

= 15 250 2

30100

××

= 30%

(b) 650

325

= 6 250 2

12100

××

= 12%

(c) 750

750

= 7 250 2

14100

××

= 14%

(d) 1050

15

= 10 250 2

20100

××

= 20%

(e) 1250

625

= 12 250 2

24100

××

= 24%

4. (a) 30% = 30

1003

10= (b) 25% =

25100

14

=

(c) 7.5% = 7 5100

751000

. = = 3

40 (d) 8

15

% =415

% =41

500

(e) 312

% =72

% =7

200

5. (a) 18

× 100 = 252

= 12.5% or 1212

% (b) 35

10020

× = 3 × 20 = 60%

(c) 820

× 100 = 8 × 5 = 40% (d) 165

× 100 = 16 × 20 = 320%

(e) 114

× 100 = 11 × 25 = 275%


6. (a) Sheena’s % = 2540

10 0125

262 5

2

5× = =% . %

(b) % of Boys = 4860

100 808

× = %

(c) % of spent money = 25075 0

100100

333

13

10

3

× = = %

(d) % of defective TV = 2406 00

100 4040

× = %

(e) % of votes polled = 40

3

212015 0

10 0 80× = %

7. (a) 0.16 = 16

100 = 16% (b) 0.025 =

251000

2 5100

= . = 2.5%

(c) 9.2 = 96210

920100

= = 920% (d) 0.08 = 8

100 = 8%

(e) 7.5 = 7510

750100

= = 750%

8. (a) 25% = 25

100= 0.25 (b) 48% = 48

100= 0.48

(c) 150% = 150100

= 1.5 (d) 6

10% =

610 100×

× 6

1000= 0.006

(e) 0.45% = 45

100 =

4510000

= 0.0045

9. (a) 0.42 = 42% (b) 1.25 = 125% (c) 30.08 = 3008% (d) 0.0625 = 6.25% (e) 31.24 = 3124%

10. (a) 35

3 205 20

60100

60= ××

= = % (b) 920

920

100 9 5 45= × = × = %

(c) 114

114

100 27525

= × = % (d) 58

58

100 5 12 5 62 5= × = × =. % . %

(e) 2925

2925

100 1164

= × = %

11. (a) Total parts = 8 Shaded parts = 2

% of shaded parts = 28

= 28

× 100 = 25%

(b) Total parts = 6

Shaded parts = 2


10013

100 3313

× = × = %


(c) Total parts = 5

Shaded parts = 2


10020

× = 40%

(d) Total parts = 6 Shaded parts = 3

% of shaded portion = 36

100× = 50%

(e) Total parts = 1

Shaded part = 14

18

18

+ + = 2 1 18

48

12

+ + = =

% of shaded part = 12

100× = 50%

(f) Total parts = 8 Shaded = 6

% of shaded part = 68

1003

2

25× = 75 %

12. (a) 40% of 150 = 40

100150× = 60

(b) 15% of 600 = 15

100600 90× = `

(c) 20% of an hour = 20% of 60 minutes = 20

10060× = 12 minute

(d) 20% of 5 kg = 20

1005

5

× = 1 kg

(e) 70% of 5 minutes = 70% of 5 × 60 sec = 70

1005 60× × = 3 minute 30 sec.

(f) 40% of 500 km = 40

100500× = 200 km.

13. (a) Let the whole quantity be x.

5% of x = 300

5

100× x = 300

x = 300 100

5

60

×

x = 6000

(b) Let the whole quantity be x

then 12% of x = 960

x × 12

100 = 960

x = 960 100

12

80

× = 800

x = ` 8000


(c) 50% of x = 700

50

100× x = 700

x = 700 100

50

2

× = 1400

(d) 70% of x = 2100

70

100× x = 2100

x = 2100 10070

30

×

x = 3000 kg

(e) 40% of x = 2 months

40

100of x = 2 months

x = 2 100

40

5

2

×

x = 5 months

(f) 25% of x = 80 litres

25

100× x = 80

x = 80 100

25

4

×

= 320 litres

14. (a) 200500

100

40

× = 40%

∴ ` 200 is 40% of 500

(b) 4860

100

8

× = 80%

48 is 80% of 60

(c) 2575

100100

33

4

× = = 3313

%

25 kg is 3313

% of 75 kg.

(d) 50

200100

5

× = 25%

∴ 50 cm is 25% of 2900 cm.

15. Daughter’s share in % = (100 – 35 – 40)% = 25%

Let the salary of a man = ` x

25% x = ` 3000

25

100× x = ` 3000

x = 3000 100

25

4

× = ` 12000

16. % of children = (100 – 35 – 30) = 35% No of children = 7000 ∴ Let the total population = x ∴ 35% of x = 7000

35

100× x = 7000

x = 7000 10035

200

× x = 20,000

Yes the number of male and children is same as both are 35%.


17. Total votes = 6,00,000

% of votes who voted = 42%

% of votes who did not vote = (100 – 42)% = 58%

∴ Number of people who did not vote = 58

1006 00 000× , , = 3,48,000

18. Saving of Raman = 40,000 % of saving = 20% Let the salary be x. 20% of x = 40,000

20

100× x = 40,000

x = 40 000 100

20

5

, × = ` 2,00,000

Roman’s solving is 2 lakh.

19. No of matches played = 25

% of matches won = 40%

No of matches won = 40% of 25 = 40100

25

4

× = 10

∴ No. of matches last = Total matches played – Total matches won = 25 – 10 = 15 20. % of copper = 60%

% of zinc = (100 – 60)% = 40%

Weight of zinc = 40% of 5 kg = 40

1005

2

× kg = 2 kg

Weight of copper = Total weight – Weight of zinc = 5 kg – 2 kg = 3 kg 21. Total students = 800 % who could not clear = 20% % who could clear the exam = (100 – 20)% = 80%

Number of students who cleared the exam = 80% of 800 = 80

100800× = 640

22. Total distance covered = 75 km + 225 km = 300 km

% of joining covered by train = Distance travelled by train

Total distance× 100

= 225300

100

75

× = 75%


23. % of Rotten fruits = 10% % of fruits reserved for guest = 30% Remaining % = (100 – 10 – 30)% = 60% ∴ % of fruits eaten by family members = 60% Number of fruits eaten by family = 30 Let the total number of fruits = x ∴ 60% of x = 30

60

100× x = 30 ⇒ x =

30 10060

50

2

× = 50

Total number of fruits in the basket was 50.

Exercise 8.3 1. Total parts required = 2 + 3

% of sugar = 25

10020

× = 40%

%offlour=35

100× = 60%

2. Total parts = 2 + 3 + 5 = 10

Sohan’s share = 2

10 of ` 600 =

210

600× = ` 120

Mohan’s share = 3

10600× ` = ` 180

Rohan’s share = 510

600× ` = ` 300

% of Sohan = 120600

100

20

× = 20%

% of Mohan = 180600

100× = 30%

% of Rohan = 300600

100× = 50%

3. Total sweets = 400

Sheena’s share = 30% of 400 = 30

100400× = 120

Meena’s share = 400 – 120 = 280 Ratio = 12 : 28 = 3 : 7 4. Let the three angles be in the ratio 1 : 3 : 5 Sum of the three angles = 180°

Ist angle is = 19

180 20× = °


IInd angle is = 39

180× ° = 60°

IIIrd angle is = 59

180× = 100°

5. (a) Shree secured 44% votes Total votes Shree got = 11484 Let the total votes be = x ∴ 44% of x = 11484

44

100× x = 11484

x = 11484 100

44×

= 26100

(b) No of votes in favour of Aman = 26% of 26100

= 26

10026100× = 6786

(c) Number of votes who did not vote = 26100 – (11484 + 6786) = 7830

6. (a) Increase % = 100024000

100× = 416

%

(b) Original price of house = 1500000 Decrease in price = 1500000 – 1200000 = 300000

Decrease % = 3000001500000

1005

20× = 20%

(c) Number of games won this year = 8 Number of games won last year = 10 Decrease in won games = 10–8 = 2

Decrease % = 2

10100× = 20%

(d) Original price = 650000 New price = 7,15,000 Change in price = 65000

Increase% = 65 000

65 0000100× = 10%

(e) Original price = ` 50/litre New price = ` 65/litre Change in price = ` 65/litre – ` 50/litre = ` 15/litre

Increase % = 1550

1002

× = 30%

(f ) Original fees = ` 1500 Increased fees = ` 1800


Increase in fee = ` 1800 – ` 1500 = ` 300

Increase % = IncreaseOriginal

× 100 = 300

1500100

5

20× = 20%

7. C.P. of steam iron = ` 1300 Profit%=15% S.P.=C.P.+Profit

Profit=15%of1300=15

1001300× = ` 195

S.P. = (1300 + 195) = ` 1495

8. C.P. of car = ` 5,00,000

S.P. of car = ` 450000

CP > SP, there is loss

loss = CP – SP = ` 5,00,000 – ` 4,50,000 = ` 50,000

loss% = 50 000

5 00 000100

,, ,

× = 10%

9. C.P. of TV = ` 10,000

Profit%=25%

Profit=25%of` 10,000

= 25

10010 000× , = 2500

S.P.=C.P.+Profit=` (10,000 + 2500) = ` 12500. 10. S.P. of item = ` 25500 Profit=50%

C.P. = 100150

25500

2

3

8500

× = ` 17,000

11. S.P. = ` 540 Loss % = 50

C.P. = 100

100 50540

−× =

10050

540× = ` 1080

12. C.P. of watch = ` 675 Cost of chain = ` 75 Actual cost price of watch = 750 S.P. = ` 850 S.P. > C.P. Thereisprofit Profit=SP–CP=` (850 – 750) = ` 100


Gain % = 100750

100

4

3

× = 403

1313

= %

13. C.P. = ` 36600 Loss = 8%

S.P. = 100

100− ×loss%

C.P. = 100 8

10036600

− ×

= 92

10036600× = ` 33672

14. SP of Sooty = ` 14495

loss = 35%

C.P. = 100

100 −×

loss%SP =

100100 −

×35

14495 = 10065

14495

20

13

1115×

= 20 × 1115 = ` 22300 15. S.P. = 1050

Profit=12%

C.P. = 100

100 +×

ProfitS.P.

% =

100112

1050× = ` 937.50

16. S.P = ` 73920

Loss = 4%

C.P. = 100

100 −×

loss%SP

= 100

100 −×

4SP =

10096

× 73920 = ` 77000

17. (a) Calcium = 9 parts Carbon = 4 parts Oxygen = 12 parts Total parts = 25 parts

% of carbon = 425

100 164

× = %

% of calcium = 925

100 364

× = %

% of oxygen = 1225

100 484

× = %

(b) Weight of carbon is 50 gm alloy is 16% of 50 gm = 16

10050

8

2

× gm = 8 gms

Weight of calcium in 50 gm = 36

10050

2

× = 18 gms


Weight of oxygen in 50 gm = 48

10050

2

× gms = 24 gms.

(c) Let the weight of alloy = x gms Weight of calcium in alloy = 45 gm 36% of x = 45

36

100× x = 45

x = 45 100

36

5 25

4

× = 125 gms.

18. (a) The ratio of students liking maths, science and social science is 5 : 3 : 2 Total parts = 5 + 3 + 2 = 10

Students who like maths is 5

10 of 70 = 35

(b) If the Number of students liking social science is 18 then let the total number of students be x.

∴ 2

10 of x = 18

x = 18 10

2×

= 90

19. (a) S.I. = P × ×r t

100 =

1200 10 3100× ×

= 360

A = P + S.I. = 1200 + 360 = ` 1560

(b) S.I. = P × ×r t

100 =

750 5 2100× ×

= ` 75

A = S.I. + P = 750 + 75 = ` 825

(c) S.I. = P × ×r t

100 =

2700 9 3100

× × = 729

A = P + S.I. = 2700 + 729 = ` 3429

(d) S.I. = P × ×r t

100

40 = 200 4

100× ×r

408

= r = 5%

r = S.I 100

P t= ×

×

20540 100

200 4 = 5%

A = P + S.I. = 200 + 40 = ` 240


(e) r = S.I.

P××

100t

= 1275 1008500 3

425

××

= 5%

A = 8500 + 1275 = ` 9775

(f) r = S.I.

P××

100t

= 1092 100

4550 2×

× = 12%

A = P + S.I. = ` (4550 + 1092) = ` 5642

(g) P = S.I. ×

×= ×

×100 300 100

5 3

10020

r t = ` 2000

A = P + S.I. = ` (2000 + 300) = ` 2300 (h) A = ` 5200 S.I. = ` 1200 P = A – S.I. = 5200 – 1200 = ` 4000

T = S.I.

P××

= ××

100 12 00 1004000 6

3 5

2r = 5 years

20. Let the principal = ` x. S.I. = ` x Time = 8 years.

S.I. = P × ×r t

100

x = x r× × 8

100

100

8

12 5.

= r

12.5% = r

Worksheet 1

1. (a) 75% = 75

10034

= (b) 83% = 83

100 = 0.83 (c)

35

100× = 60%

(d) 3.75 = 3 75100

375100

. = = 375% (e) 3

4

251216

100× = 75% (f) 2 : 5 = 25

100× = 40%

2. (a) 70% of ` 550 = 70

100550× = ` 385

(b) 30% of 5 kg = 30100

52

× = 1.5 kg

(c) 25% of 150 km = 25

100150

4

× = 37.5 km


(d) 20% of 275 litres = 20100

275

5

× = 55 litres

(e) Let the number be x 30% of x = 900

30

100× x = 900

x = 900 100

30

30

×

x = 3000 (f) Let the total no. of students = x Boys = 25% No of boys = 50 25% of x = 50

25

100× x = 50

x = 50 100

25

2

× = 200.

(g) SP = ` 260 Gain = ` 25 CP = SP – Gain = 260 – 25 = ` 235 3. (a) Let 30 be x% of 150. ∴ x% of 150 = 30

x

100150× = 30

x = 30 100

150

20

5

×

x = 20%

or 30150

100× = 10015

3100

5× = = 20%

(b) 8

24100

1003

33133

× = = % (c) 3660

1006

× = 60%

(d) 30 minutes

120 minutes× 100 =

301204

25

100× = 25%

(e) 47

100× = 4007

5717

= %

4. Total students in school = 7000

Number of students who likes to play cricket = 65% of 7000 = 65

1007000× = 4550

Number of students who play football = 7000 – 4500 = 2450


5. Let the total students be x. % of students passed = 84% % of students who failed = (100 – 84)% = 16% ∴ Number of students who failed = 1460 ∴ 16% of x = 1460

16

100× x = 1460

x = 1460 100

16×

= 9125.

∴ There are 9125 students in school.

6. Original population = 21 lakh Increase in population = 24 lakh – 21 lakh = 3 lakh

% increase = 321

100lakhlakh

× = 17

100100

714

27

× = = % .

7. C.P. of house = 60 lakh Profit=15%

S.P. = 100100

+ ×Profit%CP =

115100

60 00 000× , , = ` 69,00,000

8. SP = ` 24150

Profit=15%

CP = 70

CP = 100

100 15+×

%SP =

100115

× 24150 = ` 21,000

9. SP = ` 9900

gain % = 10%

C.P. = 100

100 +×

Gain%S.P. =

100110

9900× = ` 9000

10. S.P. = ` 1275 loss% = 15%

C.P. = 100

100 −×

lossS.P.

% =

10085

1275× = ` 1500

Gain = 10%

SP = 100

100+ ×Gain%

CP = 100 10

1001500

+ × = 110100

1500× = ` 1650

11. P = ` 35000 R = 7% T = 2 years

S.I. = P × ×r t

100 =

35000 7 2100

× × = 4900

A = P + S.I. = 35000 + 4900 = ` 39900


12. P = ` 75000 r = 5% t = 2 years

S.I. = P × × = × ×r t

10075000 5 2

100 = ` 7500.

13. P = ` 3500 A = ` 3850 R = 5% T = ? S.I. = ` 3850 – ` 3500 = ` 350

T = S.I

P. ×

×100r

= 350 1003500 5

102

××

= 2 years

14. (a) R = S.I.

P××

100t

= 5 1500 100

5000 3××

= 10%

(b) R = S.I.P

××

100t

= 168 1001050 2

××

= 8%

15. (a) T = S.IP. ×

×100r

= 800 1008000 5

2

××

= 2 years

(b) T = S.I

P. ×

×100r

= 1750 1007000 5

20

350

××

= 5 years

Worksheet 2

1. 940

100452

22122

5

× = = % 2. 3313

3313

100100

3 10013

% = =×

= = 1 : 3

3. 0.089 = 89

10008 9100

8 9= =.. % 4. 800% of ` 800 =

800100

800× = `6400

5. 1 minute1 day

× 100 = 1 minute24 60×

×3

5100 = 5

72%

6. 8% of 25 kg = 8

10025

4

× kg = 2 kg

7. 10080

100

12 5.

× = 125%.

8. Calcium 40%, Carbon 12%, Oxygen 48% Carbon = 12% of 2.5 kg

= 12

1002 5× . kg =

12100

2500× = 300 gms

Calcium = 40% of 2500 gms


40

1002500× = 1000 gm = 1 kg

9. % of lime = (100 – 55 – 33)% = 12%

Weight of lime = 12% of 800 kg = 12

100800× kg = 96 kg

10. Cost price of 24 tables = 24 × ` 450 = ` 10,800 S.P. of 16 tables = ` 600 × 16 = 9600 SP of 8 tables = ` 400 × 8 = ` 3200 Total SP = ` (3200 + 9600) = 12800 Profit=SP–CP=12800–10800=2000

Profit%=2000

10800100 18

1427

× =

11. Let the original price is x. x + 12% x = 896

x 112

100+

= 896

x112100

= 896

x = 896 100

112×

= ` 800

12. S.I. for ` 4000 = 4000 18 3

100× ×

= ` 2160

S.I. for ` 800 = 8000 15 3

100× ×

= ` 3600

Total interest = ` 2160 + 3600 = ` 5760 13. P = ` 9000 A = ` 18000 S.I. = 1800 – 9000 = ` 9000 Time = 8 years

r = S.I.

P××

100t

= 9000 100

9000 8

12 5

××

.

= 12.5%

14. Maths% = 350400

100× = 87.5

English % = 250300

100× = 8313

%

He performed batter in Maths.


Chapter 9

Exercise 9.1

1. 37

1535

614

2456

= = =

2. Positive R no are: 25

34

1712

, ,−−

−−

;NegativeRnoare:− −

−4

77

31615

, ,

3. (a) 1636

2045

2454−

−−

, , (b) −

−−12

201525

1830

, , (c) 1628

2035

2442

, ,−−

(d) 8

441055

1266

, ,

4. (a) 8

301245

1660

2075

, , , (b) −511

= − − − −1022

1533

2044

2555

, , ,

(c) 67

= 1214

1821

2428

3035− − − −

, , , (d) 78

= 1416

2124

2832

3540

, , ,

5. b, c, f, g

(a) 45

and − = ≠ − ⇒ ≠ −816

45

816

45

12

(b) 520 525 5

45

45

− ÷÷

= − = +−

(c) − × = −35

44

1220

(d) 87

78−

≠ − (e)

− ≠13

312 4

so − ≠13

14

(f) − =

−= −5

19519

519

(g) −

−=3

113

11 (h)

815

2410

125−

≠ − = −

6. (a) −45

27

−45

< 27

as −45

is negative rational

number and a negative rational number is always less than a positive number

(b) −

−7

51420

and

−

−7

575

and

− = −75

75

∴ − =

−7

51420

(c) − −87

54

− −87

54

and are in their standard

form ∴ – 8 × 4 – 5 × 7

–32 –35

–32 > –35

∴ − > −87

54

.

(d) − −15

27

− −15

27

and are in their standard

form so cross multiplying we get


–1 × 7 –2 × 5

–7 –10

–7 > –10

∴ − > −15

27

(e) 47

58

and

47

58

and are in their standard form

so multiplying both sides

47

58

4 × 8 5 × 7

32 < 35

∴ 47

58

<

(f) 29

412

−−

29

is in its standard form and

standard form of −−

412

is 13

Let us compare 29

and 13

29

13

2 × 3 1 × 9

6 9 6 < 9

∴ 29

412

< −−

7. (a) 3280

32 880 8

4 210 2

25

= ÷÷

= ÷÷

=

or LCM (32, 80) = 16

So 32 1680 16

÷÷

= 25

∴ 25 is the standard form of

3280

(b) −18108

LCM of (18, 108) = 18

∴ − ÷

÷= −18 18

108 181

6

∴ −16

is the standard form of −18108

(c) −4290

LCM of (42, 90) = 6

∴ − ÷

÷= −42 6

90 67

15

∴ −715


(d) −75125

LCM of (75, 125) = 25

∴ − ÷

÷= −75 25

125 253

5

∴ −35


(e) 40

130 LCM of (40, 130) = 10

∴ 40 10

130 104

13÷÷

=

∴ 4

13 is the S.F. of

40130

8. (a) 75

125

= (b) 47


(c) −56

(d) −= −

118

138

9. (a) (i) − ×

×= −3 7

7 721

49 (ii) − ×

×= −3 12

7 1236

84

(iii) − ×

×= −3 9

7 927

63 (iv)

− ××

= −3 217 21

63147

(b) (i) 11 55 5

5525

××

= (ii) 11 115 11

12155

××

=

(iii) 11 35 3

3315

× −× −

= −−

(iv) 11 65 6

6630

× −× −

= −−

10. (a) 35

410

715

85

, , ,

LCM (5, 10, 15) = 30

35

1830

410

1230

715

1430

85

4830

= = = =, , ,

∴ 1230

1430

1830

4830

> < <

85

35

715

410

> > >

(b) − −54

67

314

, ,

LCM (4, 7, 14) = 28

∴ − = −54

3528

, 67

2428

= , − = −314

628

∴ 2428

628

3528

> − > −

or 67

314

54

> − > −

(c) 34

23

12

56

, , ,− −

LCM (4, 3, 2, 6) = 12

34

912

23

812

12

612

56

1012

= − = − − = − =, , ,


Arranging them in descending order we get

1012

912

612

812

> > − > −

i.e., 56

34

12

23

> > − > −

(d) −

− − −1

5320

410

65

, , , or − − − −15

320

410

65

, , ,

LCM of (5, 10, 20) = 20

∴ − = − − − = − − = −15

420

320

410

820

65

2420

, , ,

Arranging them is descending order we get

− > − > − > −35

420

820

2420

i.e., − > − > − > −320

15

410

65

11. (a) 47

35

12

514

, , ,

LCM of (7, 5, 2, 14) = 70

47

4070

35

4270

12

3570

514

2570

= = = =, , ,

Arranging them in ascending order we get

2570

3570

4070

4270

< < < i.e., 5

1412

47

35

< < <

(b) + − −58

14

03

16, , ,

We know that 0 > 58

and − −14

316

and are less than 0

∴ Comparing − −14

316

and we get − −416

316

and i.e., − < − < <416

316

058

∴ − < − < <14

316

058

(c) 73

25

310

56

, , ,

LCM (3, 5, 10, 6) is 30.

∴ 73

7030

25

1230

310

930

56

2530

= = = =, , ,

Arranging them in ascending order we get

9

301230

2530

7030

< < < i.e., 3

1025

56

73

< < <


(d) 57

37

107

67− − − −

, , , or − − − −57

37

107

67

, , ,

Since the denominators are equal, so the rational number having numerator smaller is smaller.

∴ − < − < − < −107

67

57

37

.

(e) 115

113

112

117

, , ,

Since the numerators are same so the r. no having denominator smaller is greater

i.e., 117

115

113

112

< < <

12. (a) 47

25

and

LCM of (7, 5) = 35

47

2035

= and 25

1435

=

∴ 25

47

< .

∴ 25

1535

1635

1735

1835

47

< < < < <......

∴ Any two can be taken

(b) 13

34

and

LCM of (3, and 4) = 12

13

412

=

34

912

=

∴ 13

412

512

612

712

= < < <

< < =812

912

34

So any two can be taken

(c) 58

69

and

LCM (8, 9) = 72

and 69

4872

=

58

4572

=

∴ 58

4572

4872

69

= < =

∴ 58

4672

4772

69

< < < .

(d) 3 and 92

3 2

292

×and i.e.,

62

92

and

∴ 62

72

82

92

< < < .

(e) 5 1

535

×and

35

55

<

35

45

55

< <

or 3050

3150

3250

5030

< < <

There can be any other two

(f) 16

34

and

LCM of (6, 4) = 12

∴ 16

212

912

34

= =

and

∴ 2

123

124

129

12< < <...

Any two of the numbers.

13. A = 313;B=3

23;C=− 1

13;D=− 1

23


14. (a) − − − −410

1235

177

, , ,

LCM (10, 5, 7) = 70

− = − − = − − = − − = −410

2870

170

70235

32270

177

17070

, , ,

greatest = −410

;smallest=−235

(b) − − −89

511

37

, ,

LCM of numerators (+8, 5, 3) = 120

+−

+−

+−

120135

120264

120280

, ,

Since numerators are equal so the rational number having smaller denominates is greater

∴ smallest is 120135−

or −89;greatest=

120280−

= −37

(c) −

−− −3

4712

516

23

, , ,

LCM (4, 12, 16, 3) = 48

− = − − = −34

3648

712

2848

,

− = −516

1548

− = −23

3248

∴ smallest = −3648

or −34;greatest=

−516

(d) greatest = 0

− − −16

27

3, ,

LCM (6, 1, 3) = 6

− − −16

126

146

, ,

−146

is the smallest i.e., −73

is smallest.

15. (a) 2654

23

and

26 × 3 54 × 2

78 ≠ 108

So 2654

23

≠ (False)


(b) False as fraction is a positive r. no.

(c) Standard form of −−

=818

49

(False)

(d) (True) as 0 and 30 are integers and 30 ≠ 0.

(e) −

−2

554

and

− −25

54

and

cross multiplying –2 × 4 and – 5 × 5 –8 and – 25 –8 > – 25

∴ − > −25

54

(False)

(f) −49

is negative number so it lies to the left of zero (True)

(g) (True) as −−

811

811

or is positive and −811

is negative.

Exercise 9.2

1. (a) − +75

115

= − + =7 11

545

(b) 43

34

+ = 4 4 3 3

12× + ×

[LCM (3, 4) = 12]

= 16 9

122512

21

12+ = =

(c) − +910

2215

[LCM (10, 15) = 30]

− × + ×9 3 22 2

30 = − + =27 44

301730

(d) 4

115

9+ −

[LCM (11, 9) = 99]

4 9 5 11

99× + − ×( )

= 36 55

9919

99− = −

(e) 312

45

+ −

=

72

45

+ −

= 7 5 4 2

10× + − ×( )

= 35 8

102710

27

10− = =

(f) 65

23

+ −

=

6 3 2 515

× + − ×( ) =

18 1015

815

+ − =( )


2. (a) 23

57

2 7 3 521

− −

= × + × =

14 1521

2921

1821

+ = =

(b) − −613

45

= − × − ×6 5 4 13

65 =

− − = −30 5265

8265

= −11765

(c) 7

1145

711

45

− −

= + = 7 5 4 11

55× + ×

= 35 44

557955

12455

+ = =

(d) − − −

= − × + ×215

43

2 1 4 515

= − + = + = =2 20

1518

1565

115

(e) 47

35

47

35

4 5 7 335

− −

= + = × + × = 20 21

354135

1635

+ = =

(f) 214

594

51

9 5 44

− = − = − × =

9 204

114

234

− = − = −

3. (a) 7

131026

7 2 10 126

− = × − × =

14 1026

426

213

− = =

(b) 425

65

425

65

− −

= + = 4 1 6 525

× + × = 4 30

253425

1925

+ = =

(c) − − = − × − ×38

57

3 7 5 856

= − −21 40

56 =

− = −6156

15

56

(d) 87

54

8 4 5 728

32 3528

− = × − × = − =

−328

(e) 49

23

49

23

4 2 39

− −

= + = + × = 4 6

9109

119

+ = =

(f) 234

2 4 34

8 34

54

114

− = × − = − = =

4. (a) 95

74

6320

33

20× −

= − = − (b) 37

95

2735

× − = −

(c) 4 224 16

12

2

4

1

2

××

= (d) 172

517

52

212

× = =

(e) − × = −49

95

45

(f) 3

111218

211

2

6

× =

(g) 435

125

225

× = =


5. (a) − ÷ = − × = −553

535

3 (b) − ÷ = − × = − × = − = −45

23

45

32

2 35

65

115

(c) 147

1114

117

1114

÷ = ÷ = 117

1411

× = 2 (d) − ÷ = − × = −213

415

213

154

15262

(e) 45

53

45

35

1225

÷ = × = (f) 632

6 23

2

÷ = × = 2 × 2 = 4.

6. (a) 312

58

+ −

+ −

= 3 8 1 4 5 1

8× + − × + − ×( ) ( )

= 24 4 58

24 98

− − = − = 158

178

=

(b) − + − +518

924

16

= − × + − × + ×5 4 9 3 1 1272

( ) = − − + = − = −20 27 12

7035

7012

(c) 6

112

335

66− −

+ −

=

6 6 2 2 566

36 4 566

× + × + − = + −( ) =

3566

(d) − + +512

76

133

= − + × + ×5 7 2 13 4

12 =

− + + = =5 14 5212

6112

51

12

(e) − ÷

×89

43

16

= − ×

×8

934

16

2

3 =

− × = −23

16

218

1

9

= − 19

(f) 34

69

45

÷

× = 34

96

452

×

× = 98

452

× = 9

10

(g) 35

13

52

×

÷ = 15

25

225

× = .

7. Sum of 34

57

+ = 3 7 5 4

2821 20

28× + × = +

= 4128

According to question 3

144128

− = 3 2 41

286 41

2835

285

4× − = − = − = −

8. Sum of 12

34

+ = 1 2 3

454

× + =

According to question 54

95

− −

= 54

95

5 5 9 420

25 3620

+ = × + × = + = 6120

3120

= .

9. Sum of the two real numbers = 45

One of the number = − 13

Other number = 45

13

45

13

− −

= + = 4 3 1 515

12 515

1715

12

15× + × = + = =


10. Amount Sheena had = ` 50

Cost of a pen = ` 1834

Balance left = ` 50 1834

−

= 50 –

754

= 50 4 75

4200 75

4125

431

14

× − = − = = `

She has ` 3114

left with her.

11. According to question = 2 – 45

= 2 5 4

510 4

565

× − = − =

∴ 65

should be added to 45

to get 2.

12. Cost of 1 pen = ` 1534

Cost of 20 pens = ` 1534

20634

205

× = × = ` 315

∴ Cost of 20 such pens is ` 315.

13. Cost of 6 m ribbon = ` 427

Cost of 1 m ribbon = ` 427

6÷

= `

307

16

5

×

= `

57

Cost of 10 m ribbon = ` 57

10× = ` 507

717

=

14. 64

125

615

64

3

×

− ÷

185

615

46

− ×

=

185

415

18 3 415

54 415

−

= × − = − =

5015

103

313

= =

15. 34

16

65

2× × − = 3

202− = 3 2 20

203 40

2037

20− × = − = −

Worksheet 1

1. (a) 75

(b) 23

(c) −57

(d)


2. (a) − − − −25

410

615

820

, , , (b) − − − −913

1826

2739

3652

, , ,

(c) 67

1214

1821

2428

, , , (d) − − − −511

1022

1533

2044

, , ,

3. (a) 2781

39

13

= = is the S.F. of 2781

(b) − = −945

15

is S.F. of −945

(c) 3774

12−

= − is the S.F. of 37

74− (d) 75

1451529−

=−

= −1529

is the S.F. of 75145−

4. (a) −45

23

and cross multiplying we get

4 × 3 2 × 5

12 ≠ 10

∴ −45

23

and are not equivalent

(b) −

−6

71214

and lowest form of 1214

67−

= − and hence −−

67

1214

and are equivalent r.no.

(c) − −89

43

and are not equivalent as on cross multiplying we get –8 × 3 and –4 × 9

–24 ≠ –36.

(d) −311

933

and , these are not equivalent fractions as one of them is negative r. no and

the other one is positive.

20

12060

12075

12096

120< < <

16

12

58

45

< < <

5. (a) 73

is greater than −37

73

as is positive.

(b) 45

817

45

817

and or and−

− by cross multiply we get 4 × 17 8 × 5

78 > 40

∴ 45

817

>

(c) − −49

57

and

Cross multiply –4 × 7 –5 × 9 –28 – 45

∴ −49

> −57


(d) − −97

95

, < or

Cross multiplying we get –9 × 5 and –9 × 7 –45 > –63

So − > −97

95

or 97

95−

>−

Numerators are same, so the one having greater number is smaller

∴ 97

95−

>−

(e) − −47

37

, Denominators are same, so one having the greater numerator is greater.

∴ –3 >–4 so − > −37

47

(f) −93

and 4, since 4 is positive r. no so 4 > −93

or –3.

6. Express:

(a) 89

6472

= (b) 89

4045

=

7. (a) − =−

185

3610

(b) − =

−185

7220

8. (a) 12

45

16

58

, , ,

LCM of (2, 5, 6, 8) = 120

12

60120

45

96120

16

20120

58

75120

= = = =, , , , 20

12060

12075

12096

120< < < . So,

16

12

58

45

< < <

(b) − − −12

35

47

59

, , ,

35

is greatest as it is positive r. no and other three are negative real numbers, hence

let us compare

− − −12

47

59

, ,

LCM (2, 7, 9) = 126

∴ − = − − = − − = −12

63126

47

72126

59

70126

, ,


∴ − < − < −72126

70126

63126

i.e., − < − < − <47

59

12

35

9. (a) 34

410

715

35

, , ,− −

Let us compare positive real number and negatives real number separately.

Comparing negative real number we get

− − − −410

35

25

35

and or and − >25

35

The Comparing +ve real number we get

34

715

and (LCM (4, 15) = 60)

i.e., 4560

2860

and

4560

2860

> i.e., 34

715

>

∴ 34

715

410

35

> > − > −

or (LCM (4, 10, 15, 5) = 60

34

4560

=

− = −410

2460

7

152860

=

− = −35

3660

∴ 34

715

410

35

> > − > −

(b) Comparing positive real number and negative real number we get

− −54

47

and

–35 < –16

∴ − > −47

57

Comparing positive real number

67

514

and

6 × 14 5 × 7

84 > 35

67

514

>

∴ 67

514

47

57

> > − > −

or − −54

67

514

47

, , ,

LCM of (4, 7, 14) = 28

∴ − = − =54

3528

67

2428

,

5

141028

1628

47

= − = −,

∴ 2428

1028

1628

3528

> > − > −

67

54

47

57

> > − > −

10. (a) − + = − × + × = − + =37

45

3 5 4 735

15 2835

1335

(b) − + = − × + = − + = − = −31

59

3 9 59

27 59

229

249


(c) − + = − + =12

54

2 54

34

(d) − − = − × − ×37

58

3 8 5 756

= − − = − = −24 35

5659

561

356

(e) − − −

= − + = − + = − × + ×4

85

485

41

85

4 5 8 15

= − + = − = −20 8

5125

225

(f) 103

45

10 5 4 315

50 1215

3815

28

15− = × − × = − = =

(g) − ×37

146

2

2 = –1 (h)

− ×89

184

2 2

= –2 × 2 = –4

(i) − ×9109

= –10 (j) − × = −94

23

32

3

2

(k) − × = − = −87

4916

72

312

7

2 (l)

− ×−

811

4416

4

2

2

= 2

11. (a) 23

45

32

+

− = ( )2 10 4 6 3 1530

× + × − × = ( )20 24 45

3044 45

301

30+ − = − = −

(b) − ×

÷ = − ÷6

549

13

815

13

2

3 =

− × = −815

31

855

(c) 34

12

45

3 1 24

45

14

45

15

−

× = − ×

× = × =

(d) − ÷

× = − ×

× = − × = −38

69

14

38

96

14

916

14

9642

(e) 14

15

120

120

120

120

201

×

÷ = ÷ = × = 1

(f) 54

153

115

54

315

115

14

115

1605

÷

× = ×

× = × =

Worksheet 2

1. A = 215;B=2

25;C=2

35;D=2

45;P=−1

14;Q=−1

24;R=−1

34

2. A → 36

714

1020

612

510

, , , , ;B→ 46

1421

812

1218

2233

, , , , ;C→ 615

1025

2460

, ,

3. (a) No (b) Yes (c) Yes (d) Yes (e) No 4. (a) LCM of (3, 5, 7, 15) = 105

∴ − = − = =35

63105

47

60105

23

70105

, ,


4

1528

1059

7135

105= − = −

,

∴ smallest is −135105

or −97

and greatest = 23

(b) LCM of (5, 2, 15) = 30

− = − − = − − = − − = −45

2430

32

4530

135

7830

92

13530

, , ,

− = −715

1430

∴ smallest = −92

and greatest = −715

(c) 34

25

38

95

72

, , , ,

LCM of (2, 4, 5, 8) = 40

34

3040

25

1640

38

1540

95

7240

72

14040

= = = = =, , , ,

∴ Smallest = 38

and greatest = 72

5. (a) Cross multiplying –55 – 63

∴ − > −57

911

(b) Cross multiplying we get 4 × 15 6 × 13

60 78

∴ 4

136

15<

(c) − <35

53

as negative rational number is always smaller than positive rational number.

(d) − −713

965

⇒ Cross multiplying –7 × 65 –9 × 13

–441 < –117

So − < −713

965

6. (a) − =

−=

−= −11

173351

5585

88136

(b) 6

103660

35

915

= = =

7. (a) − + = − × + × = − +37

58

3 8 5 756

24 3556

= + 1156

(b) 9

1125

9 5 2 1155

45 2255

6755

11255

+ = × + × = + = =

(c) 612

214

132

94

13 2 9 14

+ = + = × + × = 26 9

4354

834

+ = =


(d) 315

213

165

73

16 3 7 515

− = − = × − × = 48 35

151315

− =

(e) 4165

4 5 165

20 165

45

− = × − = − =

8. (a) 38

87

37

× = (b) 94

98

94

89

2

÷ = × = 2 (c) 65

103

2 2

× = 4

(d) 30215

30521

10 57

507

÷ = × = × = = 717

9. (a) 25

32

103

4 1510

103

+

× = +

× = 1910

103

193

613

× = =

(b) 72

27

37

7 7 414

73

49 414

73

4514

73

15

2

−

÷ = × −

× = − × = ×( ) =

152

712

=

(c) 35

85

47

2425

74

4225

6

×

÷ = × = = 11725

(d) 34

2 4 3 312

34

8 912

÷ × − ×

= ÷ −

= 34

121

3

×−

= – 9.

10. Total distance covered = 312

45

+

km

= 72

45

+

km = 7 5 4 210

35 810

4310

43

10× + × = + = = km.

11. Cost of register = ` 13515

Cost of pen = ` 7545

Total cost = 13515

7545

+ = (135 + 75) + 15

45

+

= (210) + 55

= 211

Amount returned = ` 500 – 211 = ` 289

12. Cost of 1 litre petrol = ` 7234

2914

=

Cost of 212

litre petrol = ` 2914

52

× = ` 1455

8181

78

= `

13. Cost of 17 erasers = ` 1015

Cost of 1 eraser = ` 1015

17÷ = ` 515

17515

117

÷

= ×` = `35

Cost of 10 such erasers = ` 35

102

× = ` 6


Chapter 10

Do it yourself.

Chapter 11

Exercise 11.1 1. Area to be painted = Area of wall – Area of door = (15 × 10) m2 – 3 × 2 m2 = (150 – 6) m2 = 144 m2

Cost of painting = ` 144 × 3 = ` 432

2. (a) Area of ∆PQR = 12

base × ht. = 12

8 54× × cm2 = 20 cm2

(b) Area of ∆XYZ = 12

base × ht. = 12

4 22× × = 4 cm2

(c) Area of ∆RST = 12

base × ht. = 12

5 4 2× × = 10 cm2

(d) Area of ∆LMN = 12

base × ht. = 12

12 46

× × = 24 cm2

(e) Area of ∆XYZ = 12

6 83× × = 24 cm2

(f) Area of ∆UVW = 12

8 5× × = 20 cm2

3. Peri of rectangle = Peri. of square 2(24 + 20) = 4 × side 2 × 44 = 4 × side

2 44

4

11×

= side

22 m = side Area of rectangle = 24 × 20 = 480 m2

Area of square = 22 × 22 = 484 m2

∴ Area of square is more than area of rectangles. 4. Area of square = Area of rectangles 24 × 24 = 48 × breadth

12

2

24 2448

× = breadth

12 m = breadth Wire required to fence square = 4 × 24 = 96 m Wire required to fence rectangle = 2(12 + 48) = 2(60) = 120 m


∴ more wire is required to fence rectangle. 5. l = 400 m, b = 200 m (a) Peri of land = 2(400 + 250) = 2 × 650 = 1300 m Cost of fencing = ` 5 × 1300 = ` 6500 (b) Area of land = l × b = 400 × 250 = 100000 m2

Cost of land = ` 2050 × 100000 = ` 20,50,00000 6. Area of square = 4900 sq. m. side × side = 4900 = 49 × 10 × 10 = 7 × 10 × 7 × 10 side = 7 × 10 m = 70 m Perimeter of square = 4 × 70 m = 280 m 7. Area = l × b 1200 m2 = 40 m × b (1200 ÷ 40) m = b 30 m = b Perimeter of rectangle = 2(l + b) = 2(40 + 30) m = 2 × 70 = 140 m. 8. Area of square = Area of rectangle 40 × 40 = 160 × b

10

4

40 40160

× = b

10 m = b 9. Let the breadth = x m Length = (x + 50) m Peri = 2(l + b) 900 m = 2(x + x + 50) 900 = 2(2x + 50) 900 = 4(x + 25) 225 = x + 25 200 = x = b 250 = l Area = l × b = 200 × 250 = 50000 m2

10. Area of wall = l × b = 15 × 10 = 150 m2. Area of door and 2 windows = [3.5 × 2.5 + 2 × (1.5 × 1) m2 = (8.75 + 3) m2 = 11.75 m2

Area to be painted = Area of wall – Area of door and window = 150 m2 – 11.75 m2 = 138.25 m2

Cost of painting wall = ` 3 × 138.25 = ` 414.75 11. l = 5 x, b = 2x Peri = 2(l + b) 350 m = 2(5x + 2x) 350 m = 14 x


350 ÷ 14 = x 25 = x ∴ l = 125 m, b = 50 m Area = l × b = 125 × 50 m2 = 6250 m2

12. Areaof8flowerbeds=8×l1 × b1 = 8 × (4 × 3) = 96 m2

Area of plot = 60 × 40 = 2400 m2

Remainingarea=Areaofplot–Areaofflowerbed = (2400 – 96) m2 = 2304 m2

Cost of laying grass = ` 6 × 2304 = ` 13,824 13. Garden is in front of the house so three sides are there to fence leaving 2 m for entrance.

Hence the length of fencing required = (20 + 25 + 20 – 2) m = 63 m. ∴ Cost of fencing = ` 2.50 × 63 = ` 157.50. 14. (a) Base = 20 cm Area = 300 cm2

ht. = 2 2 300

20× = ×AreaBase

= 30 cm

(b) Ht. = 32.4 cm Area = 64.8 cm2

Base = 2 2 64 8

32 4

2× = ×Areaht.

cm.

. = 4 cm.

(c) Base = 44 cm Area = 170.5 cm2

Ht. = 2 2 170 5

44 22

× = ×AreaBase

. = 7.75 cm

(d) Ht = 8.4 m Area = 194.88 m2

Base = 2 194 88

8 4

97 44

4 2 2 1

× ..

.

. .

= 46.4 m

(e) Area = 65.52 m2

Base = 7.8 m

Ht. = 2 2 65 52

7 83 9

× = ×AreaBase

...

= 16.8 m

(f) Base = 2 2 240

60

4

× = ×AreaHt.

= 8 m


(g) Ht. = 2 2 1050

15

70

× = ×Area15

= 140 m.

15. Area of ∆ABC = 12

AC AB× = 12

48 6× = 24 cm2

Also area of ∆ABC = 12

BC × AD

24 = 12

105

× × AD

245

= AD

4.8 cm = AD

16. Area ∆PQR = 12

base ht.× = 12

QR PL×

= 12

510 12× = 60 cm2.

Also Area of ∆PQR = 12

PR QM×

60 cm2 = 12

10× ×20 QM

6010

2cmcm

= QM

6 cm = QM

17. Area of ∆ABC = 12

AB CD× = 12

15

× ×30 20 = 300 cm2

Area of ∆ABC = 12

BC AE× = 12

BC 15×

300 2

15

20

× = BC

40 cm = BC

18. Area of isosceles triangle = 12

30 3015 × = 450 cm2


A

B C30 cm

30 cm

19. Area of triangle = 12

base ht× = 12

2(2 base) ht× ( ) = 412

base ht×

Area becomes 4 times. 20. (a) Area of shaded portion = Area of rectangle ABCD – Area of ∆ECD

= 12 512

3 4× − ×

( ) m2 = 60 – 6 = 54 m2.

(b) Area of shaded rectangle = Area of rectangle ABCD – Area of ∆EDC

= 12 × 8 – 12

12 86

× = 96 – 48 = 48 m2.

(c) Area of shaded portion

= Area of ∆ABC – Area of ∆PQR

= 12

9 712

3 4 2× = × = 31.5 – 6 = 25.5 m2.

Exercise 11.2 1. (a) Area of parallelogram ABCD = base × corresponding ht. = 10 cm × 4 cm = 40 cm2

(b) Area of parallelogram UXYZ = base × corresponding ht. = 10 cm × 15 cm = 150 cm2

(c) Area of parallelogram PQRS = base × corresponding ht. = 8 cm × 12 cm = 96 m2

(d) Area of parallelogram KLMN = base × corresponding ht. = 18 cm × 10 cm = 180 cm2

2. Complete the table. (a) Area of parallelogram = Base × height 600 cm2 = 20 cm × height

60020

cm = height

30 cm = height (b) Area of parallelogram = base × height 1256 cm2 = 31.4 cm × height cm

125631 4

2cm. cm

= height

40 cm = height


(c) Area of parallelogram = base × height

170.5 = base × 11

170 5

11

2. cmcm

= base

15.5 cm = base (d) Area of parallelogram = base × height 750 cm2 = base × 15 cm

75015

2cmcm

= base

50 cm = base 3. Area of parallelogram = 540 cm2

side = 180 cm

height = Area

side or base= 540

180

2cmcm

= 3 cm

4. Area of rectangle = Area of parallelogram l × b = h × base 90 × 60 cm2 = h × 180 cm

90 60180

30

2

× = ht.

ht. = 30 cm.

5. (a) Area of shaded portion = Area of rectangle – Area of parallelogram = 10 × 10 cm2 – 8 × 5 cm2

= 100 – 40 cm2 = 60 cm2

(b) Area of shaded portion = Area of parallelogram ABCD – Area of ∆ADE

= base × corresponding ht. – 12

base × ht

= 16 × 10 cm2 – 12

(16 – 10) × 10

= 160 cm2 – 12

6 103 2× × cm

= 160 cm2 – 30 cm2 = 130 cm2

(c) Area of shaded portion = Area of parallelogram ABCD – Area of rectangle PQRS = (16 × 10) cm2 – (10 × 8) cm2

= 160 cm2 – 80 cm2 = 80 cm2

(d) Area of shaded portion = Area of circle (r = 11 cm) – (Area of 2 squares

+ Area of circle) (r = 3.5 cm)

= 227

14 14 2 2 2227

3 5 3 52

× × − × × + × ×

( ) . .


= (616 – 46.5) cm2 = 569.5 cm2

6. (a) Circumference of a circle = p × d = 227

× 10.5 = 22 × 1.5 = 33 cm

(b) Circumference of a circle = 2pr = 2 × 227

7× = 44 cm

(c) Circumference of a circle = 2pr = 2 × 227

355

× = 220 cm

(d) Distance covered in 50 rounds = 50 × circumference

= 50 × 2p × 49 = 50 × 2 × 227

497

× = 15400 cm

(e) r = Circumference

2176 72 22

4 8

π= ×

× = 28 cm

(f) Circumference = 2pr = 2 × 3.14 × 30 m = 188.4 m Cost of fencing = ` 188.4 × 15 = ` 2826

7. Perimeter of semicircle of r = 21 cm = d + pr = 42 cm + 227

213

× cm

= 42 cm + 66 cm = 108 cm

Area of semicircle of r = 21 cm = πr2

11 3

222 21 21

7 2= × ×

× = 693 cm2.

8. (a) Perimeter of the shape = 4 × Perimeter of semicircle of r = 21 cm

= 422

×

πr = 4pr = 4

227

213

× × = 264 cm

(b)Perimeterofthefigure=(2semicircleofr(7cm)+10cm+10cm

= 2227

7 20× ×

+ cm = (44 + 20) cm = 64 cm

(c) Perimeteroffigure=(2Semicircleofr = 7 cm) + 10 × 2 cm = (44 + 20) cm = 64 cm (d)Perimeteroffigure=[2(Semicircleofr = 21 cm) + 30 × 2] cm

= 2227

21 603

× ×

+

cm = 192 cm

9. (a) Area of larger circle = 227

× 14 × 14 cm2 = 616 cm2

(b) Area of smaller circle = 227

× 10.5 × 10.5 cm2 = 346.5 cm2

(c) Area of shaded portion = Area of bigger circle – Area of smaller circle = (616 – 346.5) cm2 = 269.5 cm2

(d) Difference between the circumference of two circle

= 2pR – 2pr = 2p (R – r) = 2 × 227

× (14 – 10.5)

= 2 × 227

× 3.5 = 22 cm


10. Circumference = 308 cm 2pr = 308

r = 308 72 22

147

××

= 49 cm

Area = pr2 = 227

49 497

× × = 7546 cm2

11. (a) Area of bigger circle = 3.14 × 20 × 20 = 1256 cm2

Area of smaller circle = 3.14 × 15 × 15 = 706.5 cm2

Area of remaining sheet = 1256.0 – 706.5 = 549.5 cm2

(b) 2pR : 2pr = 20 : 15 = 4 : 3

(a) Area of biggerArea of smaller

= ππ

× ×× ×

=20 2015 15

169

= 16 : 9.

12. Shaheen Circumference = 176 cm 2pr = 176

r = 176 72 22

8 4

××

= 28 cm

Area = 227

28 284

× ×

= 2464 cm2

Rohan Peri of square = 176

4 × side = 1764

Side = 176

4 = 44 cm

Area = 49 × 44 cm2

= 1936 cm2

Rosy Peri. of rectangle = 176 cm 2(x + 30) = 176 cm x + 30 = 88 cm x = 88 – 30 = 58 cm Area of rectangle = l × b = 58 × 30 cm2 = 1740 cm2

Shaheen shape i.e., circle encloses greater area. 13. Area of square = 8 × 8 cm2 = 64 cm2

Area of circle = pr2 = 3.14 × 3 × 3 = 3.14 × 9 = 28.26 cm2

Remaining area = Square Area – Circle Area = (64 – 28.26) cm2 = + 35.74 cm2

Cost of painting at the rate of 12 per cm2

= 35.74 × 12 = ` 428.88 14. Circumference = 352 cm 2pr = 352 cm

r = 352 72 22

3216

2

8

××

= 56 m


Outer circle radius = 56 m + 4 m = 60 m Circumference of outer track = 2pr = 2 × 3.14 × 60 cm = 376.8 m Cost of fencing at the rate of ` 6 per m = ` 6 × 376.8 = ` 2260.8 Area of track = Area of outer circle – Area of inner circle = p × (602 – 562) m2

= 3.14 × (116) × 4 = 1456.96 m2

15. Area that the sprinkler can cover = p × 16 m × 16 m = 3.14 × 16 × 16 = 803.84 m2

Areaofflowerbed=628m2

Yes,Areacoveredbysprinklerismorethanareaofflowerbedhencesprinklerwillwaterthewholeflowerbed.

16. Distance covered in 1 round = 227

7 71× ×. .. = 1.54 m

Number of rotation for 1.54 m = 1

Number of rotation of 1 m = 1

1 54.

Number of rotation for 462 m = 1

1 5446200

.× = 300

∴ the wheel will make 300 revolutions in going round 462 m.

Exercise 11.3

1. (a) 150 cm2 = 150

100 100× = 0.015 m2

(b) 12 hectare = 12 × 100 × 100 = 120000 m2

(c) 100 m2 = 100 × (100 × 100) cm2 = 106 cm2

(d) 600000 m2 = 600000

100 100× = 60 hectares

2. Area of path = Outer rectangle area– Inner rectangle area

= (66 × 46) m2 – 60 × 40 m2

= (3036 – 2400) m2 = 636 m2.

3. Area of path = Area of outer square– Area of inner square

= 150 × 150 – 142 × 142

= 22,500 – 20,164

= 2336 m2

4. Area of path = (100 × 6) + (80 × 6) – (6 × 6)A B

D C

80 m

100 m

6 m


= (600 + 480 – 36) m2 = 1044 m2

Cost of constructing at the rate of ` 95/m2

= ` 1044 × 95 = ` 99180 5. Area of painting = l × b = (3 – 0.4) × (2 – 0.4) m2

= (2.6 × 1.6) m2 = 4.16 m2

Area of margin = Area ABCD – Area PQRS

= 3 × 2 – 4.16

= 6 – 4.16 = 1.84 m2

Cost of painting at ` 500/m2

= 4.16 × 500 = ` 2080.

6.

600

800

(a) Area of cross roads = (800 × 15 + 600 × 15 – 15 × 15) m2

= 15 × (800 + 600 – 15)

= 15 × 1385 = 20,775 m2

(b) Area of remaining park = Area of park – Area of cross roads

= 800 × 600 – 20775 = 480000 – 20775

= 459225 m2 = 45.9225 hectares

(c) Cost of laying grass in remaining park

= 45.9225 × 50 = ` 2296.125

7. Area of square lawn = 50 × 50 m = 2500 m2

Area of cross roads = (50 × 5 + 50 × 5 – 5 × 5) m2

= (250 + 250 – 25) m2 = 475 m2

(a) Cost of constructing roads at the rate of 250/m2

= ` 475 × 250 = ` 118750 (b) Area of remaining park = Area of park – Area of cross roads = (2500 – 475) m2 = 2025 m2

Cost of leveling 1 hectare (100 × 100) m2 = ` 10000

Cost of leveling 1 m2 = 1000010 000

= ` 1

Cost of leveling 2025 m2 = ` 2025 × 1 = ` 2025 8. Length of cord required for rectangular piece = 2(16 + 12) = 2 × 28 = 56 m Length of cord required for circular piece of radius 6 m = 2 × 3.14 × 6 = 3.14 × 12

A B

CD

P Q

RS

3 m2.6 m

2m


= 37.68 m. Yes, she will have (56 – 37.68) = 18.32 m chord 9. Area of room = 8.60 × 6.25 m2

Area of tile = 0.2 m × 0.25 m = .05 m2

Number of tiles required = 8 60 6 25

05

125. .

.×

= 8.6 × 125 = 1075

Cost of tiling = ` 50 × 1075 = ` 53750. 10. (a) Shaded Area = Area of rectangle – Area of crossroads

= (17 × 13) m2 – (17 × 2 + 13 × 2 – 2 × 2)

= (221 – 56) m2 = 165 m2

(b) Shaded area = Area of rectangle – Area of two triangle

= 18 × 12 – 12

12 812

6 12× × + ×

= 192 – [48 + 36] = 108 m2

(c) Shaded area = Area ABCD – Area PQRS = 57 × 45 – (41 × 55) m2 = (2565 – 2225) m2 = 310 m2

(d) Shaded area = Area PQRS – Area ABCD = [(30 × 20) – (20 × 15)] m2 = (600 – 300) m2 = 300 m2

(e) Shaded Area = 15 × 2 + 4 × 1 = (30 + 4) m2 = 34 m2

(f) Shaded area = Area of rectangle – Area of circle

= 20 × 15 – 227

7 7× × = (300 – 154) m2 = 146 m2

(g) Shaded area = Area of circle – Area of 2 rectangles

= [p × 142 – 2 × (4 × 2)] m2 = 227

14 14 162

× × −

m2

= (616 – 16) m2 = 600 m2

(h) Area of shaded portion = (25 × 4 + 20 × 4 – 4 × 4) m2

= (100 + 80 – 16) m2 = 164 m2

Worksheet 1 1. (a) 1 hectare = 10000 m2

(b) 1 hectare = 100000000 cm2 = 108 cm2

(c) Area of square = Side × Side

(d) Area of triangle = 12

base × height

(e) Area of rectangle = l × b

(f) Circumference of circle = 2pr


(g) Area of parallelogram = base × corresponding height

2. (a) Perimeter of square = 4 × side = 4 × 16 = 64 m

(b) Perimeter of rectangle = 2(l + b) = 2(20 + 16) = 2 × 36 = 72 m.

(c) Perimeter of triangle = sum of three sides = (12.5 + 10 + 8) cm = 30.5 cm

(d) Circumference of a circle = 2pr = 2227

7× × = 44 cm

3. (a) Perimeter of square = 200 m 4 × side = 200 m

side = 200

4m

= 50 m

⇒ Area = side × side = 50 × 50 = 2500 m2

(b) Perimeter of rectangle = 2(l + b) = 300 m 2 . (80 + b) = 300 80 + b = 150 b = 70 m. Area = l × b = 80 × 70 = 5600 m2

(c) Circumference = 628 cm 2pr = 628 cm

r = 628 72 3 14

×× .

= 100 cm

Area = pr2 = 3.14 × 100 × 100 = 3,1400 cm2

4. Area of square = 4 hectare = 4 × 10000 m2

side × side = 4 × 100 × 100 = 2 × 100 × 2 × 100 = 200 × 200 m2

side = 200 m. Perimeter = 4 × side = 4 × 200 = 800 m Cost of fencing = 800 × 25 = ` 20000. 5. Area of remaining lawn = Area of lawn – Area of square Flower bed = 18 × 14 – 6 × 6 = 252 – 36 = 216 m2

Cost of laying grass in the remaining lawn = ` 6 × 216 = ` 1296 6. Area of footpath = Area of rectangle ABCD – Area of rectangle PQRS = 43 × 33 – 40 × 30 = 1419 – 1200 = 219 m2

Cost of cementing footpath at ` 150/m2

= ` 150 × 219 = ` 32850 Perimeter of outer track = 2(43 + 33) = 2 × (76) = 152 m. Cost of fencing the outer track of footpath = ` 50 × 152 = ` 7600 7. Area of cardboard needed to put at the back = Area ABCD = 170 × 110 cm2 = 18700 cm2

A B

C

P Q

RS

43 mD

40 m

33 m 30m

1.5 m


Cost of framing the painting = Perimeter of ABCD × cost/m = 2(170 + 110) m × ` 120 = 2 × 280 × 120 = ` 67200 8. Area PQRS = l × b = (30 – 4) × (20 – 4) m2

= 26 × 16 m2 = 416 m2

Cost of planting grass at the rate of 5 per m2

= ` 5 × 416 = ` 2080 Margin area where roses are planted = Area ABCD – Area PQRS = 30 × 20 – 26 × 16 = (600 – 416) m2 = 184 m2

Total number of roses in 1 m2 = 5 Total number of roses in 184 m2 = 184 × 5 = 920 roses. 9. radius = 28 cm

circumference = 2pr = 2227

284

× × = 176 cm

Number of revolution for 1.76 m = 1

Number of revolution for 1 m = 11 76.

Number of revolution for 1408 m = 1

1 76. × 1408 = 800

∴ The wheel will make 800 revolutions to cover 1408 m. 10. (a) Shaded area = Area of bigger circle – Area of smaller. = p(7)2 – p(5)2 = p(49 – 25) = p × 24 = 3.14 × 24 = 75.36 cm2

(b) Shaded area = (70 × 5 + 50 × 5 – 5 × 5) m2

= (350 + 250 – 25) m2 = 600 – 25 = 575 m2

11. Area of parallelogram = AB × DM Also Area of parallelogram = BC × DM ∴ 24 × 8 = 16 × x

12

2

24 816

× = x

x = 12 cm 12. Area used for path = 100 × 10 + 80 × 10 – 10 × 10 = 1800 – 100 = 1700 m2

Cost of leveling the path = 1700 m2 × 50 = ` 85000

Worksheet 2 1. (a) Peri. of square = 88 m 4 × side = 88 side = 22 m Area = 22 × 22 m2 = 484 m2

A B

CD

30 m

20m

P Q

RS


(b) Circumference of circle C = 2pr ⇒ 88 = 2227

× × r ⇒ r = 88 7

44

2

× = 14 m

Area = pr2 = 227

14 142

× × = 22 × 28 m2 = 616 m2.

(c) Perimeter of rectangle = 88 cm

2(l + b) = 88 = 2(30 + b) = 88 = 30 + b = 44 = b = 14.

Area = l × b = 30 × 14 cm2 = 420 cm2. (Circle has greatest area)

2. Area of triangle = Area of square

12

40base × = 40 × 40

base = 40 40

20

2×

= 80 m

3. (a) Distance covered by minute hand = 2p × 15 cm = 2 × 3.14 × 15 cm = 94.2 cm (b) Distance covered by hours hand = 2p × 12 cm = 75.36 cm Difference = 94.2 – 75.36 cm = 18.84 cm. 4. Area ungrazed = Area of square – Area of circle = (20 × 20 – p × 6 × 6) m2

= (400 – 113.04) m2 = 286.96 m2

5. Area of circle = 616 m2

pr2 = 616 m2

227

2× r = 616

r2 = 616 7

22

56

2

28

× = 28 × 7

r2 = 7 × 2 × 2 × 7

r × r = 7 × 2 × 7 × 2

r = 14 m.

Circumference = 2pr = 2227

142

× × = 88 m.

Cost of fencing = ` 12 × 88 = ` 1056 6. Circumference of length of wire = 2pr = 2 × 3.14 × 2 = 4 × 3.14 cm = 12.56 cm Peri. of square = 4 × side = 4 × 3 = 12 cm Yes, she will be able to do so as circumference of circle is more than the perimeter of

square. 7. Area of shaded portion


(a) Area of square – 4 × 14

area of circle

= 50 × 50 – 414

25 25× × × ×

π

= 2500 – 3.14 × 25 × 25 = 2500 – 1962.50 = 537.50 m2. (b) Shaded Area = Area of rectangle – 4 × Area of square = (60 × 50 – 4 × (8 × 8)) m2

= 3000 – 256 = 2744 m2

8. Area of quad. = 12

AC × BM + 12

AC × DM

= 12

25 × .5 + 12

× 25 × 8 = 6205 + 100 = 162.5 cm2

9. Area of border = Area of saree – Area of saree in side border

= 5 × 1.25 – 4.5 × .75 = 6.25 – 3.75 = 2.875 m2.

Cost of printing border 2.875 × 500 = ` 1437.5

10. Total length (outer) = 1 + 4 + 2.5 + 4 + 1 = 12.5 m. Total breadth (outer) = 1 + 3 + 3 + 1 = 8 m. (a) Area of path = outer area – Inner area = 12.5 × 8 – 10.5 × 6 = 100 m2 – 63 m2 = 37 m2

Cost of paving the path at the rate of ` 120 /m2

= ` 120 × 37 = ` 4440 (b) Area of living room = (4 × 3 + 2.5 × 4) m2 = (12 + 10) m2 = 22 m2

(c)Areaofwoodenflooring=Bedroom+Living+Kids+Kitchen = 4 × 3 m2 + (4 × 3 + 2.5 × 4) + 4 × 3 m2 + 4 × 3 m2

= 12 + (12 + 10) + 12 + 12 = 58 m2

Costofwoodenflooring=58×1200=` 69600 11. (a) Area of solar panel = 90 × 120 cm2 = 0.9 × 1. 20 m2 = 1.08 m2

(b) Area of solar cell = 0.05 × 0.1 m2 = 0.005 m2

(c) Number of solar cells in 1 panel = 1 080 005

..

= 216.

(d) Utilisation of natural resource and renewable sources.

12. (a) Distance covered in one round = 2 2227

14 882

πr = × × = m (b) Time taken to cover 8 m = 1 min

Time taken to cover 1 m = 18

min

Time taken to cover 88 m = 18

88 11× = min

(c) Respect of elders or senior citizens.


Chapter 12

Exercise 12.1

1. (a) 2x – p (b) 13

(p – q) (c) px (d) 15

(2x + 3y)

(e) 50 – xy (f) 5 + 3xy (g) 7z + (x + y) (h) 3y – 5

2. (a) 2ab + 9 (b) x2y2 – 3 Terms = 2ab and 9 Term = x2y2 and –3 Factor of 2ab = 2, a, b Factor of x2y2 are x, x, y, y Factors of 9 = 9 Factor of –3 are = –1, 3

2 + 9ab

2ab

2b

9

a

(c) 6m2 - 4xy (d) 14

(x + y) → 14

x + 14

, y

Terms are 6 m2 and - 4xy Terms are = 14

x and 14

y

Factors of 6 m2 are 6, m, m Factors of 14

x are 14

, x

Factors of –4xy are –1, 4, x, y Factors of 14

y are 14

, y

x

( + )x y14

x14

y14

14

14

y

(e) 10 + 2xy (f) y3 + 2x2 Terms are 10 and 2xy Term are y3 and 2x2

Factor of 10 = 10 Factor of y3 are 1, y, y, y Factors of 2xy = 2, x, y Factor of 2x2 are 1, 2, x, x

10 + 2xy

10 2xy

2x

y

y x3 2+ 2

y3

yy

2y x

x

2x2


(g) 8a2b + ab (h) 3xy +4x2y + 3y2x Terms are 8a2b and ab Term are 3xy +4x2y and 3y2x Factor of 8a2b are 8, a, a, b Factor of 4x2y are 4, x, x, y Factors of ab are a, b Factor of 3y2x are 3, y, y, x

3 + 4 + 3xy x y y x2 2

3xy

x

x1y

y

3y2

y3

4x y2

xx

y14

x

3. (a)Coefficientofx in 4x2yz = 4xyz (b) Coefficientofx2 in 9x2yz = 9yz (c)Coefficientofy in 9x2y2z = 9x2yz (d) Coefficientofz of –14xyz = –14xy (e)Coefficientxz in 4x2z = 4x (f) Coefficientofx3 in x3yz = yz (g)Coefficientofy2 in 3xy3z is 3xyz (h) Coefficientofz in 4xy2z = 4xy2

(i) k in k2mn is kmn. 4. Term other than constant Numeral coefficient (a) a, 2ab, –3c, 5d 1, 2, –3, 5 (b) 5m2 –81m 5, –81 (c) 0.5x2, –0.8y2, 0.6z2 0.5, –0.8, 0.6

(d) 225

x2y, 225

(e) 2l2, 2b2 2, 2 (f) –x4, y3, –7x2 – 1, 1, 7 5. Term containing x Coefficient of x (a) 6x 6 (b) –8yx, 7x –8y, 7 (c) xy y (d) 5xy 5y (e) z2x z2

(f) 7x, xy2 7, y2

6. Term containing y2 Coefficient of y2 (a) –x2y2 –x2

(b) 5y2, 7xy2 5, 7x (c) –15xy2, 7y2 –15x, 7 (d) 12x2y2, 7xy2 12x2, 7x (e) 4x2y2, 5x2y2 4x2, 5x2

(f) 94

xy2, 6y2, 7y2z 94

x, 6, 7z

7. (a) 6x – 13y2 — Binomial (b) –5m3 + m2 + 4m — Trinomial (c) m2 + 81m + y — Trinomial (d) 18xyz + 9 — Monomial (e) x3 + 5x2 — Binomial (f) –21x + 8y2 – 3xy — Trinomial

8. (a) (b)


(c) x x y z3 2+ 5 + 7

x3

x x1

7z

1

5x y2

xx

y15

zx

7

(d) – 21 + 4xy x2

–21xy

21 x4

4x2

–1 xy

x

(e) –7 + 4 + 7xy x y2

–7xy

7 x

1

7

–1

4x y2

xx

y

4y 7

9. (a) 2xy2, –3xy2, 7xy2, 8y2x, are like terms (b) 5x2yz, –5x2zy, 4yzx2, , are like terms (c) 7xyz, 9xyz, 16xyz , are like terms 8yz, –3yz, 5zy, –11yz are like terms –6xy2, 7y2x , are like terms (d) 10p2y, 7p2y are like terms 9pq, 11pq are like terms 8p, 9p are like terms 8y2p, 4py2, 11py2 , are like terms 10. (a) x2 and 8x2 are like terms in x2 + yz + 8x2

(b) –9x2y and 8yx2 are like terms in7xy – 9x2y + 8yx (c) 4ba and 8ab are like terms in 2a2b + 4ba + 6b2 + 8ab

(d) abc and 7abc are like terms in abc + a2bc + abc2 + 7abc

(e) –8xy2 and 2xy2 are like terms in 6x2y – 8x2y + 5xy + 2xy2

(f) t, –3t, 8t and 4t2, –7t2 are like terms t – 3t + 4t2 – 7t2 + 8t

Exercise 12.2 1. (a) 2xy2 + (–4)y2x + 6xy2 + 8y2x = (2 + (–4) + 6 + 8) xy2 = 12xy2

(b) (p – 8qr) + (7qr + p) + (–p + 9qr) + (2p – 3qr) = (p + p + (–p) + 2p) + (–8qr + 7qr + 9qr – 3qr) = 3p + 5qr

(c) (5x2 + 3xy – 4) + (6x2 – 6xy + 7) = (6x2 + 5x2) + (3xy + (–6)xy) + (–4 + 7) = 11x2 + (–3)xy + 3

(d) (4p2q2 + 3pq) + (6pq – 7p2q2) + (8pq + 15p2q2) = (4p2q2 – 7p2q2 + 15p2q2) + (3pq + 6 pq + 8 pq) = 12 p2q2 + 17pq

(e) (p2 + p2q + q2p) + (–3p2 + 4p2q – 7q2p) = (p2 + (–3p2)) + (p2q + 4p2q) + (q2p – 7q2p) = –2p2 + 5p2q2 – 6q2p


(f) 2a + 3b + 4x2y, 4a – 4b + 7x2y = (2a + 4a) + (3b – 4b) + (4x2y + 7x2y) = 6a – b + 11x2y

2. (a) 4x2y – 3xy + 7

8x2y + 9xy + 5

+ 4x2y – 6xy – 9

16x2y + 0xy + 3

(b) 15pq – 8p2q – 4

– 30pq + 9p2q + 8

– 4pq + 9p2q + 3

– 19pq + 10p2q + 7

(c) 30xy – 12x2y + 40

– 60xy + 24x2y – 60

4xy + 7x2y – 40

– 26xy + 19x2y – 60

(d) 8ab + 14bc – 15ca

3ab + 7bc – 4ca

6ab + 4bc – 4ca

17ab + 25bc – 23ca

(e) 3xy2 – 2x2y + 15z2y

– 8xy2 + 4x2y + 6z2y

– 4xy2 – 6x2y + 6yz2

– 9xy2 – 4x2y + 27z2y

(f) 11x + 13y + 15z

16x – 14y – z

12x + y + 17z

39x + 0y + 31z

(g) 7a – 9b + 4c – 3

– 4a + 6b + 8c – 6

– 4a – 4b + 9c + 5

– a –7b + 21c – 4

(h) 3x2 – 5y2 + 2z + 1

–2x2 – y2 + 3z + 0

0 + 2y2 – 7z + 9

x2 + 4y2 + 2z + 10

(i) 8x2 – 17x + 12– 19x2 + 11x – 5 4x2 + 18x – 11– 7x2 + 12x – 4

(j) 9x2 + 6xy + 2y2 + 0

0 + –7xy + 4y2 + 0

– x2 + 0 + 0 + 6

8x2 – xy + 6y2 + 6

3. (a) –6x2y – 8x2y = –14x2y (b) 19xy2z – 13xy2z = 6xy2z (c) (–2a + b + c) – (a + b + c) = –3a + 0b +0c (d) (15x + 4y) – (8x + 2y) = 7x + 2y (e) (9x – 14y + 7c) – (6x – 4y + c) = 3x – 10y + 6c (f) (23xy – 18x2 + 10y2) – (16x2 – 7xy – 28y2) = 30xy – 34x2 + 38y2

(g) (16a + 5b) – (5a + b + 16c) = 11a + 4b – 16c (h) (16b – 7a + 11c) – (6a – 10b + 6c) = –13a + 26b + 5c (i) (9ab – 2a2 – 2b2) – (5a2 – 8ab + 6b2) = 17ab –7a2 – 8b2

(j) (8q2 – 5p2 + 9pq) – (4pq – 5q2 + 8p2) = 13q2 –13p2 + 5pq

4. (a) 16x + 9+ 15x – 1– –

x + 8

(b) 7m – 8p – 12x

15m + 12p + 0 – 16n

– – +

– 8m – 20p – 12x + 16n

(c) 19a + 10b – 15c32a + b – 18c

– – +– 13a – 9b + 3c

(d) 8x2 – 17x + 12– 5x2 + 18x – 12+ – +

13x2 – 35x + 24


(e) 8x y – 14yz + 6abc– 8xy – 12yz + 3abc+ + –

16xy – 2yz + 3abc

(f) 15ab2 + 13a2b – 5ab2ab2 – 16a2b + 3ab

– + –13ab2 + 29a2b – 8ab

(g) 12ab + 18bc + 10ca+ 10ab – 12bc – 15ac

22ab + 6bc – 5ac

(h) 17xy + 8yz + 6p2

– 18xy – 9yz – 3p2

– xy – yz – 3p2

(i) 15x – 14x – 1 + 13x + 2 = 14x – 1

(j) 15

12

p q r− +

–

25

12

12

p q r+ −

15

25

12

12

12

p p q q r r−

− − −

+ +

= – 15

p – q + 32

r

5. (a) x2 + 4xy + 3y2

2x2 + 3xy + 4– – –

–x2 + xy + 3y2 – 4

(b) 30a2b+ 25b – 1725a2b – 31b + 15

– + –5a2b + 56b – 32

∴ –x2 + xy + 3y2 – 4 should be added to 5a2b + 56b – 32 should be added to 2x2 + 3xy + 4 to get x2 + 4xy + 3y2 25a2b – 31b + 15 to get 30a2b + 25b – 17

(c) 7x – 5y– 3x + 5y – 2z+ – +

10x– 10y + 2z

(d) 5x2 + 7y – 3z– 4x2 + 5y + 7z+ – –

9x2 + 2y – 10z 10x – 10y + 2z should be added to ∴ 9x2 + 2y –10z should be added to –3x + 5y – 2z to get 7x – 5y. –4x2 + 5y +7y to get 5x2 + 7y – 3z. 6. (a) Let A should be subtracted from 3x2 – 4y2 + 8xy to get –4x2 + 8y2 – 7xy ∴ (3x2 – 4y2 + 8xy) – A = –4x2 + 8y2 – 7xy (3x2 – 4y2 + 8xy) – (–4x2 + 8y2 – 7xy) = A. 3x2 – 4y2 + 8xy + 4x2 – 8y2 + 7xy = A.

7x2 –12y2 + 15xy = A. (b) Let A be subtract from 8a3 – 10a2 – 3a + 4 to get 11a3 + 6a – 10. ∴ (8a3 – 10a2 – 3a + 4) – A = 11a3 + 6a – 10 (8a3 – 10a2 – 3a + 4) – (11a3 + 6a – 10) = A. (8a3 – 11a3)+ (–10a2) + (–3a – 6a) + (4 + 10) = A

–3a3 – 10a2 – 9a + 14 = A (c) Let A be added to 12x2 – 3x + 18 to get 17x2 + 4x – 23. (12x2 – 3x + 18) + A = 17x2 + 4x – 23

A = (17x2 + 4x – 23) – (12x2 – 3x + 18) = (17x2 – 12x2) + (4x + 3x) + (–23 – 18) = 5x2 + 7x – 41


∴ 5x2 + 7x – 41 must be added. (d) Let A must be subtracted from 18xyz + 4yz – 16 to get 20xyz – 5yz + 17. (18xyz + 4yz – 16) – A = 20xyz – 5yz + 17 (18xyz + 4yz – 16) – (20xyz – 5yz + 17) = A

– 2xyz + 9yz – 33 = A ∴ 2xyz + 9yz – 33 must be subtracted. 7. (a) 14a + 12b + 12c

12a – 14b + 16c26a – 2b + 28c26a – 2b + 28c16a – 18b – 12c

– + +10a + 16b + 40c

(b) 5a3 + 2b3 – 6ab– 2a3 – 3b3 + 9ab

3a3 – b3 + 3ab3a3 – b3 + 3aba3 + b3 – 3ab

– – +2a3 – 2b3 + 6ab

(c) 2x2 – 4xy + 7y2 – 8x2 + 8xy – 3y2 + 7

– + + –x2 – 12xy + 10y2 – 15 x2 – 12xy + 10y2 – 15

+ x2 – 3xy + 7y2 – 22x2 – 15xy + 17y2 – 17

(d) 4x2y + 7xy + 9– 2x2y – 9xy + 16+

2x2y – 2xy + 2525x2y + 9xy + 122x2y – 2xy + 25

– + –23x2y + 11xy – 13

8. (a) Let A should be taken away from 4x2 + 3y2 + 8xy – 17 to get –5x2 – 6y2 + 8xy – 20. (4x2 + 3y2 + 8xy – 17) – A = –5x2 – 6y2 + 8xy – 20. (4x2 + 3y2 + 8xy – 17) – (–5x2 – 6y2 + 8xy – 20) = A 4x2 + 3y2 + 8xy – 17 + 5x2 + 6y2 – 8xy + 20 = A (4x2 + 5x2) + (3y2 + 6y2) + (8xy – 8xy) + (–17 + 20) = A 9x2 + 9y2 + 0 + 3 = A 9x2 + 9y2 + 3 must be taken away. (b) Let A should be added to 4x2 + 3y2 + 8xy – 15 to get 9x2 – 8y2 + 9xy + 30. (4x2 + 3y2 + 8yx – 15) + A = 9x2 + 8y2 + 9xy + 30. \ 9x2 + 8y2 + 9xy + 30

+ 4x2 + 3y2 + 8yx – 15– – – +

5x2 + 5y2 + yx + 45

∴ A = 5x2 + 5y2 + yx + 45

(c) 9a2 + 7a – 2– 2a3 + 4a2 + 3a + 1+ – – –

2a3 + 5a2 + 4a – 3

\ 9a2 + 7a – 2 exceeds – 2a3 + 4a2 + 3a + 1 by 2a3 + 5a2 + 4a – 3


(d) 13x2 – 7y2 + 56x2 – 9y2 + 7

– + –7x2 + 2y2 – 2

∴ 13x2 –7y2 + 5 is 7x2 + 2y2 – 2 greater than 6x2 – 9y2 + 7

(e) Let A should be added. ∴ A= 4x2 – x + 2 – (2x3 – 3x + 6) = 4x2 – x + 2 – 2x3 + 3x – 6 = 4x2 – 2x3 + 2x – 4.

9. (a) A + B + C5x2 + 7x + 84x2 – 7x + 32x2 + 9x – 7

11x2 + 9x + 4

(b) 5x2 + 7x + 8+ 4x2 – 7x + 3

9x2 + 0x + 112x2 + 9x – 7

– – +7x2 – 9x + 18

(c) A + (B – C)4x2 – 7x + 32x2 + 9x – 76x2 + 2x – 45x2 + 7x + 86x2 + 2x – 4– – +

– x2 + 5x + 12

(d) 4x2 + 7x + 3+ 2x2 + 9x – 7– – +

2x2 – 16x + 105x2 + 7x + 8

+ 2x2 – 16x + 10– + –

3x2 + 23x – 2

(e) 5x2 + 7x + 82x2 + 9x – 7

– – +3x2 – 2x + 154x2 – 7x + 33x2 – 2x + 15

– + –x2 – 5x – 12

(f) 5x2 + 7x + 82x2 + 9x – 7

+7x2 + 16x + 14x2 – 7x + 37x2 + 16x + 1

– – –– 3x2 – 23x + 2

Exercise 12.3 1. (a) 2x + 7 = 2 × (–2) + 7 = – 4 + 7 = 3 (b) 4x – 5 = 4 × (–2) – 5 = – 8 – 5 = –13 (c) x2 + 4x + 5 = (–2)2 + 4 × ( –2) + 5 = +4 – 8 + 5 = 1 (d) 50 – 2x2 = 50 – 2(–2)2 = 50 – 8 = 42 (e) 5x2 + 5x + 1 = 5 × (–2)2 + 5 × (–2) + 1 = 20 – 10 + 1 = 11 (f) 2x2 – 5 = 2 × (–2)2 – 5 = 8 – 5 = 3 (g) 2x2 + 3x + 4 = 2(–2)2 + 3 (–2) + 4 = 8 – 6 + 4 = 6 (h) –3x + 7 = –3 (–2) + 7 = 6 + 7 = 13. 2. (a) 5x2 + 1 = 5(5)2 + 1 = 125 + 1 = 126 (b) –2x2 – 7x + 5 = – 2 × (5)2 – 7(5) + 5 = –2 × 25 – 35 + 5 = – 50 – 35 + 5 = – 80 (c) 3x2 – 2x + 7 = 3(5)2 – 2 × 5 + 7 = 3 × 25 – 10 + 7 = 75 – 10 + 7 = 72 (d) –4x + 9 = –4 × 5 + 9 = –20 + 9 = –11 (e) x2 – 9 = (5)2 – 9 = 25 – 9 = 16 (f) 9x2 – 6x + 4 = 9 × (5)2 – 6 × 5 + 4 = 225 – 30 + 4 = 199 3. (a) a2 + 2ab = (1)2 + 2(1) (– 2) = 1 + (–4) = –3 (b) a2 + ab – b2 = (1)2 + (1) (–2) – (2)2 = 1 – 2 – 4 = –5


(c) a2 + 4ab + b2 = (1)2 + 4(1) (–2) + (–2)2 = 1 – 8 + 4 = –3 (d) a3 + b3 + 3ab = (1)3 + (–2)3 + 3(1)(–2) = 1 – 8 – 6 = –13 (e) b2 – a2 – 2ab = (–2)2 – (1)2 – 2(+1) (–2) = 4 – 1 + 4 = 7 (f) – 2a2 + b2 + 4ab = –2(1)2 + (–2)2 + 4(1)(–2) = –2 + 4 – 8 = –6 4. (a) a3 + b3 – 3ab = (–1)3 + (3)3 – 3 (–1)(3) = – 1 + 27 + 9 = 35 (b) a2 + b2 + 2a + 3b = (–1)2 + (3)2 + 2(–1) + 3(3) = 1 + 9 – 2 + 9 = 17 (c) 3a + 4b – b2 = 3(–1) + 4(3) – (3)2 = –3 + 12 – 9 = 0 (d) a3 – b3 = (–1)3 – (3)3 = –1 – 27 = –28 (e) 3ab – a2 + b2 = 3(–1) (3) – (–1)2 + 32 = –9 – 1 + 9 = –1 (f) –4(–1)(3) + 5(–1)2 + (3)2 = 12 + 5 +9 = 26 5. (a) 2a+ b + 3c – 3a = –a + b + 3c = –(–1) + 2 + 3(1) = + 1 + 2 + 3 = 6 (b) 2a + a – b – 7 = 2a – b + 2 = 2 × (–1) –2 + 2 = –2 (c) 6a + 5(b + a) – 4b = 6a + 5b + 10 – 4b = 6a + b + 10 = 6 × (–1) + (2) + 10 = –6 + 12 = 6. (d) (8a + 4b – 3c) + (5a – 2b + 3c) = 13a + 2b = 13 × (–1) + 2 × 2 = –13 + 4 = –9 (e) 6(a + 5) – 3 (a – 2) = 6a + 30 – 3a + 6 = 3a + 36 = 3 × (–1) + 36 = – 3 + 36 = 33 (f) 4(2b – 3) + (6b – 11) = 8b – 12 + 6b – 11 = 14b – 23 = 14 × 2 –23 = 28 – 23 = 5 (g) 2(a2 + ab) + 7 – ab = 2a2 + 2ab + 7 – ab = 2a2 + ab + 7

= 2(–1)2 + (–1)(2) + 7 = 2 – 2 + 7 = 7 (h) a2 – 3(b2 – 3a2) = a2 – 3b2 + 9a2 = 10a2 – 3b2 = 10(–1)2 – (3) × (2)2 = 10 – 12 = –2 (i) (c2 – 2ac) – (2c2 + 2ac) = c2 – 2ac – 2c2 – 2ac = – c2 – 4ac = – (1)2 – 4(–1)(1) = – 1 + 4 = 3 (j) (a + b) + (b + c) + (c – a) = 2b + 2c = 2(2) + 2(1) = 6 6. (a) x2 + y2 + z2 = (1)2 + (–1)2 + (3)2 = 1 + 1 + 9 = 11 (b) x2 – xy + 3y – y2 = (1)2 – (1)(–1) + (–1) (3) – (–1)2 = 1 + 1 – 3 – 1 = –2 (c) y2 + z2 – 3xyz = (–1)2 + (3)2 – 3 (1) (–1) (3) = 1 + 9 + 9 = 19 (d) x + y2 –3z2 = 1 + (–1)2 – 3 (3)2 = 1 + 1 – 27 = –25 (e) z3 – y2 – x = (3)3 – (–1)2 – (1) = 27 – 1 – 1 = 25 (f) x3 – 2y2 – 3z2 = (1)3 – 2(–1)2 – 3(3)2 = 1 – 2 – 27 = –28 7. (a) z3 – 4z2 + 3z + 30 = (5)3 – 4(5)2 + 3(5) + 30 = 125 – 100 + 15 + 30 = 70 (b) p2 –3(p + 1) – 2 = (–5)2 – 3(–5 + 1) – 2 = +25 + 12 – 2 = 35 (c) q3 – 2(q + 1) + 3q2 = (10)3 – 2(10 + 1) + 3(10)2 = 1000 – 22 + 300 = 1278 (d) x3 + y3 – 3x2 – 4y2

(1)3 + (–1)3 – 3(1)2 – 4 (–1)2 = 1 – 1 – 3 – 4 = –7 8. (a) 2(1)2 + 5(a) – 9 = 3

5a – 7 = 3 5a = 10

a = 105

= 2

(b) 3x2 + 2(x + 52) – a = 5 3(–2)2 + 2(–2 + 52) – a = 5 +12 + 2 × 50 – a = 5


+12 + 100 – a – 5 112 – a = 5 112 – 5 = a 107 = a

(c) 4y2 + 4(y – 7) + b = 20 4(–1)2 + (4) (–1–7) + b = 20 4 + 4 (–8) + b = 20 4 – 32 + b = 20 b = 20 + 28 = 48 9. (a) 2(a2 + b2 + c2) –3(ab + bc + ca)

= 2(52 + (–3)2 + (–2)2) – 3[(5 × (–3) + (–3)(–2) + (2)(5)] = 2 (25 + 9 + 4) – 3 (–15 + 6 – 10) = 2 × 38 – 3 (–19) = 76 + 57 = 133

(b) 3ab + 3bc + ab + bc = 4ab + 4bc – 2ca = 4(5) (–3) + 4(–3) (–2) – 2(–2)(5) = –60 + 24 + 20 = –16

10. No of pattern (a) is 6, 11, 16... ∴ 5n + 1 is the rule (b) 4 no of sticks required are in the pattern of 4, 7, 10 rule is 3n + 1 (c) The no of line segment required are in the pattern of 5, 10, 15... rule is 5n. 11. (a) Area = s × s (b) Perimeter = 4 × side = 4a (c) Area of rect. = l × b (d) Perimeter = 2(l × b) (e) Circumference of circle = 2pr (f) Area = pa2

Worksheet 1

1. (a) (b)

(c)

2. (a) 2x + 8 (b) 12

(x + y) + 7 (c) 20 – xy (d) 7 – x2

3. (a) a2x, 4a2x; 2ax, 5ax (b) 5abc2, –4abc2;8acb2, –3ab2c, 4ab2c 4. (a) (5x – 3x) + (7h + 4h) (b) 3x2 + 2x – 2x2 + 3x + x2

= 2x + 11h = (3x2 – 2x2 + x2) + (2x + 3x)


= 2x2 + 5x (c) 6ab + 3bc – 4ab + bc – 3ab (d) 3m + 2q – 5q – 4m – 7q = (6ab – 4ab – 3ab) + (3bc + bc) = (3m – 4m) + (2q –5q – 7q) = – ab + 4bc = – m – 10q

5. (a) – 5x6 + 3x4 – 4x3

– 9x6 + 4x4 + 20x3

– 14x6 + 7x4 + 16x3

(b) + 7x2 – 2x + 5– 2x2 – 15x + 4– 3x2 – 5x + 4

2x2 – 22x + 13

(c) 5x2 + 3xy + y2

+ 2x2 – 5xy – 3y2

– 4x2 – 0 + 5y2

3x2 – 2xy + 3y2

(d) – 8a2b + 3ab2 – ab– 6a2b + 2ab2 –2ab– 14a2b + 5ab2 –3ab

6. (a) 3a + 7b + 86a + 4b – 6

– – +– 3a + 3b + 14

(b) – 4a + 54b – 7– 9a – 7b + 9+ + –

5a + 61b – 16

(c) 15m – 15n + 4mn16m + 12n – 3mn

– – +– m – 27n + 7mn

(d) – 8a2b + 3ab2 – ab6a2b + 2ab2 –3ab

– – +– 4a2b + ab2 +2ab

7. (a) Let A be added x2 + 2xy + y2 to get 3x2 + 4xy + 5y2

Then A = 3x2 + 4xy + 5y2

x2 + 2xy + y2

– – –2x2 + 2xy + 4y2

∴ A = 2x2 + 2xy + 4y2

(b) Let A should be subtracted from 2ab + ab2+ 10a2 to get –4a2 + 8ab – 7b2

∴ A = 2ab + 9b2 + 10a2

+ 8ab – 7b2 – 4a2

– + +– 6ab + 16b2 + 14a2

8. (a) Let A should be taken Away then 3x2 – 4y2 + 7xy + 40 – A = 5x2 + 7y2 – 8xy – 10

∴ A = 3x2 – 4y2 + 7xy + 40– 5x2 + 7y2 – 8xy – 10+ – + +

8x2 – 11y2 + 15xy + 50


(b) 5ab + 3b + 2c4ab + 6b – 3c

– – +1ab – 3b + 5c

9. (a) 3x – 2y + 11– 3y + 7

3x – 5y + 18+ 4x + 7y – 15– – +– x – 12y + 33

(b) 4x2 + 5y – 9x3x2 + 0 – 8x7x2 + 5y – 17x9x2 + 8y + 15x7x2 + 5y – 17x

– – +2x2 + 3y + 32x

10. (a) 2p – 3q + 7 (2 × – 1) – (3 × 2) + 7 = – 2 – 6 + 7 = – 8 +7 = –1

(b) (m – n) + m2 + n2

(2 + 1) + (2)2 + (–1)2

3 + 4 + 1 = 8 (c) 7x2 + 3xy + y2

7 × 22 + 3(2) (3) + (3)2 = 28 + 18 + 9 = 55 (d) 3x4 – 2x2 + 5x – 6 3(–1)4 – 2(–1)2 + 5 × (–1) – 6 = 3 – 2 – 5 – 6 = 10 (e) 7abc + 3a2bc – 4a2b2c 7 × 1 × (–1) × (2) + 3(1)2 (–1) (2) – 4 × (1)2 × (–1)2 × (2) = –14 – 6 – 8 = –28

11. (a) P = 4 × s (b) A = P + I (c) S = dt

Worksheet 2 1. (a) p (b) like (c) unlike (d) r (e) one (f) n and 6n (g) Monomial (h) 55y km (i) 2x2 (j) (b + c) 2. (a) True (b) False (c) True (d) False (e) True (f) False (g) False (h) False (i) True (j) False 3. (a) x2 + xy — Binomial (b) r – 3p × 2q = r – 6pq — Binomial 4. (a) 1 (b) –2 (c) 3 (d) 0 5. (a) 3x2yz2 – 3xy2z + x2yz2 + 7xy2z = (3x2yz2 – x2yz2) + (–3xy2z + 7xy2z) = 4x2yz2 + 4xy2z Binomial (b) x4 + 3x3y + 3x2y2 – 3x3y – 3xy3 + y4 – 3x2y2

= x4 + (3x3y – 3x3y) + (3x2y2 – 3x2y2) – 3xy3 + y4 = x4 + 0 + 0 – 3x3y + y4

Trinomial (c) p3q2r + pq2r3 + 3p2qr2 – 9p2qr2 = p3q2r + pq2r3 + (–6) p2qr2

Trinomial (d) 2a + 2b + 2c – 2a – 2b – 2c – 2b + 2c + 2a = 2a + (–2)b + 2c


Trinomial (e) 50x3 – 21x + 107 + 41x3 – x + 1 – 93 + 71x – 31x3

= 60x3 + (–21 – 1 + 71)x + (107 + 1 – 93) = 60x3 + 49x + 15 Trinomial 6. (a) (y4 – 12y2 + y + 14) – (17y3 + 34y2 – 51y + 68)

= y4 – 12y2 + y + 14 – 17y3 – 34y2 + 51y – 68 = y4 – 17y3 – 46y2 + 52y – 54

(b) (93p2 – 55p + 4) – (13p3 – 5p2 + 17p – 90) 93p2 – 55p + 4 – 13p3 + 5p2 – 17p + 90 = + 98p2 – 72p – 13p3 + 94

(c) –99x3 + 33x2 + 13x + 41 must be added to 99x3 – 33x2 – 13x – 41 to make the sum zero.

7. (a) m + n + p = 1 + (–1) + 2 = 2 (b) m2 + n2 + p2 = (1)2 + (–1)2 + (2)2 = 1 + 1 + 4 = 6 (c) m3 + n3 + p3 = (1)3 + (–1)3 + (2)3 = 1 – 1 + 8 = 8 (d) mn + np + pm = 1 × (–1) + (–1) × 2 + 2 × 1 = –1 – 2 + 2 = –1 (e) m3 + n3 + p3 – 3mnp = (1)3 + (–1)3 + (2)3 – 3 (1) (–1) (2) = 1 – 1 + 8 + 6 = 14 (f) m2n2 + n2p2 + p2m2 = (1)2 (–1)2 + (–1)2 (2)2 + (2)2 (1)2 = 1 + 4 + 4 = 9 8. P + Q + R = – (x – 2) + (–2 (y + 1)) + (–x + 2y)

ax = –x + 2 – 2y – 2 – x + 2y ax = –2x a = –2.

9. (i) (b) (ii) (c) (iii) (d) (iv) (a) (v) (f) (vi) (e) 10. We know the sum of numbers up to 10 times of table 1 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 = 55 or 55 × 1 (a) Sum of number 7 + 14 + 21 + ...... + 70 = 55 × 7 = 385 (b) Sum of number 10 + 20 + 30 + ...... + 100 = 55 × 10 = 550

(c) Sum of number 19 + 38 + 57 + ...... + 190 = 55 × 19 = 1045

Chapter 13

Exercise 13.1 1. (a) 25 = 2 × 2 × 2 ×2 × 2 = 32 (b) a3 = 9 × 9 × 9 = 729. (c) 83 = 8 × 8 × 8 = 512 (d) 172 = 289 (e) 54 = 5 × 5 × 5 × 5 = 625 (f) 94 = 6561 (g) 36 = 3 × 3 × 3 × 3 × 3 × 3 = 729 (h) 73 = 7 × 7 × 7 = 343 (i) 84 = 8 × 8 × 8 × 8 = 4096 (j) 56 = 5 × 5 × 5 × 5 × 5 × 5 = 15625 2. (a) 7 × 7 × 7 × 7 = 74 (b) t × t × t = t3

(c) a × a × a × a × a = a5 (d) 5 × 5 × 7 × 5 × 7 × 5 = 54 × 72

(e) 2 × a × 2 × a × 2 = 23a2 (f) a × a × a × c × c × c × d = a3c3d 3. (a) 600 = 2 × 2 × 2 × 3 × 5 × 5 = 23 × 3 × 52

(b) 1125 = 5 × 5 × 5 × 3 × 3 = 32 × 53

(c) 5040 = 2 × 2 × 2 × 2 × 3 × 3 × 5 × 7 = 24 × 32 × 5 × 7


(d) 1440 = 2 × 2 × 2 × 2 × 2 × 3 × 3 × 5 = 25 × 32 × 5 (e) 800 = 2 × 2 × 2 × 2 × 2 × 5 × 5 = 25 × 52

(f) 1600 = 2 × 2 × 2 × 2 × 2 × 2 × 5 × 5 = 26 × 52

(g) 1728 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3 = 26 × 33

4. (a) 34 or 43

34 = 3 × 3 × 3 × 3 = 81 43 = 4 × 4 × 4 = 64 34 > 43

(b) 26 = 2 × 2 × 2 × 2 × 2 × 2 = 64 62 = 6 × 6 = 36 26 > 62

(c) 35 = 3 × 3 × 3 × 3 × 3 = 243 53 = 5 × 5 × 5 = 125 35 > 53

(d) 25 = 2 × 2 × 2 × 2 × 2 = 32 52 = 25 25 > 52

(e) 210 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 1024 102 = 10 × 10 = 100 210 > 102

(f) 45 = 1024 54 = 625 ∴ 45 > 54

5. (a) 2 12962 6482 3242 1623 813 273 9

3

(b) 2 16202 8105 4053 813 273 93 3

1 ∴ 1296 = 2 × 2 × 2 × 2 × 3 × 3 × 3 × 3 ∴ 1620 = 2 × 2 × 3 × 3 × 3 × 3 × 5

(c) 2 10802 5402 2705 1353 273 9

3

(d) 2 72002 36002 18002 9002 4505 2255 453 9

3


∴ 1080 = 2 × 2 × 2 × 3 × 3 × 3

7200 = 2 × 2 × 2 × 2 × 2 × 3 × 3 × 5 × 5 (e) 2 9600

2 48002 24002 12002 6002 3002 1505 755 15

3

(f) 2 28802 14402 7202 3602 1802 903 453 15

5

9600 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 × 5 × 5 2880 = 2 × 2 × 2 × 2 × 2 × 2× 3 × 3 × 5

(g) 5 12255 2457 49

7

(h) 3 21873 7293 2433 813 273 9

3

1225 = 5 × 5 × 7 × 7

2187 = 3 × 3 × 3 × 3 × 3 × 3 × 3

(i) 7 24017 3437 49

7

(j) 2 24303 12153 4053 1353 453 15

5

2401 = 7 × 7 × 7 × 7

2430 = 2 × 3 × 3 × 3 × 3 × 3 × 5 6. (a) 4 × 103 = 4 × 10 × 10 × 10 = 4000 (b) 6 × 24 = 6 × 2 × 2 × 2 × 2 = 96 (c) 3 × 43 = 3 × 4 × 4 × 4 = 192 (d) 52 × 32 = 25 × 9 = 225 (e) (–1)5 × 73 = –1 × 7 × 7 × 7 = –343 (f) (–3)2 × 43 = –3 × – 3 × 4 × 4 × 4 = 576 (g) 42 × (–2)5 = 4 × 4 × (–2) × (–2) × (–2) × (–2) × (–2) = –512 (h) 33 × 52 = 3 × 3 × 3 × 5 × 5 = 675 7. (a) (–3)2 × (–5) = –3 × –3 × – 5 = –45 (b) (–2)3 × (3)2 = (–2) × (–2) × (–2) × 3 × 3 = –72 (c) (–5)2 × (–2)4 = –5 × –5 × –2 × –2 × –2 × –2 = 400 (d) (–7)2 × (–3)3 = –7 × –7 × –3 × –3 × –3 = – 1323


(e) (–2)4 × (–3)3 = –2 × – 2 × –2 × –2 × –3 × –3 × –3 = –432 (f) (–3)4 × (–2)3 = –3 × –3 × –3 × –3 × –2 × –2 × –2 = –648 8. (a) 1.9 × 1012 : 2.3 × 1015

Digits in 2.3 × 1015 is more as compared to 1.9 × 1012 80 2.3 × 1015 > 1.9 × 1012. (b) 2.7 × 1013 and 1.3 × 1015

Digits in 1.3 × 1015 is more as compared to 2.7 × 1013 so 1.3 × 1015 is greater. (c) 2.8 × 1020 and 3.5 × 1018

Digits in 2.8 × 1020 is more as compared to 3.5 × 1018 so 2.8 × 1020 is more greater. (d) 2.034 × 1015 and 1.089 × 1017

Digits in 2.034 × 1015 is less as compared to 1.089 × 1017 so 2.034 × 1015 is greater

9. (a) 1000000 = 106 (b) 625 = 5 × 5 × 5 × 5 = 54

(c) 64486

2 2 2 2 23 3 3 3 3

23

32

243

5

= × × × ×× × × ×

=

(d) 2187128

3 3 3 3 3 3 32 2 2 2 2 2 2

32

7

= × × × × × ×× × × × × ×

=

(e) 9004900

949

37

2

= =

10. (a) (12 × 3)5 = (12)5 × (3)5 = (2 × 2 × 3)5 × 35 = 25 × 25 × 35 × 35 = 210 × 310

(b) (49)3 = (72)3 = 76

(c) (–6 m)3 = (–6)3m3

(d) (–625a)5 = (–625)5a5 =(–54)5a5 = – 520a5

(e) (3mn)3 = 33m3n3

(f ) 4 4 2 26 6

6

2 6

6

12

6m m m m

= = =( )

11. (a) 85 = (2 × 2 × 2)5 = (23)5 = 2 3 × 5 = 215

(b) (16)4 = (4 × 4)4 = (42)4 = 48

(c) 25 × 25 × 25 = (25)3 = (5 × 5)3 = (52)3 = 56

12. (a) 4x × 3x = (4 × 3)x = 12x

(b) (52)5 ÷ 58 = 510 ÷ 58 = 510 – 8 = 52

(c) (62 × 63) ÷ 65 = (62+3) ÷ 65 = 65 ÷ 65 = 65–5 = 60 = 1 (d) 23 × 53 × 24 = 23 × 24 × 53 = 23+4 × 53 = 27 × 53

(e) (23)2 × 36 × 56 = 26 × 36 × 56 = (2 × 3 × 5)6 = (30)6

(f) 44

48

35

× = (48 –3) × 45

xx

xa

ba b=

−

= 45 × 45 = 45+5 = 410 (xm × xn = xm+n) (g) 23 × a3 × 2 × a5 × 24 ( xm × ym = (x + y)m

23+4+1 × a3+5 × 5 = 28 × a8 = (2a)8 ( xm × xn = xm + n)

2 1282 642 322 162 82 4

2

3 21873 7293 2433 813 273 9

2


(h) [(42)2 × 54] × 64 = [44 × 54] × 64 = (4 × 5 × 6)4 = (120)4

(i) (215 ÷ 210) ÷ 23 = (215–10) ÷ 23 (xm ÷ xn = xm–n) = 25 ÷ 23 = 25 – 3 = 22

13. (a) ( )

( )2 74 7

2 74 7

2 72 7

2 72 7

4 3 5

2 2

12 5

2 2

12 5

2 2 2

12 5

4 2×

×= ×

×= ×

×= ×

× = 212 – 4 × 75 – 2 = 28 × 73

(b) 13 5169 25

13 513 5

3 7 3 7

2 2××

= ××( ) ( )

= 133–2 × 57 – 2 = 131 × 55

(c) (3° + 2°) ÷ 2° = (1 + 1) ÷ 1 = 2

(d) 5 3 153 125

5 3 3 53 5

5 33 5

5 33 5

3 8

4

3 8

4 3

3 1 8 1

4 3

4 9

4 3× ×

×= × × ×

×=

×= ×

×

+ + = 54–3 × 39–4 = 51 × 35

(e) 3 22 81

3 22 3

8 5

3

8 5

3 4××

= ××

= 38–4 × 25–3 = 22 × 34

(f) 3 7 1111 3 7

4 3 5

3 2× ×

× × = 34–2 × 73–1 × 115–3 = 32 × 72 × 112

(g) [(52)3 ÷ 54] × 53 = (56 ÷ 54) × 53 = (56–4) × 53 = 52+3 = 55

(h) aa

8

5

× a4 = (a8–5) × a4 = a3 × a4 = a7

(i) 44

8 9 5

5 3 4××

a ba b

= 48 – 5 × a9 – 3 × b5 – 4 = 43a6b1

(j) 3 10 355 6

3 2 5 7 55 2 3

3 2 5 7 55 2 3

5 5

7 5

5 2

7 5

5 2 2

7 5 5× ×

×= × × × ×

× ×= × × × ×

× ×( )

( )

= 3 2 5 7

5 2 33 7

5 23 75 2

5 2 3

7 5 5

5 5

7 3 5 2

0

4 3× × ×

× ×= ×

×= ×

×

−

− − = 7

5 24 3×

(k) 27

9 33

3 3

3 7 5

6 3

3 3 7 5

2 6 3× − ×

× ×= × − ×

× ×( ) ( ) ( )

( )m n

nm n

n

= 3

3 33

3 3

9 7 5

12 3

9 7 5 1

12 3

7

15 94× − ×

× ×= × − × = − ×

−

+ −( ) ( ) ( )m n

nm n m

n = ( )− ×m n7 4

63

(l) 343 6456

7 22 7

7 5

4 3

3 6 7 5

3 4 3×

× ×= × ×

× × ×a b

a ba b

a b( ) ( )

= 73 – 1 × 26 – 3 × a7 – 4 b5 – 3 = 72 × 23 × a3 × b2

(m) 125 510

5 52 5

52 5

3 5

3 3

3 3 5

3 3

6 5

3 3 3× ×

×= × ×

× ×= ×

× ×t

ttt

tt( )

= 5

2

6 3 5 3

3

− −× t = 52

3 2

3× t

(n) 3 10 1255 6

3 2 5 55 2 3

2 5

7 3

2 5 3

7 3× ×

×= × × ×

× ×( )

( ) =

3 2 5 55 2 3

3 2 55 2 3

2 5 5 3

7 3 3

2 5 8

7 3 3× × ×

× ×= × ×

× ×

= 2 53

5 3 8 7

3 2

− −

−× = 2 5

3

2 ×


14. (a) True (b) False (c) False (d) False (e) False (f) True 15. (a)2°+3°+4°=1+1+1=3≠1 (b) 2° × 3° × 4° = 1 × 1 × 1 = 1 (c) (3° – 2°) × 4° = 0 × 1 = 0 (d) (3° – 2°) × (3° + 2°) = 0 × 2 = 0 So b is equal to 1.

16. (a) 711

711

711

3 4 3 4 12

=

=

×

(b) 23

23

23

23

10 2 5 10 10 0

÷

=

=

−

(c) a6 × a5 × a4 = a6+5+4 = a15

(d) 1 Lakh = 105

(e) 432 = 24 × 33

(f) 32 < 15

(g) 25

52

13 3

÷

=−

17. 23

23

23

3 6 2 1

×

=

−m

23

23

9 2 1

=−m

∴ 9 = 2m – 1 (since bases are same so power are also equal) 9 + 1 = 2m

5102

= = m

18. (a) 12

18

84 1

88

38

83

23

23

32−

× = −

× = × = 33 × 82–3 = 33 × 8–1 = 38

278

3

=

(b) (12 + 22 – 32) ÷ (–4)0

(1 + 4 – 9) ÷ (–4)0

(–4) ÷ (–4)0 = –4

(c) (34 – 23) ÷ 52 = (81 – 8) ÷ 52 = 65 ÷ 25 = 13 × 5 ÷ 5 × 5 = 135 19. No. of rows 4, No of soilder in each row = 8 × 4 = 32

(a) Total no. of soldiers in marching troop = 32 × 4 = 128

(b) No of soilder is 16 such troops =16 × 128 = 24 × 27 = 211

(c) On Republic Day our constitution was implemented. It is celebrated on 26th January every year.


Exercise 13.2 1. (a) 5.89564 × 104 (b) 6.95380 × 105 (c) 3.456 × 107

(d) 7.0048007 × 108 (e) 4.85 × 106 (f) 5.6 × 107

(g) 1.089763 × 103 (h) 403.8563 = 4.038563 × 102

(i) 9.27586 × 108

2. (a) 289563 = 2 × 105 + 8 × 104 + 9 × 103 + 5 × 102 + 6 ×101 + 3 (b) 3056897 = 3 × 106 + 5 × 104 + 6 × 103 + 8 × 102 + 9 × 101 + 7 (c) 20823467 = 2 × 107 + 8 × 105 + 2 × 104 + 3 × 103 + 4 × 102 + 6 × 101 + 7 (d) 10897936 = 1 × 107 + 8 × 105 + 9 × 104 + 7 × 103 + 9 × 102 + 3 × 101 + 6 3. (a) 904245 (b) 6347009 (c) 906050347 (d) 84007462 (e) 5473025 (f ) 30450349 4. (a) 3.7 × 105 km (b) 3.288 × 106 sq. km (c) 6.6 × 1020 meteric tonnes (d) 5.102 × 108 km2 (e) 3.0 × 108 m/s (f) 3.61419 × 108 sq. km (g) 1.48647 × 107 sq km

Worksheet 1

1. (a) (6)5 (b) (–2)5 (c) −

15

4 (d) −

1513

6

2. (a) −

= −125

1728125

3 (b) –216 (c) 2 × 9 × 1 = 18

(d) +1625

(e) 4 × 9 × 25 = 900 (f) 0

3. (a) 23

23

6

6

6

=

(b) − = −

1512

12

9 (c) −

110

5 (d) 2

7

4

(e) − = −

25

25

5

5

5 (f) 2880 = 2 × 2 × 2 × 2 × 2 × 3 × 3 × 2 × 5 = 26 × 32 × 5

4. (a) (152)3 = 152 × 3 = 156 (b) −

× −

= −

= −

+59

59

59

59

3 8 3 8 11

(c) 12

12

12

123

4

12

12

12

12

× = = =

(d) −

=1

31

3 0

(e) 12

12

12

12

3 2 4 3 2 4

×

=

=

+ 55 4 2012

×

=

(f) 8 88

88

85 4

3

9

39 3× = = − = 86 = (23)6 = 218


(g) 15 3 5

15

15 3 5

15

15 15

15

15

151

7 2 2

9

7 2

9

7 2

9

9

9

× × = × × = × = =( )

(h) 13

993

9

3

3

333

22 4

2

4

2

2 4

2

8

2× −

= − =( )

=( )

=( )( ) ( )

= 38 – 2 = 36

5. (a) 3.8 × 1015 and 0.9 × 1018

Digits in 0.9 × 1018 is more as compared to Digits in 3.8 × 1015. So 0.9 ×1018 is greater

(b) 4 × 1014 : 3 × 1017

Digits in 3 × 1017 is more as compared to digits in 4 × 1014. (c) 2.7 × 1012 and 3.9 × 109

Digits in 2.7 × 1012 are more as compared to digits in 3.9 × 109. So 2.7 × 1012 is greater.

(d) 5.2 × 1017 and 2.5 × 1019

Digits in 2.5 × 1019 is more as compared to 5.2 × 1017. So 2.5 × 1019 is greater.

6. (a) 2 100002 50002 25002 12505 6255 1255 25

5

(b) 2 6250002 3125002 1562505 781255 156255 31255 6255 1255 25

5 10000 = 2 × 2 × 2 × 2 × 5 × 5 × 5 × 5 625000 = 2 × 2 × 2 × 5 × 5 × 5 × 5 × 5 × 5 × 5 = 24 × 54 = 23 × 57

(c) 5 729005 145802 29162 14583 7293 2433 813 273 9

3

(d) 2 18002 9002 4505 2255 453 93 3

1

72900 = 5 × 5 × 2 × 2 × 3 × 3 × 3 × 3 × 3 × 3 1800 = 2 × 2 × 2 × 5 × 5 × 3 × 3 = 22 × 36 × 52 = 23 × 32 × 52


(e) 2 28802 14402 7202 3602 1802 903 453 15

5

2880 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 5 = 26 × 32 × 5

7. (a) 9 6 12

54 12

3 2 3 2 3

2 3 2 3

3 2 3 23 8

5

2 3 2 8

3 2 5

2 3 3 16× ××

= × × × ×× × ×

= × × × ×( ) ( )

( )

33

2 3 2 3

8

3 10 5× × ×

= 3 2

2 3

2 3 8 3 16

1 10 3 5

+ + +

+ +××

= 3 2

2 3

13 19

11 8

××

= 313 – 8 × 219–11 = 35 ×28

(b) 2

4

2

2

8 5 6

3 2 2

8 5 2 6 2

3

× ×× ×

= × ×− −a t

a t

a t

( ) = 28–6 × a3 × t4 = 22 × 93 × t4

(c) 2 7

8 72 72 7

2 72 7

5 2 3

3 2

10 3

3 3 2

10 3

9 2

( ) ×

×= ×

×= ×

×( ) = 210–9 × 71 = 2 × 7 = 14

(d) 125 2

10

5 2

2 5

3 7

3 4

3 3 7

3 3 4× ×

×= ×

× ×t

t

t

t = 53–3 × 23–3 t7–4 = 50 × 20 × t3 = t3

(e) 3 10 125

5 6

3 10 5

5 2 3

3 2 5 5

5 2 3

8 5

7 3

8 5 3

7 3 3

8 5 5 3

7 3 3× ×

×= × ×

× ×= × × ×

× ×

= 38 – 3 × 25 – 3 × 58 – 7 = 51 × 22 × 35 = 22 × 35 × 5

(f) 2 4 53 64

2 2 5

3 2

4 3 7 4 6 7

6× ×

×= × ×

× = 2 5

3 2

2 53

2 53

10 7

6

10 6 7 4 7××

= × = ×−

8. (a) 2 × 106 + 4 × 105 + 9 × 104 × 3 × 103 + 7 × 102 × 8 × 101 + 6

(b) 3 × 106 × 8 × 103 + 5 × 101 + 6

(c) 1 × 108 + 8 × 106 + 5 × 105 × 9 × 102 + 7 × 101 + 2

(d) 4 × 106 + 1 × 104 + 5 × 103 + 6 × 102 + 2 × 101 + 3

9. (a) 3.85 × 108 (b) 9.86 × 107 (c) 1.005078 × 106

(d) 8.23 × 106 (e) 1.48536 × 102 (f) 2.46315 × 103

10. (a) 1.28 × 1011 years (b) 6.023 × 1014 (c) 1.27 × 108 m

(d) 1.027 × 109 (e) 8.68 × 1020 tonnes



1. (a) (b) (c) (d)

(e) (f) No line of symmetry (g)

(h) No line of symmetry (i) (j)

2. (a) Scalene triangle (b) parallelogram (c) non-isosceles trapezium 3. (a) perpendicular bisector (b) angle bisector (c) Diameter 4. (a) F, G, J, N, L, P, Q, R, S, Z (b) A, B, C, D, E, K, M, T, U, V, W, Y

(c) H, I, X (d) O

6. (a) one (b) one (c) 5 (d) 8

7. (a) 4 (b) 2 (c) 2 (d) 1 (e) 1 (f) 4

8. (a) (b) (c) (d)

(e) (f)

9. (a) 3 lines of symmetry (d) 2 lines of symmetry (e) 4 lines of symmetry (f) 4 lines of symmetry 10. (a) False (b) False (c) True (d) True (e) False (f) True (g) False (h) False


Exercise 14.2 1. H, I, N, S, Z 2. From the key point. 3. (a) 2 (b) 1 (c) 4 (d) 4 (e) 6 (g) 8 (h) 3 (i) 6 4. Equilateral triangle and square have both line symmetry and rotational symmetry. 5. Parallelogram 6. Perpendicular bisector of diameter, No 7. Name of Figure Centre of rotational symmetry Angle of rotation (a) Rectangle Point of intersection of diagonals 180° (b) Parallelogram Point of intersection of diagonals 180° (c) Square Point of intersection of diagonals 90° (d) Regular pentagon Point of intersection of diagonals 72° (e) Isosceles triangle Mid-point of perpendicular from vertex to base. 360° (f) Equilateral triangle Orthocentre 120° (g) Hexagon Centre 60° (h) Circle Centre Not defined (i) Semicircle Mid-point of diameter 360°

8. Capital alphabets

No. of lines of symmetry

Rotational symmetry

Order of rotational symmetry

A 1 No 1B 1 No 1C 1 No 1H 2 Yes 2E 1 No 1I 2 Yes 2N 0 Yes 2S 0 Yes 2

9. Parallelogram 10. At 240° it will look exactly the same

11. (a) Yes (b) Yes (c) Yes (d) Yes (e) No

12. (a) sides (b) rotational (c) angle of rotation (d) no (e) line symmetry 13. More than 1

Worksheet 1. (a) B, H, O, W, X (b) All have line symmetry 2. (a) 2 (b) 5 (c) 4 (d) 2 3.


4. I, O, N, Z, S 5. (a) 90° (b) 90° (c) 120° (d) 0 (e) 180° 6. Rectangle

7. (a) (b) (c)

(d) (e) No line of symmetry

8. 88.

Chapter 15

Exercise 15.1 1. (a) 6, 12, 8 (b) 6, 12, 8 (c) 3, 2, 0 (d) 2, 1, 1 (e) 1, 0, 0 (f) 5, 9, 6 (g) 5, 8, 5 (h) 9, 16, 9 (i) 7, 15, 10 2. (a) Cylinder (b) Cone (c) Cuboid (d) Cube (e) Prism (f) pyramid (triangular) 3. Cuboid – juice can, pencil box, eraser etc. Sphere – ball, tomato, globe etc. Cone – Joker cap, ice cream cone, etc. Cylinder – powder can, glass, bottle etc. Cube – Dice, ice cube 4. (a) a = 5, b = 3, c = 6 (b) a = 4, b = 5, c = 1 5. (a) Yes (b) Yes 6. (a) Yes (b) Yes (c) Yes (d) Yes (e) Yes (f) No 7. (a) Cuboid (b) Cuboid (c) Cube (d) Cube (e) Cylinder (f) Cube (g) Cone (h) Cuboid (i) Sphere 8. (a) Cube (b) Cone (c) Cuboid (d) Cube/Cuboid (e) Cylinder (f) Sphere 9. Cylinder and cone 10. Cuboid

Exercise 15.2

1. (a) (b)


(c)

2. (i) (ii)

(iii)

3.


4. (a)

(b)

5.

6. (a)


(b) (c)

7. (a) 4 + 1 (b) 2 + 3 (c) 5 + 4

8.

9. (a) 9 (b) 16 (c) 29

10. (a) 15 (b) 18 (c) 15

11. (a) rectangle (b) circle (c) circle (d) rectangle

(e) triangle (vertically), circle (horizontally) (f) circle

(g) rectangle (vertically), circle (horizontally) (h) square

12. (a) book (b) dice (c) birthday caps (d) ball


13.

14. (a) 6

(b) 8

(c) 7

(d) 8

(e) 6

(f) 8

(g) 6


(h) 8

Worksheet 1. (a) triangular. (b) Cone (c) sphere (d) 2 (e) Prism (f) 12 × 4 × 4 (g) Triangle 2. (a) Cuboid (b) Cylinder (c) Cube (d) Cylinder (e) Cuboid (f) Cylinder (g) Sphere (h) Cuboid 3. (a) 5 × 5 × 5 = 125 (b) 6 (c) 20 4. (a) Circle (b) Circle (c) Triangle (d) Rectangle 5. Refer to key point

6. (a) (b) (c) (d)

7. Do it yourself.

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