CHAPTER-3 UNIT-1

ADDIIONAL PROBLEMS ON “AXIOMS<POSTULATES”

1. Choose the correct option:

i. If a=60 and b=a, then b=60 by....

a) Axiom 1 b) Axiom 2

c) Axiom 3 d) Axiom 4 [a]

2. What is the angle between the hour`s hand and minute`s hand of a

clock at (i) 1.40 hours, (ii) 2.15 hours? (Use 10=60 minutes)

1 hour=3600

m 1 hour the minuts and complete 1 revolutions=1 circle

1 circle=3600

12 hours=3600

1 hours=360/12=300

5 minutes is also =300

1 munites=30/5=60

1.40 hrs =32 minuts -22×6=1920

2.15 hrs= 4 minuts 4×6=240

3. How much would hour`s hand have moved from its position ta 12

noon when the time is 4.24 p.m.?

12.00 noon at 4.24 pm

Number of minutes the hour hand moved = 24 minutes

1 minute = 60

∴24 minuts = 24×6 = 1440

4. Let 𝐀𝐁 be a line segment and let C be the midpoint of 𝐀𝐁 to D such

that B lies between A and D. Prove that AD+BD=2CD.

A a C a B D

In the above figure AC=BC=a AB s extended to D

Now AD+BD = AC+CB+BD+BD

= a + a +BD+BD

= 2a + 2BD

= 2(a+BD)

= 2(CB+BD)

= 2CD

5. Let 𝐀𝐁 and 𝐂𝐃 be two lines intersecting at a point O. Prove that the

ray bisecting ∟𝐁𝐎𝐃. Prove that the extension of 𝐎𝐗 to the left of O

bisects ∟𝐀𝐎𝐂.

A D

a

Y O a X

C B

The lines AB and CD intersect at O

Let OX is the bisects of ∟BOD

Let ∟BOX = ∟OBX = a0

∴ ∟BOX = ∟AOY = a0

And ∟DOX = ∟COY = a0

∴ ∟AOY = ∟COY

∴ ∟ OY bisects ∟AOC

6. Let 𝐎𝐗 be a ray and let 𝐎𝐀 and 𝐎𝐁 be two rays on the same side of

𝐎𝐗 and 𝐎𝐁 . Let 𝐎𝐂 be the bisects of ∟AOB. Prove that ∟AOX +

∟XOB = 2 ∟XOC.

In the above figure OC is the angle bisecter of ∟AOB

∴ ∟AOC =∟BOC

Now XOA + XOB

=∟ XOA +∟XOA + ∟AOC + ∟BOC

= 2 ∟XOA + ∟AOC + ∟AOC

= 2 ∟XOA + 2 ∟AOC

= 2 (∟XOA + 2 ∟AOC)

∴ ∟XOA + ∟XOB = 2 (XOC)

7. Let 𝐎𝐀 and 𝐎𝐁 be two rays and let 𝐎𝐗 be a ray between 𝐎𝐀 and 𝐎𝐁

such that ∟AOX > ∟XOB. Let 𝐎𝐂 be the bisector of ∟AOB. Prove

that ∟AOB – ∟XOB = 2 ∟COX

B

X

C

O

In the along side figure ∟AOX > ∟XOB

Ray OC the angle bisecting of ∟AOB

∴ ∟AOC = ∟BOC

Now ∟AOX = ∟XOB

= ∟AOC + ∟COX -( ∟BOC - ∟COX)

= ∟AOC + ∟COX – (∟BOC + ∟COX)

= ∟COX + ∟COX

= 2 ∟COX

∴ ∟AOX – ∟XOB = 2 ∟COX

8. Let 𝐎𝐀 , 𝐎𝐁 , 𝐎𝐂 , be three rays such that 𝐎𝐂 lies between 𝐎𝐀 and 𝐎𝐁 .

Suppose the bisectors of ∟AOC and ∟COB are perpendicular to each

other. Prove that B, O, A are collinear.

C

Y X

b a

B b a A

In the figure OC lies in between the rays OA and OB.

OX and OY are the bisectors of ∟AOC and ∟COB

∟AOX = ∟COX = a0 and

∟BOY = ∟COY = b0

It is given OX ⊥ OY

∴ ∟COX + ∟COY = 900

a0 + b

0 = 90

0

∴ ∟AOX = ∟BOY a0 + b

0 = 90

0

∟AOY + ∟COX + ∟COY + ∟BOY = 900 + 90

0 +=180

0

∴ ∟BOA is straight angle

∴ BOA is a straight line

Hence B, O, A are collinear.

9. In the adjoining figure, 𝐀𝐁 ∥ 𝐃𝐄 . Prove that ∟ABC + ∟DCB + = 1800

B

A D E

C

Given: In the given figure AB || DE

To Prove: ∟ABC – ∟DCB + ∟CDE = 180

Construction: Draw a PQ through C parallel to AB and DE.

Proof: AB || PC and BC is transversal

∴ ∟ABC + ∟BCP = 1800 .................................(1)

(Seen of interior angles on its side)

lllly ∟CDE + ∟DCQ = 1800 ........................(2)

Seen of interior angles on its side)

Adding (1) and (2) we get

∟ABC + ∟CDE + ∟BCP +∟ DCQ = 360 ..........(3)

Now ∟BCP + ∟DCQ = 180 – ∟DCB ..............(4)

Substituting in (4) and (3)

∟ABC + ∟CDE + 180 – ∟DCB = 3600

∟ABC + ∟CDE - ∟DCB = 3600-180

0

∟ABC + ∟CDE – ∟DCB = 1800

10. Consider two parallel lines and a traversal. Among the measurement

of triangles formed, how many distinct numbers are there?

P

a

L b

A b a B

a H b

C b a D

Q

In the figure AB || CD and PQ is Traversal PQ intersects

AB in L and CD in H respectively

∴ ∟ALP = ∟BLM = a0 V.O.A

∟LHC = ∟DMQ = a0 V.O.A

PLB = ALM = b0 V.O.A

∟PLB = ∟LMD = b0 Corresponding angles

∴ ∟LMD = ∟CHQ = b0 V.O.A

There are only two distinct angles a0 and b

0 are these.

CHAPTER-1 UNIT-2

ADDITIONAL PROBLEM ON SQUARES, SQUARE ROOTS, CUBES

AND CUBE ROOTS

1. Match the numbers in the column A with their squares in the column

B.

A B Answers

(1) 5 (2) 8

(3) 2 (4) -6 (5) -22

(6) 12

(a) 25 (b) 144

(c) 36 (d) 484 (e) 64

(f) 4 (g) 121

(1) (a) (2) (e)

(3) (f) (4) (c) (5) (d)

(6) (b)

2. Choose the correct option.

a) The number of perfect squares from 1 to 500 is

a) 1 b) 16 c) 22 d) 25 [c]

b) The last digit of a perfect square can never be

a) 1 b)3 c) 5 d) 9 [b]

c) If a number ends in 5 zeros, its square ends in

a) 5 zeros b) 8 zeros

c) 10 zeros d) 12 zeros [c]

d) Which could be the remainder among the following when a perfect

square is divided by 8?

a) 1 b) 3 c) 5 d) 7 [a]

e) The 6th

triangular number is

a) 6 b) 10 c) 21 d) 28 [c]

3. Consider all integers from -10 to 5 and the square of each of them.

How many distance numbers do you get?

(-10)2 =100, -9

2=81.................................................Answer: 11

4. Write the digit in units place when the following numbers are

squared.

4, 5, 9, 24, 17, 76, 34, 52, 33, 2319, 18, 3458, 3453.

Sl No. n n2

Digit in the Unit place

1 2

3 4 5

6 7

8 9

10 11

12 13

4 5

9 24 17

76 34

52 33

2319 18

3458 3453

16 25

81 576 289

5776 1156

2704 1089

5377761 324

11957764 119232209

6 5

1 6 9

6 6

4 9

1 4

4 9

5. Write all numbers from 400 to 425 which end in 2, 3, 7, 8. Check if

any of these is a perfect square.

402

403 407 408

412

413

417 418 422

423

None of the above numbers are perfect squares.

6. Find the sum of digits of (111111111)2.

Answer: 81

7. Suppose x2+y

2=z

2.

If x=4 and y=3 find z substituting 42+32=16+9=25 32=25 3=5

If x=5 and z=13 find y substituting 52+y2=132 25+y2=169 y2=169-25=144 y=12

If y=15 and z=17 find x substituting x2+152=172 x2+225=289 x2=289-225=64 x=8

8. A sum of Rs. 2304 is equally distributed among several people. Each

gets as many ropees as the number of persons. How much does each

one get?

2304 is a perfect square of 48 therefore 48 get RS 48 each.

9. Define a new addition * on the set of all natural numbers by

m*n=m2+n

2

1) Is N closed under *? Ans:Yes it is closed

2) Is * Commutative on N? Yes it is commutative

3) Is * Associative ? Ans: Yes it is associative.

4) Is there an identity element in N with respect to *?

Ans: No.

10. Find all perfect squares from 1 to 500, each of which is a sum of two

perfect squares:32+4

210. Find all perfect squares from 1 to 500, each

of which is a sum of perfect squares.

32+4

2=9+16=25 25 is a perfect square.

9+16+25

36+64=100

81+144=225

25+144=169

11. Suppose the area of a square field is 7396 m2 Find its perimeter:

Area of a square field =a2 =7396

Each side = √7396 =86m

Perimeter =4a=4x86=344m.

12. Can 101 be written as a difference of two perfect squares?

Ans:1010=a2-b

2 for some integers of a and b. So possibilities are 1. Both a

and b are odd 2. Both a and b are even that is a2-b

2 is divisible by 4. But

1010 is not therefore it cannot be expressed as the difference of two squares.

13. What are the remainders when a percfect cube is divided by 7?

Ans: 0,1,6 Divide 27 by 7, 64 by 7 34 by 7 and can confirm.

14. What is the least perfect square which leaves the remainder 1 when

divided by 7 as well as 11?

Ans: (34) 2=1156 when divided by 7 and 11 leaves the remainder 1.

15. Find two smallest perfect squares whose product is a perfect cube

42X16

2=16

2

CHAPTER-2 UNIT-2

ADDITIONAL PROBLEMS ON FACTORISATION

1. Choose the correct answer:

a) 4a+12b is equal to

a) 4a b) 12b c) 4(a+3b) d)3a [c]

b) The product of two numbers is positive and their sum negative only when

a) both are positive

b) both are negative

c) one positive the other negative

d)one them is equal to zero [b]

c) Factorising x2+6x+8, we get

a) (x+1)(x+8) b) (x+8)(x+2)

c) (x+10)(x-2) d) (x+4)(x+2) [d]

d) The denominator of an algebraic fraction should not be

a) 1 b) 0 c) 4 d) 7 [b]

e) If the sum of two integers is -2 and their product is -24, the numbers

are

a) 6 and 4 b) -6 and 4

c) -6 and -4 d) 6 and -4

f) The difference (0.7) 2-(0.3)

2 simplifies to

a) 0.4 b) 0.04 c) 0.49 d) 0.56 [a]

2. Factorise the following:

i) x2+6x+9=(x+3)

2

|(a+b)2=a

2+b

2+2ab|

ii) 1-8x+16x2

rearranging

16x2-8x+1=(4x-1)

2 |(a+b)

2=a

2+b

2+2ab |

iii) 4x2-81y

2

= (2x+9y)(2x-9y) |a2-b

2=(a+b)(a-b)|

iv) 4a2+4ab+b

2

=(2a+b) 2 |(a+b)

2=a

2+b

2+2ab |

v) a2b

2+c

2d

2-a

2c

2-b

2d

2

rearranging

a2b

2- a

2c

2+ c

2d

2- b

2d

2= a

2(b

2-c

2)-d

2(b

2-c

2)=(a

2-d

2)(b

2-c

2)

3. Factorise the following: (splitting the middle term)

(i) x2+7x+12 (ii) x

2+x-12

(iii) x2-3x-18 (iv) x

2+4x-21

(v) x2-4x-192 (vi) x

4-5x

2+4

(vii) x4-13x

2y

2+36y

4

4. Factorise the following: (i) 2x

2+7x+6

6×2=12,

12 can be split as 4×3=12 and 4+3=7 Therefore the Ans: is (2x+3)(x+2)

(ii) 3x

2-17x+20

3×20=60 Factors required 12×5=60 12+5=17

Ans: (3x-5)(x-4)

(iii) 6x

2-5x-14

6×14=-84 Factors required 7×12=84 12-7=5

Ans: (x-2)(6x+7)

(iv) 4x2+12xy=5y

2

4×5=20 Factors required 10×2=20 10+2=12

Ans: (2x+y)(2x+5)

(v) 4x

4-5x

2+1

4×1=4 Factors required 4×1=4 4+1=5

Ans: 4x4-4x

2-x

2+1

4x2(x

2-1)-1(x

2-1)

(x2-1)(4x

21)

(x+1)(x-1)(2x-11)(2x-1)

5. Factorise the following:

(i) x

8-y

8

this can be written as (x4)2-(y

4)2 applying a

2-b

2=(a+b)(a-b)

We get (x

4+y

4)(x

4-y

4)

(x4-y

4)=(x

2)

2-(y

2)2

=(x2-y

2)(x

2+y

2)

=(x+y)(x-y)(x2+y

2)

Ans: (x4+y

4)(x

2+y

2)(x+y)(x-y)

(ii) a

12x

4-a

4x

12=a

4x

4(a

8-x

8)

(a8-x

8)=(a

4+x

4)(a

2+x

2)(a+x)(a-x)

Ans: =a4x

4(a

4+x

4)(a

2+x

2)(a+x)(a-x)

(iii) x4+x

2+1=(x

2+x+1) (x

2-x+1) this is of the form x

2+px+q=(x+a)(x+b)

Where a.b=q a+b=p

(iv) x

4+5x

2+9=(x

2+x+3)(x

2-x+3)

(as the previous problem)

6. Factorise x4+4y

4. Use this to prove that 2011

4+64 is a composite

number.

X4+4y

4=(x

2+2xy+2y

2)(x

2-2xy+2y

2) Applying to 2011

4+64=2011

4+4(2)

4=

(20112+2×2011×2+2×2

2)(2×2011×2+2×2

2) simplification shows that it is

a composite number.

CHAPTER-3 UNIT-2

ADDITIONAL PROBLEMS ON THEOREMS ON TRIANGLES

1. Fill in the blacks to make the following statement true:

a) Sum of the angles a triangle is .......(1800)

b) An exterior angle of a triangle is equal to the sum of ........ opposite

angles. (interior)

c) An exterior angle of a triangle is always ..... than either of the interior

opposite angles. (greater)

d) A triangle cannot have more than ........ right angle. (one)

e) A triangle cannot have more than ........ obtuse angle. (one)

2. Choose the correct answer from the given alternatives:

a) In a triangle ABC, ∠A = 800 and AB = AC, then ∠B is ........

a. 500 b. 60

0 c. 40

0 d. 70

0 [a]

b) In right angled triangle, ∠A is right angle and ∠B = 350, then ∠C is

...........

a. 650 b. 55

0 c. 75

0 d. 45

0 [b]

c) In a triangle, ABC, ∠B = ∠C = 450, then the triangle is ...........

a. right triangle

b. acute angled triangle

c. obtuse angle triangle

d. equilateral triangle [a]

d) In an equilateral triangle, each exterior angle is ..........

a. 600 b. 90

0 c. 120

0 d. 150

0 [c]

e) Sum of the three exterior angle of a triangle is ..........

a. two right angles

b. three right angles

c. one right angles

d. four right angles [d]

3. In a triangle ABC, ∠B = 700. Find ∠A + ∠C?

A

∠𝐴 + ∠B + ∠C = 180 110

∠A + ∠C + ∠B =180

∠A + ∠C + 70 = 180 x x

∠A + ∠C = 180-70 = 1100 B C

4. In atriangle ABC, ∠𝑨 = 1100 and AB = AC. Find Fing. ∠B and ∠C.

∠A + ∠B + ∠C =1800

∠A + x + x = 180

∠A + 2x =180

110 + 2x = 180-110 = 70

x = 70

2 = 35

0

∠B = 350 ∠C = 35

0

5. If three angles of triangle are in the ratio 2:3:4, determine three

angles.

2x + 3x + 5x =180

10x = 180

x = 180

10 =18

6. The angles of a triangle are RRnged in ascending order of

magnitude. If the difference between two consecutive angles is 150,

find the three angles?

x + x 15 + x + 30= 180

3x + 45 =180

3x = 18045 = 135

x = 135

3 = 45

0

x = 450

x + 15 = 45 + 15 = 600

x + 30 = 45 + 30 = 750

7. The sum of two angles of a triangle is equal to its third angle.

Determine the measure of the third angle.

∠A + ∠B = ∠C

∠A + ∠B + ∠C = 180

∠C + ∠C =180

2∠C =180

∠C= 180

2 = 90

0

8. In a triangle ABC, if 2∠A = 3 ∠B=6 ∠C, Determine ∠A, ∠B and ∠C.

2∠A = 6∠C

∠A = 3∠C

3∠B = 6∠C

∠B = 2∠C

∠A + ∠B = ∠C=180

3∠C + 2∠C + ∠C =180

6∠C =180

9. The angles of a triangle x – 400, x – 20

0 and

𝟏

𝟐 x + 15

0. Find the value

of x.

A + C + C = 180

x-40+x-20+x+15=180

21

2x-45=180

5𝑥 = 225

∴x=225

5×2=90

10. In a triangle ABC, ∠A-∠B=150 and ∠B-∠C=30

0, find ∠A, ∠B and ∠C.

∠A+∠B=15

∠A=15+∠B

∠B-∠C=30

∠B-30=C

∠A+∠B=∠C=180

15+C+B+B-30=180

3B-15=180

3B=180+15=195

B=135

3=65

0

11. In a triangle ABC, ∠A-∠B=150 and ∠B-∠C=30

0,find ∠A, ∠B and ∠C.

∠A=15+B=15+65=800

∠C=∠B-30=65-30=350

∠A=800, ∠B=65

0, ∠C=35

0

12. The sum of two angles of a triangle is 800 and their difference is 20

0.

Find the angles of the triangle.

A+B=80

A-B=20

2A=100

A=50

B=30

∠C=100

13. In a triangle ABC, ∠B=600 and ∠C=80

0. Suppose the bisector of ∠B-

and ∠C meet at I. Find BIC?

A

40

I

110

60

30 4080

B C

14. In a triangle, each of the smaller angles is half the largest angle. Find

the angles.

∠B=1

2 ∠A and ∠C=

1

2 ∠A

∠A+∠B+∠C=180

∠A+1

2 ∠A+

1

2 ∠A=180

2A=180

A=180

2 =90

0

A=900 B=45

0 C=45

0

15. In a triangle, each of the bigger is twice the third angle, Find the

angles.

∠Q=2∠P and ∠R=2∠P

∠P+∠Q+∠R=1800

∠P+2∠P+2∠Q=1800

5∠P=1800

∠P=180

5 =36

0

∠P=360 ∠Q=72

0 ∠R=72

0

16. In atriangle ABC, ∠B=500 and ∠A=60

0. Suppose BC is extended to

D. Find ∠ACD.

A

60

50o

B C D

Ext.: ∠ACD = Sum of interior opposite angles

Ext.: ∠ACD = ∠A+∠B=600+50

0

Ext.: ∠ACD = 1100

17. In an Isosceles triangle, the vertex angle is twice the sum of the base

angles. Find the angles of the tringle.

A

4a

a a

B C

∠A+∠B+∠C=180

4a+a+a=180

6a=180

a=300

∠A=4a=4×30=1200

∠B=a=300

∠C=a=300

CHAPTER-1 UNIT-3

RATIONAL NUMBERS

EXERCISE 1.3.1

1. Identify the property in the following statements:

(i) 2+(3+4)=(2+3)+4

Ans: Associative property of addition.

(ii) 2.8=8.2

Ans: Commutative property of multiplication.

(iii) 8.(6+5)=(8.6)+(8.5)

Ans: Distributive property.

2. Find the additive inverses of the following integers: (i) 6

Ans: „-6‟ is the additive inverse of 6. (ii) 9

Ans: „-9‟ is the additive inverse of 9. (iii) 123 Ans: „-123‟ is the additive inverse of 123.

(iv) -76 Ans: „76‟ is the additive inverse of -76.

(v) -85 Ans: „-85‟ is the additive inverse of 85.

(vi) 1000 Ans: „-1000‟ is the additive inverse of 1000.

3. Find the integer m in the following:

(i) m+6=8

Ans: m=6-8 m=2

(ii) m+25=15 Ans: m=15-25

m=-10

(iii) m-40=26 Ans: m=-26+40

m=+14

(iv) m+28=-49 Ans: m=-49-28

m = -77

4. Write in the following in increasing order: 21,-8,26,85,38,-333,-210,0,2011

Ans: -333,-210,-26,-8,0,21,33,85,2011

5. Write the following in decreasing order: 85,210,-58,2011,-1024,528,364,-10000,12

Ans: 2011,528,364,210,85,12,-58,-1024,-10000

EXERCISE 1.3.2

1. Write down ten rational numbers which are equivalent to 𝟓

𝟕 and the

denominator not exceeding 80.

Ans: 5×2

7×2 =

𝟏𝟎

𝟏𝟒

5×3

7×3 =

𝟏𝟓

𝟐𝟏

5×4

7×4 =

𝟐𝟎

𝟐𝟖

5×5

7×5 =

𝟐𝟓

𝟑𝟓

5×6

7×6 =

𝟑𝟎

𝟒𝟐

5×7

7×7 =

𝟑𝟓

𝟒𝟗

5×8

7×8 =

𝟒𝟎

𝟓𝟔

5×9

7×9 =

𝟒𝟓

𝟔𝟑

5×10

7×10 =

𝟓𝟎

𝟕𝟎

5×11

7×11 =

𝟓𝟓

𝟕𝟕

2. Write down 15 rational numbers which are equivalent to 𝟏𝟏

𝟓 and the

numerator not exceeding 180.

Ans: 11 ×2

5×2=

𝟐𝟐

𝟏𝟎 ,

11 ×3

5×3=

𝟑𝟑

𝟏𝟓 ,

11 ×4

5×4=

𝟒𝟒

𝟐𝟎 ,

11 ×5

5×5=

𝟓𝟓

𝟐𝟓 ,

11 ×6

5×6=

𝟔𝟎

𝟑𝟎 ,

11 ×7

5×7=

𝟕𝟕

𝟑𝟓 ,

11 ×8

5×8=

𝟖𝟖

𝟒𝟎,

11 ×9

5 ×9=

𝟗𝟗

𝟒𝟓,

11 ×10

5 ×10=

𝟏𝟏𝟎

𝟓𝟎,

11 ×11

5 ×11=

𝟏𝟐𝟏

𝟓𝟓,

11 ×12

5 ×12=

𝟏𝟑𝟐

𝟔𝟎,

11 ×13

5 ×13=

𝟏𝟒𝟑

𝟔𝟓,

11 ×14

5 ×14=

𝟏𝟓𝟒

𝟕𝟎,

11 ×15

5 ×15=

𝟏𝟔𝟓

𝟕𝟓,

11 ×16

5 ×16=

𝟏𝟕𝟔

𝟖𝟎

3. Write down the ten positive numbers such that the sum of numerator and denominator of each is 11. Write them in decreasing order.

Ans: Number : 10

1,

9

2,

8

2,

7

4,

6

5,

5

6,

4

7,

3

8,

2

9,

1

10

Decreasing order : 10

1,

9

2,

8

3,

7

4,

6

5,

5

6,

4

7,

3

8,

2

9,

1

10

4. Write down the ten positive numbers such that the numerator and

denominator for each them is -2. Write them in increasing order.

Ans: Increasing order : 10

12 = 0.833

9

11 = 0.811,

8

10 = 0.80,

3

5 = 0.6,

1

3 = 0.33

5. Is 𝟑

−𝟐 a rational number? If so, how do you write it in a form

conforming to the definition of a rational number (that is, the denominator as a positive integer)?

Ans: 3

−2 is not a rational number. It should be written as

−3

2 to be rational

number.

6. Earlier you have studied decimals 0.9,0.8. Can you write these as rational numbers?

Ans: Yes, we can write decimals like 0.9, 0.8 as rational numbers.

Ex.: 0.9 = 9

10, 0.8 =

8

10, 1.8 =

18

10

EXERCISE 1.3.3

1. Name the property indicated in the following:

(i) 315+115 = 430

Ans: Closure property of addition.

(ii) 𝟑

𝟒×

𝟗

𝟓=

𝟐𝟕

𝟐𝟎

Ans: Closure property of multiplication.

(iii) 5+0 = 0+5 =5

Ans: 0 is the additive identity.

(iv) 𝟖

𝟗× 𝟏 =

𝟖

𝟗

Ans: 1 is the multiplicative identity.

(v) 𝟖

𝟏𝟕+

−𝟖

𝟏𝟕= 𝟎

Ans: Additive inverse.

(vi) 𝟐𝟐

𝟐𝟑×

𝟐𝟑

𝟐𝟐= 𝟏

Ans: Multiplication inverse.

2. Check the commutative property of addition for the following pairs:

(i) 𝟏𝟎𝟐

𝟐𝟎𝟏,𝟑

𝟒

Ans: a+b = b+a

102

201+

3

4

3

4+

102

201

408 +603

804

603 +408

804

1011

804

1011

804

∴102

201+

3

4=

3

4+

102

201

(ii) −𝟖

𝟏𝟑,𝟐𝟑

𝟐𝟕

a + b =b+a

−8

13+

23

27=

23

27+

−8

13

−216 +299

351=

299−216

351

83

351=

83

351 ∴Proved

(iii) −𝟕

𝟗,−𝟏𝟖

𝟏𝟗

a+b=b+a −7

9+

18

19=

18

19+

−7

9

−133 +(−162 )

171=

−162 +(−133 )

171

−295

171=

−295

171 ∴Proved

3. Check the commutative property of multiplication for the following pairs:

(i) 𝟐𝟐

𝟒𝟓,𝟑

𝟒

Ans: a×b = b×a

22

45×

3

4=

3

4×

22

45

66

180=

66

180

11

30=

11

30 ∴Proved

(ii) −𝟕

𝟏𝟑,𝟐𝟓

𝟐𝟕

a×b = b×a

−7

13×

25

27=

25

27×

−7

13

−175

351=

−175

351 ∴ Proved

(iii) −𝟖

𝟗,−𝟏𝟕

𝟏𝟗

Ans: a×b = b×a

−8

9×

−17

19=

−17

19×

−8

9

136

171=

136

171 ∴Proved

4. Check the distributive property for the following triples of rational numbers:

(i) 𝟏

𝟖,𝟏

𝟗,𝟏

𝟏𝟎

Ans:a(b+c) = ab+ac

1

8

1

9+

1

10 =

1

8×

1

9+

1

8×

1

9

1

8

10 +9

90 =

1

72+

1

80

1

8×

19

90=

80 +72

5760

19

720=

152

5760

19

720=

19

720

(ii) −𝟒

𝟗,𝟔

𝟓,𝟏𝟏

𝟏𝟎

Ans: a = −𝟒

𝟗 b =

𝟔

𝟓 c =

𝟏𝟏

𝟏𝟎

a(b+c) = ab + ac

−4

9×

6

5+

11

10 =

−4

9×

6

5 +

−4

9×

11

10

−4

9×

12 +11

10 =

−24

45+

−44

90

−4

9×

23

10=

−8

15×

−22

45

−46

45=

−24−22

45

−46

45=

−46

45

(iii) 𝟑

𝟖,𝟎,

𝟏𝟑

𝟕

Ans: = 𝟑

𝟖× 𝟎 +

𝟏𝟑

𝟕 =

𝟑

𝟖× 𝟎 +

𝟑

𝟖×

𝟏𝟑

𝟕

𝟑

𝟖×

𝟏𝟑

𝟕= 𝟎 +

𝟑𝟗

𝟓𝟔

=𝟑𝟗

𝟓𝟔 ∴ Proved

5. Find the additive inverse of each of the following numbers:

(i) 𝟖

𝟓,𝟔

𝟏𝟎,−𝟑

𝟑,−𝟏𝟔

𝟑,−𝟒

𝟏

Ans: 8

5=

−8

5

6

10=

−6

10

−3

8=

3

8

−16

3=

16

3

−4

1=

4

1

6. Find the multiplicative inverse of each of the following numbers:

2, 𝟔

𝟏𝟏,−𝟖

𝟏𝟓,𝟏𝟗

𝟏𝟖,

𝟏

𝟏𝟎𝟎𝟎

Ans: 2

1=

1

2

6

11=

11

6

8

−15=

−15

8

19

18=

18

19

1

1000=

1000

1

EXERCISE 1.3.4

1. Represent the following rational numbers on the number line:

(i) −8

5

B A

-2 −8

5 -1 0

∴ AB represents −8

5

(ii) 3

8

A B

0 3

8 1

∴ AB represents 3

8

(iii) 2

7

A B

0 2

7 1

∴ AB represents 2

7+

(iv) 12

5

A B

0 1 2 12

5 3

∴ AB represents 12

5

(v) 45

13

A B

0 1 2 3 45

13 4

∴ AB represents 45

13

2. Wrtie the following rational numbers in ascending order: 3

4,

7

12,15

11,22

19,101

100,−4

5,−102

81,−13

7

Ans: Ascending order −13

7,−102

81,−4

5,

7

12,3

4,101

100,22

19,15

11

Method:

3

4,

7

12,15

11,22

19,101

100,−4

5,−13

7,−102

81

0.75 0.5 1.3 1.1 1.01 -0.3 -1.8 -1.2

3. Write 5 rational number between 𝟐

𝟓 and

𝟑

𝟓 , having the same

denominators.

Ans:

2

5 0.42 0.44 0.57 0.59

3

5

0.4 0.41 0.43 0.45 0.56 0.58 0.6

∴ Numbers between 2

5 and

3

5 are

41

100,

42

100,

43

100,

44

100,

45

100

Method II:

13

30

15

30

17

30

0 1 2 3 4 5

14

30

16

30

Totally there are 30 divisions,

Each division is 1

30

Between 2

5 and

3

5 there are

13

30,

14

30,

15

30,

16

30,

17

30

4. How many positive rational numbers less than 1 are there such that the sum of the numerator and denominator does not exceed 10?

Ans: 1

9,

1

8,

1

7,

1

6,

1

5,

1

4,

1

3,

1

2,

2

7,

2

5,

2

3,

3

7,

3

5,

3

4,

4

5

∴ Only 15 positive rational number are possible such that they are

less then 1 and the sum of the numerator and denominator does not exceed 10.

5. Suppose m/n and p/q are two positive rational numbers. Where does

𝐦+𝐩

𝐧+𝐪 lie, with respect to m/n and p/q?

Ans: 𝑚

𝑛=

1

2 𝑎𝑛𝑑

𝑝

𝑞=

3

4

𝑚+𝑝

𝑛+𝑞=

1+3

2+4=

4

6=

2

3= 0.66

0.5

m+p

n+q lies between

m

n and

p

q

6. How many rational numbers are there strictly between 0 and 1 such that the denominator of the rational number is 80?

Ans: 1

80,

2

80,

3

80,

4

80,……… . .

77

80,

78

80,

79

80

∴ There are 79 positive rational numbers.

7. How many rational numbers are there strictly between 0 and 1 with

the property that the sum of the numerator and denominator is 70?

Ans: 1

69,

2

68,−3

67,

4

66,……… .

33

37,

34

36

∴ There are 34 rational numbers.

ADDITIONAL PROBLEMS ON “RATIONAL NUMBERS”

1. Fill in the blanks:

0.66 0.75

(a) The number 0 is not in the set of ________ (Natural numbers)

(b) The least number in the set of all whole number is _________ (0)

(c) The least number in the set of all even natural numbers is ________ (2)

(d) The successor of 8 in the set of all natural numbers is _________ (9)

(e) The sum of two odd integers is ________ (even)

(f) The product of two odd integers is _________ (odd)

2. State whether the following statements are true or false.

(a) The set of all even natural numbers is a finite set. [False]

(b) Every non-empty subset of Z jas the smallest element. [False]

(c) Every integer can be identified with a rational number. [True]

(d) For each rational number, one can find the next rational number. [False]

(e) There is the largest rational number. [False]

(f) Every integer is either even or odd. [false]

(g) Between any two rational numbers, there is an integer. [False]

3. Simplify

(i) 100(100-3)-(100×100-3)

=100× (97)-(10000-3)

=9700-9997

=-297

(ii) [20-(2011-201)]+[2011-(201-20)]

=[20-1810]+[2011-181]

=-1790+1830

=40

4. Suppose m is an integer such that m≠-1 and m≠-2. Which is larger 𝒎

𝒎+𝟏 or

𝒎+𝟏

𝒎+𝟐? State your reasons.

Ans- Let m=2 ∈ z then

𝑚

𝑚+1=

2

2+1 =

2

3 [

2

3 = 0.66]

𝑚+1

𝑚+2 =

2+1

2+2 =

3

4 [

3

4 = 0.75]

3

4 >

2

3

5. Define an operation * on the set of all rational numbers Q as follows:

r*s = r+s-(r×s) for any two rational numbers r,s. Answer the

following with justification:

(i) Is Q closed under the operation *?

Ans- Let r=2, s=3

2*3 = 2+3 – (2*3)

2*3 = 5-6

2*3 = -1 ∈ Q

∴ Q is closed under *

(ii) Is * an associative operator on Q?

Let r,2 and t be three integers

∴(r*t)*t=r*(s*t)

Assume r=1, s=2 and t=3 (1*2)*3=1*(2*3)

Now (1*2)*3=1*(2*3) (1*2)=1+2-1×2=3-2=1

(1*2)*3=1*3

(1*2)*=1+3-1×3=4-3=1

3=1

r*(s*t)=1*(2*3) 2*3=2×3-2×3=5-6=-1

1*(2*3) = 1*-1

1*(2*3) = 1*-1 = 1+(-1)-(1×-1)

= 0+1=1

1*(2*3) = 1

∴(1*2)*3 = 1*(2*3) ∴ is associative on Q.

(iii) Is * a commutative operation on Q?

Assume 4, 5∈Q

4*5=5*4

4*5=4+5-(4×5) = 9-20 = -11 5*4 = 5+4-(5×4) =9-20 = -11

∴ 4*5 = 5*4

∴ is commutative on Q.

(iv) What is a*1 for any a in Q ?

Now r * s =r×s-(r×2)

a*1 = a+1-(a×1)=a+1-a=1

a*1 = 1

(v) Find two integers a ≠ 0 and b ≠ 0 such that a*b=0.

Let a*b = a+b-(a×b)

A*b +a×b-(a×b)

2×2=2+2-(2×2)

=4 – 4=0

2*2=0

∴a=2 and b=2

6. Find the multiplicative inverse of the following rational numbers: 𝟖

𝟏𝟑,𝟏𝟐

𝟏𝟕,𝟐𝟔

𝟐𝟑,−𝟏𝟑

𝟏𝟏,−𝟏𝟎𝟏

𝟏𝟎𝟎

Ans:

Rational numbers Multiplicative inverse

8

13

12

17

26

23

−13

11

−101

100

13

8

17

12

23

26

−11

13

−100

101

7. Wtrite the following in increasing order:

𝟏𝟎

𝟏𝟑,𝟐𝟎

𝟐𝟑,𝟓

𝟔,𝟒𝟎

𝟒𝟑,𝟐𝟓

𝟐𝟖,𝟏𝟎

𝟏𝟏

Ans: Taking the LCM of the denominator and computing the numerators

we observe. 𝟏𝟎

𝟏𝟑,𝟓

𝟔,𝟐𝟎

𝟐𝟑,𝟐𝟓

𝟐𝟖,𝟏𝟎

𝟏𝟏and

𝟒𝟎

𝟒𝟑 are in order of increasing order.

8. Write the following in decreasing order:

𝟐𝟏

𝟏𝟕,𝟑𝟏

𝟐𝟕,𝟏𝟑

𝟏𝟏,𝟒𝟏

𝟑𝟕,𝟓𝟏

𝟒𝟕and

𝟗

𝟖

Ans: Taking the LCM of the denominators and arranging the numerators

in the decreasing order.

21

17,

13

11,

31

27,

9

8,

41

37 and

51

47 are in decreasing order.

9. a. What is the additive inverse of 0?

Additive inverse of 0 is 0.

b. What is the multiplicative inverse of 1?

Additive inverse of 1 is 1?

c. Which integers have multiplicative inverses?

Every integer have multiplicative inverse.

10. In the set of all rational numbers, give 5 examples each illustrating

the following properties.

(i) associativity

(ii) commutativity

(iii) distributivity of multiplication over addition.

Associativity Commutativity Distributivity

1. Let 1,2,3 ∈ Q

a+(b+c)=(a+b)+c

1+5=3+3

6=6

1,2 ∈ Q

a+b=b+a

1+2=2+1

3=3

1,2,3 ∈ Q

a(b+c)=ab+ac

1(2+3)=1×2+1×3

1(5)=2+3

5=5

2. Let 1,2,3 ∈ Q

a×(b×c)=(a×b) ×c

2×(3×4)=(2×3) ×4

2×12=(6) ×4

24=24

(a×b)=b×a

2×3=3×2

6=6

a(b+c)=ab+ac

2(2+3)=2×3+2×4

2×7=6×8

14=14

11. Simplify the following using distributive property:

(i) 𝟐

𝟓×

𝟏

𝟗+

𝟐

𝟓

=2

5×

1

9+

2

5×

2

5

1

9

= 2

45+

4

25=

10 +36

275=

46

225

(ii) 𝟓

𝟏𝟐×

𝟐𝟓

𝟗+

𝟑𝟐

𝟓

= 5

12×

25

9+

5

12×

32

5

= 125

108+

8

3=

125 +228

108=

353

108

(iii) 𝟖

𝟗×

𝟏𝟏

𝟐+

𝟐

𝟗

= 8

9×

11

2+

8

9×

2

9=

44

9+

16

81=

396 +16

81

= 412

81

12. Simplify the following:

(i) 𝟐𝟓

𝟗+

𝟏𝟐

𝟑 +

𝟑

𝟓

= 35 +36

9+

3

5=

306 +27

45=

332

45

(ii) 𝟐𝟐

𝟕+

𝟑𝟔

𝟓 ×

𝟔

𝟕

= 110 +252

35 ×

6

7=

362

35×

6

7=

2177

245

(iii) 𝟓𝟏

𝟐×

𝟕

𝟔 ÷

𝟑

𝟓

= 153 +7

6 ×

5

3=

160 ×5

6×3=

400

9= 44

4

9

(iv) 𝟏𝟔

𝟕+

𝟐𝟏

𝟖 ×

𝟏𝟓

𝟑−

𝟐

𝟗

= 128 +147

56 ×

45−2

9 =

275

56×

23

9=

11825

504

13. Which is the property that is there in the set of all rationals but

which is not in the set of all integers?

Every non-zero number is inversible but only 1 is not inversible.

14. What is the value of 𝟏 +𝟏

𝟏+𝟏

𝟏+𝟏

?

= 1 +1

1+1

1+1

= 1 +1

1+1

1+2

= 1 +1

2+1

2

= 1 +13

2

= 1 +2

3=

3+2

3=

5

3

15. Find the value of

1

3−

1

4

1

2−

1

3

.

= 4−3

12 ÷

3−2

6

= 1

12÷

1

6=

1

12×

1

6=

6

12=

1

2

16. Find all rational numbers each of which is equal to its reciprocal.

+1 and -1 are the rational numbers equal to their reciprocal.

17. A bus shuttles between two neighbouring towns every two hours. It

starts from 8am in the morning and last trip is at 6 pm. On one day

the driver observed that the first trip had 30 passengers and each

subsequent trip has one passenger less than the previous trip. How

many passengers travelled on that day?

1st trip 8 am -30 passengers

2nd

trip 10 am -29 passengers

3rd

trip 12 noon -28 passengers

4th

trip 2 pm -27 passengers

5th

trip 4 pm -26 passengers

6th

trip 6 pm -25 passengers

165 passengers

Total number of passengers travelled on that day =165

18. How many rational numbers p/q are there 0 and 1 for which q<p?

Not a single rational number lie between 0 and 1with q < p. When p > p

then the number is greater than 1.

19. Find all integers such that 𝟑𝒏+𝟒

𝒏+𝟐 is also integer.

When n = 0, 3𝑛+4

𝑛+2=

3 0 +4

𝑛+2=

4

2= 2

2 is an integer

When n = -1, 3 −1 +4

−1+2=

−3+4

−1+2=

1

1= 1

1 is an integer.

20. By inserting parenthesis (that is brackets), you can get several values

2×3+4× 𝟓. (For example ((2× 𝟑) + 𝟒) × 𝟓 one way of inserting

parenthesis). How many such values are there?

(i) (2×3) + (4×5) = 6 + 20 = 26

(ii) 2×(3+4) ×5=2×7×5=70

(iii) (2×3+4) ×5=10×5=50

(iv) 2× (3+4×5)=2×23=46

There are four values.

21. Suppose p/q is a positive rational in its lowest form. Prove that 𝟏

𝒒+

𝟏

𝒑+𝒒 is also in its lowest form.

Let 𝒑

𝒒=

𝟑

𝟓 i.e., p=3 and q=5

Then 1

𝑞+

1

𝑝+𝑞=

1

5+

1

3+5=

1

5+

1

8

=8+5

40=

13

40

13

40 is in the lowest form.

22. Show that for each natural number n, the fraction 𝟏𝟒𝒏+𝟑

𝟐𝟏𝒏+𝟒 is in its

lowest form.

Let n=5 be a natural number

Then 14𝑛+3

21𝑛+4=

14 5 +3

21 5 +4=

70 +3

105 +4=

73

109

∴73

109 is also in the lowest form.

23. Find all integers n for which the number (n+3)(n-1) is also an

integer.

Let n=-2 is an integer

(n+3)(n-1)=(-2+3)(2-1)

= (+1)(-3)

=-3

-3 is an integer.

CHAPTER-2 UNIT-3

LINEAR EQUATIONS IN ONE VARIABLE

EXERCISE 2.3.2

1. Solve the following

(i) x+3=11 (ii) y-9=21 x=11-3 y=21+9

x=8 y=30

(iii) 10=z+3 (iv) 3

11+ x =

9

11

10-3=z x=9

11−

3

11

7=z z=𝟔

𝟏𝟏

z=7

(v) 10x=z+3 (vi) s

7 = 4

x=30

10 s=4×7

x=3 s=28

(vii) 3x

6 = 10 (viii) 1.6=

x

1.5

3x=10×6 1.6×1.5=x

x=10 ×6

3 2.4=x

x=20 x=2.4

(ix) 8x=48+8 (x) x

3+ 1 =

7

15

8x=48+8 x

3=

7

15−

1

1

x=56

8

x

3=

−8

15

x=7 x=−8

15× 3

x=−8

5

(xi) x

5= 12 (xii)

3x

5 = 15

x=12×5 3x=15×5

x=60 x=15 ×5

3

x=25

(xii) 3(x+6)=24 (xiv) x

4 - 8=1

3x+18=24 x

4 = 1+8

3x=24-18 x

4 = 9

x=2 x=9×4

x=36

(xv) 3(x+2)-2(x-1)=7

3x+6-2x+2=7 x+8=7

x=7-8 x=1

2. Solve the equations

(i) 5x=3x+24

5x-3x=24 2x=24

x=24

2

x=12

(ii) 8t+5=2t-31

8t-2t=-31-5

6t=-36

t= −36

6

t = -6

(iii) 7x-10=4x+11

7x-4x=10+11

3x=21

x = 21

3

x = 7

(iv) 4z+3=6+2z

4z-2z=6-3 2z = 3

Z= 3

2

(v) 2x-1=14-x

2x+x=14+1

3x=15

x = 15

3

x = 5 (vi) 6x+1=3(x-1)+7

6x+1=3x-3+7

6x-3x=-3+7-1

3x= 3

3

x = 1

(vii) 2x

5−

3

2=

x

2+ 1

x

2−

2x

5=

−3

2−

1

1

5x−4x

10=

−5

2

x

10=

−5

2

10= −5

2×10

2= -25

x = -25

(viii) x−3

5− 2 =

2x

5

x−3−10

5=

2x

5

x−3−10

5× 5 = 2x

x-13 = 2x

x-2x = 13

-x=13

x = -13

(ix) 3(x+1) = 12+4(x-1)

3x+3 = 12+4x-4 3x-4x = 12-4-3 -x=+5

x = -5

(x) 2x-5 = 3(x-5) 2x-5 = 3x-15

2x-3x=-15+5 -x=-10 x

x = +10

(xi) 6(1-4x)+7(2+5x) = 53 6-24x+14+35x = 53

-24x+35x = 53

+11x= 33

11

x=3

(xii) 3(x+6)+2(x+3) = 64 3x+18+2x+6 = 64

3x+2x = 64-19-6 5x=40

x = 40

5

x = 8

(xiii) 2m

3+ 8 =

m

2− 1

2m

3−

m

2= -1-8

4m−3m

6= −9

m

6= −9

m = -9×6 m = -54

(xiv) 3

4 (x-1) = x-3

3x

4−

3

4= x-3

3x

4−

x

1=

−3

1+

3

4

3x−4x

4=

−12 +3

4

−x

4=

−9

4

-x = −9

4× 4

x = 9

EXERCISE 2.3.3

1. If 4 is added to a number and the sum is multiplied by 3, the result is

30. Find the number.

Ans: (x+4) = 30 3x+12 =30

3x = 30-12

x = 18

3

x = 6

2. Find three consecutive odd numbers whose sum is 219.

Ans: Let the numbers be x+x+2+x+4

x+x+2+x+x4=219

3x+6=219

3x=219-6

x= 219

3

∴ x = 71

∴ x+2 = 71+2 = =73

∴ x+4 = 71+4 = 75

∴ The numbers are 71,73 and 75.

3. A rectangle has length which is 5cm less than twice its breadth. If the length is decreased by 5cm and breadth is increased by 2cm. The

perimeter of the resulting rectangle will be 74cm. Find the length and breadth of the original rectangle.

Ans: Length is 5 less than 2 × breadth

Let the length be ℓ

Let the breadth be b

ℓ = 2b-5

New length = ℓ - 5

= (2b-5)-5

L = 2b-10

New breadth = b+2

B=b+2

Perimeter of rectangle = 74

2(L+B) = 74

2(2b-10+b+2) = 74

2(3b-8) = 74

6b = 74+16

b = 90

6

b = 15cm

ℓ = 2b-5

= 2(15)-5

= 30-5= 25cm

∴ Original length = 25cm

∴ Original breadth = 15cm

4. A number subtracted by 30 gives 14 subtracted by 3 times the

number. Find the number.

Let the number be x

30-x = 3x-14

-x-3x=-14-30

-4x=-44

x = −44

−4

x = 11

5. Sristi’s salary is same as 4 times Azar’s salary. If together they earn

Rs. 3,750 a month. Find their individual salaries.

Ans: Let Azar‟s salary be x

∴Sristi‟s salary = 4x

∴Sum of their salaries = 3,750

x+4x =3,750

5x = 3,750

x = 3750

5

x = 750

∴Azar‟s salary = 750

∴Sristi‟s salary = 4x

= 4×750

=3,000

6. Prakri’s age is 6 times Sahil’s age. After 15 years, Prakruthi will be 3

times as old as Sahil. Find their age.

Ans: Let Sahil‟s age be x

∴ Prakruthi‟s age = 6x Sahil

= 6x

After 15 years = Sahil‟s age = (x+15)

Prakruthi‟s age = (6x+15)

Prakruthis age = 3× Sahil‟s age

(6x+15) = 3× (x+15)

6x+15 = 3x+45

6x-3x = 45-15

3x = 30

x = 30

3

x = 10

∴Sahil‟s age = 10 years

∴Prakruthi‟s age = 6x

= 6×10

= 60 years

7. In the figure, AB is a straight line. Find.

C D

x+40

x+20 x A B

Ans: x+20+x+40+x = 1800

3x+60 = 1800

3x = 1800-60

0

x = 120 0

3

x = 400

8. If 5 is subtracted from three times a number, the result is 16. Find

the number.

Ans: Let the number be x 3x-5 = 16

3x = 16+5

x = 21

3

x = 7

9. Find two numbers such that one of them exceeds the other by 9 and their sum is 81.

Ans: Let the no. be x, the other no. be x+9

x+x+9 = 81 2x+9 = 81

2x = 81-9

x = 72

2

x = 36 ∴x+9=36+9=45

∴The numbers are 36 and 45

10. The length of a rectangular field is twice its breadth. If the perimeter

of the field is 288m. Find the dimensions of the field.

Ans: Let the length of the field be ℓ

Let the breadth of the field be b

Length is twice its breadth → ℓ = 26

Perimeter of the field = 288m

2(2b+b) = 288 2(ℓ+b) = 288

2b+2b+b+b=288

6b = 288

b = 288

6

b = 48m

∴ b = 48m

∴ ℓ = 2b

= 2× 48

= 96m

11. Ahmed’s father is thrice as old as Ahmed. After 12 years, his age will

be twice that of his son. Find their present age.

Ans: Let Ahmed‟s age be „x‟ years

Let Ahmed‟s father age be „3x‟ years

After 12 years, Ahmed‟s father‟s age = (3x+12)

Ahmed‟s age = (x+12)

Ahmed‟s father‟s age = 2× Ahmed‟s age

(3x+12) = 2(x+12)

3x+12 = 2x+24

3x-2x = 24-12

x = 12

∴ Ahmed‟s father‟s age = 12 years

∴ Ahmed‟s father‟s age = 3x = 3×12 = 36 years

12. Sajnu is 6 years older than his brother Nishu. If the sum of their ages is 28 years, what are their present age?

Ans: Let Nishu‟s age be „x‟

Let Sanju‟s age be (x+6)

Sum of their ages = 28 years

∴Present age = x+6+x = 28

2x+6 = 28

2x = 28-6

x = 22

2

x = 11

∴ Nishu‟s age = 11 years

∴ Sanju‟s age = x+6

= 11+6

= 17 years

13. Viji is twice as old as his brother Deepu. If the difference of their

ages is 11 years. Find theirpresent age.

Ans: Let Deepu‟s age be „x‟

Let Viji‟s age be „2x‟

Difference between their ages = 11 years

2x-x = 11

x = 11

∴ Deepu‟s age = 11 years

∴ Viji‟s age = 2x

= 2x×11

= 22 years

14. Mrs. Joseph is 27 years ilder than her daughter Bindu. After 8 years she will be twice as old as Bindu. Find their present age.

Ans: Let Bindu‟s age be „x‟

Let Mrs. Joseph‟s age be = (x+2x)

After 8 years, Bindu‟s age = (x+8)

Mrs. Joseph age = (x+27+8)

Mrs. Joseph‟s age = 2×BIndu‟s age

(x+27+8) = 2(x+8)

x+27+8 = 17-27-8

-x = -19

x = 19

x = 11

∴ Bindu‟s age = 19 years

∴ Mrs. Joseph‟s age = x+27

= 19+27

= 46 years

15. After 16 years, Leena will be three times as old as she is now. Find

her present age

Ans: Let Leena‟s age be „x‟

After 16 years, Leena‟s age will be x+16

x+16 = 3x

16 = 3x-x

16 = 2x

2x = 16

x = 16

2

x = 8 years

ADDITIONAL PROBLEM ON “LINEAR EQUATIONS IN ONE VARIABLE”

1. Choose the correct answer.

a. The value of x in the equation 5x-35 = 0is: a) 2 b) 7 c) 8 d) 11 [b]

b. If 14 is taken away from one fifth of a number, the result is 20.

The equation expressing this statement is

a) (x/5)-14 = 20 b) x-(14/5) = (20/5) c) x-14 = (20/5) d) x+(14/5) = 20 [a]

c. If five times a number increased by 8 is 53, the number is: a) 12 b) 9 c) 11 d) 2 [b]

d. The value of x in the equation 5(x-2) = 3(x-3) is:

a) 2 b) ½ c) ¾ d) 0 [b]

e. If the sum of two numbers is 84 and their difference is 30, the numbers are:

a) -57 and 27 b) 57 and 27

c) 57 and -27 d) -57 and -27 [b]

f. If the area of a rectangle whose length is its breadth is 800m2,

then length and breadth of the rectangle are:

a) 60m and 20m b) 40m and 20m c) 80m and 10m d) 100m and 8m [b]

g. If the sum of three consecutive odd numbers is 249, the

numbers are: a) 81,83,85 b) 79,81,83

c) 103,105,107 d) 95,97,99 [a]

h. If the (x+0.7x)/2 = 0.85, then x equals: a) 2 b) 1 c) -1 d) 0 [b]

i. If 2x-(3x-4)=3x-5 then x equals:

a) 14/9 b) 9/4 c) 3/2 d) 2/3 [b]

2. Solve: (i) (3x+24)/(2x+7)=2

3x+24

2x+7= 2

Multiply both sides by (2x+7)

3x + 24

2x + 7 x 2x + 7 = 2 2x + 7

3x+4x = 4x+14

3x-4x = 14-24

-x = -10

x = 10

∴ x = 10

(ii) (1-9y)/(11-3y)=(5/8)

𝟏−𝟗𝐲

𝟏𝟏−𝟑𝐲=

𝟓

𝟖

By cross multiplication

8(1-9y)=5(11-3y)

8-72y=55-15y

-72y+15y=55-8

-57y=47

-y=𝟒𝟕

𝟓𝟕

∴ y=𝟒𝟕

𝟓𝟕

3. The sum of two numbers 45 and their ratio is 7:8. Find the numbers.

Ans: Let the numbers be x and y

x+y=45

x=45-y ............ (1)

Now 𝐱

𝐲=

𝟕

𝟖

Cross multiplication 8x=7y ................. (2)

Substituting (1) in (2)

8(45-y) =7y

360-8y=7y

-8y-7y=-360

-15y= -360

y= −360

−15 = 24

Now x=45-y

x=45-24

x=21

The numbers are 21 and 24.

4. Shona’s mother is four times as old as Shona. After five years, her

mother will be three times as old as Shona (at that time). What are their present ages?

Ans: Let the present age of Shone be x years.

∴ her mother‟s age 4x years.

After 5 years Shona will be (4x+5) years

Her mother will be three times as old as Shona

∴ (4x+5)=3(x+5)

4x+5=3x+15

4x+3x=15-5

x=10

Present age of Shona = 10 years

5. The sum of three consecutive even numbers is 336. Find the?

Ans: Let the 3 consecutive even numbers x, 2+2 and x+4

Sum of three consecutive even numbers

=x=x+x+x+4

3x+6=336

3x=336-6

3x=330

x = 330

3 =110

The three consecutive even numbers are 110,112,114.

6. Two friends A and B start a joint business with a capital Rs. 60,000.

If A’s share is twice that of B, how much have each invested? Ans: Let the share of B Rs. X

∴ Share of A = Rs. 2x

x+2x=60,000

3x=60,000

x= 60 ,000

3 = 20,000

A‟s share = 2x = 2×20,000 = Rs.40,000

B‟s share = x = 1×20,000=Rs.20,000

7. Which is the number when 40 is subtracted gives one third of the original number?

Ans: Let the number be x

x-40= 1

3 x

3(x-40)=x 3x-120=x

Cross multiplication 3x-x=120 2x=120

∴ x = 12 0

2 = 60

The original number = 60

8. Find the number whose sixth exceeds its eight parts by 3.

Ans: Let the number be x

∴ Sixth part of the number = 𝐱

𝟔

Eighth part of the number =𝐱

𝟖

X

6−

x

8= 3

4x−3x

24= 3 LCM=24

x=3×24 x=72

∴ The number = 72

9. A house and a garden together coast Rs.8,40,000. The price of the

garden is 𝟓

𝟏𝟐 times the price of the house. Find the price of the house

and the garden.

Ans: Price of the garden = 5

12× 8,40,000

= Rs. 3,50,000

∴ Price of the house = 8,40,000-3,50,000

= Rs. 4,90,000

10. Two farmers A and B together own a stock of grocery. They agree to

divide it by its value. Farmer A takes 72 bags while B takes 92 bags and Rs. 8,000 to A. What is the cost of each bag?

Ans: Let the cost of each bag be Rs. X

Cost of 72 bags = Rs. 72x

Cost of 92 bags = Rs. 92x

72x+8000=92x-8000

72x-92x= -8000 – 8000

-20x= -16000

x = −16000

−20= 800

∴ Cost of each bag = Rs. 800

11. A father’s age is four times that of his son. After 5 years, it will be three times that of his son. How many more years will takes if

father’s age is to be twice that of his son?

Ans: Let the age of the son be x years

The father age = (4x) years

After 5 years son‟s age = (x+5) years

After 5 years faher‟s age = (4x+5) years

(4x+5)=3(x+5)

4x+5=3x+15

4x-3x=15-5

x=10

Present age of the son = 10 years

Present age of the father = 4x=4×10=40 years

Let after 4 years father will be twice of his son

∴ 40+y=2(10+y)

40+y=20+2y

y-2y=20-40

-y=-20

y=20

After 20 years father will be twice of his son. How many more

years father age is twice that of his son?

y=-5=20-5=15 years

12. Find a number which when multiplied by 7 is as much above 132 as

it was originally below it.

Ans: Let the number be x

7 twice of the number =7x

∴ 7x is as much above 132 is 7x-132 originally it was (132-x)

below it

Now both are equal

7x-132=132-x

7x+x=132+132

8x=264

x = 264

8 =33

∴ The original number =33

13. A person buys 25 pens worth Rs. 250 each of equal cost. He wants to

keep 5 pens for himself and sell the remaining to recover his money. What should be price of each pen?

Ans: Number of pens bought = 25

Cost price of 25 pens = Rs.250

Price of each pen = Rs. 250

25 = Rs. 10

Number of pens he sold = 25-5=20

Let the selling price of each pen = Rs. X

∴ Selling price of 20 pens = Rs. 20x

20x=250

x = 250

20=

25

2= 12.50

∴ Selling price of each pen = Rs. 12.50

14. The sum of the digits of a two-digiy number is 12. If the new number formed by reversing the digits is greater than the original

number. Check your solution.

Ans: Let the digit in the unit place is y and tens place is x

x+y=12

x =(12-y)

∴ The original number = (10x+y)

The reversed number = (10y+x)

10y+x=10x+y+18

10y-y+x-10y=18

9y-9x=18

9(y-x)=18

y-x=2

-x+y=2

x+y=12

-x+y=12

2y=14 adding both equation

y = 7

x+y=12

x+7=12

x = 12-7=5

Ten‟s digit = 5 units digit =7

∴ The original number = 57

15. The distance between two stations is 340km. Two trains start simultaneously from these stations on parallel tracks and cross each

her. The speed of one of them is greater than that if the other by 5km/hr. If the distance between two trains after 2 hours of their start

is 30km. Find the speed of each train.

Ans: Let the speed of one train be x km/hr.

The speed of the other train = (x+5) km/hr.

The distance travelled by first train in 2 hours = 2x km

The distance travelled by second train in 2 hours = 2(x+5) km

Distance between the two trains is 2 hours = 30 km

Total distance travelled = 340+30=370 km

2x+2x+10=370

4x = 370-10=360

x = 360

4 = 90

Speed of the first train = 90 km/hr.

Speed of the second train = (90+5) = 95 km/hr.

16. A steamer goes down stream and covers the distance between two in

4 hours while it covers the same distance up stream in 5 hours. If the speed of the steamer up stream is 2 km/hour. Find the of streamer in still water.

Ans: Speed of steamer up stream = 2 km/hr.

Time taken by the steamer upstream = 5 hrs.

Distance travelled = 2×5 = 10 km.

Speed of steamer down steam = 10

4

= 5

2= 2

Let the speed of streamer in still water be =x km/hr.

The speed of steam = y km/hr.

Speed of steamer down steam = (x+y) km/hr.

x+y = 2.5

x-y = 2.0

2x = 4.5

x = 2.25

Speed of steamer in still water = 2.25 km/hr.

17. The numerator of the rational number is less than its denominator by 3. If the numerator becomes three times and the denominator is

increased by 20, the new number becomes 1/8. Find the original number.

Ans: Let the original number be x

y

x = y-3

3x

20 +y=

1

8

3x+8=y+20

24x=y+20 24(y-3)=y+20

24y-72=y+20 24y-y=20+72

23y=92

y= 92

23 =4

x=y-3 x=4-3=1

∴ The original number = 1

4

18. The digit at the tens place of a two digit number is three times the digit at the units place. If the sum of this number and the number

formed by reversing its digits is 88. Find the numbers.

Ans: The digit in the units place be x

The digit in the ten‟s place = 3x

∴ The reversed number = 10x+3x=13x

Sum of the numbers = 31x+13x=44x

44x=88

x= 88

44 =2

Digit in the units place = 2

Digit in the ten‟s place = 3x=3×2=6

The number = 62

19. The altitude of a triangle is five-thirds. The length of its

corresponding base. If the altitude is increased by 4cm and the base decreased by 2cm, the area of the triangle would remain the same.

Find the base and altitude of the triangle.

Ans: Let the base of the triangle be x cms.

∴ its altitude = 5

3x cms

Area of the triangle =1

2× b × h =

1

2× x ×

5x

3

= 5x

6

2

cm2

Altitude is increased by 4

It becomes = 5x

3+ 4 cms

Base is decreased by 2, it becomes (x-2) cms

Area of the triangle = 1

2× b × h

= 1

2 x − 2 ×

5x

2+ 4

= 1

2 x − 2 ×

5x+12

3

= x−2 5x +12

6

5x

6

2 =

5x2 +2x−24

6

5x2 =

5x2+2x-24

∴ 2x-24=0

2x=24

x=12

base = x =12 cms

Altitude = 5x

3=

5×12

3=

60

3= 20cms.

Base of the triangle = 12 cm. Altitude of the triangle = 2cm.

20. One of the angle of a triangle is equal to the sum of the other two angles. If the ratio of the other two angles of the triangle is 4:5, find

the angles of the triangle.

Ans: Let the angle of the triangle be y

Other two angles are 4:5, i.e., 4x, 5x

y+9x=180

y=9x

y+9x=180

9x+9x=180

18x= 180

18 = 10

The three angle are 4x=400

5x=500

9x=900

∴ Angles of the triangle 40, 50 and 900

CHAPTER-3 UNIT-3

CONGRUENCY OF TRIANGLES

EXERCISE 3.3.1

1. Identify the corresponding sides and corresponding angles in the

following congruent triangles.

(a)

P X

B R Y Z

Corresponding angles Corresponding sides

1. ∟Q = ∟Y 1. PR=XZ

2. ∟R = ∟Z 2. PQ = XY

3. ∟P = ∟X 3. QR = YZ

(b)

R A B

Q P C

Corresponding angles Corresponding sides

1. ∟R = ∟C 1. PQ=AB

2. ∟Q = ∟B 2. PR = AC

3. ∟A = ∟P 3. RQ = BC

2. Pair of congruent triangle and incomplete statements related to them

are given below. Observe the figures carefully and fill up the blanks:

(a) In the adjoining figure if ∟C = ∟F, then AB=DE and BC=EF

C D E

A B F

(b) In the adjoining figure if BC=EF, then

∟C = ∟F and ∟A = ∟D

A D

B C E F

(c) In the adjoining figure, if AC = CE and ∆ ABC = ∆ DEC, then

∟D=∟B and ∟A=∟E

D A

C

E B

EXERCISE 3.3.2

1. In the adjoining figure PQRS in a triangle. Identify the congruent

triangles formed by the diagonals.

P Q

S R

Ans: Data: PQRS is a rectangle

O is the mid point of PR and SQ

Proof: In ∆ PQR and ∆ SOR

1. ∆ PSQ ≅ ∆ RSQ

2. ∆ POQ ≅ ∆ SOR

3. ∆ POS ≅ ∆ ROQ

4. ∆ PQR ≅ ∆ PSR

∴ ∆ POS ≅ ∆ QOS (SAS postulate)

∴The congruent triangles formed by the diagonals are ∆ POQ,

∆ SOQ, ∆ POS and ∆ QOR.

2. In the figure ABCD in a square, M,N,O and P rae the midpoints of

sides AB,BC,CD and DA respectively. Identify the congruent

triangles.

A M B

P N

D O C

O

Ans: Data: ABCD is a square.

M, N, O and P are the midpoints of sides

AB , BC ,CD and DA respectively.

Proof: In ∆ APM and ∆ PDO

1. ∆ AP< ≅ ∆ BNM

2. ∆ DPO ≅ ∆ ONO

3. ∟ B = ∟C (900)

∴ ∆ MBN ∆ NOC (SAS postulate)

∴ The congruent triangles are ∆ APM, ∆ PDO, ∆ MBN and

∆ NOC.

3. In a triangle ABC, AB = AC. Points E on AB and D on AC are such

that AE = AD. Prove that triangles BCD and CBE are congruent.

A

E D

B C

Ans: Data: AB = AC

AE = AD

To Prove: ∆ BCD and ∆ CBE

Proof: In ∆ BCD and ∆ CBE

1. BC = BC (Common side)

2. ∟B = ∟C (AB = AC → data)

AB = AC → data

AE = AD → data

3. BE = DC

From (1), (2) and (3)

In ∆ BCD ≅ ∆ CBE (SAS postulate)

4. In the adjoining figure, the idea BA and CA have been produced

such that BA = AD and CA = AE. Prove that DE || BC. [Hint: Use the

cincept of alternate angles]

E D

B C

Ans: BA = AD and CA = AE

To Prove: DE || BC

Proof: In ∆ EAD and ∆ BAC.

1. BC = BC (Common side)

2. ∟ EAD = ∟ BAC (V.O.A)

3. BS = AD = (Data)

∴ ∆ EAD = ∆ BAC (SAS postulate)

∴ ABC = ADE (Congruent property)

But they are alternate angle

∴ DE || BC

A