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1

2 LINEAR EQUATIONS IN ONE VARIABLE

Exercise 2.1

Q.1. Solve the following equations. 1. x – 2 = 7 2. y + 3 = 10 3. 6 = z + 2

4. 37

+ x = 177

5. 6x = 12 6. 5t = 10

7. 23x = 18 8. 1.6 =

1.5y 9. 7x – 9 = 16

10. 14y – 8 = 13 11. 17 + 6p = 9 12. 3x + 1 = 7

15

Ans. 1. x – 2 = 7 Adding 2 to both sides, we get x – 2 + 2 = 7 + 2 ∴ x = 9 2. y + 3 = 10 Subtracting 3 from both sides, we get y + 3 – 3 = 10 – 3 ∴ y = 7 3. 6 = z + 2 Subtracting 2 from both sides, we get 6 – 2 = z + 2 – 2 4 = 2 ∴ z = 4



2

4. 37

+ x = 177

Subtracting 37

from both sides, we get

⇒ 3 – 7 7

3 + x = 17 3– 7 7

⇒ x = 17 – 37

= 147

= 2

∴ x = 2 5. 6x = 12 Dividing both sides by 6, we get

66x = 12

6

∴ x = 2

6. 5t = 10

Multiplying both sides by 5, we get

× 55

t = 10 × 5

∴ t = 50

7. 23x = 18


⇒ 2 × 33

x = 18 × 3

⇒ 2x = 54



3

Dividing both sides by 2, we get

22x = 54

2

∴ x = 27

8. 1.6 = 1.5y

Multiplying both sides by 1.5, we get

1.6 × 1.5 = 1.5y × 1.5

∴ y = 2.4 9. 7x – 9 = 16 Adding 9 to both sides, we get ⇒ 7x – 9 + 9 = 16 + 9 ⇒ 7x = 25

⇒ x = 257

∴ x = 257

10. 14y – 8 = 13 Adding 8 to both sides, we get ⇒ 14y – 8 + 8 = 13 + 8 ⇒ 14y = 21

∴ y = 2114

= 32

11. 17 + 6p = 9 Subtracting 17 from both sides, we get ⇒ 17 – 17 + 6p = 9 – 17



4

⇒ 6p = – 8

⇒ p = –86

∴ p = – 43

12. 3x + 1 = 7

15

Subtracting 1 from both sides, we get

3x + 1 – 1 = 7

15 – 1

⇒ 3x = 7 1–

15 1

⇒ 3x = 7 – 15

15 = –8

15


⇒ 3x × 3 = –8

15 × 3

∴ x = –85

Exercise 2.2

Q.1. If you subtract 12

from a number and multiply the result

by 12

, you get 18

. What is the number?

Ans. Let x be the number. As per condition :



5

1 – 2 2

x⎛ ⎞⎜ ⎟⎝ ⎠

1 = 18

⇒ 1– 2 4x = 1

8

Adding 14

to both sides, we get

⇒ 1 – + 2 4x 1

4 = 1 1+

8 4

⇒ 2x = 1 1+

8 4

⇒ 2x = 1 + 2

8

⇒ 2x = 3

8


⇒ 2x × 2 = 3

8 × 2

∴ x = 34

Hence, the required number is 34

.

Q.2. The perimeter of a rectangular swimming pool is 154 m. Its length is 2 m more than twice its breadth. What are the length and the breadth of the pool?

Ans. Let breadth of the rectangle be x metre. We know, perimeter of rectangle = 2(l + b) As per condition :



6

Length = 2x + 2 Perimeter = 2(length + breadth) 154 = 2(2x + 2 + x) ⇒ 154 = 2(3x + 2) ⇒ 154 = 6x + 4 ⇒ 154 – 4 = 6x + 4 – 4 ⇒ 150 = 6x

⇒ 1506

= x

⇒ 25 = x; ∴ x = 25 m ⇒ Breadth of rectangle is 25 m. ∴ Length = 2x + 2 (put x = 25) ⇒ Length = 2 × 25 + 2 = 50 + 2 = 52 m. Hence, length of the pool is 52 m, and breadth of

pool is 25 m.

Q.3. The base of an isosceles triangle is 43

cm. The perimeter

of the triangle is 2415

cm. What is the length of either of

the remaining equal sides?

Ans. Let length of remaining equal sides be x cm. As per condition

Perimeter of triangle = 2415

cm = 6215

cm



7

∴ (x + x + 43

) cm = 6215

cm

⇒ (2x + 43

) cm = 6215

cm

⇒ 2x + 43

= 6215

⇒ 2x = 62 4– 15 3

⇒ 2x = 62 4– 15 3

⇒ 2x = 62 – 2015

= 4215

⇒ x = 422 × 15

⇒ x = 2115

= 75

∴ x = 215

cm

Hence, length of remaining equal sides is 215

cm.

Q.4. Sum of two numbers is 95. If one exceeds the other by 15, find the numbers.

Ans. Let one number be x then other number is x + 15. As per condition, (x + x + 15) = 95 ⇒ 2x + 15 = 95



8

⇒ 2x = 95 – 15 ⇒ 2x = 80

⇒ 22x = 80

2

∴ x = 40 ∴ Other number = x + 15 = 40 + 15 = 55 Hence, the numbers are 40 and 55.

Q.5. Two numbers are in the ratio 5:3. If they differ by 18, what are the numbers?

Ans. Since the ratio of two numbers is 5:3 we can take numbers as,

1st number = 5x 2nd number = 3x As per condition, 5x – 3x = 18 ∴ 2x = 18

x = 182

;

∴ x = 9 ∴ 1st number = 5x = 5 × 9 = 45 2nd number = 3x = 3 × 9 = 27 Hence, the numbers are 45 and 27.

Q.6. Three consecutive integers add up to 51. What are these integers?

Ans. Three consecutive integers are x, x + 1 and x + 2. As per condition,



9

Sum of three consecutive integers = 51 ⇒ x + x + 1 + x + 2 = 51 ⇒ 3x + 3 = 51 ⇒ 3x = 51 – 3 ⇒ 3x = 48

⇒ x = 483

∴ x = 16 Hence, three consecutive integers are 16, 17 and 18.

Q.7. The sum of three consecutive multiples of 8 is 888. Find the multiples.

Ans. If x is the multiple of 8, the next multiple is x + 8. The next to this is x + 8 + 8 or x + 16

So, three consecutive multiples of 8 as x, x + 8 and x + 16. As per condition, x + x + 8 + x + 16 = 888 ⇒ 3x + 24 = 888 ⇒ 3x = 888 – 24 3x = 864

⇒ 33x = 864

3

∴ x = 288 Hence, three consecutive multiples of 8 are 288, 296

and 304 whose sum is 888.

Q.8. Three consecutive integers are such that when they are taken in increasing order and multiplied by 2, 3 and 4 respectively, they add up to 74. Find these numbers.



10

Ans. Three consecutive integers are : x, x + 1 and x + 2 As per condition, 2x + 3(x + 1) + 4(x + 2) = 74 ⇒ 2x + 3x + 3 + 4x + 8 = 74 ⇒ 9x + 11 = 74 ⇒ 9x = 74 – 11 ⇒ 9x = 63

⇒ 99x = 63

9

∴ x = 7 Hence, three consecutive integers are 7, 8 and 9.

Q.9. The ages of Rahul and Haroon are in the ratio 5:7. Four years later the sum of their ages will be 56 years. What are their present ages?

Ans. Ratio of age of Rahul and Haroon = 5:7. So, we can take Rahul’s age = 5x Haroon’s age = 7x As per condition, 5x + 4 + 7x + 4 = 56 ⇒ (5x + 7x + 8) = 56 ⇒ 12x + 8 = 56 ⇒ 12x + 8 – 8 = 56 – 8 ⇒ 12x = 48

⇒ x = 4812

∴ x = 4 years



11

Hence, Rahul’s age = 5x = 5 × 4 = 20 years Haroon’s age = 7x = 7 × 4 = 28 years

Q.10. The number of boys and girls in a class are in the ratio 7:5. The number of boys is 8 more than the number of girls. What is the total class strength?

Ans. Number of boys in the class = 7x Number of girls in the class = 5x As per condition, 7x = 5x + 8 ⇒ 7x – 5x = 8 ⇒ 2x = 8

⇒ x = 82

∴ x = 4 ∴ Number of boys = 7 × 4 = 28 Number of girls = 5 × 4 = 20 ∴ Total Number of students in the class is 28 + 20 = 48.

Q.11. Baichung’s father is 26 years younger than Baichung’s grandfather and 29 years older than Baichung. The sum of the ages of all the three is 135 years. What is the age of each one of them?

Ans. Let Baichung’s father’s age = x years So, Baichung’s age = x – 29 years ∴ Baichung’s grandfather’s age = x + 26 years As per condition, x + x – 29 + x + 26 = 135 ⇒ 3x – 3 = 135



12

⇒ 3x = 135 + 3 ⇒ 3x = 138

⇒ x = 1383

∴ x = 46 Hence, Baichung’s father’s age = 46 years, Baichung’s age = 46 – 29 = 17 years, and Baichung’s grand father’s age = x + 26 = 46 + 26 = 72 years.

Q.12. Fifteen years from now Ravi’s age will be four times his present age. What is Ravi’s present age?

Ans. Let Ravi’s present age = x years As per condition, 4x = x + 15 ⇒ 4x – x = 15 ⇒ 3x = 15

⇒ x = 153

∴ x = 5 years Hence, the present age of Ravi is 5 years.

Q.13. A rational number is such that when you multiply it by 52

and add 23

to the product you get 7–12

. What is the

number?

Ans. Let x be the rational number. As per condition,



13

5 + 2 3x 2 = –7

12

⇒ 52x = –7 2–

12 3

⇒ 52x = –7 – 8

12 = –15

12

⇒ x = –15 2× 12 5

∴ x = –12

Hence, the required rational number is –12

.

Q.14. Lakshmi is a cashier in a bank. She has currency notes of denominations Rs 100, Rs 50 and Rs 10, respectively. The ratio of the number of these notes is 2:3:5. The total cash with Lakshmi is Rs 4,00,000. How many notes of each denomination does she have?

Ans. Let the number of Rs 100 notes, Rs 50 notes and Rs 10 notes be 2x, 3x and 5x respectively.

The amount she has from Rs 100 notes : 2x × 100 = Rs 200x from Rs 50 notes : 3x × 50 = Rs 150x from Rs 10 notes : 5x × 10 = Rs 50x hence, the total money she has = 200x + 150x + 50x = 4,00000 (by question) ⇒ 400x = 4,00000



14

x = 400,000400

⇒ x = 1,000 ∴ x = 1,000 Number of Rs 100 notes she has = 2 × 1,000 = 2,000 notes Number of Rs 50 notes she has = 3 × 1,000 = 3,000 notes Number of Rs 10 notes she has = 5 × 1,000 = 5,000 notes

Q.15. I have a total of Rs 300 in coins of denomination Re 1, Rs 2 and Rs 5. The number of Rs 2 coins is 3 times the number of Rs 5 coins. The total number of coins is 160. How many coins of each denomination are with me?

Ans. Let the number of Rs 2 coins and Rs 5 coins be 3x and x respectively.

Therefore the number of Rs 1 coin = 160 – (3x + x) = 160 – 4x

The amount I have from Rs 5 coins = Rs 5x From Rs 2 coins = Rs 6x From Rs 1 coins = 160 – 4x Hence, the total amount I have 5x + 6x + (160 – 4x) = 300 ⇒ 7x + 160 = 300 ⇒ 7x = 300 – 160 ⇒ 7x = 140

⇒ 77x = 140

7

⇒ x = 20 The number of Rs 5 coins I have = 20



15

The number of Rs 2 coins I have = 60 The number of Rs 1 coins I have = 80

Q.16. The organisers of an essay competition decide that a winner in the competition gets a prize of Rs 100 and a participant who does not win gets a prize of Rs 25. The total prize money distributed is Rs 3,000. Find the number of winners, if the total number of participants is 63.

Ans. Let the number of winners be x. As per condition, 100 × x + (63 – x) × 25 = 3000 ⇒ 100x + 1575 – 25x = 3000 ⇒ 75x = 3000 – 1575 75x = 1425

⇒ x = 142575

∴ x = 19 Hence, the number of winners is 19.

Exercise 2.3

Q.1. Solve the following equations and check your results. 1. 3x = 2x + 18 2. 5t – 3 = 3t – 5 3. 5x + 9 = 5 + 3x 4. 4z + 3 = 6 + 2z 5. 2x – 1 = 14 – x 6. 8x + 4 = 3(x – 1) + 7

7. x = 45

(x + 10) 8. 23x + 1 = 7

15x + 3

9. 2y + 53

= 263

– y 10. 3m = 5m – 85



16

Ans. 1. 3x = 2x + 18 ------------- (i) ⇒ 3x – 2x = 18 ∴ x = 18, It is the Checking the answer : put x = 18 in eqn. (i) we have 3x = 2x + 18 ⇒ 3 × 18 = 2 × 18 + 18 ⇒ 54 = 36 + 18 ⇒ 54 = 54 L.H.S = R.H.S Hence, x = 18 is the required solution. 2. 5t – 3 = 3t – 5 ------------------ (i) ⇒ 5t – 3t – 3 = – 5 ⇒ 2t – 3 = – 5 ⇒ 2t = – 5 + 3 ⇒ 2t = – 2

⇒ t = –22

∴ t = –1 It is the Checking the answer : put t = –1 in eqn. (i) we have, 5 × (–1) – 3 = 3 (–1) – 5 ⇒ – 5 – 3 = – 3 – 5 ⇒ – 8 = – 8 ∴ L.H.S = R.H.S Hence, t = –1 is the required solution.



17

3. 5x + 9 = 5 + 3x ------------- (i) We get, ⇒ 5x – 3x + 9 = 5 ⇒ 2x + 9 = 5 ⇒ 2x = 5 – 9 ⇒ 2x = – 4

⇒ x = –42

∴ x = – 2 Checking the answer : put x = – 2 in eqn. (i) we have ⇒ 5 (– 2) + 9 = 5 + 3 (– 2) ⇒ – 10 + 9 = 5 – 6 ⇒ – 1 = – 1 L.H.S = R.H.S Hence, x = – 2 is the required solution. 4. 4z + 3 = 6 + 2z -------------- (i) ⇒ 4z + 3 – 2z = 6 + 2z – 2z ⇒ 2z + 3 = 6 ⇒ 2z = 6 – 3 ⇒ 2z = 3

⇒ z = 32

∴ z = 32

Checking the answer :



18

put z = 32

in equation (i) we have,

⇒ 4 32

⎛ ⎞⎜ ⎟⎝ ⎠

+ 3 = 6 + 2 32

⎛ ⎞⎜ ⎟⎝ ⎠

⇒ 6 + 3 = 6 + 3 ⇒ 9 = 9 L.H.S = R.H.S

Hence, z = 32

is the required solution.

5. 2x – 1 = 14 – x ------------- (i) ⇒ 2x – 1 + x = 14 ⇒ 3x – 1 = 14 ⇒ 3x = 14 + 1 ⇒ 3x = 15

⇒ x = 153

∴ x = 5 Checking the answer : put x = 5 in eqn. (i) we have, ⇒ 2 × 5 – 1 = 14 – 5 ⇒ 10 – 1 = 14 – 5 ⇒ 9 = 9 ∴ L.H.S = R.H.S Hence, x = 5 is the required solution.

6. 8x + 4 = 3 (x – 1) + 7 ----------- (i)



19

⇒ 8x + 4 = 3x – 3 + 7 ⇒ 8x + 4 = 3x + 4 ⇒ 8x – 3x + 4 = 4 ⇒ 5x + 4 = 4 ⇒ 5x = 4 – 4 ⇒ 5x = 0

⇒ x = 05

∴ x = 0 Checking the answer : put x = 0 in eqn. (i) we have 8x + 4 = 3x + 4 8 × 0 + 4 = 3(0 – 1) + 7 ⇒ 0 + 4 = 0 – 3 + 7 ∴ 4 = 4 Hence, x = 0, is the required solution.

7. x = 45

(x + 10) ------------- (i)

⇒ x = 45

x + 405

⇒ x = 45

x + 8

⇒ x – 45

x = 8

⇒ 5 – 45

x x = 8



20

⇒ 5x = 8

⇒ x = 8 × 5 ∴ x = 40 Checking the answer : put x = 40 in eqn. (i) we have

⇒ 40 = 45

(40 + 10)

⇒ 40 = 4 × 505

∴ 40 = 40 L.H.S = R.H.S Hence, x = 40 is the required solution.

8. 23x + 1 = 7

15x + 3 ---------------- (i)

⇒ 23x + 1 – –7

15x = 3

⇒ 23x – 7

15x + 1 = 3

⇒ 10 – 715x x + 1 = 3

⇒ 315

x + 1 = 3

⇒ 315

x + 1 – 1 = 3 – 1

⇒ 315

x = 2

⇒ 3x = 2 × 15 ⇒ 3x = 30



21

⇒ x = 303

∴ x = 10 Checking the answer : put x = 10 in eqn. (i) we have

⇒ 2 × 103

+ 1 = 7 × 1015

+ 3

⇒ 20 1+ 3 1

= 70 3+ 15 1

⇒ 20 + 33

= 70 + 4515

⇒ 233

= 11515

⇒ 233

= 233

∴ L.H.S = R.H.S Hence, x = 10 is the required solution.

9. 2y + 53

= 263

– y ------------- (i)

⇒ 2y + y + 53

= 263

⇒ 3y + 53

= 263

⇒ 3y + 53

= 263

⇒ 3y = 263

– 53

⇒ 3y = 26 – 53

= 213

⇒ 3y = 7

y = 73



22

∴ y = 73

Check :

put y = 73

in eqn. (i) we have

2 × 73

+ 53

= 263

– 73

⇒ 14 5+ 3 3

= 26 7– 3 3

⇒ 193

= 26 – 73

= 193

∴ L.H.S = R.H.S

Hence, y = 73


10. 3m = 5m – 85

---------------- (i)

⇒ 3m – 5m = 85

⇒ – 2m = – 85

⇒ m = – 8– 2 × 5

∴ m = 45

(required solution)

Check :

put m = 45

in eqn. (i) we have

⇒ 3 × 45

= 5 × 45

– 85

⇒ 125

= 20 8– 5 5



23

⇒ 125

= 20 – 85

⇒ 125

= 125

∴ L.H.S = R.H.S

Hence, m = 45


Exercise 2.4

Q.1. Amina thinks of a number and subtracts 52

from it. She

multiplies the result by 8. The result now obtained is 3 times the same number she thought of. What is the number?

Ans. Let the number be x. As per condition,

8 (x – 52

) = 3x

⇒ 8x – 8 × 52

= 3x

⇒ 8x – 20 = 3x ⇒ 8x – 3x – 20 = 3x – 3x ⇒ 5x – 20 = 0 ⇒ 5x = 20

⇒ x = 205

∴ x = 4 Hence, 4 is the required number.



24

Q.2. A positive number is 5 times another number. If 21 is added to both the numbers, then one of the new numbers becomes twice the other new number. What are the numbers?

Ans. Let the number be x and another number be 5x. As per condition, 2(x + 21) = 5x + 21 ⇒ 2x + 42 = 5x + 21 ⇒ 5x + 21 – 2x – 42 = 0 ⇒ 3x – 21 = 0 ⇒ 3x – 21 + 21 = 0 + 21 ⇒ 3x = 21

⇒ x = 213

⇒ x = 7 ∴ So, first number = 7 another number = 5 × 7 = 35

Q.3. Sum of the digits of a two-digit number is 9. When we interchange the digits, it is found that the resulting new number is greater than the original number by 27. What is the two-digit number?

Ans. Let us take the two digit number be such that the digit in the units place is x, and digit in tens place be (9 – x.)

Now, original number = 10(9 – x) + x = 90 – 10x + x = 90 – 9x

Number after interchanging the digits = 9 – x + 10x = 9 + 9x

As per condition :



25

9 + 9x = 90 – 9x + 27

or, 18x = 90 + 27 – 9 = 108

∴ x = 10818

= 6

So, digit in units place = 6

and digit in tens place = 9 – 6 = 3

So, required number is 36.

Q.4. One of the two digits of a two digit number is three times the other digit. If you interchange the digits of this two-digit number and add the resulting number to the original number, you get 88. What is the original number?

Ans. Let the digit in tens place of the number is x. Digit in units place is 3x ∴ Number = 10x + 3x = 13x Number after interchanging the digits = 10(3x) + x = 31x As per condition, 13x + 31x = 88 ⇒ 44x = 88

4444

x = 8844

∴ x = 2 ∴ Number = 13 × 2 = 26 or, another possible number is = 31x = 31 × 2 = 62



26

Hence, the original number may be 26 or 62.

Q.5. Shobo’s mother’s present age is six times Shobo’s present age. Shobo’s age five years from now will be one third of his mother’s present age. What are their present ages?

Ans. Let Shobo’s present age be x years and Shobo’s mother’s age is 6x years. As per condition,

(x + 5) = 13

(6x)

⇒ x + 5 = 2x x + 5 – x = 2x – x ∴ 5 = x Hence, Shobo’s present age is 5 years. Mother’s age is 6 × 5 = 30 years.

Q.6. There is a narrow rectangular plot, reserved for a school, in Mahuli village. The length and breadth of the plot are in the ratio 11:4. At the rate Rs 100 per metre it will cost the village panchayat Rs 75000 to fence the plot. What are the dimensions of the plot?

Ans. Let, length of the plot be 11x and breadth be 4 x. We know perimeter of rectangle = 2(length + breadth) ⇒ 2(11x + 4x) = 30x

At the rate Rs 100 per metre it will cost the village panchayat Rs 75000 to fence the plot.

So, perimeter of plot = 75000100

= 750 metre

As per condition,



27

30x = 750

⇒ x = 75030

= 25

Hence, length of the plot = 11x = 11 × 25 = 275 m and, breadth of the plot = 4x = 4 × 25 = 100 m

Q.7. Hasan buys two kinds of cloth materials for school uniforms, shirt material that costs him Rs 50 per metre and trouser material that costs him Rs 90 per metre. For every 2 meters of the trouser material he buys 3 metres of the shirt material. He sells the materials at 12% and 10% profit respectively. His total sale is Rs 36,660. How much trouser material did he buy?

Ans. Let cloth materials of trouser be x metres and

cloth materials of shirt be 32

x metres.

Cost price of the cloth material of trouser = x × 90 He sells it with 12% profit, so its selling price

= 90 × x + 90 × x × 12100

= 100.8x

Cost price of the cloth material of shirt = 32

x × 50 = 75x

He sells it with 10% profit, so its selling price

= 75x + 75x × 10100

= 82.5x As per condition, 100.8x + 82.5x = 36660 ⇒ 183.3x = 36660



28

∴ x = 36660183.3

= 200 metres

Hence, he purchased 200 metres for trouser.

Q.8. Half of a herd of deer are grazing in the field and three fourths of the remaining are playing nearby. The rest 9 are drinking water from the pond. Find the number of deer in the herd.

Ans. Let the number of deer in the herd be x. As per condition,

3 + × + 92 2 4x x = x

⇒ 3+ + 92 8x x = x

L.C.M. of 2, 4 and 8 is 8. ⇒ 4x + 3x + 72 = 8x ⇒ 7x + 72 = 8x ⇒ 7x – 8x = –72 ⇒ –x = –72

⇒ x = –72–1

= 72 Hence, the number of deer in the herd is 72.

Q.9. A grandfather is ten times older than his granddaughter. He is also 54 years older than her. Find their present ages.

Ans. Let the age ofgranddaughter be x years. So, grandfather’s age be 10x years.



29

As per condition, 10x = x + 54 ⇒ 10x – x = 54 ⇒ 9x = 54

⇒ 99x = 54

9

∴ x = 6 Hence, granddaughter’s age is 6 years and grandfather’s age

is 60 years.

Q.10. Aman’s age is three times his son’s age. Ten years ago he was five times his son’s age. Find their present ages.

Ans. Let Aman’s son’s age be x years and Aman’s age be 3x years. As per condition, 5(x – 10) = 3x – 10 ⇒ 5x – 50 = 3x – 10 ⇒ 5x – 3x – 50 = – 10 ⇒ 2x – 50 = – 10 ⇒ 2x = – 10 + 50 ⇒ 2x = 40

⇒ x = 402

∴ x = 20 Hence, Aman’s sons age is 20 years and

Aman’s age is 60 years.

Exercise 2.5



30

Q. Solve the following linear equations.

1. 1 – 2 5x = 1 +

3 4x 2. 3 5– +

2 4 6n n n = 21

3. x + 7 – 83x = 17 5–

6 2x 4. – 5

3x = – 3

5x

5. 3 – 2 2 + 3 – 4 3

t t = 2 – 3

t

6. m – – 12

m = 1 – – 23

m

Simplify and solve the following linear equations. 7. 3(t – 3) = 5(2t + 1) 8. 15(y – 4) –2(y – 9) + 5(y + 6) = 0 9. 3(5z – 7) –2(9z – 11) = 4(8z – 13) – 17 10. 0.25(4f – 3) = 0.05(10f – 9)

Ans. 1. 1 – 2 5x = 1+

3 4x

1 – – 2 3x x

5 = 1

4

⇒ 3 – 2 1– 6 5

x x = 14

⇒ 1– 6 5x = 1

4

⇒ 6x = 1 1

4 5+

⇒ 6x = 9

20

⇒ 6x × 6 = 9

20 × 6

∴ x = 5420

= 2710



31

2. 3 5 – + 2 4 6n n n = 21

⇒ 6 – 9 + 1012

n n n = 21 [L.C.M of 2, 4, 6 = 12]

⇒ 712

n = 21

Multiply both sides by 127

7 1× 12 7

n 2 = 21 × 127

∴ n = 36

3. x + 7 – 83x = 17 5–

6 2x

⇒ x – 8 + 73x = 17 5–

6 2x

⇒ 3 – 83

x x + 7 = 17 5– 6 2

x

⇒ –5 73

x+ = 17 5–

6 2x

–5 5 + 73 2

x x+ = 17

6

⇒ –10 + 156

x x + 7 = 176

⇒ 56x + 7 = 17

6

⇒ 56x = 17

6 – 7



32

⇒ 56x = 17 – 42

6

⇒ 56x = –25

6

⇒ 5 6 6 5

x × = –25 6 6 5

×

⇒ x = –255

= – 5 ∴ x = – 5

4. – 53

x = – 35

x

⇒ 5x – 25 = 3x – 9 (Cross multiplying) ⇒ 5x – 3x = – 9 + 25 ⇒ 2x = 16

⇒ 22x = 16

2

∴ x = 8

5. 3 – 2 2t + 3 – 4 3

t = 2 – 3

t

⇒ 3 – 2 2 + 3 – 4 3

t t⎛ ⎞⎜ ⎟⎝ ⎠

= 2 – 33

t

⇒ 3(3 – 2) – 4(2 – 3)12

t t = 2 – 33

t

⇒ 9 – 6 – 8 – 12 12

t t = 2 – 33

t



33

⇒ – 1812

t = 2 – 33

t

⇒ 3(t – 18) = 12(2 – 3t) ⇒ 3t – 54 = 24 – 36t ⇒ 3t + 36t = 24 + 54 ⇒ 39t = 78

∴ t = 7839

= 2

∴ t = 2

6. m – ( )– 12

m = 1 – ( )– 2

3m

⇒ 2 – + 12

m m = ( )3 – – 23m

⇒ 2 – + 12

m m = 3 – + 23m

⇒ + 12

m = 5 – 3

m

⇒ 3m + 3 = 10 – 2m (Cross multiplying) ⇒ 3m + 2m = 10 – 3 ⇒ 5m = 7

∴ m = 75

7. 3(t – 3) = 5(2t + 1) ⇒ 3t – 9 = 10t + 5 ⇒ 3t – 10t = 5 + 9 ∴ t = – 2



34

8. 15(y – 4) – 2(y – 9) + 5(y + 6) = 0 ⇒ 15y – 60 – 2y + 18 + 5y + 30 = 0 ⇒ 18y – 12 = 0 ⇒ 18y = 12

⇒ y = 1218

∴ y = 23

9. 3(5z – 7) – 2(9z – 11) = 4(8z – 13) – 17 ⇒ 15z – 21 – 18z + 22 = 32z – 52 – 17 ⇒ – 3z + 1 = 32z – 69 ⇒ –3z – 32z = –69 – 1 ⇒ –35z = –70

⇒ z = –70–35

⇒ z = 2 10. 0.25(4f – 3) = 0.05(10f – 9) ⇒ 1.0f – 0.75 = 0.5f – 0.45 ⇒ 1.0f – 0.5f = 0.45 + 0.75

∴ f = 0.30 3= 0.5 5

Exercise 2.6

Q.1. Solve the following equations.

1. 8 – 33x

x = 2 2. 9

7 – 6x

x = 15 3.

+ 15z

z = 4

9



35

4. 3 + 42 – 6

yy

= –25

5. 7 + 4+ 2

yy

= – 43

Ans. 1. 8 – 33x

x = 2

⇒ 8x – 3 = 2 × 3x (Cross multiplication) ⇒ 8x – 6x = 3 ⇒ 2x = 3

∴ x = 32

2. 97 – 6

xx

= 15

⇒ 9x = 15 (1 – 6x) (Cross multiplication) ⇒ 9x = 15 × 7 – 15 × 6x 9x + 90x = 105 99x = 105

∴ x = 10599

= 3533

3. + 15

zz

= 49

⇒ z × 9 = 4 (x + 15) (Cross multiplication) ⇒ 9z = 4z + 60 ⇒ 9z – 4z = 60 ⇒ 5z = 60



36

∴ z = 605

= 12

4. 3 42 – 6y

y+ = –2

5

⇒ 5(3y + 4) = (–2) (2 – 6y) ⇒ 15y + 20 = – 4 + 12y ⇒ 15y – 12y = – 4 – 20 ⇒ 3y = – 24

∴ y = –243

= –8

5. 7 + 4+ 2

yy

= – 43

⇒ 3 (7y + 4) = (– 4) (y + 2) ⇒ 21y + 12 = – 4y – 8 ⇒ 21y + 4y = – 8 – 12 ⇒ 25y = – 20

∴ y = –20 – 4= 25 5

Q.6. The ages of Hari and Harry are in the ratio 5:7. Four years from now the ratio of their ages will be 3:4. Find their present ages.

Ans. Let present ages of Hari and Harry be 5x years and 7x years respectively.

After four years Hari’s age = (5x + 4) years Harry’s age = (7x + 4) years As per condition,



37

5 + 47 + 4

xx

= 34

or 4(5x + 4) = 3(7x + 4) or 20x + 16 = 21x + 12 or 21x + 12 = 20x + 16 or 21x – 20x = 16 – 12 ∴ x = 4 Therefore, Hari’s present age = 5x = 5 × 4 = 20 years and Harry’s present age = 7x = 7 × 4 = 28 years.

Q.7. The denominator of a rational number is greater than its numerator by 8. If the numerator is increased by 17 and the denominator is decreased by 1, the number obtained

is 32

. Find the rational number.

Ans. Let numerator be x then denominator should be x + 8

So, rational number = + 8x

x

As per condition,

+ 17+ 8 – 1

xx

= 32

⇒ 2(x + 17) = 3(x + 7) (Cross multiplication) ⇒ 2x + 34 = 3x + 21 ⇒ 3x + 21 – 2x – 34 = 0 ⇒ x – 13 = 0 ∴ x = 13 Hence, numerator = 13, and



38

denominator = 13 + 8 = 21

∴ Rational number = + 8x

x

= 1313 + 8

= 1321

linear equations in one variable - testlabz ratio of the number of these notes is 2:3:5. the total...

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