Saturday, March 25, 2023

# Linear Equations ICSE Class-8th Concise Selina Maths

Linear Equations ICSE Class-8th Concise Selina Mathematics Solutions of Chapter-14. We provide step by step Solutions of Exercise / lesson-14 Linear Equations in one Variable ( with Problems Based on Linear Equations ) for ICSE Class-8 Concise Selina Mathematics.

Our Solutions contain all type Questions with Exe-14 AExe-14 B  and Exe-14 C to develop skill and confidence . Visit official Website for detail information about ICSE Board Class-8.

## Linear Equations in one Variable ICSE Class-8th Concise Selina Mathematics Solutions of Chapter-14

–: Select Topics :–

Exe-14 A,

Exe-14 B,

Exe-14 C,

### Solved Questions of Exercise – 14 A, Linear Equations in one Variable ICSE Class-8th

Solve the following equations:

20 = 6 + 2x

20 = 6 + 2x
20 – 6 = 2x
14 = 2x
7 = x
x = 7

15 + x = 5x + 3

15 – 3 = 5x – x
12 = 4x
3 = x
x = 3

#### Question 3 :-

$\frac { 3x+2 }{ x-6 }$ = -7

3x + 2 = -7 (x – 6) (by cross multiplying)
3x + 2 = -7x + 42
3x + 7x = 42 – 2
10x = 40
x = 4

#### Question 4 :-

3a – 4 = 2(4 – a)

3a – 4 = 8 – 2a
3a + 2a = 8+4
5a = 12
a = 2.4

#### Question 5 :-

3(b – 4) = 2(4 – b)

⇒ 3b – 12 = 8 – 2b

⇒ 3b + 2b = 8 + 12

⇒ 5b = 20

⇒ b = 20/5

⇒ b = 4

#### Question 6 :-

$\frac { x+2 }{ 9 } =\frac { x+4 }{ 11 }$

⇒ 11(x + 2) = 9(x + 4)  ..(by cross multiplying)

⇒ 11x + 22 = 9x + 36

⇒ 11x – 9x = 36 – 22

⇒ 2x = 14

⇒ x = 14/2

⇒ x = 7

#### Question 7 :-

$\frac { x-8 }{ 5 } =\frac { x-12 }{ 9 }$

⇒ 9(x – 8) = 5(x – 12) …(by cross multiplying)

⇒ 9x – 72 = 5x – 60

⇒ 9x – 5x = -60 + 72

⇒ 4x = 12

⇒ x = 12/4

⇒ x = 3

#### Question 8 :-

5(8x + 3) = 9(4x + 7)

⇒ 40x + 15 = 36x + 63

⇒ 40x – 36x = 63 – 15

⇒ 4x = 48

⇒ x = 48/4

⇒ x = 12

#### Question 9 :-

3(x + 1) = 12 + 4(x – 1)

3(x + 1) = 12 + 4(x – 1)
3x + 3 = 12 + 4x – 4
3x – 4x = 12 – 4 – 3
-x = 5
x = -5

#### Question 10 :-

$\frac { 3x }{ 4 } -\frac { 1 }{ 4 } \left( x-20 \right) =\frac { x }{ 4 } +32$

⇒ x = 27 × 4

⇒ x = 108

#### Question 11 :-

$3a-\frac { 1 }{ 5 } =\frac { a }{ 5 } +5\frac { 2 }{ 5 }$

…..(Multiplying each term by 5)

⇒ 15a – a = 27 + 1

⇒ 14a = 28

⇒ a = 28/14

⇒ a = 2

#### Question 12 :-

$\frac { x }{ 3 } -2\frac { 1 }{ 2 } =\frac { 4x }{ 9 } -\frac { 2x }{ 3 }$

⇒  (x/3) – (5/2) = (4x/9) – (2x/3)

Since, L.C.M. of denominators 3, 2, 9 and 3 = 18

⇒ (x/3) × 18 – (5/2) × 18 = (4x/9) × 18 – (2x/3) ×18
…..[Multiplying each term by 18]

⇒ 6x – 45 = 8x – 12x

⇒ 6x + 12x – 8x = 45

⇒ 18x – 8x = 45

⇒ 10x = 45

⇒ x = 45/10

⇒ x = 4.5

#### Question 13 :-

$\frac { 4\left( y+2 \right) }{ 5 } =7+\frac { 5y }{ 13 }$

…(by cross multiplying)

⇒ 13 (4y + 8) = 5(91 + 5y)

⇒ 52y + 104 = 455 + 25y

⇒ 52y – 25y = 455 – 104

⇒ 27y = 351

⇒ y = 351/27

⇒ y = 13

#### Question 14 :-

$\frac { a+5 }{ 6 } -\frac { a+1 }{ 9 } =\frac { a+3 }{ 4 }$

Since, L.C.M. of denominators 6,9 and 4 = 36

…(Multiplying each term by 36)

⇒ 6(a + 5) – 4(a + 1) = 9(a + 3)

⇒ 6a + 30 – 4a – 4 = 9a + 27

⇒ 6a – 4a – 9a = 27 – 30 + 4

⇒ 6a – 13a = 1

⇒ -7a = 1

⇒ a = -1/7

#### Question 15 :-

$\frac { 2x-13 }{ 5 } -\frac { x-3 }{ 11 } =\frac { x-9 }{ 5 } +1$

⇒ 11(2x – 13)-5 (x – 3) = 11(x – 9) + 55

⇒ 22x – 143 – 5x + 15 = 11x – 99 + 55

⇒ 22x – 5x – 11x = -99 + 55 + 143 – 15

⇒ 6x = 198 – 114

⇒ 6x = 84

⇒ x = 84/6

⇒ x = 14

#### Question 16 :-

6(6x – 5) – 5 (7x – 8) = 12 (4 – x) + 1

6(6x – 5) – 5(7x – 8) = 12(4 – x) + 1

36x – 30 – 35x + 40 = 48 – 12x + 1

⇒ x + 12x = 49 – 10

⇒ 13x = 39

⇒ x = 19/13

⇒ x = 3

#### Question 17 :-

(x – 5) (x + 3) = (x – 7) (x + 4)

(x – 5) (x + 3) = (x – 7)(x + 4)

⇒ x2 + 3x – 5x – 15 = x2 + 4x – 7x – 28

⇒ -2x – 15 = – 3x – 28

⇒ 3x – 2x = 15 – 28

⇒ x = -13

#### Question 18 :-

(x – 5)2 – (x + 2)2 = -2

(x – 5)2 – (x + 2)2 = -2

⇒ (x2 – 10x + 25) – (x2 + 4x + 4) = -2

⇒ x2 – 10x + 25 – x2 – 4x – 4 = -2

⇒ -10x – 4x + 25 – 4 = -2

⇒ -14x = 4 – 2 – 25 = -23

⇒ x = (-23/-14) = (23/14) = 1 (9/14)

#### Question 19 :-

(x – 1) (x + 6) – (x – 2) (x – 3) = 3

(x – 1 )(x + 6) – (x – 2)(x – 3) = 3

x2 – x + 6x – 6 – (x2 – 3x – 2x + 6) = 3

⇒ x2 – x + 6x – 6 – x2 + 3x + 2x – 6 = 3

⇒ – x + 6x + 3x + 2x – 6 – 6 = 3

⇒ x + 11x – 6 – 6 = 3

10x = 15

x = 15/10 = 3/2 = 1 (1/2)

#### Question 20 :-

$\frac { 3x }{ x+6 } -\frac { x }{ x+5 } =2$

⇒ 2x2 + 9x = 2 (x2 + 11x + 30)

⇒ 2x2 + 9x = 2x2 + 22x + 60

⇒ 2x2 – 2x2 + 9x – 22x = 60

⇒ -13 x = 60

x = (-60/13) = -4 (8/13)

#### Question 21 :-

$\frac { 1 }{ x-1 } +\frac { 2 }{ x-2 } =\frac { 3 }{ x-3 }$

⇒ (x – 3)(3x – 4) = (x2 – 3x + 2)

⇒ 3x2 – 4x – 9x + 12 = 3x2 – 9x + 6

⇒ 3x2 – 13x – 3x2 + 9x = 6 – 12

⇒ -4x = -6

x = (-6/-4) = 3/2 =1(1/2)

#### Question 22 :-

⇒ (x – 1)(7x – 26) = (7x – 14)(x – 3)

⇒ 7x2 – 7x – 26x + 26 = 7x2 – 14x – 21x + 42

⇒ -33x + 26 = -35x + 42

⇒ 35x – 33x = 42 – 26

⇒ 2x = 16

⇒ x = 8

#### Question 23 :-

$\frac { 1 }{ x-1 } -\frac { 1 }{ x } =\frac { 1 }{ x+3 } -\frac { 1 }{ x+4 }$

= (x + 3)(x + 4) = x(x – 1)

⇒ x2 + 4x + 3x + 12 = x2 – x

⇒ x2 + 7x – x2 + x = -12

8x = -12

x = -12/8 = -3/2 = -1(1/2)

#### Question 24 :-

Solve: $\frac { 2x }{ 3 } -\frac { x-1 }{ 6 } +\frac { 7x-1 }{ 4 } =2\frac { 1 }{ 6 }$
Hence, find the value of ‘a’, if $\frac { 1 }{ a }$ + 5x = 8.

⇒ 27x – 1 = 26

⇒ 27x = 26 + 1

⇒ x = 27/27 = 1

Now, 1/a+5x=8

⇒ 1/a + 5×1 = 8

⇒1a+5 = 8

⇒1/a =8-5 = 3

∵ 3a = 1

⇒ a = 1/3

∴ x = 1 and a = 1/3

#### Question 25 :-

Solve: $\frac { 4-3x }{ 5 } +\frac { 7-x }{ 3 } +4\frac { 1 }{ 3 } =0$
Hence find the value of ‘p’ if 2p – 2x + 1 = 0

…(L.C.M. of 5, 3, 3 = 15)

-14x + 112 = 0

⇒ -14x = -112

⇒ x = (-112/-14) = 8

Hence x = 8

Now, 3p – 2x + 1=0

⇒ 3p – 2 x 8 + 1 = 0

⇒ 3p – 16 + 1 =0

⇒ 3p – 15 = 0.

⇒ 3p=15

⇒ p = 5

#### Question 26 :-

Solve: $0.25+\frac { 1.95 }{ x } =0.9$

0.25 + (1.95/x) =0.9

0.25x + 1.95 = 0.9x

⇒ 0.9x – 0.25x = 1.95

⇒ 0.65x = 1.95

⇒ x = (1.95/0.65) =3

Hence, x = 3

#### Question 27 :-

Solve: $5x-\left( 4x+\frac { 5x-4 }{ 7 } \right) =\frac { 4x-14 }{ 3 }$

6x + 12 = 28x – 98

22x = 98 + 12

x = 110/22 = 5

### Exe -14 B, Solved Questions of Linear Equations in one Variable Concise Selina ICSE Maths for Class-8th

#### Question 1 :-

Fifteen less than 4 times a number is 9. Find the number.

the required number be x

4 times the number = 4x

15 less than 4 times the number = 4x-15

According to the statement :

4x – 15 = 9

⇒ 4x = 9 + 15

⇒ 4x = 24

⇒ x = 6

#### Question 2 :-

If Megha’s age is increased by three times her age, the result is 60 years. Find her age

Megha’s age = x years

Three times Megha’s age = 3x years

According to the statement :

x + 3x = 60

=> 4x = 60

=> x = 15

Megha’s age = 15 years

#### Question 3 :-

28 is 12 less than 4 times a number. Find the number.

the required number be x

4 times the number = 4x

12 less than 4 times the number = 4x – 12

According to the statement

4x – 12 = 28

=> 4x = 28 + 12

=> 4x = 40

x = 10

Required number = 10

#### Question 4 :-

Five less than 3 times a number is -20. Find the number.

the required number = x

3 times the number = 3x

5 less than 3 times the number = 3x – 5

According to statement :

3x – 5 = -20

⇒ 3x = -20 + 5

⇒ 3x = -15

⇒ x = -5

Required number = -5

#### Question 5 :-

Fifteen more than 3 times Neetu’s age is the same as 4 times her age. How old is she?

Neetu’s age = x years

3 times Neetu’s age = 3x years

Fifteen more than 3 times Neetu’s age = (3x + 15) years

4 times Neetu’s age = 4x

According to the statement :

4x = 3x + 15

⇒ 4x – 3x = 15

⇒ x = 15

Neetu’s age = 15 years

#### Question 6 :-

A number decreased by 30 is the same as 14 decreased by 3 times the number; Find the number.

the required number = x

The number decreased by 30 = x – 30

14 decreased by 3 times the number = 14 – 3x

According to the statement :

x – 30 = 14 – 3x

⇒ x + 3x = 14 + 30

⇒ 4x = 44

x = 11

Required number =11

#### Question 7 :-

A’s salary is same as 4 times B’s salary. If together they earn Rs.3,750 a month, find the salary of each.

B’s salary = Rs. x

A’s salary = Rs. 4x

According to the statement :

x + 4x = 3750

⇒ 5x = 3750

⇒ x = 750

4x = 750 x 4 = 3000

A’s salary = Rs. 3000

B’s salary = Rs. 750

#### Question 8 :-

Separate 178 into two parts so that the first part is 8 less than twice the second part.

first part = x

Second part = 178 – x

According to the problem :

First Part = 8 less than twice the second part

x = 2(178 – x) – 8

⇒ x = 356 – 2x – 8

⇒ x+2x = 356 – 8

⇒ 3x = 348

⇒ x = 116

First Part = 116

⇒ Second Part = 178 – x = 178 – 116 = 62

First Part = 116

⇒ Second Part = 62

#### Question 9 :-

Six more than one-fourth of a number is two-fifth of the number. Find the number.

the required number = x

∴ One-fourth of the number = x/4

Two-fifth of the number = 2x/5

According to the statement:

….[Multiplying each term by 20 because L.C.M. of 5,4 and 1 = 20]

⇒ 8x- 5x = 120

⇒ 3x = 120

⇒ x = 120/3

x = 40

Required number = 40

#### Question 10 :-

The length of a rectangle is twice its width. If its perimeter is 54 cm; find its length.

width of the rectangle = x cm

Length of the rectangle = 2x cm

Perimeter of the rectangle = 2 [Length + Width] = 2 [2x + x] = 2 x 3x = 6x cm

Given perimeter = 54 cm

6x = 54

⇒ x = 9

Length = 2x = 2 x 9 = 18 cm

#### Question 11 :-

A rectangle’s length is 5 cm less than twice its width. If the length is decreased by 5 cm and width is increased by 2 cm; the perimeter of the resulting rectangle will be 74 cm. Find the length and the width of the original rectangle.

width of the original rectangle = x cm

Length of the original rectangle = (2x – 5)cm

Now, new length of the rectangle = 2x – 5 – 5 = (2x – 10) cm

New width of the rectangle = (x + 2) cm

New perimeter = 2[Length+Width] = 2[2x – 10 + x + 2] = 2[3x – 8] = (6x – 16) cm

Given; new perimeter = 74 cm

6x – 16 = 74

⇒ 6x = 74 + 16

⇒ 6x = 90

⇒ x = 15

Length of the original rectangle = 2x – 5 = 2 x 15 – 5 = 30 – 5 = 25 cm

Width of the original rectangle = x = 15 cm

#### Question 12 :-

The sum of three consecutive odd numbers is 57. Find the numbers.

the three consecutive odd numbers be x, x+2, x+4.

According to the statement :

x + x + 2 + x + 4 = 57

⇒ x + x + x = 57 – 2 – 4

⇒ 3x = 51

⇒ x = 17

Three consecutive odd numbers are 17, 19, 21

#### Question 13 :-

A man’s age is three times that of his son, and in twelve years he will be twice as old as his son would be. What are their present ages.

present age of the son = x years

present age of the man = 3x years

In 12 years :

Son’s age will be = (x + 12) years

The man’s age will be = (3x + 12) years

According to the statement :

3x + 12 = 2(x + 12)

⇒ 3x + 12 = 2x + 24

⇒ 3x – 2x = 24 – 12

⇒ x = 12

3x = 3 x 12 = 36

Hence, present age of the man = 36 years
Present age of the son = 12 years.

#### Question 14 :-

A man is 42 years old and his son is 12 years old. In how many years will the age of the son be half the age of the man at that time?

Man’s age = 42 years

Son’s age = 12 years

after x years the age of the son will be half the age of the man.

Man’s age after x years = (42 + x) years

Son’s age after x years = (12 + x) years

According to the statement :

12 + x = (42 + x)/2

⇒ 2(12 + x) = 42 + x   …(by cross multiplying)

⇒ 24 + 2x = 42 + x

⇒ 2x – x = 42 – 24

⇒ x = 18

Hence after 18 years, the age of the son will be half the age of the man

#### Question 15 :-

A man completed a trip of 136 km in 8 hours. Some part of the trip was covered at 15 km/hr and the remaining at 18 km/hr. Find the part of the trip covered at 18 km/hr.

Total distance of the trip = 136 km

part of the trip covered at 18 km/hr = x km

∴ Distance of the trip covered at 15 km/hr = (136 – x)km

Time taken by the man to cover x km = (Distance/speed) = x/18 hours

Time taken by the man to cover (136 – x) km = (136-x)/15 hours

Time taken by the man to cover a trip of 136 km = 8 hours

….[Multiplying each term by 90 because L.C.M. of denominators = 90]

⇒ 5x + 6 (136 – x) = 720

⇒ 5x + 816 – 6x = 720

⇒ 5x – 6x = 720 – 816

⇒ -x = -96

⇒ x = 96

∴ Part of the trip covered at 18 km/hr = 96 km

#### Question 16 :-

The difference of two numbers is 3 and the difference of their squares is 69. Find the numbers.

one number = x

Second number = x + 3 [Difference of two numbers is 3]

According to the statement :

(x + 3)2 – (x)2 = 69

⇒ (x)2 + (3)2 + 2 × x × 3 – x2 = 69

⇒ x2 + 9 + 6x – x2 = 69

⇒ 6x = 69 – 9

⇒ 6x = 60

⇒ x = 60/6

⇒ x = 10

One number = 10

Second number = x + 3 = 10 + 3 = 13

#### Question 17 :-

Two consecutive natural numbers are such that one-fourth of the smaller exceeds one-fifth of the greater by 1. Find the numbers.

two consecutive natural numbers = x, x+1

∴ One-fourth of the smaller = x/4

one-fifth of the greater = (x+1)/5

According to the statement:

⇒ x – 4 = 20  …(Cross multiplying)

⇒ x = 20 + 4 ⇒ x = 24

∴ x + 1 = 24 + 1 = 25

Two consecutive numbers are 24 and 25

#### Question 18 :-

Three consecutive whole numbers are such that if they are divided by 5, 3 and 4 respectively; the sum of the quotients is 40. Find the numbers.

The three consecutive whole numbers be x, x + 1 and x + 2

[Multiplying each term by 60 because L.C.M. of denominators = 60]

⇒ 12x + 20(x + 1) + 15(x + 2) = 2400

⇒ 12x +20x + 20 + 15x + 30 = 2400

⇒ 12x + 20x + 15x = 2400 – 20 -30

⇒ 47x = 2350

⇒ x = 2350/47

x = 50

x + 1 = 50+1 = 51

x + 2 = 50 + 2 = 52

Three consecutive whole numbers are 50, 51 and 52

#### Question 19 :-

If the same number be added to the numbers 5, 11, 15 and 31, the resulting numbers are in proportion. Find the number.

Let x be added to each number, then the numbers will be 5 + x, 11 + x, 15 + x and 31 + x

By cross multiplication,

(5 + x)(31 + x) = (15 + x)(11 + x)

⇒ 155 + 5x + 31x + x2 = 165 + 11x + 15x + x2

⇒ 155 + 36x + x2 = 165 + 26x + x2

⇒ 36x + x2 – 26x – x2 = 165 – 155

⇒ 10x = 10 ⇒ x = 1010=1

#### Question 20 :-

The present age of a man is twice that of his son. Eight years hence, their ages will be in the ratio 7 : 4. Find their present ages.

Present age of son = x year

Then age of his father = 2x

8 years hence,

Age of son = (x + 8) years and age of father = (2x + 8) years

⇒ 8x + 32 = 7x + 56

⇒ 8x – 7x = 56 – 32

⇒ x = 24

Present age of son = 24 years

and age of father = 2x = 2 x 24 = 48 years

Hence age of man = 48 years and age of his son = 24 years

### Selina Solutions of concise Mathematics Exe-14 C Linear Equations in one Variable

#### Question 1 :-

Solve:

(i) $\frac { 1 }{ 3 } x-6=\frac { 5 }{ 2 }$
(ii) $\frac { 2x }{ 3 } -\frac { 3x }{ 8 } =\frac { 7 }{ 12 }$
(iii) (x + 2)(x + 3) + (x – 3)(x – 2) – 2x(x + 1) = 0
(iv) $\frac { 1 }{ 10 } -\frac { 7 }{ x } =35$
(v) 13(x – 4) – 3(x – 9) – 5(x + 4) = 0
(vi) x + 7 – $\frac { 8x }{ 3 } =\frac { 17x }{ 6 } -\frac { 5x }{ 8 }$
(vii) $\frac { 3x-2 }{ 4 } -\frac { 2x+3 }{ 3 } =\frac { 2 }{ 3 } -x$
(viii) $\frac { x+2 }{ 6 } -\left( \frac { 11-x }{ 3 } -\frac { 1 }{ 4 } \right) =\frac { 3x-4 }{ 12 }$
(ix) $\frac { 2 }{ 5x } -\frac { 5 }{ 3x } =\frac { 1 }{ 15 }$
(x) $\frac { x+2 }{ 3 } -\frac { x+1 }{ 5 } =\frac { x-3 }{ 4 } -1$
(xi) $\frac { 3x-2 }{ 3 } +\frac { 2x+3 }{ 2 } =x+\frac { 7 }{ 6 }$
(xii) $x-\frac { x-1 }{ 2 } =1-\frac { x-2 }{ 3 }$
(xiii) $\frac { 9x+7 }{ 2 } -\left( x-\frac { x-2 }{ 7 } \right) =36$
(xiv) $\frac { 6x+1 }{ 2 } +1=\frac { 7x-3 }{ 3 }$

(i)

(ii)

L.C.M. of 3 and 8 = 2 × 2 × 2 × 3 = 24

∴ x = 2

(iii)

(x+2) (x+3) +( x−3) (x−2) − 2x (x+1 ) = 0
x²+ 5x +6 + x² − 5x + 6 − 2x² − 2x = 0
12−2x=0
x=122=6
verify :
L . H . S . = (6+2) (6+3) + (6−3) (6−2) − 2×6 (6+1)
= 72 + 12 − 84

= 0

= R . H . S .

(iv)

(v)

13(x – 4) – 3(x – 9) – 5(x + 4) = 0

⇒ 13x – 52 – 3x + 27 – 5x – 20 = 0

⇒ 13x – 3x – 5x – 52 + 27 – 20 = 0

⇒ 13x – 8x – 72 + 27 = 0

⇒ 5x – 45 = 0

by Dividing 5,

5x/5 – 45/5 = 0

⇒ x – 9 = 0

⇒ x = 9

Verify:

L.H.S. = 13(x – 4) -3(x – 9) – 5(x + 4)

= 13(9 – 4) – 3(9 – 9) – 5(9 + 4)

= 13 × 5 – 3 × 0 – 5 × 13

= 65 – 0 – 65 = 0 = R.H.S.

(vi)

⇒ 3 × 53x = 24 (- 5x + 21)

⇒ 159x = -120x + 504

⇒ 159x + 120 x = 504

⇒ 279 x = 504

(vii)

= 3(x – 18) = 12(2 – 3x)

= 3x – 54 = 24 – 36x

= 3x + 36x = 24 + 54

= 39x = 78

x= 78/39

=2

∴ x = 2

(viii)

⇒ 12(6x – 37) = 12(3x – 4)

⇒ 72x – 444 = 36x – 48

⇒ 72x – 36x = -48 + 444

⇒ 36x = 396

⇒ x = 396/36

=11

∴ x = 11

(ix)

∴ x = -19

(x)

⇒ 4(2x + 7) = 15(x – 7)

⇒ 8x + 28 = 15x – 105

⇒ 8x – 15x = -105 – 28

⇒ -7x = – 133

x = (-133/-7)

∴ x = 19

(xi)

⇒ 6(12x + 5) = 6(6x + 7)

⇒ 72x + 30 = 36x – 42

⇒ 72x – 36x = 42 – 30

⇒ 36x = 12

x = 12/36

∴ x = 1/3

(xii)

⇒ 3(x + 1) = 2(5 – x)

⇒ 3x + 3 = 10 – 2x

⇒ 3x + 2x = 10 – 3

⇒ 5x = 7

∴ x = 7/5

(xiii)

⇒ 51x + 53 = 14 × 36

⇒ 51x = 504 – 53

⇒ 51x = 459

⇒ x = 459/51

∴ x = 9

(xiv)

⇒ 3(6x + 3) = 2(7x – 3)

⇒ 18x + 9 = 14x – 6

⇒ 18x – 14x = – 6 – 9

⇒ 4x = -15

∴ x = -15/4

#### Question 2 :-

After 12 years, I shall be 3 times as old as 1 was 4 years ago. Find my present age.

Present age = x years

According to question,

(x + 12) = 3(x – 4)

x + 12 = 3x – 12

2x = 24

⇒ x = 12 years

Present age = 12 years

#### Question 3 :-

A man sold an article for 7396 and gained 10% on it. Find the cost price of the article

S.P. of article = ₹ 396

Gain = 10%

Let cost price = ₹ x

Cost price of an article = ₹ 360

#### Question 4 :-

The sum of two numbers is 4500. If 10% of one number is 12.5% of the other, find the numbers.

The first number = x

and the second number = y

x + y = 4500      …(i)

and 10% x = 12.5% y

i.e. 10x = 12.5y

x = (12.5/10)y         …(ii)

Substitute value of x in equation (i),

(12.5/10)y + y = 45000

12.5y + 10y = 45000

22.5y = 45000

y = (45000/22.5) = 2000

Now, put the value of y in equation (ii)

x = (12.5/10) × 2000

x = 2500

Hence,

The numbers are 2500 and 2000

#### Question 5 :-

The sum of two numbers is 405 and their ratio is 8 : 7. Find the numbers.

The first number = x

and the second number = 7

x + y = 405 ……..(i)

and the numbers are in the ratio

8 : 7

i.e. 8 x/7y = 1

⇒ 8x = 7y

⇒ x = 78y

Substitute value of x in equation (i)

(7/8y) + y = 405

7y + 8y = 405 × 8

15y = 3240

y = 3240/15

y = 216

Now, put the value of y in equation (ii)

x = 7/8 × 216

x = 189

Hence,

The numbers are 189 and 216

#### Question 6 :-

The ages of A and B are in the ratio 7 : 5. Ten years hence, the ratio of their ages will be 9 : 7. Find their present ages.

Ratio in the present ages of A and B = 7 : 5

Age of A = 7x years

Age of B = 5x years

10 years hence,

Then age of A = 7x + 10 years

and age of B = 5x + 10 years

According to the condition,

By crossing multiplication

7(7x + 10)

= 9(5x + 10)

⇒ 49x + 70

= 45x + 90

⇒ 49x – 45x

= 90 – 70

⇒ 4x = 20

⇒ x = 5

Present age of A = 7x = 7 x 5 = 35 years

and present age of B = 5x = 5 x 5 = 25 years

#### Question 7 :-

Find the number whose double is 45 greater than its half.

The number be x .

or 3x = 90  [After cross multiplication]
or x = 90/3
or x = 30
Thus, the number is 30.

#### Question 8 :-

The difference between the squares of two consecutive numbers is 31. Find the numbers.

The numbers be x and x + 1 .

x = 15
Thus, the numbers are 15 and 16 .

#### Question 9 :-

Find a number such that when 5 is subtracted from 5 times the number, the result is 4 more than twice the number.

The number be x.

x = 3
Thus, the number is 3.

#### Question 10 :-

The numerator of a fraction is 5 less than its denominator. If 3 is added to the numerator, and denominator both, the fraction becomes $\frac { 2 }{ 3 }$. Find the original fraction.

denominator of the original fraction = x

Then numerator = x – 5

and fraction = x-5/x

Condition According

⇒ 5(x – 2) = 4x + 12

⇒ 5x – 10 = 4x + 12

⇒ x = 22

— End of Linear Equations in one Variable Solutions :–

Thanks

RELATED ARTICLES