WORD PROBLEMS INVOLVING PAIR OF LINEAR EQUATIONS

Problem 1 :

Sum of two numbers is 50 and their difference is 10, then the numbers are

(a) 30 and 20     (b) 24 and 14     (c) 12 and 2     (d) none of these

Solution:

Let the two numbers are x and y.

x + y = 50 --- > (1)

x - y = 10 --- > (2)

Adding (1) and (2),

2x = 60

x = 60/2

x = 30

By applying x = 30 in equation (1)

30 + y = 50

y = 50 - 30

y = 20

Therefore, the two numbers are 30 and 20.

So, option (a) is correct.

Problem 2 :

The sum of the digits of a two-digit number is 12. The number obtained by interchanging its digit exceeds the given number by 18, then the number is

(a) 72     (b) 75     (c) 57     (d) none of these

Solution:

Let the tens digit of the required number be x and the units digits be y.

x + y = 12 --- > (1)

Required number = (10x + y)

Number obtained on reversing the digits = (10y + x)

(10y + x) - (10x + y) = 18

9y - 9x = 18

y - x = 2 --- > (2)

Adding (1) and (2), we get

2y = 14

y = 14/2

y = 7

By applying y = 7 in (1),

x + 7 = 12

x = 12 - 7

x = 5

Hence, the required number is 57.

So, option (c) is correct.

Problem 3 :

The sum of a two-digit number and the number obtained by interchanging its digit is 99. If the digits differ by 3, then the number is

(a) 36     (b) 33     (c) 66     (d) none of these

Solution:

Let the two digits number be x and y.

Number = 10y + x

(10x + y) + (10y + x) = 99

11x + 11y = 99

x + y = 9 --- > (1)

x - y = 3 --- > (2)

Adding (1) and (2),

2x = 12

x = 12/2

x = 6

By applying x = 6 in equation (1),

6 + y = 9

y = 9 - 6

y = 3

Therefore, number = 10y + x

= 10(3) + 6

= 36

Hence, the required number is 63 or 36.

So, Option (a) is correct.

Problem 4 :

Seven times a two-digit number is equal to four times the number obtained by reversing the order of its digit. If the difference between the digits is 3, then find the number.

(a) 36     (b) 33     (c) 66     (d) none of these

Solution:

Let numbers be x at onces place and y at tens place.

Reversed digit = 10x + y

7(10y + x) = 4(10x + y)

70y + 7x = 40x + 4y

70y - 4y = 40x - 7x

66y = 33x

x = 2y --- > (1)

Now, 

Given, x - y = 3

Substitute x = 2y in equation (1)

2y - y = 3

y = 3

So, x = 2y

x = 2(3) 

x = 6

So, required original number = 10y + x

= 10(3) + 6

= 36

So, option (a) is correct.

Problem 5 :

A two-digit number is 4 more than 6 times the sum of its digit. If 18 is subtracted from the number, the digits are reversed. Find the number.

(a) 36     (b) 46     (c) 64     (d) none of these

Solution:

Let, the ones digit number be x and tens digit number be y.

x + 10y = 6(x + y) + 4

x + 10y = 6x + 6y + 4

10y - 6y = 6x - x + 4

4y = 5x + 4 ---> (1)

And, from if 18 is subtracted from the number, the digits are reversed.

x + 10y - 18 = y + 10x

10y - 18 = y + 9x

9y - 18 = 9x

x = y - 2 --- > (2)

Substitute x = y - 2 in (1)

4y = 5(y - 2) + 4

4y = 5y - 10 + 4

4y - 5y = -6

-y = -6

y = 6

By applying y = 6 in (2)

x = 6 - 2

x = 4

Therefore, the number is 64.

So, option (c) is correct.

Problem 6 :

The sum of two numbers is 1000 and the difference between their squares is 256000, then find the numbers.

(a) 616 and 384     (b) 628 and 372     (c) 564 and 436     (d) none of these

Solution:

Let the numbers are x and y.

x + y = 1000 --- > (1)

x² - y² = 256000

x² - y² = (x + y) (x - y)

1000 × (x - y) = 256000

x - y = 256 --- > (2)

Adding (1) and (2),

2x = 1256

x = 1256/2

x = 628

By applying x = 628 in (1),

628 + y = 1000

y = 1000 - 628

y = 372

So, option (b) is correct.

Problem 7 :

Five years ago, A was thrice as old as B and ten years later A shall be twice as old as B, then the present age of A is

(a) 20     (b) 50     (c) 30     (d) none of these

Solution:

Let the age of A is x years and the age of B is y years.

Five years ago,

A's age = (x - 5) years

B's age = (y - 5) years

x - 5 = 3y - 15

x = 3y - 10 --- > (1)

Ten years later,

A's age = (x + 10)

B's age = (y + 10)

x + 10 = 2(y + 10)

x + 10 = 2y + 20

x = 2y + 10 --- > (2)

From equation (1) and (2),

3y - 10 = 2y + 10

3y - 2y = 10 + 10

y = 20

By applying y = 20 in equation (1),

x = 3(20) - 10

x = 60 - 10

x = 50

Hence, the age of person A is 50.

So, option (b) is correct.

Problem 8 :

The sum of thrice the first and the second is 142 and four times the first exceeds the second by 138, then the numbers are

(a) 40 and 20     (b) 40 and 22     (c) 12 and 22     (d) none of these

Solution:

Let the numbers be x and y.

3x + y = 142 --- > (1)

4x - y = 138

y = 4x - 138 --- > (2)

Substitute the value of y in (1)

3x + 4x - 138 = 142

7x = 142 + 138

7x = 280

x = 280/7

x = 40

By applying x = 40 in equation (2)

y = 4(40) - 138

y = 160 - 138

y = 22

Therefore, the numbers are 40 and 22.

So, option (b) is correct.

Problem 9 :

The sum of twice the first and thrice the second is 92 and four times the first exceeds seven times the second by 2, then find the numbers.

(a) 25 and 20     (b) 25 and 14     (c) 14 and 22     (d) none of these

Solution:

Let us consider, first number = x and second number = y

2x + 3y = 92 --- > (1)

4x - 7y = 2 --- > (2)

(1) × 2 ==> 4x + 6y = 184

Subtract (1) from (2)

13y = 182

y = 182/13

y = 14

By applying y = 14 in (1),

2x + 3(14) = 92

2x + 42 = 92

2x = 92 - 42

2x = 50

x = 50/2

x = 25

Therefore, the numbers are 25 and 14.

So, option (b) is correct.

Problem 10 :

The difference between two numbers is 14 and the difference between their squares is 448, then find the numbers.

(a) 25 and 9     (b) 22 and 9     (c) 23 and 9     (d) none of these

Solution:

Let us consider, first number = x and second number = y

x - y = 14 --- > (1)

x² - y² = 448

(x + y) (x - y) = 448

(x + y) × 14 = 448

x + y = 448/14

x + y = 32 --- > (2)

Adding (1) and (2)

2x = 46

x = 46/2

x = 23

By applying x = 23 in (1)

23 - y = 14

-y = 14 - 23

y = 9

Therefore, the numbers are 23 and 9.

So, option (c) is correct.