Chapter 3: Pair of Linear Equations in Two Variables Exercise – 3.7

Question: 1

The sum of two numbers is 8. If their sum is four times their difference, find the numbers.

Solution:

Let the numbers are x and y. One of them must be greater than or equal to the other. Let us assume that x is greater than or equal to y.

The sum of the two numbers is 8. Thus, we have x + y = 8

The sum of the two numbers is four times their difference. Thus, we have

x + y = 4(x — y)

⟹ x +y = 4x - 4y

⟹ 4x – 4y – x – y = 0

⟹ 3x – 5y = 0

So, we have two equations

x + y = 8

3x - 5y = 0

Here x and y are unknowns.

We have to solve the above equations for x and y.

Multiplying the first equation by 5 and then adding with the second equation, we have

5(x + y)+(3x – 5y) = 5 × 8 + 0

⟹ 5x + 5y + 3x - 5y = 40

⟹ 8x = 40

⟹ x = 5

x = Substituting the value of x in the first equation, we have

5 + y = 8

⟹ y = 8 - 5

⟹ y = 3

Hence, the numbers are 5 and 3.

Question: 2

The sum of digits of a two digit number is 13. If the number is subtracted from the one obtained by interchanging the digits, the result is 45. What is the number?

Solution:

Let the digits at units and tens place of the given number be x and y respectively. Thus, the number is 10y + x.

The sum of the digits of the number is 13. Thus, we have x + y = 13

After interchanging the digits, the number becomes10x + y.

The difference between the number obtained by interchanging the digits and the original number is 45. Thus, we have

(10x + y) — (10y + x) = 45

⟹ 110x + y-10y — x = 45

⟹ 9x - 9y = 45

⟹ 9(x – y) = 45

⟹ x - y = 5

So, we have two equations

x + y = 13

x - y = 5

Here x and y are unknowns. We have to solve the above equations for x and y.

Adding the two equations, we have

(x + y) + (x — y) = 13 + 5

⟹ x + y + x - y = 18

⟹ 2x = 18

⟹ x = 9

Substituting the value of x in the first equation, we have

9 + y = 13

⟹ y = 13 – 9

⟹ y = 4

Hence, the number is 10 × 4 + 9 = 49

Question: 3

A number consists of two digits whose sum is five. When the digits are reversed, the number becomes greater by nine. Find the number.

Solution:

Let the digits at units and tens place of the given number be x and y respectively. Thus, the number is 10y + x.

The sum of the digits of the number is 5. Thus, we have x + y = 5

After interchanging the digits, the number becomes 10x + y.

The number obtained by interchanging the digits is greater by 9 from the original number. Thus, we have

10x + y = 10y + x + 9

⟹ 10x + y – 10y – x = 9

⟹ 9x – 9y = 9

⟹ 9(x – y) = 9

⟹ x – y = 1

So, we have two equations

x + y = 5

x - y = 1

Here x and y are unknowns. We have to solve the above equations for x and y.

Adding the two equations, we have

(x + y) + (x – y) = 5 + 1

⟹ x + y + x – y = 5 + 1

⟹ 2x = 6

⟹ x = 6/2

⟹ x = 3

Substituting the value of x in the first equation, we have

3 + y = 5

⟹ y = 5 - 3

⟹ y = 2

Hence, the number is 10 × 2 + 3 = 23

Question: 4

The sum of digits of a two digit number is 15. The number obtained by reversing the order of digits of the given number exceeds the given number by 9. Find the given number.

Solution:

Let the digits at units and tens place of the given number be x and y respectively. Thus, the number is 10y + x.

The sum of the digits of the number is 15. Thus, we have x + y = 15

After interchanging the digits, the number becomes 10x + y.

The number obtained by interchanging the digits is exceeding by 9 from the original number. Thus, we have

10x + y = 10y + x + 9

⟹ 10x + y – 10y – x = 9

⟹ 9x - 9y = 9

⟹ 9(x - y) = 9

⟹ x - y = 9/9

⟹ x - y = I

So, we have two equations

x + y = 15

x - y = I

Here x and y are unknowns. We have to solve the above equations for x and y. Adding the two equations, we have

(x + y) + (x – y) = 15 + 1

⟹ x + y + x - y = 16

⟹ 2x = 16

⟹ x = 16/2

⟹ x = 8

Substituting the value of x in the first equation, we have

8 + y = 5

⟹ y = 15 - 8

⟹ y = 7

Hence, the number is 10 × 7 + 8 = 78

Question: 5

The sum of two- digit number and the number formed by reversing the order of digits is 66.If the two digits differ by 2, find the number. How many such numbers are there?

Solution:

Let the digits at units and tens place of the given number be x and y respectively. Thus, the number is 10y + x.

The two digits of the number are differing by 2. Thus, we have x - y = ±2

After interchanging the digits, the number becomes 10x + y.

The sum of the numbers obtained by interchanging the digits and the original number is 66. Thus, we have

(10x +  y) + (10y + x) = 66

⟹ 10x + y + 10y + x = 66

⟹ 11x + 11y = 66

⟹ 11(x + y) = 66

⟹ x + y = 66/11

⟹ x + y = 6

So, we have two systems of simultaneous equations

x - y = 2,

x + y = 6

x –  y = -2,

x + y = 6

Here x and y are unknowns. We have to solve the above systems of equations for x and y.

(i) First, we solve the system

x - y = 2,

x + y = 6

Adding the two equations, we have

(x — y) + (x + y) = 2 + 6

⟹ x - y + x + y = 8

⟹ 2x = 8

⟹ x = 8/2

⟹ x = 4

Substituting the value of x in the first equation, we have

4 - y = 2

⟹ y = 4 - 2

⟹ y = 2

Hence, the number is 10 × 2 + 4 = 24

(ii) Now, we solve the system

x - y = -2,

x + y = 6

Adding the two equations, we have

(x – y) + (x + y) = -2 + 6

⟹ x - y + x + y = 4

⟹ 2x = 4

⟹ x = 4/2

⟹ x = 2

Substituting the value of x in the first equation, we have

2 - y = -2

⟹ y = 2 + 2

⟹ y = 4

Hence, the number is 10 × 4 + 2 = 42

There are two such numbers.

Question: 6

 The sum of two numbers is 1000 and the difference between their square is 256000. Find the numbers.

Solution:

Let the numbers are x and y. One of them must be greater than or equal to the other. Let us assume that x is greater than or equal to y.

The sum of the two numbers is 1000. Thus, we have x + y = 1000

The difference between the squares of the two numbers is 256000. Thus, we have

x² –y² = 256000

⟹ (x + y)(x - y) = 256000

⟹ 1000(x - y) = 256000

⟹ x - y = 256000/1000

⟹ x - y = 256

So, we have two equations

x + y = 1000

x - y = 256

Here x and y are unknowns. We have to solve the above equations for x and y. Adding the two equations, we have

(x + y) + (x - y) = 1000 + 256

⟹ x + y + x - y = 1256

⟹ 2x = 1256

⟹ x = 1256/2

x = 628

Substituting the value of x in the first equation. We have

628 + y = 1000

⟹ y = 1000 - 628

⟹ y = 372

Hence, the numbers are 628 and 372

Question: 7

The sum of a two digit number and the number obtained by reversing the order of its digits is 99.If the digits differ by 3, find the number.

Solution:

Let the digits at units and tens place of the given number be x and y respectively. Thus, the number is 10y + x.

The two digits of the number are differing by 3. Thus, we have x – y = ±3

After interchanging the digits, the number becomes 10x + y.

The sum of the numbers obtained by interchanging the digits and the original number is 99. Thus, we have

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

⟹ 10x + y + 10y + x = 99

⟹ 11x + 11y = 99

⟹ 11(x + y)= 99

⟹x + y = 99/11

⟹ x = 9

So, we have two systems of simultaneous equations

x - y = 3,

x+ y = 9

x - y = -3,

x + y = 9

Adding the two equations, we have

(x – y) + (x + y) = 3 + 9

⟹ x – y + x + y = 12

⟹ 2x = 12

⟹ x = 12/2

⟹x = 6

Substituting the value of x in the first equation, we have

6 – y = 3

⟹ y = 6 - 3

⟹ y = 3

Hence, the number is 10 × 3 + 6 =36

(ii) Now, we solve the system

x – y = – 3,

x + y = 9

Adding the two equations we have

(x – y)+(x + y) = –3 + 9

⟹ x – y + x + y = 6

⟹ 2x = 6

⟹ x = 3

Substituting the value of x in the first equation, we have

3 - y = -3

⟹ y = 3 + 3

⟹ y = 6

Hence, the number is 10 × 6 + 3 = 63

Note that there are two such numbers.

Question: 8

A two- digit number is 4 times the sum of its digits. If 18 is added to the number, the digits are reversed. Find the number.

Solution:

Let the digits at units and tens place of the given number be x and y respectively. Thus, the number is 10y + x.

The number is 4 times the sum of the two digits. Thus, we have

10y + x = 4(x + y)

⟹ 10y + x = 4x + 4y

⟹ 4x + 4y - 10y - x = 0

⟹ 3x - 6y = 0

⟹ 3(x — 2y) = 0

⟹ x - 2y = 0

After interchanging the digits, the number becomes10x + y.

If 18 is added to the number, the digits are reversed. Thus, we have

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

⟹ 10x + y - 10y - x = 18

⟹ 9x - 9y = 18

⟹ 9(x - y) = 18

⟹ x - y = 18/9

⟹ x - y = 2

So, we have the systems of equations

x - 2y = 0,

x - y = 2

Here x and y are unknowns. We have to solve the above systems of equations for x and y. Subtracting the first equation from the second, we have

(x - y) - (x – 2y) = 2 - 0

⟹ x - y - x + 2y = 2

⟹ y = 2

Substituting the value of y in the first equation, we have

x - 2 × 2 = 0

⟹ x -4 = 0

⟹ x = 4

Hence, the number is 10 × 2 + 4 = 24

Question: 9

A two-digit number is 3 more than 4 times the sum of its digits. If 18 is added to the number, the digits are reversed. Find the number.

Solution:

Let the digits at units and tens place of the given number be x and y respectively. Thus, the number is 10y + x.

The number is 3 more than 4 times the sum of the two digits. Thus, we have

10y + x = 4(x + y) + 3

⟹ 10y + x = 4x + 4y + 3

⟹ 4x + 4y - 10y - x = -3

⟹ 3x – 6y = -3

⟹ 3(x – 2y) = -3

⟹ x - 2y = -3/3

⟹ x - 2y = -1

Alter interchanging the digits, the number becomes 10x + y.

If 18 is added to the number, the digits are reversed. Thus, we have

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

⟹ 10x + y - 10y - x = 18

⟹ 9x - 9y = 18

⟹ 9(x — y) = 18

⟹ x - y = 18/9

⟹ x - y = 2

So, we have the systems of equations x - 2y = -1,

x - y = 2

Here x and y are unknowns. We have to solve the above systems of equations for x and y. Subtracting the first equation from the second, we have

(x - y) - (x - 2y) = 2 – (-1)

⟹ x- y - x + 2y = 3

⟹ y = 3

Substituting the value of y in the first equation, we have

x - 2 × 3 = -1

⟹ x - 6 = -1

⟹ x = -1 + 6

⟹ x = 5

Hence the number is 10 × 3 + 5 = 35

Question: 10

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

Solution:

Let the digits at units and tens place of the given number be x and y respectively. Thus, the number is 10y + x.

The number is 4 more than 6 times the sum of the two digits. Thus, we have

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

⟹ 10y + x = 6x + 6y + 4

⟹ 6x + 6y - 10y – x = -4

⟹ 5x - 4y = -4

After interchanging the digits, the number becomes 10x + y.

If 18 is subtracted from the number, the digits are reversed. Thus, we have

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

⟹ 10x + y - 10y – x = – 18

⟹ 9x - 9y = – 18

⟹ 9(x - y) =  – 18

⟹ x - y = – 18/9

⟹ x - y = – 2

So, we have the systems of equations

5x – 4y = – 4,

x – y = – 2

Here x and y are unknowns. We have to solve the above systems of equations for x and y. Multiplying the second equation by 5 and then subtracting from the first, we have

(5x – 4y) – (5x – 5y) = – 4 – (– 2 × 5)

⟹ 5x – 4y – 5x + 5y= – 4 +10

⟹ y = 6

Substituting the value of y in the second equation, we have

x – 6 = – 2

⟹ x = 6 – 2

⟹ x = 4

Hence, the number is 10 × 6 + 4 = 64

Question: 11

A two-digit number is 4 times the sum of its digits and twice the product of the digits. Find the number.

Solution:

Let the digits at units and tens place of the given number be x and y respectively. Thus, the number is 10y + x.

The number is 4 times the sum of the two digits. Thus, we have

10y + x = 4(x + y)

⟹ 10y + x = 4x + 4y

⟹ 4x + 4y - 10y – x = 0

⟹ 3x – 6y = 0

3(x – 2y) = 0

⟹ x – 2y = 0

⟹ x = 2y

After interchanging the digits, the number becomes 10x + y.

The number is twice the product of the digits. Thus, we have I0y + x = 2xy

So, we have the systems of equations

x = 2y,

10y + x = 2xy

Here x and y are unknowns. We have to solve the above systems of equations for x and y. Substituting x = 2y in the second equation, we get

10y + 2y = 2x × 2xy

⟹ 12y = 4y²

⟹ 4y² – 12y = 0

⟹ 4y(y – 3) = 0

⟹ y(y – 3) = 0

⟹ y = 0 or y = 3

Substituting the value of y in the first equation, we have

Hence, the number is 10 x 3 + 6 = 36

Note that the first pair of solution does not give a two digit number

Question: 12

A two-digit number is such that the product of its digits is 20. If 9 is added to the number, the digits interchange their places. Find the number.

Solution:

Let the digits at units and tens place of the given number be x and y respectively. Thus, the number is 10y + x.

The product of the two digits of the number is 20. Thus, we have xy = 20

After interchanging the digits, the number becomes 10x+ y

If 9 is added to the number, the digits interchange their places. Thus, we have

(10y + x) + 9 = 10x + y

⟹ 10y + x + 9= 10x + y

⟹ 10x + y - 10y - x = 9

⟹ 9x - 9y = 9

⟹ 9(x – y) = 9

⟹ x - y = 9/9

⟹ x - y = 1

So, we have the systems of equations

xy = 20,

x - y = 1

Here x and y are unknowns. We have to solve the above systems of equations for x and y.

Substituting x = I + y from the second equation to the first equation, we get

(1 + y) y = 20

⟹ y + y² = 20

⟹ y² + y - 20 = 0

⟹ y² + 5y - 4y – 20 = 0

⟹ y(y + 5)- 4(y +  5) = 0

⟹ (y + 5)(y - 4) = 0

⟹ y = – 5 or y = 4

Substituting the value of y in the second equation, we have

Hence, the number is 10 × 4 + 5 = 45

Note that in the first pair of solution the values of x and y are both negative. But the digits of the number can’t be negative. So, we must remove this pair.

Question: 13

The difference between two numbers is 26 and one number is three times the other. Find them.

Solution:

Let the numbers are x and y. One of them must be greater than or equal to the other. Let us assume that x is greater than or equal to y.

The difference between the two numbers is 26. Thus, we have x -y = 26

One of the two numbers is three times the other number. Here, we are assuming that x is greater than or equal to y. Thus, we have x = 3y

So, we have two equations

x -y = 26

x = 3y

Here x and y are unknowns. We have to solve the above equations for x and y.

Substituting x = 3y from the second equation in the first equation, we get

3y – y = 26

⟹ 2y = 26

⟹ y = 13

Substituting the value of y in the first equation, we have

x- 13 = 26

⟹ x = 13 + 26

⟹ x = 39

Hence the numbers are 39 and 13.

Question: 14

The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.

Solution:

Let the digits at units and tens place of the given number be x and y respectively Thus, the number is 10y + x.

The sum of the two digits of the number is 9.

Thus, we have x + y = 9

After interchanging the digits, the number becomes 10x + y.

Also, 9 times the number is equal to twice the number obtained by reversing the order of the digits. Thus, we have

9(10y + x) = 2(10x + y)

⟹ 90y + 9x = 20x + 2y

⟹ 20x + 2y – 90y – 9x = 0

⟹11x - 88y = 0

⟹11(x -8y) = 0

⟹ x - 8y = 0

So, we have the systems of equations

x + y = 9,

x – 8y = 0

Here x and y are unknowns. We have to solve the above systems of equations for x and y. Substituting x = 8y from the second equation to the first equation, we get

8y + y = 9

⟹ 9y = 9

⟹ y = 9/9

⟹ y = 1

Substituting the value of y in the second equation, we have

x - 8 × 1 = 0

⟹ x - 8 = 0

⟹ x = 8

Hence, the number is 10 x 1 + 8 = 18

Question: 15

Seven times a two-digit number is equal to four times the number obtained by reversing the digits. If the difference between the digits is 3. Find the number.

Solution:

Let the digits at units and tens place of the given number be x and y respectively. Thus, the number is 10y + x.

The difference between the two digits of the number is 3. Thus, we have x - y = ±3

After interchanging the digits, the number becomes 10x + y.

Seven times the number is equal to four times the number obtained by reversing the order of the digits. Thus, we have

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

⟹ 70y + 7x = 40x + 4y

⟹ 40x + 4y – 70y-7x = 0

⟹ 33x – 66y = 0

⟹33(x — 2y) = 0

⟹ x – 2y = 0

So, we have two systems of simultaneous equations

x – y = 3,

x - 2y = 0

x – y = - 3,

x – 2y = 0

Here x and y are unknowns. We have to solve the above systems of equations for x and y.

(i) First, we solve the system

x – y = 3,

x – 2y = 0

Multiplying the first equation by 2 and then subtracting from the second equation, we have

(x  - 2y) - 2(x – y) = 0 – 2 × 3

⟹ x – 2y – 2x + 2y = -6

⟹ -x = -6

⟹ x = 6

Substituting the value of x in the first equation, we have

6 - y = 3

⟹ y = 6 – 3

⟹ y = 3

Hence, the number is 10 x 3 + 6 = 36

(ii) Now, we solve the system

x – y = -3,

x - 2y = 0

Multiplying the first equation by 2 and then subtracting from the second equation we have

(x - 2y) - 2(x – y) = 0 – (-3 × 2)

⟹ x- 2y – 2x + 2y = 6

⟹ -x = 6

⟹ x = -6

Substituting the value of x in the first equation, we have

- 6 – y = – 3

⟹ y = – 6 + 3

⟹ y = – 3

But, the digits of the number can’t be negative. Hence, the second case must be removed.

Question: 16

Two numbers are in the ratio 5: 6. If 8 is subtracted from each of the numbers the ratio becomes 4: 5. Find the numbers.

Solution:

Let the numbers be 5x and 6x

Now subtracting 8 we get the numbers as

5x – 8 and 6x – 8

Thus, (5x – 8)/(6x – 8) = 4: 5

By cross multiplying we get,

5(5x – 8) = 4(6x – 8)

⟹ 25x – 40 = 24x – 32

⟹ x = 8

Hence, the numbers are

5x = 5 x 8 = 40

6x = 6 x 8 = 48

Question: 17

A two-digit number is obtained by either multiplying the sum of the digits by 8 and then subtracting 5 or by multiplying the difference of the digits by 16 and then adding 3. Find the number.

Solution:

Let the unit digit and ten’s digit of the number be x and y respectively.

Therefore the number = 10y + x

Sum of digits = x + y

10y +x = 8(x + y) – 5

10y + x = 8x + 8y – 5

7x – 2y = 5        ……(1)

Difference of the digits = y – x [if x < y]

10y + x = 16(y – x) + 3

10y + x = 16y – 16x +3

17x – 6y = 3       ……(2)

Multiply equation (1) and (2) and subtracting equation (2)

21x – 6y = 15

17x – 6y = 3

4x = 12

x = 12/4 = 3

Putting the value of x = 3 in equation (1)

7 × 3 – 2y = 5

2y = 21 – 5

2y = 16

y = 16/2 = 8

Thus the unit digit of the number is 3 and ten’s digit is 8.

Therefore the number is 83.