# R.D. Sharma Solutions Class 10th: Ch 3 Pair of Linear Equations in Two Variables Exercise 3.7

## Chapter 3 Pair of Linear Equations in Two Variables R.D. Sharma Solutions for Class 10th Math Exercise 3.7

**Exercise 3.7**

**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 = 40/8

⇒ x = 5

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

5 + y = 8

⇒ y = 8 – 5

⇒ y = 3

Hence, the number are 5 and 3 .

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**

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

After interchanging the digits, the number becomes 10x + 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

⇒ 10x + y – 10y – x = 45

⇒ 9x – 9y = 45

⇒ 9(x-y) = 45

⇒ x-y = 45/9

⇒ 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

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 .

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

⇒ y = 4

Hence, the number is 10 × 4 + 9 = 49 .

3. A number consist 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 = 9/9

⇒ 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 = 6

⇒ 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 .

4. The sum of digits of a two 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

⇒ 9x – 9y = 9

⇒ 9(x-y) = 9

⇒ x – y = 9/9

So, we have two equations

x+y = 15

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) = 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 = 15

⇒ y = 15 – 8

⇒ y = 7

Hence, the number is 10 × 7 + 8 = 78 .

5. The sum of a two-digit number and the number formed by reversing the order of digit 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 unknows. 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 = 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 .

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

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

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.

Let the digits at units and tens plac e 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 + y = 9

So, we have two systems of simultaneous equations

x-y=3,

x+y=9

x-y = -3

x+y=9

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+y = 9

Adding the two equations, we have

(x-y) + (x+y) = 3 + 9

⇒ x – y + x + y = 12

⇒ 2x = 12

⇒ x = 12/2

⇒ 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.

⇒ 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 = 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 .

6. The sum of two numbers is 1000 and the difference between their squares 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

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 plac e 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 + y = 9

So, we have two systems of simultaneous equations

x-y=3,

x+y=9

x-y = -3

x+y=9

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+y = 9

Adding the two equations, we have

(x-y) + (x+y) = 3 + 9

⇒ x – y + x + y = 12

⇒ 2x = 12

⇒ x = 12/2

⇒ 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

⇒ x = 6/2

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 .

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.

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 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 = 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 .

9. A two-digit number is 3 more than 4 times the sum of its digits. If 8 is added to the number, the digits are reversed. Find the number.

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

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

If 8 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 systems of equations

x – y = -1,

x – y = 2

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

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.

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

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

x = 6 = -2

⇒ x = 6 – 2

x– y = -3

x + y = 9

Adding the two equations, we have

(x-y) + (x+y) = -3+9

⇒ x – y + x + y = 6

⇒ x = 6/2

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 .

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 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 = 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 .

9. A two-digit number is 3 more than 4 times the sum of its digits. If 8 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

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

If 8 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 systems of equations

x – y = -1,

x – y = 2

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

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

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 .

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

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 10y + 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 = 2 × 2y × y

⇒ 12y = 4y

⇒ 4y

⇒ 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 × 3 + 6 = 36 .

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

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.

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 system of equations

xy = 20

x-y = 1

Here, x and y are unknows . We have to solve the above systems of equations for x and y. Substituting x = 1 + y from the second equation to the first equation, we get

(1+y)y = 20

⇒ y + y

⇒ y

⇒ y

⇒ 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 cant be negative . So, we must remove this pair .

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

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 unknows . 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 = 26/2

⇒ 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 .

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.

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 unknows . We have to solve the abpve systems of equations for x and y substituting x = 8y from the second 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 × 1 + 8 = 18 .

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.

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 × 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.

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

Let the two numbers be x and y

So,

x/y = 5/6

⇒ 6x = 5y

⇒ 6x – 5y = 0 ….(i)

Now, when 8 is subtracted from each of the numbers, then

x-8/y-8 = 4/5

⇒ 5x – 40 = 4y – 32

⇒ 5x – 4y = 8 …. (ii)

Multiplying (i) with 4 and (ii) with 5, we get

24x − 20y = 0 .....(iii)

25x − 20y = 40 .....(iv)

Subtracting (iii) from (iv), we get

x = 40

Putting x = 40 in (i), we get

240 − 5y = 0

⇒ y = 48

Thus, the two numbers are 40 and 48 .

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 .

Let the digits of the two digit number be x and y. So, the two digit number will be 10x + y.

Now, according to the given condition the number is obtained in two ways.

Case I: 8(x + y) − 5 = 10x + y

⇒ 8x + 8y − 5 = 10x + y

⇒ 2x −7y = −5 .....(1)

Case II: 16(x − y) + 3 = 10x + y

⇒ 16x − 16y + 3 = 10x + y

⇒ 6x − 17y = −3 .....(2)

Multiplying (1) by 3, we get

6x − 21y = −15 .....(3)

Subtracting (2) from (3), we get

− 4y = −12

⇒ y = 3

Putting y = 3 in (1), we get

2x − 21 = −5

⇒ 2x = 16

⇒ x = 8

So, the required number is 10x + y = 10 × 8 + 3 = 83.

Hence, the number is 10 × 6 + 4 = 64 .

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 10y + 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 = 2 × 2y × y

⇒ 12y = 4y

^{2}⇒ 4y

^{2}– 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 × 3 + 6 = 36 .

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

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 system of equations

xy = 20

x-y = 1

Here, x and y are unknows . We have to solve the above systems of equations for x and y. Substituting x = 1 + y from the second equation to the first equation, we get

(1+y)y = 20

⇒ y + y

^{2}= 20⇒ y

^{2}+ y – 20 = 0⇒ y

^{2}+ 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 cant be negative . So, we must remove this pair .

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 unknows . 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 = 26/2

⇒ 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 .

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 unknows . We have to solve the abpve systems of equations for x and y substituting x = 8y from the second 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 × 1 + 8 = 18 .

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 × 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.

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

**Solution**Let the two numbers be x and y

So,

x/y = 5/6

⇒ 6x = 5y

⇒ 6x – 5y = 0 ….(i)

Now, when 8 is subtracted from each of the numbers, then

x-8/y-8 = 4/5

⇒ 5x – 40 = 4y – 32

⇒ 5x – 4y = 8 …. (ii)

Multiplying (i) with 4 and (ii) with 5, we get

24x − 20y = 0 .....(iii)

25x − 20y = 40 .....(iv)

Subtracting (iii) from (iv), we get

x = 40

Putting x = 40 in (i), we get

240 − 5y = 0

⇒ y = 48

Thus, the two numbers are 40 and 48 .

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 digits of the two digit number be x and y. So, the two digit number will be 10x + y.

Now, according to the given condition the number is obtained in two ways.

Case I: 8(x + y) − 5 = 10x + y

⇒ 8x + 8y − 5 = 10x + y

⇒ 2x −7y = −5 .....(1)

Case II: 16(x − y) + 3 = 10x + y

⇒ 16x − 16y + 3 = 10x + y

⇒ 6x − 17y = −3 .....(2)

Multiplying (1) by 3, we get

6x − 21y = −15 .....(3)

Subtracting (2) from (3), we get

− 4y = −12

⇒ y = 3

Putting y = 3 in (1), we get

2x − 21 = −5

⇒ 2x = 16

⇒ x = 8

So, the required number is 10x + y = 10 × 8 + 3 = 83.