# Class 10 RD Sharma Solutions – 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 us assume the first number and second number to be x and y respectively.

Now, according to the given conditions, we have,

x + y = 8 ….(i)

and x + y = 4 (x – y)

⇒ 4 (x – y) = 8

⇒ x – y = 2 ….(ii)

On adding eq(i) and (ii), we get

2x = 10 ⇒ x = 5

On subtracting eq(ii) from (i), we get

2y = 6 ⇒ y = 3

Therefore, the required numbers are 5 and 3 respectively.

### 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 us assume the unit’s digit to be x and ten’s digit to be y respectively.

Therefore, the number = x + 10y

Now, as per the specified constraints,

x + y = 13 ….(i)

Also,

The same number after interchanging their digits = y + 10x

Now, we have,

y + 10x – x – 10y = 45

On solving, we get,

9x – 9y = 45

⇒ x – y = 5

x – y = 5 ….(ii)

On adding eq(i) and (ii), we get

2x = 18 ⇒ x = 9

2y = 8 ⇒ y = 4

Substituting the value, we get,

Number = x + 10y = 9 + 4 x 10 = 9 + 40 = 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 us assume the unit’s digit to be x and ten’s digit to be y respectively.and ten’s digit = y

Therefore, the number = x + 10y

The same number after interchanging their digits = y + 10x

Now, we have,

x + y = 5 ….(i)

and y + 10x = x + 10y + 9

On solving, we get,

⇒ y + 10x – x – 10y = 9

⇒ 9x – 9y = 9

On dividing by 9, we get,

⇒ x – y = 1 ….(ii)

On adding eq(i) and (ii), we get

2x = 6 ⇒ x = 3

On subtracting eq(i) and (ii), we get

2y = 4 ⇒ y = 2

Therefore, the number = x + 10y

= 3 + 10 x 2

= 3 + 20 = 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 us assume the unit’s digit to be x and ten’s digit to be y respectively.and ten’s digit = y

Therefore, the number = x + 10y

The same number after interchanging their digits = y + 10x

Now, we have,

x + y = 15 ….(i)

y + 10x = x + 10y + 9

⇒ y + 10x – x – 10y = 9

⇒ 9x – 9y = 9

On dividing by 9, we get,

⇒ x – y = 1 ……..(ii)

Now adding eq(i) and (ii) equations together, we get,

2x = 16

=> x = 8

On subtracting eq(i) and (ii),

2y = 14 ⇒ y = 7

Therefore, the number = x + 10y

= 8 + 10 x 7

= 8 + 70 = 78

### Question 5. The sum of a 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:**

According to the given constraints, sum of two-digit number and

number formed by reversing its digits = 66

Let us assume the units digit to be x and the tens digit to be x + 2

Therefore, the number = x + 10 (x + 2)

= x + 10x + 20 = 11x + 20

The number obtained by reversing its digits, is,

assuming the units digit to be x + 2 and the tens digit to be x.

Now, number = x + 2 + 10x = 11x + 2

11x + 20 + 11x + 2 = 66

⇒ 22x + 22 = 66

⇒ 22x = 66 – 22 = 44

Solving, we get,

⇒ x = 2

The required number = 11x + 20 = 11 x 2 + 20 = 22 + 20 = 42

Also, the number by reversing its digits will be 11x + 2 = 11 x 2 + 2 = 22 + 2 = 24

Therefore, the numbers are 42 and 24 respectively.

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

**Solution:**

Let us assume the first number and second number to be x and y respectively.

Now,

x + y = 1000 ……..(i)

and x

^{2 }– y^{2}= 256000Dividing by eq(i), we obtain,

We get, x – y = 265 … (ii)

On solving eq(i) and (ii), we get,

2x = 1256

x = 628

Now substitute the value of x in eq(ii), we get the value of y

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

Therefore, the number = x + 10y

Upon reversing the digits, the number = y + 10x

Given,

x + 10y + y + 10x = 99

⇒ 11x + 11y = 99

⇒ x + y = 9 ….(i)

and x – y = 3 ….(ii)

On adding eq(i) and (ii), we get,

2x = 12

x = 6

On subtracting eq(i) and (ii), we get,

2y = 6

y = 3

Therefore,

The required number = x + 10y = 6 + 10 x 3

= 6 + 30 = 36

### 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 unit digit and tens digit of the number be x and y respectively.

Therefore, the number = x + 10y

The same number after interchanging their digits = y + 10x

Now, we have,

x + 10y = 4 (x + y)

⇒ x + 10y = 4x + 4y

⇒ 4x + 4y – x – 10y = 0

⇒ 3x – 6y = 0

⇒ x – 2y = 0

⇒ x = 2y …. (ii)

and x + 10y + 18 = y + 10x

⇒ x + 10y – y – 10x = -18

⇒ – 9x + 9y = -18

⇒ x – y = 2 ….(ii)

From eq(i) put the value of x in eq(ii), we get

⇒ 2y – y = 2

⇒ y = 2

x = 2y = 2 x 2 = 4

Therefore,

The required number = x + 10y = 4 + 10 x 2

= 4 + 20 = 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 us considered the unit digit and tens digit of the number be x and y.

Therefore, the number = x + 10y

The same number after interchanging their digits = y + 10x

Now, we have,

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

⇒ x + 10y = 4x + 4y + 3

⇒ x + 10y – 4x – 4y = 3

⇒ -3x + 6y = 3

⇒ x – 2y = -1 ….(i)

and x + 10y + 18 = y + 10x

⇒ x + 10y – y – 10x = -18

⇒ -9x + 9y = -18

⇒x – y = 2 ….(ii)

On subtracting eq(i) from (ii), we get,

y = 3

x – 3 = 2

⇒ x = 2 + 3 = 5 [From eq(ii)]

Therefore,

The required number = x + 10y = 5 + 10 x 3

= 5 + 30 = 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 us considered the unit digit and tens digit of the number be x and y.

Therefore, the number = x + 10y

The same number after interchanging their digits = y + 10x

Now, we have,

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

⇒ x + 10y = 6x + 6y + 4

⇒ x + 10y – 6x – 6y = 4

⇒ -5x + 4y = 4 ….(i)

and x + 10y – 18 = y + 10x

⇒ x + 10y – y – 10x = 18

⇒ -9x + 9y = 18

⇒ x – y = -2 ….(ii)

⇒ x = y – 2

On substituting the value in eq(i),

-5 (y – 2) + 4y = 4

-5y + 10 + 4y = 4

-y = 4 – 10 = – 6

On solving, we get,

y = 6

The required number = x + 10y = 4 + 10 x 6

= 4 + 60 = 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 unit’s digit of the number and ten’s digit number x and y respectively.

Therefore, the number = x + 10y

Also, the number obtained by reversing the order of the digits = y + 10x

As per the specified constraints :

x + 10y = 4(x + y)

⇒ x + 10y = 4x + 4y

⇒ x + 10y – 4x – 4y = 0

⇒ -3x + 6y = 0

⇒ x = 2y …(i)

Also,

x + 10y = 2xy …(ii)

Substituting the value of x, in eq(ii)

2y + 10y = 2 * 2y * y ⇒ 12y = 4y

^{2}⇒ 3y = y

^{2}⇒ y(y – 3) = 0

y = 0 is not possible. Then y = 3.

Substituting y = 3, we obtain, x = 6

Therefore,

The required number = x + 10y = 36

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

Therefore, the number = x + 10y

Also, the number obtained by reversing the order of the digits = y + 10x

As per the specified constraints :

xy = 20

x + 10y + 9 = y + 10x

=> -9x + 9y = -9

x- y +1

On substituting in eq (i), we get

(1 + y)y = 20

y

^{2 }+ y – 20 = 0y

^{2}+ 5y – 4y – 20 = 0y(y + 5) – 4(y + 5) = 0

(y – 4)(y + 5) = 0

y = -5 is not possible. Therefore, y = 4

Therefore,

x = 1 + y = 1 + 4 = 5

Therefore, the required number is x + 10y = 5 + 40 = 45

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

**Solution:**

Let us assume the first number and second number to be x and y respectively.

Now, according to the given constraints,

x – y = 26 ….(i)

x = 3y ….(ii)

On substituting the value of x in eq.(i), we get,

3y – y = 26

⇒ 2y = 26

⇒ y = 13

On substituting the value of y, in eq(ii) we get,

x = 3y = 3 x 13 = 39

Therefore, the numbers are 39 and 13 respectively.

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

Therefore, the number = x + 10y

Also, the number obtained by reversing the order of the digits = y + 10x

As per the specified constraints :

x + y = 9 …..(i)

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

⇒ 9x + 90y = 2y + 20x

⇒ 9x + 90y – 2y – 20x = 0

⇒ -11x + 88y = 0

On dividing the equation by -11, we get,

⇒ x – 8y = 0

⇒ x = 8y

On substituting the value of x in eq(i)

8y + y = 9

⇒ 9y = 9

⇒ y= 1

x = 8y = 1 x 8 = 8

Therefore,

The required number = x + 10y = 8 + 10 x 1

= 8 + 10 = 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 unit’s digit of the number and ten’s digit number x and y respectively.

Therefore, the number = x + 10y

Also, the number obtained by reversing the order of the digits = y + 10x

As per the specified constraints :

x – y = 3 ….(i)

Also,

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

⇒ 7x + 70y = 4y + 40x

⇒ 7x + 70y – 4y – 40x = 0

⇒ -33x + 66y = 0

⇒ x – 2y = 0 (Dividing by -33)

⇒ x = 2y

On substituting the value of x in eq(i),

2y – y = 3 ⇒ y = 3

x = 2y = 2 x 3 = 6

Therefore,

The required number = x + 10y = 6 + 10 x 3

= 6 + 30 = 36

### 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 two numbers be x and y respectively.

Ratio of these two numbers = 5 : 6

that is, x : y = 5 : 6

⇒ x/y = 5/6

⇒ y = 6x/5

If 8 is subtracted from both the numbers, the ratio becomes 4:5

That is, x – 8 /(y – 8) = 4/5

Now,

⇒ 5x – 40 = 4y – 32

⇒ 5x – 4y = 8

Now, substituting the value,

5x – 4(6x/5) = 8

x = 40

Substituting x = 40, we get,

y = 48.

Therefore, the numbers are 40 and 48 respectively.

### 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 us assume the two-digit number to be 10x + y

Case I : Multiplying the sum of the digits by 8 and then subtract 5

⇒ 8 x (x + y) – 5 = 10x + y

⇒ 8x + 8y – 5 = 10x + y

⇒ 2x – 7y = – 5

Case II : Multiplying the difference of the digits by 16 and then add 3

16 * (x – y) + 3 = 10x + y

⇒ 6x – 17y = -3 …(i)

Multiplying eq(i) by 3, we obtain,

6x – 21y = -15 … (ii)

On subtracting eq(ii) from (i),

4y = 12

y = 3

On substituting the value of y, in eq(i) we get

x = 8

Hence,

The required number = 10x + y

⇒ 10 * 8 + 3

⇒ 83