Class 10 RD Sharma Solutions - Chapter 3 Pair of Linear Equations in Two Variables - Exercise 3.7

In Class 10 mathematics, solving linear equations is a fundamental concept that underpins many advanced topics. Chapter 3 of RD Sharma's textbook titled "Pair of Linear Equations in Two Variables" explores methods to solve systems of equations involving the two variables. Understanding these concepts is crucial for solving real-world problems that can be modeled using linear equations.

Pair of Linear Equations in Two Variables

A pair of linear equations in the two variables consists of the two equations with the two unknowns. These equations can be represented as:

ax+by=c

dx+ey=f

To solve such a system one can use various methods including:

• Graphical Method: Plotting both equations on a graph and identifying the intersection point.
• Substitution Method: Solving one equation for one variable and substituting it into the other equation.
• Elimination Method: Adding or subtracting equations to eliminate one variable and solve for the other.

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  x²  - y ² = 256000

Dividing 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 = 4y²

⇒ 3y = y²

⇒ 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

y² + y - 20 = 0

y² + 5y - 4y - 20 = 0 

y(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

Conclusion

Mastering the techniques for solving pairs of the linear equations in the two variables is essential for the students to tackle more complex algebraic problems. Exercise 3.7 in RD Sharma’s Class 10 textbook provides the valuable practice to reinforce these concepts and ensure a solid foundation in the solving linear systems.