Chapter 3 - Pair Of Linear Equations In Two Variables, RD Sharma Solutions - (Part-19) | RD Sharma Solutions for Class 10 Mathematics PDF Download

Page No 3.86

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

Q.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.
Ans. 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 = 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 = 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 + 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 x 6 + 3 = 63
Note that there are two such numbers.

Q.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.
Ans. Let the digits at units and tens place of the given number be x and y respectively. Thus, the number is.
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 - 2 = 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 x 2 = 0
⇒ x - 4 = 0
⇒ x = 4
Hence, the number is 10 x 2 + 4 = 24

Page No 3.86:

Q.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.
Ans. Let the digits at units and tens place of the given number be x and y respectively. Thus, the number is.
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 = - 1
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 = 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 x 3 = - 1
⇒ x - 6 = - 1
⇒ x = -1 + 6
⇒ x = 5
Hence, the number is 10 x 3 + 5 = 35

Q.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.
Ans. Let the digits at units and tens place of the given number be x and y respectively. Thus, the number is 10 y + 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 = - 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 x 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 x 6 + 4 = 64

Q.11. A two-digit number is 4 times the sum of its digits and twice the product of the digits. Find the number.
Ans. 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 x 2y x y
⇒ 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.

Q.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.
Ans. Let the digits at units and tens place of the given number be x and y respectively. Thus, the number is 10 y + 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 = 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 = 1 + 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 x 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.

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