RD Sharma Solutions Chapter 8 Quadratic Equations Exercise 8.7 Class 10 Maths
Chapter Name  RD Sharma Chapter 8 Quadratic Equations 
Book Name  RD Sharma Mathematics for Class 10 
Other Exercises 

Related Study  NCERT Solutions for Class 10 Maths 
Exercise 8.7 Solutions
1. Find the consecutive numbers whose squares have the sum 85.
Solution
Let the two consecutive natural numbers be 'x' and 'x + 1'
Given that the sum of their squares is 85.
Then by hypothesis, we get
x^{2} + (x + 1)^{2} = 85
⇒ x^{2} + x^{2} + 2x + 1 = 85
⇒ 2x^{2} + 2x + 1  85 = 0
⇒ 2x^{2} + 2x + 84 = 0 ⇒ 2[x^{2} + x  42] = 0
⇒ x^{2} + 7x  6x  42 = 0 [by the method of factorisation]
⇒ x(x + 7)  6(x + 7) = 0
⇒ (x  6)(x + 7) = 0
⇒ x = 6 or x = 7
case i : if x = 6x + 1 = 6 + 1 = 7
case ii: if x = 7x + 1 =  7 + 1 =  6
∴ The consecutive numbers that the sum of this squares be 85 are 6, 7 and 6,  7.
2. Divide 29 into two parts so that the sum of the squares of the parts is 425.
Solution
Case ii: x = 16; 29  x = 29  16 = 13∴ The two parts that the sum of the squares of the parts is 425 are 13, 16.
The sum of their areas = 656 cm^{2} .
The area of the square = side × side
∴ Area of the square = x(x + 4) cm^{2}
⇒ Given that sum of the areas is 656 cm^{2} .
Hence by hypothesis, we have
⇒ x(x + 4) + x(x + 4) = 656
⇒2x(x + 4) = 656
⇒ x^{2} + 4x = 328 [dividing both sides by 2]
⇒ x^{2} + 4c  328 = 0
⇒ x^{2} + 20x  16x  328 = 0 [∵ by the method of factorisation]
⇒ x(x + 20)  16(x + 20) = 0
Case i: If x = 16; x + 4 = 20
∴ The sides of the squares are 16 cm and 20cm.
Note : No negative value is considered as the sides will never be measured negatively.
Let the two numbers be x and 48  x also given their product is 432.
Hence x(48  x) = 432
⇒ 48x  x^{2} = 432
⇒ 48x  x^{2}  432 = 0
⇒ x^{2}  48x + 432 = 0
⇒ x^{2}  36x  12x + 432 = 0 [By method of factorisation]
⇒ x(x  36)  12(x  36) = 0
⇒ (x  36)(x  12) = 0
⇒ x = 36 or x = 12
∴ The two numbers are 12, 36.
Given that if an integer is added to its square, the sum is 90.
⇒ x + x^{2} = 90
⇒ x + x^{2}  90 = 0
⇒ x^{2} + 10x  9x  90 = 0
⇒ x(x + 10)  9(x + 10) = 0
⇒ x =  10 or x = 9
⇒ Given that the product of the natural numbers is 20
Hence ⇒ x(x + 1) = 20
⇒ x^{2} + x = 20
⇒ x^{2} + x  20 = 0
⇒ x^{2} + 5x  4x  20 = 0
⇒ x(x + 5)  4(x + 5) = 0
⇒ x =  5 or x = 4
Considering positive value of x as x ∈ N
For r = 4, x + 1 = 4 + 1 = 5
∴ The two consecutive natural numbers are 4 as 5.
8. The sum of the squares of the two consecutive odd positive integers as 394. Find them.
Solution
Let the consecutive odd positive integers are 2x  1 and 2x + 1
Given that the sum of the squares is 394.
(2x  1)^{2} + (2x + 1)^{2} = 394
4x^{2} + 1  4x + 4x^{2} + 1 + 4x = 394
8x^{2} + 2 = 394
4x^{2} = 392
x^{2} = 36
x = 6
As x = 6,
2x  1 = 2×6  1 = 11
2x + 1 = 2×6 + 1 = 13
∴ The two consecutive odd positive numbers are 11 and 13.
9. The sum of two numbers is 8 and 15 times the sum of their reciprocals is also 8. Find the numbers.
Solution
Let the numbers be x and 8 x
Given that the sum of these numbers is 8
And 15 times the sum of their reciprocals as 8
By the hypothesis we have
13. The product of two successive integral multiples of 5 is 300. Determine the multiples.
Let the integers be 5x, and 5(x + 1)
Then, by the integers be 5x and 5(x + 1)
Then, by the hypothesis, we have
5x . 5(x + 1) = 300
Then the other number = 2x  3
According to the given hypothesis,
Then according to the given hypothesis,
Given that the difference of two numbers is 4.
By the given hypothesis, we have 1/(x  4)  1/x = 4/21
By the given hypothesis,
By the given hypothesis, we have
Given that their product is 270
According to the given hypothesis
22. A twodigit number is such that the product of its digits is 8. When 18 is subtracted from the number, the digits interchange their places. Find the number ?
Let the two digits be x and x  2
24. A two digit number is such that the product of the digits is 16. When 54 is subtracted from the number the digits are interchanged. Find the number .
Given that their product is 192
given that = 10x + y
⇒ 10x + y = 4 (sum of digits) and 2xy
⇒ 10x + y = 4(x + y) and 10x + y = 3xy
⇒ 10x + y = 4x + 4y and 10x + y = 3xy
⇒ 6x  3y = 0 and 10x + y  3xy = 0
⇒ y = 2x and 10x + 2x = 2xy(2x)
⇒ 12x = 4x^{2}
⇒ 4x(x  3) = 0
⇒ 4x = 0 or x = 3
⇒ here we have y = 2x ⇒ 2×3 = 6
∴ x = 3 and y = 6
Hence 10x + y = 10×3+6 = 36
∴ The required two digit number is 36.
Also, square of the smaller number = x^{2}
It is given that the sum of the square of the integers is 208.
∴ square of the larger number = 18x = 18 × 8 = 144
⇒ larger number are 8 and 18.
⇒ according to the given hypothesis
By the given hypothesis, we have
According to the hypothesis, we have
It is given that the difference of the square of the number is 88.
By the given hypothesis, we have
Square of larger number 8x =  80 here no perfect square exist, hence the numbers are 18, 12.