# Class 10 Maths MCQ – Quadratic Equations

This set of Class 10 Maths Chapter 4 Multiple Choice Questions & Answers (MCQs) focuses on “Quadratic Equations”. These MCQs are created based on the latest CBSE syllabus and the NCERT curriculum, offering valuable assistance for exam preparation.

1. The sum of a number and its reciprocal is $$\frac {65}{8}$$. What is the number?
a) 8
b) 4
c) 2
d) 6

Explanation: Let the number be x
x+$$\frac {1}{x}=\frac {65}{8}$$
$$\frac {x^2+1}{x}=\frac {65}{8}$$
8(x2+1)=65x
8x2+8=65x
8x2-65x+8=0
8x2-64x-x+8=0
8x(x-8)-1(x-8)=0
(x-8)(8x-1)=0
x=8, $$\frac {1}{8}$$
The number is 8 or $$\frac {1}{8}$$.

2. Find two numbers such that the sum of the numbers is 12 and the sum of their squares is 74.
a) 84
b) 75
c) 66
d) 48

Explanation: Sum of the numbers is 12.
Let one number be x. Other number is 12-x.
Sum of their squares = 74
x2+(12-x)2=74
x2+144+x2-24x=74
2x2-24x+144-74=0
2x2-24x+70=0
x2-12x+35=0
x2-7x-5x+35=0
x(x-7)-5(x-7)=0
(x-7)(x-5)=0
x=7, 5
The number is 57 or 75.

3. The sum of two numbers is 13 and the sum of their reciprocals is $$\frac {13}{40}$$. What are the two numbers?
a) 76
b) 49
c) 58
d) 94

Explanation: Sum of the numbers is 13.
Let one number be x. Other number is 13-x.
Sum of their reciprocals = $$\frac {13}{40}$$
$$\frac {1}{x} + \frac {1}{13-x}=\frac {13}{40}$$
$$\frac {13-x+x}{x(13-x)}=\frac {13}{40}$$
$$\frac {13}{13x-x^2}=\frac {13}{40}$$
$$\frac {1}{13x-x^2}=\frac {1}{40}$$
40=13x-x2
x2-13x+40=0
x2-5x-8x+40=0
x(x-5)-8(x-5)=0
(x-8)(x-5)=0
x=8, 5
The number is 58 or 85.

4. The sum of the squares of the left and right pages of a book is 481. What are the page numbers?
a) 11, 12
b) 12, 13
c) 17, 18
d) 15, 16

Explanation: Since the pages of books are consecutive numbers, so let the left page number be x. The right page number will be x+1.
Sum of the squares of the pages is 481
x2+(x+1)2=481
x2+x2+1+2x=481
2x2+2x-480=0
x2+x-240=0
x2+16x-15x-240=0
x(x+16)-15(x+16)=0
(x-15)(x+16)=0
x=15, -16
Since, page number cannot be negative, so x=15
The two page numbers are 15 and 16.

5. The sum of the squares of two consecutive positive even numbers is 3364. What are the two numbers?
a) 40, 42
b) 38, 40
c) 42, 44
d) 44, 46

Explanation: Let one number be x. The other number is x+2
Sum of the squares of the numbers is 3364.
x2+(x+2)2=3364
x2+x2+4x+4=3364
2x2+4x-3360=0
x2+2x-1680=0
x2+42x-40x-1680=0
x(x+42)-40(x+42)=0
(x+42)(x-40)=0
x=40, -42
Since we only need positive numbers.
Hence, x=40
The two numbers are 40, 42.
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6. The sum of the length and the breadth of a rectangle are 97 and the area of the rectangle is 1752. What will be the value of the length and breadth of the rectangle?
a) 42, 76
b) 73, 24
c) 45, 73
d) 22, 77

Explanation: Let the length of the rectangle be x. The sum of length and breadth is 97.
Area of rectangle = 1752.
x(97-x)=1752
97x-x2=1752
x2-97x+1752=0
x2-73x-24x+1752=0
x(x-73)-24(x-73)=0
(x-73)(x-24)=0
x=73, 24
Hence, the length is 73 and breadth 24 or length 24, breadth 73.

7. The product of digits of a two digit number is 21 and when 36 is subtracted from the number, the digits interchange their places. What is the number?
a) -24
b) 42
c) 73
d) -37

Explanation: Let the units place of the two digit number be x and the tens place be y.
Product of the digits of the places = 21
xy=21
y=$$\frac {21}{x}$$
Now, the number will be 10y+x
If 36 is subtracted from the number the digits interchange their places.
New number = 10x+y
10y+x-36=10x+y
9y-9x=36
y-x=4
Now, y=$$\frac {21}{x}$$
$$\frac {21}{x}$$-x=4
21-x2=4x
x2+4x-21=0
x2+7x-3x-21=0
x(x+7)-3(x+7)=0
(x+7)(x-3)=0
x = -7, 3
The number is 73.

8. If the nth term of the AP is 5n+2. What will be the value of n so that the sum of the first n terms is 295?
a) 9
b) 10
c) 11
d) 4

Explanation: The nth term of the AP is 5n+2.
T1=a=5+2=7
T2=5(2)+2=12
d=T2-T1=12-7=5
Sn=295
Sn=$$\frac {n}{2}$$(2a+(n-1)d)
295=$$\frac {n}{2}$$(2(7)+(n-1)5)
590=14n+5n2-5n
5n2+9n-590=0
5n2+59n-50n-590=0
n(5n+59)-10(5n+59)=0
(n-10)(5n+59)=0
n=10, $$\frac {-59}{5}$$

9. The denominator of a fraction is 1 more than 9 times the numerator. If the sum of the fraction and its reciprocal is $$\frac {101}{10}$$ then, what will be the fraction?
a) $$\frac {89}{10}$$
b) $$\frac {10}{89}$$
c) $$\frac {1}{10}$$
d) 10

Explanation: Let the numerator be x.
Denominator of a fraction is 1 more than 9 times the numerator.
Denominator = 1+9x
The fraction is $$\frac {x}{11+9x}$$
Fraction + Reciprocal = $$\frac {101}{10}$$
$$\frac {x}{1+9x}+\frac {1+9x}{x}=\frac {101}{10}$$
$$\frac {x^2+(1+9x)^2}{x(1+9x)}=\frac {101}{10}$$
10(x2+1+81x2+18x)=101x(1+9x)
10(82x2+18x+1)=101x+909x2
820x2+180x+10=101x+909x2
89x2-79x-10=0
89x2-89x+10x-10x=0
89x(x-1)+10(x-1)=0
(x-1)(89x+10)=0
x=1, $$\frac {-10}{89}$$
The fraction is $$\frac {1}{10}$$

10. If the sides of the right angled triangle is x+2, x+1, x then what is the value of x?
a) 1
b) 2
c) 3
d) 4

Explanation: Sides of the right angled triangle is x+2, x+1, x
From Pythagoras theorem,
(x+2)2=(x+1)2+(x)2
x2+4x+4=x2+1+2x+x2
4x+4=x2+1+2x
x2-2x-3=0
x2+3x-x-3=0
x(x+3)-1(x+3)=0
(x-1)(x+3)=0
x=1, -3
Since, sides of triangle cannot be negative.
Hence, x=1

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