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

View Answer

Explanation: Let the number be x

x+\(\frac {1}{x}=\frac {65}{8}\)

\(\frac {x^2+1}{x}=\frac {65}{8}\)

8(x

^{2}+1)=65x

8x

^{2}+8=65x

8x

^{2}-65x+8=0

8x

^{2}-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

View Answer

Explanation: Sum of the numbers is 12.

Let one number be x. Other number is 12-x.

Sum of their squares = 74

x

^{2}+(12-x)

^{2}=74

x

^{2}+144+x

^{2}-24x=74

2x

^{2}-24x+144-74=0

2x

^{2}-24x+70=0

x

^{2}-12x+35=0

x

^{2}-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

View Answer

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-x

^{2}

x

^{2}-13x+40=0

x

^{2}-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

View Answer

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

x

^{2}+(x+1)

^{2}=481

x

^{2}+x

^{2}+1+2x=481

2x

^{2}+2x-480=0

x

^{2}+x-240=0

x

^{2}+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

View Answer

Explanation: Let one number be x. The other number is x+2

Sum of the squares of the numbers is 3364.

x

^{2}+(x+2)

^{2}=3364

x

^{2}+x

^{2}+4x+4=3364

2x

^{2}+4x-3360=0

x

^{2}+2x-1680=0

x

^{2}+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.

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

View Answer

Explanation: Let the length of the rectangle be x. The sum of length and breadth is 97.

Breadth will be 97-x.

Area of rectangle = 1752.

length × breadth = 1752

x(97-x)=1752

97x-x

^{2}=1752

x

^{2}-97x+1752=0

x

^{2}-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

View Answer

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-x

^{2}=4x

x

^{2}+4x-21=0

x

^{2}+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 n^{th} 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

View Answer

Explanation: The n

^{th}term of the AP is 5n+2.

T

_{1}=a=5+2=7

T

_{2}=5(2)+2=12

d=T

_{2}-T

_{1}=12-7=5

S

_{n}=295

S

_{n}=\(\frac {n}{2}\)(2a+(n-1)d)

295=\(\frac {n}{2}\)(2(7)+(n-1)5)

590=14n+5n

^{2}-5n

5n

^{2}+9n-590=0

5n

^{2}+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

View Answer

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(x

^{2}+1+81x

^{2}+18x)=101x(1+9x)

10(82x

^{2}+18x+1)=101x+909x

^{2}

820x

^{2}+180x+10=101x+909x

^{2}

89x

^{2}-79x-10=0

89x

^{2}-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

View Answer

Explanation: Sides of the right angled triangle is x+2, x+1, x

From Pythagoras theorem,

hyp

^{2}=adj

^{2}+base

^{2}

(x+2)

^{2}=(x+1)

^{2}+(x)

^{2}

x

^{2}+4x+4=x

^{2}+1+2x+x

^{2}

4x+4=x

^{2}+1+2x

x

^{2}-2x-3=0

x

^{2}+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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