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Class 10-Quadratic Equations MCQs Multiple Choice Questions with Answers
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Class 10 Maths MCQs Quadratic Equations
1. The sum of a number and its reciprocal is \(\frac {65}{8}\). What is the number? Let the number be x
a) 8
b) 4
c) 2
d) 6
Answer
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
Answer
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? Sum of the numbers is 13.
a) 76
b) 49
c) 58
d) 94
Answer
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? Since the pages of books are consecutive numbers, so let the left page number be x. The right page number will be x+1.
a) 11, 12
b) 12, 13
c) 17, 18
d) 15, 16
Answer
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? Let one number be x. The other number is x+2
a) 40, 42
b) 38, 40
c) 42, 44
d) 44, 46
Answer
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.
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? Let the length of the rectangle be x. The sum of length and breadth is 97.
a) 42, 76
b) 73, 24
c) 45, 73
d) 22, 77
Answer
Breadth will be 97-x.
Area of rectangle = 1752.
length × breadth = 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? Let the units place of the two digit number be x and the tens place be y.
a) -24
b) 42
c) 73
d) -37
Answer
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? The nth term of the AP is 5n+2.
a) 9
b) 10
c) 11
d) 4
Answer
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? Let the numerator be x.
a) \(\frac {89}{10}\)
b) \(\frac {10}{89}\)
c) \(\frac {1}{10}\)
d) 10
Answer
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? Sides of the right angled triangle is x+2, x+1, x
a) 1
b) 2
c) 3
d) 4
Answer
From Pythagoras theorem,
hyp2=adj2+base2
(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
11. For the equation x2 + 5x – 1, which of the following statements is correct? Roots are real. ∴ b2 – 4ac ≥ 0
a) The roots of the equation are equal
b) The discriminant of the equation is negative
c) The roots of the equation are real, distinct and irrational
d) The discriminant is equal to zero
Answer
52 – 4(1)(1)
25 – 4 = 21 which is greater than 0. Hence, the discriminant of the equation is greater than zero, so roots are real.
12. If the roots of the equation ax2 + bx + c are real and equal, what will be the relation between a, b, c? Roots are real and equal. ∴ b2 – 4ac = 0
a) b = ±\(\sqrt {ac}\)
b) b = ±\(\sqrt {4c}\)
c) b = ±\(\sqrt {-4ac}\)
d) b = ±\(\sqrt {4ac}\)
Answer
b2 = 4ac
b = ±\(\sqrt {4ac}\)
13. What will be the value of k, so that the roots of the equation are x2 + 2kx + 9 are imaginary? Roots are imaginary. ∴ b2 – 4ac < 0
a) -5 < k < 5
b) -3 < k < 3
c) 3 < k < -3
d) -5 < k < 3
Answer
(2k)2 – 4(9)(1) < 0
4k2 – 36 < 0
k2 – 9 < 0
k2 < 9
k < ±3
-3 < k < 3
14. What will be the nature of the roots of the quadratic equation 5x2 – 11x + 13? To check the nature of the roots, the discriminant must be either equal to zero, less than zero or greater than zero.
a) Imaginary
b) Real
c) Irrational
d) Equal
Answer
Discriminant = b2 – 4ac = – 112 – 4 × 5 × 13 = 121 – 260 = – 139
Since discriminant is less than zero, the roots of the equation are imaginary.
15. What will be the nature of the roots of the quadratic equation x2 + 10x + 25? To check the nature of the roots, the discriminant must be either equal to zero, less than zero or greater than zero.
a) Imaginary
b) Real
c) Irrational
d) Equal
Answer
Discriminant = b2 – 4ac = 102 – 4 × 25 × 1 = 100 – 100 = 0
Since discriminant is equal to zero, the roots of the equation are equal.
16. What will be the nature of the roots of the quadratic equation 2x2 + 10x + 9? To check the nature of the roots, the discriminant must be either equal to zero, less than zero or greater than zero.
a) Imaginary
b) Real
c) Irrational
d) Equal
Answer
Discriminant = b2 – 4ac = 102 – 4 × 2 × 9 = 100 – 72 = 28
Since discriminant is greater than zero, the roots of the equation are real and distinct.
17. What will be the value of a, for which the equation 5x2 + ax + 5 and x2 – 12x + a will have real roots? The roots of both the equations are real.
a) a = 37
b) 10 < a < 36
c) 36 < a < 10
d) a = 9
Answer
Discriminant of 5x2 + ax + 5 : b2 – 4ac = a2 – 4 × 5 × 5 = a2 – 100
Since, roots are real; discriminant will be greater than 0.
a2 ≥ 100
a ≥ ±10
Now, discriminant of x2 – 12x + a : b2 – 4ac = -122 – 4 × 1 × a = 144 – 4a
Since, roots are real; discriminant will be greater than 0.
144 ≥ 4a
a ≤ \(\frac {144}{4}\) = 36
For both the equations to have real roots the value of a must lie between 36 and 10.
18. What will be the value of k, if the roots of the equation (k – 4)x2 – 2kx + (k + 5) = 0 are equal? Roots are equal. ∴ b2 – 4ac = 0
a) 18
b) 19
c) 20
d) 21
Answer
-(2k)2 – 4(k – 4)(k + 5) = 0
4k2 – 4(k2 – 4k + 5k – 20) = 0
4k2 – 4(k2 + k – 20) = 0
4k2 – 4k2 – 4k + 80 = 0
-4k = – 80
k = \(\frac {-80}{-4}\) = 20
19.One year ago, a man was 8 times as old as his son. Now his age is equal to the square of his son’s age. Their present ages are
(a) 7 years, 49 years
(b) 5 years, 25 years
(c) 1 years, 50 years
(d) 6 years, 49 years
Answer
Answer: (a) 7 years, 49 years
20.If -5 is a root of the quadratic equation 2x2 + px – 15 = 0, then
(a) p = 3
(b) p = 5
(c) p = 7
(d) p = 1
Answer
Answer: (c) p = 7
21.The two consecutive odd positive integers, sum of whose squares is 290 are
(a) 13, 15
(b) 11, 13
(c) 7, 9
(d) 5, 7
Answer
Answer: (b) 11, 13
22.The value of b2 – 4ac for equation 3x2 – 7x – 2 = 0 is
(a) 49
(b) 0
(c) 25
(d) 73
Answer
Answer: (d) 73
23.Value(s) of k for which the quadratic equation 2x2 -kx + k = 0 has equal roots is
(a) 0
(b) 4
(c) 8
(d) 0 and 8
Answer
Answer: (d) 0 and 8
24.Which of the following are the roots of the quadratic equation, x2 – 9x + 20 = 0 by factorisation?
(a) 3,4
(b) 4, 5
(c) 5, 6
(d) 6, 7
Answer
Answer: (b) 4, 5
24.The equation (x – 2)2 + 1 = 2x – 3 is a
(a) linear equation
(b) quadratic equation
(c) cubic equation
(d) bi-quadratic equation
Answer
Answer: (b) quadratic equation
25.Two candidates attempt to solve a quadratic equation of the form x2 + px + q = 0. One starts with a wrong value of p and finds the roots to be 2 and 6. The other starts with a wrong value of q and finds the roots to be 2 and – 9. Find the correct roots of the equation :
(a) 3, 4
(b) – 3, – 4
(c) 3, – 4
(d) – 3, 4
Answer
Answer: (b) – 3, – 4
26.If (x – a) is one of the factors of the polynomial ax2 + bx + c, then one of the roots of ax2 + bx + c = 0 is
(a) 1
(b) c
(c) a
(d) none of these
Answer
Answer: (c) a
27.The equation x2 – px + q = 0 p, q e R has no real roots if :
(a) p2 > 4q
(b) p2 < 4q
(c) p2 = 4q
(d) None of these
Answer
Answer: (b) p2 < 4q
28.If the roots of px2 + qx + 2 = 0 are reciprocal of each other, then
(a) P = 0
(b) p = -2
(c) p = ±2
(d) p = 2
Answer
Answer: (d) p = 2
29.If a, p are the roots of the equation (x – a) (x – b) + c = 0, then the roots of the equation (x – a) (x – P) = c are
(a) a, b
(b) a, c
(c) b, c
(d) none of these
Answer
Answer: (a) a, b
30.The sum of the roots of the quadratic equation 3x2 – 9x + 5 = 0 is
(a) 3
(b) 6
(c) -3
(d) 2
Answer
Answer: (c) -3
31.The equation 2x2 + kx + 3 = 0 has two equal roots, then the value of k is
(a) ±√6
(b) ± 4
(c) ±3√2
(d) ±2√6
Answer
Answer: (d) ±2v6
32.The quadratic equation whose one rational root is 3 + v2 is
(a) x2 – 7x + 5 = 0
(b) x2 + 7x + 6 = 0
(c) x2 – 7x + 6 = 0
(d) x2 – 6x + 7 = 0
Answer
Answer: (d) x2 – 6x + 7 = 0
33The quadratic equation 2x2 – 3x + 5 = 0 has?
(a) Real and distinct roots
(b) Real and equal roots
(c) Imaginary roots
(d) All of the above
Answer
Answer: (c) Imaginary roots