Mathematics Mcqs

A. If a < b
B. If a ≤ b
C. If the relationship between a and b cannot be established
D. If a ≥ b
Explanation:
I. (a – 5)(a – 4) = 0
=> a = 5, 4
II. (2b + 3)(b – 4) = 0
=> b = 4, -3/2 => a ≥ b

A. if a < b
B. if a ≤ b
C. if the relationship between a and b cannot be established.
D. if a > b
Explanation:
I.(a – 3)(a – 4) = 0
=> a = 3, 4
II. (b – 2)(b – 1) = 0
=> b = 1, 2
=> a > b

A. 10
B. 8
C. 15
D. 7.50
Explanation:
Let the price of each note book be Rs.x.
Let the number of note books which can be brought for Rs.300 each at a price of Rs.x be y.
Hence xy = 300
=> y = 300/x
(x + 5)(y – 10) = 300 => xy + 5y – 10x – 50 = xy
=>5(300/x) – 10x – 50 = 0 => -150 + x2 + 5x = 0
multiplying both sides by -1/10x
=> x2 + 15x – 10x – 150 = 0
=> x(x + 15) – 10(x + 15) = 0
=> x = 10 or -15
As x>0, x = 10

A. 3×2 + 5x – 2 = 0
B. 3×2 + 5x + 2 = 0
C. 3×2 – 5x + 2 = 0
D. 3×2 – 5x – 2 = 0

The quadratic equation whose roots are reciprocal of 2×2 + 5x + 3 = 0 can be obtained by replacing x by 1/x.
Hence, 2(1/x)2 + 5(1/x) + 3 = 0
=> 3×2 + 5x + 2 = 0

A. 15
B. 14
C. 24
D. 26
Explanation:
a/b + b/a = (a2 + b2)/ab = (a2 + b2 + a + b)/ab
= [(a + b)2 – 2ab]/ab
a + b = -8/1 = -8
ab = 4/1 = 4
Hence a/b + b/a = [(-8)2 – 2(4)]/4 = 56/4 = 14.

A. -21
B. 1
C. 61
D. 21
Explanation:
a2 + b2 + ab = a2 + b2 + 2ab – ab
i.e., (a + b)2 – ab
from x2 – 9x + 20 = 0, we have
a + b = 9 and ab = 20. Hence the value of required expression (9)2 – 20 = 61.

A. 18, 20, 22
B. 20, 22, 24
C. 22, 24, 26
D. 24, 26, 28
Explanation:
Three consecutive even natural numbers be 2x – 2, 2x and 2x + 2.
(2x – 2)2 + (2x)2 + (2x + 2)2 = 1460
4×2 – 8x + 4 + 4×2 + 8x + 4 = 1460
12×2 = 1452 => x2 = 121 => x = ± 11
As the numbers are positive, 2x > 0. Hence x > 0. Hence x = 11.
Required numbers are 20, 22, 24.

A. 29
B. 27
C. 28
D. 7
Explanation:
Let the roots of the quadratic equation be x and 3x.
Sum of roots = -(-12) = 12
a + 3a = 4a = 12 => a = 3
Product of the roots = 3a2 = 3(3)2 = 27.

A. 9, 10
B. 10, 11
C. 11, 12
D. 12, 13
Explanation:
Let the two consecutive positive integers be x and x + 1
x2 + (x + 1)2 – x(x + 1) = 91
x2 + x – 90 = 0
(x + 10)(x – 9) = 0 => x = -10 or 9.
As x is positive x = 9
Hence the two consecutive positive integers are 9 and 10.

A. 3
B. 4
C. 5
D. 6
Let the roots of the equation 2a and 3a respectively.
2a + 3a = 5a = -(- 5/2) = 5/2 => a = 1/2
Product of the roots: 6a2 = b/2 => b = 12a2
a = 1/2, b = 3.