# Algebra Practice Questions for CAT with Solutions

Question 1:
Let M = x2 – 12x + 21 and N = x2 + x – 20. Find the sum of all the values of x which satisfy the equation MN = 1.
[1] –1
[2] 12
[3] 11
[4] 1
MN = 1 is possible if M = 1 or N = 0

M = x2 – 12x + 21 = 1

x2 – 12x + 20 = 0
x = 2, 10

N = x2 + x – 20 = 0

x = -5, 4

Sum of all valid values of x = 2 + 10 – 5 + 4 = 11. Option C

Question 2:
For what values of p does the equation 4x2 + 4px + 4 –3p = 0 have two distinct real roots?
[1] p < -4 or p > 1
[2] -1 < p < 4
[3] p < -1 or p > 4
[4] –4 < p < 1
For the roots to be distinct and real

b2 – 4ac > 0

(4p)2 – 4×4×(4 – 3p) > 0
p2 – (4 – 3p) > 0
p2 + 3p – 4 > 0
(p + 4)(p – 1) > 0
p < -4 or p > 1. Option A

Question 3:
If x2 + 4x + n > 13 for all real number x, then which of the following conditions is necessarily true?
[1] n > 17
[2] n = 20
[3] n > -17
[4] n < 11
x2 + 4x + n > 13
x2 + 4x + 4 + n > 13 + 4 {Adding 4 to both sides}
(x + 2)2 + n > 17
Minimum value (x + 2)2 can take is 0 when x = – 2
For this to be true for all real values of x, n > 17. Option A

Question 4:
If (x + 1)×(x – 2)×(x + 3)×(x – 4)×(x + 5)…(x – 100) = a0 + a1x + a2x2… + a100x100 then the value of a99 is equal to:
[1] 50
[2] 0
[3] -50
[4] -100
a100 = 1

Sum of the roots = $-\frac{{{a}_{99}}}{{{a}_{100}}}=-\ {{a}_{99}}$

a99 = - (– 1 + 2 – 3 + 4 – 5 + 6 … – 99 + 100) = - 50. Option C

Question 5:
If a, b, and c are the solutions of the equation x3 – 3x2 – 4x + 5 = 0, find the value of $\frac{1}{{ab}} + \frac{1}{{bc}} + \frac{1}{{ca}}$
[1] 3/5
[2] -3/5
[3] -4/5
[4] 4/5
$\frac{1}{{ab}} + \frac{1}{{bc}} + \frac{1}{{ca}} = \frac{{a + \;b + c}}{{abc}} = \frac{{ - ( - 3)}}{{ - 5}} = - \;\frac{3}{5}$

Option B

Algebra Practice Questions for CAT with Solutions