# Algebra Practice Questions for CAT with Solutions

Question 1:
The solution(s) to $4{x^2} + 8x - 2\sqrt {4{x^2} + 8x - 3} = 6$ are/is in the interval:
[1] (-10, -5)
[2] (-6, 0)
[3] (0, 6)
[4] (5, 10)
Let $\sqrt {4{x^2} + 8x - 3}$= y

$4{x^2} + 8x - 2\sqrt {4{x^2} + 8x - 3} = 6$

$4{x^2} + 8x - 3 + 2\sqrt {4{x^2} + 8x - 3} = 6 - 3$

y2 + 2y – 3 = 0

y = -3, 1

If y = 1

4x2 + 8x – 3 = 1
4x2 + 8x – 4 = 0
x2 + 2x – 1 = 0
x = (-2 +- sqrt(8))/2 = -1 +- sqrt(2)

If y = 3

4x2 + 8x – 3 = 9
4x2 + 8x – 12 = 0
x2 + 2x – 3 = 0
x = -3, 1

Question 2:
The sum of the integers in the solution set of |x2-5x|<6 is:
[1] 10
[2] 15
[3] 20
[4] 0
|x2-5x|<6
x2 - 5x – 6 < 0
(x – 6)(x + 1) < 0
- 1 < x < 6
x = {0, 1, 2, 3, 4, 5}

And

|x2-5x|<6

-(x2 - 5x) – 6 < 0
x2 – 5x + 6 > 0
(x – 2)(x - 3) > 0
x < 2 or x > 3

Values of x common to both {0, 1, 4, 5}

Sum of values of x = 0 + 1 + 4 + 5 = 10. Option A

Question 3:
Find abc if a+b+c = 0 and a3+ b3+ c3=216
[1] 48
[2] 72
[3] 24
[4] 216
a3 + b3 + c3 - 3abc = (a + b + c)(a2 + b2 + c2 - ab - bc - ca)
216 – 3abc = 0
abc = 72. Option B

Question 4:
Solve for x: $\sqrt {x + \sqrt {x + \sqrt x + ....} } = \frac{3}{2}$
[1] Empty Set
[2] 3/2
[3] 3/4
[4] 3/16
$\begin{array}{l}\sqrt {x + \sqrt {x + \sqrt x + ....} } = \frac{3}{2}\\\sqrt {x + \frac{3}{2}} = \frac{3}{2}\\x + \frac{3}{2} = \frac{9}{4}\\x = \frac{3}{4}\end{array}$

Option C

Question 5:
Solve for x $\sqrt {\frac{3}{2} + \sqrt {\frac{3}{2} + \sqrt {\frac{3}{2}} + ....} } = x$
[1] $\frac{{1 \pm \sqrt 7 }}{2}$
[2] $\frac{{1 + \sqrt 7 }}{2}$
[3] $\frac{{\sqrt 7 }}{2}$
[4] $\frac{3}{2}$
$\begin{array}{l}\sqrt {\frac{3}{2} + \sqrt {\frac{3}{2} + \sqrt {\frac{3}{2}} + ....} } = x\\\sqrt {\frac{3}{2} + x} = x\end{array}$

3/2 + x = x2

2x2 – 2x – 3 = 0

x = $\frac{{1 + \sqrt 7 }}{2}$Option B

Algebra Practice Questions for CAT with Solutions
5 (100%) 8 vote[s]