# Algebra Practice Questions for CAT with Solutions

Question 1:
If $\sqrt {x + \sqrt {x + \sqrt {x + ....} } } = 10.$What is the value of x?
[1] 80
[2] 90
[3] 100
[4] 110
\begin{align}& \sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+.....}}}}=10 \\;& \sqrt{x+10}=10 \\ \end{align}

x + 10 = 100

x = 90. Option B

Question 2:
Solve for x over the real numbers: ${x^{\frac{4}{3}}} + 14{x^{\frac{2}{3}}} - 51 = 0$
[1] 3√3
[2] 3√3, 17√17
[3] -17, 3
[4] 3
$\begin{array}{l}{x^{\frac{4}{3}}} + 14{x^{\frac{2}{3}}} - 51 = 0\\{x^{\frac{2}{3}}} = - 17,3\end{array}$

x = 3√3. Option A

Question 3:
Find the value of $\sqrt {4 - \sqrt {4 + \sqrt {4 - \sqrt {4 + ...} } } }$
[1] $\frac{{\sqrt {13} - 1}}{2}$
[2] $\frac{{\sqrt {13} + 1}}{2}$
[3] $\frac{{\sqrt {11} + 1}}{2}$
[4] $\frac{{\sqrt {15} - 1}}{2}$
$\begin{array}{l}\sqrt {4 - \sqrt {4 + \sqrt {4 - \sqrt {4 + ...} } } } = x\\\sqrt {4 - \sqrt {4 + x} } = x\\4 - \sqrt {4 + x} = {x^2}\\4 - {x^2} = \sqrt {4 + x} \\16 - 8{x^2} + {x^4} = 4 + x\\{x^4} - 8{x^2} - x + 12 = 0\end{array}$

We can say that the answer is a bit bigger than 2 which eliminates option A and option D.

Using the options we can say the answer is $\frac{{\sqrt {13} + 1}}{2}$. Option B

Question 4:
If the roots of the equation x3ax2 + bx c =0 are three consecutive integers, then what is the smallest possible value of b?
[1] -1/√3
[2] -1
[3] 0
[4] 1/√3
b is sum of product of the roots taken 2 at a time which will be minimum when the roots are -1, 0 & 1

b = -1×0 + 0×1 + (-1)×1 = -1. Option B

Question 5:
Three consecutive positive integers are raised to the first, second and third powers respectively and then added. The sum so obtained is a perfect square whose square root equals the total of the three original integers. Which of the following best describes the minimum, say m, of these three integers?
[1] 1 ≤ m ≤ 3
[2] 4 ≤ m ≤ 6
[3] 7 ≤ m ≤ 9
[4] 10 ≤ m ≤ 12
We are given

m + (m+1)2 + (m+2)3 = (3m+3)2 = 9(m+1)2

m + (m+2)3 = 8(m+1)2

Using options,

3 + 42 + 53 = 3 + 16 + 125 = 144 = 122 = (3 + 4 + 5)2

So, m = 3. Option A

Alternatively

Let us take the three numbers as a – 1, a and a+1

(a-1) + (a+1)3 = 8a2
a – 1 + a3 + 3a2 + 3a + 1 = 8a2
a3 – 5a2 + 4a = 0
a(a2 – 5a + 4) = 0
a = 0, 1, 4
0 and 1 are invalid values because a – 1 should be a positive integer
a = 4
m = a – 1 = 3. Option A

Algebra Practice Questions for CAT with Solutions
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