Algebra Practice Questions for CAT with Solutions

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Question 1:
What is/are the value(s) of x if \(\sqrt {{x^2} + \sqrt {{x^2} + \sqrt {{x^2} + ...} } } = 9 \)
[1] 6√2
[2] 3√10
[3] ±3√10
[4] ±6√2
\(\begin{array}{l}\sqrt {{x^2} + \sqrt {{x^2} + \sqrt {{x^2} + ...} } }  = 9\\\sqrt {{x^2} + 9}  = 9\\{x^2} + 9 = 81\\{x^2} = 72\\x =  \pm 6\sqrt 2 \end{array}\)     

Option D


Question 2:
For x ≠ 1 and x ≠ -1, simplify the following expression: \(\frac{{{\rm{(}}{{\rm{x}}^{\rm{3}}} + 1)({{\rm{x}}^3} - 1)}}{{({{\rm{x}}^2} - 1)}} \)
[1] x4 + x2 + 1
[2] x4 + x3 + x + 1
[3] x6 – 1
[4] x6 + 1
Put x = 2, we get 9×7/3 = 21

We get 21 from x4 + x2 + 1. Option A


Question 3:
If √x + √y = 6 and xy = 4 then for: x>0, y>0 give the value of x+y
[1] 2
[2] 28
[3] 32
[4] 34
\(\begin{array}{l}\sqrt x  + \sqrt y  = 6\\x + y + 2\sqrt {xy}  = 36\\x + y + 4 = 36\\x + y = 32\end{array}\)

Option C


Question 4:
Find a for which a<b and \(\sqrt {1 + \sqrt {21 + 12\sqrt 3 } } = \sqrt a + \sqrt b \)
[1] 1
[2] 3
[3] 4
[4] None of these
\(\begin{array}{l}\sqrt {1 + \sqrt {21 + 12\sqrt 3 } }  = \sqrt {1 + \sqrt {9 + 12 + 2 \times 3 \times 2\sqrt 3 } } \\ = \sqrt {1 + \sqrt {9 + 12 + 2 \times 3 \times 2\sqrt 3 } }  = \sqrt {1 + \sqrt {{{(3 + 2\sqrt 3 )}^2}} }  = \sqrt {1 + 3 + 2\sqrt 3 } \\ = \sqrt {{{(1 + \sqrt 3 )}^2}}  = 1 + \sqrt 3 \end{array}\)

Since a < b, b = 3 & a = 1. Option A


Question 5:
One root of the following given equation \(2{x^5} - 14{x^4} + 31{x^3} - 64{x^2} + 19x + 130 = 0 \) is
[1] 1
[2] 3
[3] 5
[4] 7
f(x) = \(2{x^5} - 14{x^4} + 31{x^3} - 64{x^2} + 19x + 130 = 0\)

f(1) = 2-1+31-64+19+130 = 117

f(3)= 2(3)5 – 14(3)4 + 31(3)-64(3)2 + 19(3) + 130 = -200

f(5) = 2(5)5 -14(5)4 + 31(5)-64(5)2 + 19(5) + 130 =  0

f(7) = 2(7)5 -14(7)4 + 31(7)-64(7)2 + 19(7) + 130 = 7760

Therefore the root in the above equation is 5. Option C


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Algebra Practice Questions for CAT with Solutions
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