Algebra Practice Questions for CAT with Solutions

Question 1:
The price of Darjeeling tea (in rupees per kilogram) is 100 + 0.10 n, on the nth day of 2007 (n = 1, 2, ..., 100), and then remains constant. On the other hand, the price of Ooty tea (in rupees per kilogram) is 89 + 0.15n, on the nth day of 2007 (n = 1, 2, ..., 365). On which date in 2007 will the prices of these two varieties of tea be equal?
[1] May 21
[2] April 11
[3] May 20
[4] April 10
100 + 0.1n = 89 + 0.15n
0.05n = 11
n = 220

But for Darjeeling tea n cannot be more than 100.

Maximum price of Darjeeling tea = 100 + 0.1×100 = 110

Price of Ooty tea should also be 110

89 + 0.15n = 110
n = 140

On the 140th day of 2007, the prices of the Darjeeling tea and Ooty tea will be equal

140 = 31 (Jan) + 28 (Feb) + 31 (March) + 30 (April) + 20 (May)

So, price will be equal on 20th May. Option C


Question 2:
The polynomial f(x)=x2-12x+c has two real roots, one of which is the square of the other. Find the sum of all possible value of c.
[1] -37
[2] -12
[3] 25
[4] 91
Let the roots be r and r2

Sum of the roots = r2 + r = 12

r2 + r – 12 = 0
r = -4, 3

Product of the roots = c = r3 = -64 or 27

Sum of values of c = -64 + 27 = -37. Option A


Question 3:
Two sides of a triangle have lengths 10 and 20. How many integers can take the value of the third side length:
[1] 18
[2] 19
[3] 20
[4] 21
Let the third side be x

 20 – 10 < x < 10 + 20

10 < x < 30

No. of integral values of x = 19. Option B


Question 4:
Which of the following is a solution to: \(6{\left( {x + \frac{1}{x}} \right)^2} - 35\left( {x + \frac{1}{x}} \right) + 50 = 0 \)
[1] 1
[2] 1/3
[3] 4
[4] 6
Let \(x + \frac{1}{x} = y\)
6y2 – 35y + 50 = 0
(3y – 10)(2y – 5) = 0
y = 10/3, 5/2

From the options only valid value of x = 1/3. Option B


Question 5:
Find x if \(\frac{5}{{3 + \frac{5}{{3 + \frac{5}{{3 + ...}}}}}} = x. \) \( \)
[1] \(\frac{{ - 3 + \sqrt {29} }}{2} \)
[2] \(\frac{{3 + \sqrt {29} }}{2} \)
[3] \(\frac{{ - 1 + \sqrt 5 }}{2} \)
[4] \(\frac{{1 + \sqrt 5 }}{2} \)
\(\begin{array}{l}\frac{5}{{3 + \frac{5}{{3 + \frac{5}{{3 + ...}}}}}} = x\\ \Rightarrow \frac{5}{{3 + x}} = x\\ \Rightarrow 5 = 3x + {x^2}\\ \Rightarrow {x^2} + 3x - 5 = 0\\ \Rightarrow x = \frac{{ - 3\; \pm \;\sqrt {{3^2} - 4 \times 1 \times ( - 5)} }}{2}\\ \Rightarrow x = \frac{{ - 3\; \pm \;\sqrt {29} }}{2}\\ \Rightarrow x = \frac{{ - 3\; + \;\sqrt {29} }}{2}\end{array}\)

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Algebra Practice Questions for CAT with Solutions
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