# CAT Algebra Questions

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CAT Algebra questions will be mostly based on Linear equations, Quadratic equations and functions for CAT. This is one of the practice set on CAT Algebra topic.

CAT Algebra Questions:

Question 1:

A change-making machine contains one-rupee, two-rupee and five-rupee coins. The total number of coins is 300. The amount is Rs. 960. If the numbers of one-rupee coins and two-rupee coins are interchanged, the value comes down by Rs. 40. The total number of five-rupee coins is

A. 100

B. 140

C. 60

D. 150

Question 2:

The owner of a local jewellery store hired three watchmen to guard his diamonds, but a thief still got in and stole some diamonds. On the way out, the thief met each watchman, one at a time. To each he gave 1/2 of the diamonds he had then, and 2 more besides. He escaped with one diamond. How many did he steal originally?

A. 40

B. 36

C. 25

D. None of these

Question 3:

Ujakar and Keshab attempted to solve a quadratic equation. Ujakar made a mistake in writing down the constant term. He ended up with the roots (4, 3). Keshab made a mistake in writing down the coefficient of x. He got the roots as (3, 2). What will be the exact roots of the original quadratic equation?

A. (6, 1)

B. (–3, –4)

C. (4, 3)

D. (–4, –3)

Question 4:

If | r – 6 | = 11 and | 2q – 12 | = 8, what is the minimum possible value of q / r?

A. -2/5

B. 2/17

C. 10/17

D. None of these

Question 5:

The number of solutions of the equation 2x + y = 40 where both x and y are positive integers and x <= y is:

A. 7

B. 13

C. 14

D. 18

Solutions:

1) Answer (B)

Let the number of coins of the three denominations be x, y and z respectively.
x+y+z = 300
x+2y+5z = 960
2x+y+5z = 920
=> 3(x+y) + 10z = 1880
=> 3(300 – z) + 10z = 1880
=> 900 + 7z = 1880 => z = 980/7 = 140
So, the number of 5 rupee coins is 140

2) Answer (B)

Suppose the thief stole ‘x’ diamonds.
After giving the share to the first watchman, the thief has (x/2)-2 diamonds.
After giving to the second watchman, the thief has (x/4)-3 diamonds.
After giving to the third watchman, the thief has (x/8)-(7/2) diamonds.
This is equal to 1. So, (x/8) – 7/2 = 1
Solving this equation, we get x = 36

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3) Answer (A)

e know that quadratic equation can be written as $x^2$-(sum of roots)*x+(product of the roots)=0.
Ujakar ended up with the roots (4, 3) so the equation is $x^2$-(7)*x+(12)=0 where the constant term is wrong.
Keshab got the roots as (3, 2) so the equation is $x^2$-(5)*x+(6)=0 where the coefficient of x is wrong.
So the correct equation is $x^2$-(7)*x+(6)=0. The roots of above equations are (6,1).

4) Answer (D)

| r-6 | = 11 => r = -5 or 17
| 2q – 12 | = 8 => q = 10 or 2
So, the minimum possible value of q/r = 10/(-5) = -2

5) Answer (B)

y = 38 => x = 1
y = 36 => x = 2
y = 14 => x = 13
y = 12 => x = 14 => Cases from here are not valid as x > y.
Hence, there are 13 solutions.

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