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:**

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**Question 1:Â **

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:Â **

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.**Â (6, 1)

**B.**Â (â€“3, â€“4)

**C.**Â (4, 3)

**D.**Â (â€“4, â€“3)

**Question 4:**

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

A.Â -2/5

B.Â 2/17

C.Â 10/17

D.Â None of these

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**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 thirdÂ 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.

Hope you find this CAT Algebra Questions useful. You can also download all our CAT Quantitative Aptitude Questions PDF.