# Simple Equations Problems for CAT PDF

0
7803

## Simple Equations Problems for CAT PDF:

Download important CAT Simple Equations Questions PDF based on previous asked questions in CAT and other MBA exams. Top 25 Simple Equation questions for CAT quantitative aptitude.

Question 1: A box contains 5 apples, 7 oranges and 11 pineapples. How many fruits should one pick to have atleast 4 fruits of the same kind?

a) 9

b) 10

c) 11

d) 12

Question 2: What is the total number of natural number solutions to 3X+2Y = 50 if X < Y?

a) 10

b) 6

c) 4

d) 2

Question 3: Ajay has 131 coins consisting of one rupee, 50 paisa and 25 paisa coins. The total value of the coins is Rs. 65. If the 50 paisa and 25 paisa coins are interchanged, the value comes down by Rs. 3. Find the number of 1 rupee coins he has?

a) 23

b) 31

c) 19

d) 27

Question 4: If the boys of a class wear one shirt each, 4 shirts remain extra. If, each boy wears two shirts each, two boys are left shirtless. How many boys are there in the class?

a) 6

b) 7

c) 8

d) 10

Question 5: The cost of 5 oranges, 4 apples and 6 tomatoes is Rs. 100 while the cost of 10 oranges, 6 apples and 9 tomatoes is Rs. 180. What is the cost of 8 oranges?

a) Rs. 96

b) Rs. 64

c) Rs. 80

d) Can’t be determined

Question 6: 20 teams play in a hockey tournament and get 3 points for a win and 2 points each for a draw. If each team plays the other exactly once and the total number of points won by the teams is 530, how many games were drawn?

a) 30

b) 40

c) 50

d) 60

Question 7: How many solutions does the equation |3x – 5| = |x-2| + x have?

a) 0

b) 1

c) 2

d) 3

Question 8: For what value of k do the following equations have no solution: 3x + 4y = 24 and 10x + ky = 75?

a) 12

b) 9

c) 13.33

d) None of these

Question 9: A carrot, costing Rs. 30, has 50 units of Vitamin A and a beetroot, costing Rs. 40, has 60 units of Vitamin A. How should Ram spend Rs. 1200 to gain the maximum amount of Vitamin A?

a) 20 carrots and 15 beetroots

b) 0 carrots and 30 beetroots

c) 40 carrots and 0 beetroots

d) None of these

Question 10: 5 Cabbages, 12 Coconuts and 19 Brinjals cost Rs. 275. 23 Cabbages, 19 Coconutes and 15 Brinjals cost Rs.360. How much do 1 Coconut, 1 Brinjal and 1 Cabbage cost?

a) 17

b) 18

c) 20

d) Can’t be determined

Question 11: A man can make 30 bricks a day while a woman can make 20 bricks a day. The fixed pay for a man and woman is Rs. 300 and Rs. 250 per day respectively. The variable pay per brick made for each is Rs. 10 and Rs. 12. If a total of 600 bricks should be made in a day, how many men should the firm employ to minimize cost?

a) 30

b) 20

c) 29

d) 19

Question 12: How many integral solutions are there for the equation X+Y+X*Y=0

a) 0

b) 1

c) 2

d) 3

Question 13: Which of the following conditions should a, b and c solve for the set of equations given below to have atleast one solution?
X+Y+Z=a;
2X-5Y+6Z=b and
11X-3Y+19Z=c

a) 7a +2b – c =0

b) 2a+7b – c =0

c) 7a + 2b + c =0

d) 2a + b -7c =0

Question 14: The number of integral solutions to the equation 3X+5Y = 216 given that 20 < X < 216 and 3 < Y < 216 is?

a) 9

b) 10

c) 11

d) 12

Question 15: The cost of 7 apples, 10 oranges and 13 pineapples is Rs. 25. The cost of 17 apples, 21 oranges and 25 pineapples is Rs.41. What is the cost of one apple, one orange and one pineapple?

a) 1.0

b) 0.80

c) 1.5

d) 0.75

Question 16: What is the maximum number which cannot be expressed as 3X + 11Y for any two whole numbers X and Y?

a) 19

b) 27

c) 41

d) 37

Question 17: A box contains 3 apples, 5 oranges and 13 pineapples. Atleast how many fruits should one pick to have atleast 6 fruits of the same kind?

a) 9

b) 11

c) 14

d) None of these

Question 18: Ajay and Vijay have Rs. 68 each in coins. They have coins of denomination 10, 5 and 1 only. If Ajay has the maximum number of coins possible and Vijay has the minimum number of coins possible, what is the total number of coins both have?

a) 68

b) 78

c) 72

d) None of these

Question 19: For what value of ‘m’, does the following system of equations not have a unique solution? X + 2Y + 3Z = 10
4X + 5Y -6Z = 12
6X + 8Y -m*Z = 23

a) 12

b) -6

c) 6

d) -4

Question 20: After the impending birth of identical twins, the percentage of boy grandchildren of Old Johnny will be either 33.33% or 44.44%. Find the total number of grandchildren Old Johnny has right now (excluding twins).

a) 16

b) 20

c) 24

d) Can’t be determined

Question 21: In olden days, there were only 4 paisa, 7 paisa and 11 paisa coins. What is the maximum amount you can’t exactly pay using just the three sets of coins?

a) 23 paisa

b) 17 paisa

c) 19 paisa

d) 29 paisa

Question 22: In an examination containing 100 questions, a student gets 1 mark for a right answer, -1/3 for a wrong answer and -1/4 for an unattempted answer. If a student scores 80 marks in the examination, how many questions did he answer wrong?

a) 4

b) 6

c) 0

d) Can’t be determined

Question 23: In a football team having ‘n’ teams each, all the teams play each other twice and get 3 points for a win and 1 point each for a draw. If the total number of points earned by the teams is 217, how many games were drawn?

a) 37

b) 53

c) 71

d) Can’t be determined

Question 24: While multiplying three natural numbers, Rajdeep used 75 as one of the numbers instead of 57. As a result, the product went up by 360. What is the value of the correct product?

a) 1040

b) 1140

c) 1500

d) Can’t be determined

Instructions

An airline has a certain free luggage allowance and charges for excess luggage at a fixed rate per kg. Two passengers, Raja and Praja have 60 kg of luggage between them, and are charged Rs 1200 and Rs 2400 respectively for excess luggage. Had the entire luggage belonged to one of them, the excess luggage charge would have been Rs 5400.

Question 25: What is the free luggage allowance?[CAT 2004]

a) 10 kg

b) 15 kg

c) 20 kg

d) 25 kg

e) 30 kg

The maximum number of fruits one can pick without having 4 fruits of the same kind equals 9 when x=y=z=3.
So, if one picks 10 fruits, there will be at least 4 fruits of the same kind.

The only solutions for the equation are (2,22); (4,19);(6,16) and (8,13).
So total number of solutions is 4.

Let X be the number o f 1 rupee coin, Y be the 50 paise coin and Z be the 25 paise coin.
$\text{X+Y+Z} = 131$,
$\text{X}+\frac{\text{Y}}{2}$+$\frac{\text{Z}}{4} = 65$,
$\text{X} + \frac{\text{Y}}{4}$ +$\frac{\text{Z}}{2} = 62$.
So, $\text{X} = 23$

2*(x-2) = x+4. So, x = 8

5a + 4b + 6c = 100 –> (1)
and 10a+6b+9c = 180 –> (2)
(2) – 1.5*(1) gives a = 12

Total number of games is 20C2 = 190. Let x be the number of drawn games, so 2*x + 3*(190-x) = 530. So, x =40

If x> 2, the equation becomes, 3x-5 = x-2 + x or x=3
If 5/3 < x < 2, the equation becomes 3x-5 = 2-x+x or x = 7/3. However, 7/3 > 2 and hence is not a valid solution
If x < 5/3, the equation becomes 5 – 3x = 2 or x = 1.
So, valid solutions for x are 1 and 3.

For the system to have no solutions, the determinant of the matrix should be zero. So, 4*10 = 3*k. So, k = 13.33

1 rupee spent on a carrot gives 1.66 units of Vitamin A, while 1 rupee spent on a beetroot gives 1.5 units of Vitamin A.
So, 40 carrots and 0 beetroots give the maximum amount of Vitamin A

5X+12Y+19Z=275 –> (1)
23X+19Y+15Z = 360 –>(2)

We note that the coefficients of the first equation differ by 7 and the coefficients of the second equation differ by 4. So, multiplying the first by 4 and the second by 7, 4*(1) + 7*(2) gives X+Y+Z =20

A man makes 30 bricks a day for Rs. 600 while a woman makes 20 bricks a day for Rs. 490. So, it is economical to hire a man. Hence, the total number of men needed is 600/30=20

$X = – \frac{Y}{Y+1}$. This is an integer only when Y = 0 or Y = -2. Hence, the only solutions are (0,0) and (-2,-2)

For the set of equations to have a consistent solution, the determinant of the below matrix should be zero. The determinant of $\begin{bmatrix}a & 1 & 1 \\ b & -5 & 6 \\ c & -3 & 19 \end{bmatrix}$ is zero.
So, 7a+2b=c

The given equation 3X+5Y = 216 can be written as 3X=216-5Y => X=72-(5/3)Y.
For X to be an integer Y should be a multiple of 3. So let Y be ‘3k’ => X=72-5k.
Thus for all X of the form 72 – 5k and Y of the form 3k, the equation above has a solution.
Based on the conditions given, k>1 and k<11.
So, number of integral solutions is 9.

7X + 10Y + 13Z = 25 -> (1) 17X + 21Y+25Z = 41 -> (2) The difference in the coefficients of equation (1) is 3 and the difference in the coefficients of equation (2) is 4. So, 4*(1) – 3*(2) gives X+Y+Z =1

11 = 2 (mod 3). So, any number greater than 11 and of the form 3k+2 can be expressed as 3X + 11Y. Similarly, 22 = 1 (mod 3), so any number greater than 22 which is of the form 3k+1 can be expressed as 3X+11Y. Hence, all numbers greater than 22 can be expressed as 3X+11Y. The largest number less than 22 which can’t be expressed as 3X+11Y is 19 which is the answer.

6 fruits of the same kind can be only pineapples. So, the minimum number of fruits to be pick to be certain of having 6 pineapples is 3 (apples) + 5 (oranges) + 6 (pineapples) = 14.

Maximum number of coins possible is 68 (all 1 rupee coins) Minimum number of coins possible is 6 (10 rupee coins) + 1 (5 rupee) + 3 (1 rupee) = 10 So total is 78

The system will have no unique solution if the determinant of the matrix $\begin{bmatrix}1&2&3\\4&5&-6 \\6&8&-m \end{bmatrix}$ is zero, $\implies 6*(-12-15) -8(-6-12) -m(5-8) = 0$ $\implies m = 6$

Let Old Johnny have x boy and y girl grandchildren right now. So, $\frac{x+2}{x+y+2} = \frac{4}{9}$ and $\frac{x}{x+y+2} = \frac{1}{3}$. So, x = 6 and y =10.

4 = 0(mod 4) 7 = 3 (mod 4) 14 = 2(mod 4) and 21 = 1(mod 4)

Any amount greater than 21 paisa, can be paid using these three coins. For example, if the amount leaves a remainder of 0 when divided by 4, we can pay using 4 paisa coins only. If it leaves a remainder of 1 when divided by 4, we can pay three 7 paisa coins and the rest by 4 paisa coins. Similarly if it leaves a remainder of 2 when divided by 4, we can pay two 7 paisa coins and the rest by 4 paisa coins. Similarly if it leaves a remainder of 3 when divided by 4, we can pay one 7 paisa coin and the rest by 4 paisa coins.

Of the numbers less than 21 paisa, the maximum we can’t pay using these coins is 17 paisa coins.

100 – x -y – x/3 – y/4 = 80. 15x + 16y = 240. This has two integer solutions (0,15) and (16,0). Hence, we can’t determine the number of answers the student got wrong.

Let x represent the number of wins and y represent the number of draws.

For every win, 3 point will be added to 1 team and 0 to other. Hence, net increase in points = 3x.

For every draw, 1 point will be added to both the teams. So, net increase = 2y.
3x+2y = 217 and x+y = n*(n-1) for some n. The only integral solutions to the above equation are x=37 and y = 53 and n=10