CAT Previous Year Algebra Questions

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1420 CAT Previous Year Algebra Questions

Download important CAT Algebra Problems with Solutions PDF based on previously asked questions in CAT exam. Practice Algebra Problems with Solutions for CAT exam.

Question 1: Every 10 years the Indian Government counts all the people living in the country. Suppose that the director of the census has reported the following data on two neighbouring villages Chota Hazri and Mota Hazri.
Chota Hazri has 4,522 fewer males than Mota Hazri.
Mota Hazri has 4,020 more females than males.
Chota Hazri has twice as many females as males.
Chota Hazri has 2,910 fewer females than Mota Hazri.
What is the total number of males in Chota Hazri?

a) 11,264

b) 14,174

c) 5,632

d) 10,154

Question 2: Three travellers are sitting around a fire, and are about to eat a meal. One of them has 5 small loaves of bread, the second has 3 small loaves of bread. The third has no food, but has 8 coins. He offers to pay for some bread. They agree to share the 8 loaves equally among the three travellers, and the third traveller will pay 8 coins for his share of the 8 loaves. All loaves were the same size. The second traveller (who had 3 loaves) suggests that he will be paid 3 coins, and that the first traveller be paid 5 coins. The first traveller says that he should get more than 5 coins. How much should the first traveller get?

a) 5

b) 7

c) 1

d) None of these

Question 3: I bought 5 pens, 7 pencils and 4 erasers. Rajan bought 6 pens, 8 erasers and 14 pencils for an amount which was half more what I had paid. What per cent of the total amount paid by me was paid for the pens?

a) 37.5%

b) 62.5%

c) 50%

d) None of these

Question 4: Mrs. Sonia buys Rs. 249.00 worth of candies for the children of a school. For each girl she gets a strawberry flavoured candy priced at Rs. 3.30 per candy; each boy receives a chocolate flavoured candy priced at Rs. 2.90 per candy. How many candies of each type did she buy?

a) 21, 57

b) 57, 21

c) 37, 51

d) 27, 51

Question 5: Mother Dairy sells milk packets in boxes of different sizes to its vendors. The vendors are charged Rs. 20 per packet up to 2000 packets in a box. Additions can be made only in a lot size of 200 packets. Each addition of one lot to the box results in a discount of one rupee an all the packets in the box. What should be the maximum size of the box that would maximize the revenue per box for Mother Dairy?

a) 2400 packets

b) 3000 packets

c) 4000 packets

d) None of the above

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Question 6: The number of solutions $(x, y, z)$ to the equation $x – y – z = 25$, where x, y, and z are positive integers such that $x\leq40,y\leq12$, and $z\leq12$ is

a) 101

b) 99

c) 87

d) 105

Question 7: In a school, students were called for the Flag Hoisting ceremony on August 15. After the ceremony, small boxes of sweets were distributed among the students. In each class, the student with roll no. 1 got one box of sweets, student with roll number 2 got 2 boxes of sweets, student with roll no. 3 got 3 boxes of sweets and so on. In class III, a total of 1200 boxes of sweets were distributed. By mistake one of the students of class III got double the sweets he was entitled to get. Identify the roll number of the student who got twice as many boxes of sweets as compared to his entitlement.

a) 22

b) 24

c) 28

d) 30

Question 8: A survey shows that 61%, 46% and 29% of the people watched “3 idiots”, “Rajneeti” and “Avatar” respectively. 25% people watched exactly two of the three movies and 3% watched none. What percentage of people watched all the three movies?

a) 39%

b) 11%

c) 14%

d) 7%

Question 9: A milk vendor sells 10 litres of milk from a can containing 40 litres of pure milk to the 1st customer. He then adds 10 litres of water to the milk can. He again sells 10 litres of mixture to the 2nd customer and then adds 10 litres of water to the can. Again he sells 10 litres of mixture to the 3rd customer and then adds 10 litres of water to the can and so on. What amount of pure milk will the 5th customer receive?

a) 510/128 litres

b) 505/128 litres

c) 410/128 litres

d) 405/128 litres

Question 10: A car rental agency has the following terms. If a car is rented for 5 hr or less, then, the charge is Rs. 60 per hour or Rs. 12 per kilometre whichever is more. On the other hand, if the car is rented for more than 5 hr, the charge is Rs. 50 per hour or Rs. 7.50 per kilometre whichever is more. Akil rented a car from this agency, drove it for 30 km and ended up playing Rs. 300. For how many hours did he rent the car?

a) 4 hr

b) 5 hr

c) 6 hr

d) None of these

Let the number of males in Mota Hazri = x
No. of males in Chota Hazri = x – 4522
Let the number of females in Mota Hazri = y
No. of females in Chota Hazri = y – 2910
(y – 2910) = 2(x – 4522) => y = 2x – 9044 + 2910 = 2x – 6134
Also y = x + 4020
So, x + 4020 = 2x – 6134 => x = 10154
So, number of males in Chota Hazri = 10154 – 4522 = 5632

Suppose A, B and C have 5 pieces of bread, 3 pieces of bread and 8 coins respectively. Since in total there are 8 pieces of bread, each one should get around 2.66 bread. So A must give 2.33 part of his bread to C and B must give 0.33. Distributing the amount in the same ratio of bread contribution, A must get 7 coins and B must get 1 coin.

Let the cost of pen, pencil and eraer be x,y,z respectively

5x+7y+4z = A

6x+8z+14y = 3A/2

4x + 16/3 z + 28/3 y = A

Comparing two equations

5x+7y+4z = 4x+16/3 z + 28/3 y

x = 7/3 y + 4/3 z

3x = 7y+4z

Now required percentage = $\frac{5x}{5x+7y+4z}\times100=\frac{5x}{5x+3x}=62.5$%

Let the number of strawberry flavoured candy = x and the number of chocolate flavoured candy = y
Thus, as given in the question:-
3.3x + 2.9y = 249
Using options,
Option A = 3.3*21 + 2.9*57 = 69.3+165.3 = 234.6 Rs $\neq$ 249
Hence, option A cannot be the answer.
Option B = 3.3*57 + 2.9*21 = 188.1+60.9 = 249
Thus, option B can be the answer.
Option C = 3.3*37 + 2.9*57 = 241.8 $\neq$ 249
Hence, option C cannot be the answer.
Option D = 3.3*27 + 2.9*51 = 237 $\neq$ 249
Thus, option D cannot be the answer.

Hence, option B is the correct answer.

We are given that, The vendors are charged Rs. 20 per packet up to 2000 packets in a box. Additions can be made only in a lot size of 200 packets. Each addition of one lot to the box results in a discount of one rupee an all the packets in the box.
Let x be number of additional lots.
Thus,
$(20-x)(2000+200x) = 40000 – 2000x + 4000x -200x^2$
=> $-200x^2+2000x+40000$
We need to maximize this
The minimum/maximum value of a quadratic equation is when $x = -b/2a$
Thus, the maximum value = $-2000/400 = 5$
Thus, the maximum size of the box that would maximize the revenue per box for Mother Dairy = 2000+200*5 = 3000

x – y – z = 25 and $x\leq40,y\leq12$, $z\leq12$
If x = 40 then y + z = 15. Now since both y and z are natural numbers less than 12, so y can range from 3 to 12 giving us a total of 10 solutions.Similarly, if x = 39, then y + z = 14. Now y can range from 2 to 12 giving us a total of 11 solutions.
If x = 38, then y + z = 13. Now y can range from 1 to 12 giving us a total of 12 solutions.
If x = 37 then y + z = 12 which will give 11 solutions.
Similarly on proceeding in the same manner the number of solutions will be 10, 9, 8, 7 and so on till 1.
Hence, required number of solutions will be (1 + 2 + 3 + 4 . . . . + 12) + 10 + 11
= 12*13/2 + 21
78 + 21 = 99

1+2+3+4+………………+n < 1200

and

1+2+3+4+………………+n+ (n+1) > 1200

We need to calculate value of n.

Using first and second inequality,

$\frac{n (n+1)}{2}$ < 1200

N = 48

Total number of boxes of sweets distributed is: 48*49/2 = 1176

So, roll no 24 got doubles number of boxes.

Let the number of people who watched exactly one movie be ‘a’, the number of people who watched exactly two movies be ‘b’ and the number of people who watched exactly three movies be ‘c’
Then, a + b + c = 100 – 3 = 97 ……………(i)
also, a + 2b + 3c = 61 + 46 + 29 = 136…………………..(ii)
On subtracting (i) from (ii), we get
b + 2c = 39
It is given that, b = 25
So, c = 7
Therefore, 7% people watched all the three movies.
Hence, option D is the correct answer.

The amount of pure milk received by 5th customer = amount of pure milk after 4 replacements – amount of milk after 5 replacements

Formula for amount of milk after n replacements is : 40 $(1 – \frac{10}{40})^{n}$

Amount of milk received by 5th customer = 40 $(1 – \frac{10}{40})^{4}$ – 40 $(1 – \frac{10}{40})^{5}$

=> 40 * $(\frac{3}{4})^{4} (1 – \frac{3}{4})$

=> $\frac{405}{128}$ lts.