CAT 2024 Arithmetic Questions with Answers

CAT 2024 Arithmetic Questions with Answers: Arithmetic is a very important part of the quant section of the CAT 2024 exam, carrying significant weightage. Quantitative Aptitude section is the final section, which you should aim to complete in 40 minutes. This section has 22 questions on advanced-level mathematics topics from Class 9 and 10. Out of the total questions, students can expect around 9-10 arithmetic questions. The difficulty level of the questions can be quite high covering subjects like arithmetic progressions, averages, ratios, time & work, profit & loss, simple interest, and percentage. Candidates can check here 10+ sample CAT 2024 arithmetic questions and answers, based on previous year question papers.

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CAT 2024 Arithmetic Questions with Answers

Check out some of the top CAT 2024 Arithmetic questions and answers:
1. The price of a precious stone is directly proportional to the square of its weight. Sita has a precious stone weighing 18 units. If she breaks it into four pieces with each piece having a distinct integer weight, then the difference between the highest and lowest possible values of the total price of the four pieces will be 288000. Then, the price of the original precious stone is

(a) 1944000
(b) 972000
(c) 1620000
(d) 1296000

Solution

It is given that the price of a precious stone is directly proportional to the square of its weight. Let the price be denoted by C and the weight is denoted by W.

Hence,

C ∝ W2=> C =kw2  (where k is the proportional constant)

Now, Sita has a precious stone weighing 18 units.

Therefore, C = kw2= k . 182 = 324

If she breaks it into four pieces with each piece having a distinct integer weight, then the difference between the highest and lowest possible values of the total price of the four pieces will be 288000.

To get the lowest possible value of C, we will get the weight of the four-piece as close as possible (3,4,5,6). To get the highest value we will try to take three pieces as low as possible, and one is as high as possible (1, 2, 3, 12).

Hence, the maximum cost = k(122+12+22+32) = 158k2, and the minimum cost is k(32+42+52+62) = 86k2

Hence, the difference is (158k2−86k2)=72k2, which is equal to 288000.

=> 72k2 = 288000

=> k2 = 4000

Hence, the price of the original stone is 324k2=324× 4000 = 1296000

The correct option is D

2. Working together, Rahul, Rakshita, and Gurmeet would have taken more than 7 days to finish a job. On the other hand, Rahul and Gurmeet, working together, would have taken less than 15 days to finish the job. However, they all worked together for 6 days, followed by Rakshita, who worked alone for 3 more days to finish the job. If Rakshita had worked alone on the job then the number of days she would have taken to finish the job, cannot be

(a) 20
(b) 17
(c) 16
(d) 21

Solution

Let the work done by Rahul, Rakshita, and Gurmeet be a, b, and c units per day, respectively, and the total units of work are W.

Hence, we can say that 7(a+b+c) < W ( Rahul, Rakshita, and Gurmeet, working together, would have taken more than 7 days to finish a job).

Similarly, we can say that 15(a+c) > W ( Rahul and Gurmeet, working together would have taken less than 15 days to finish the job)

Now, comparing these two inequalities, we get: 7(a+b+c) < W < 15(a+c)

It is also known that they all worked together for 6 days, followed by Rakshita, who worked alone for 3 more days to finish the job. Therefore, the total units of work done is: W = 6(a+b+c)+3b

Hence, we can say that 7(a+b+c) < 6(a+b+c)+3b < 15(a+c)

Therefore, (a+b+c) < 3b => a+c < 2b, and 9b < 9(a+c) => b < a+c

=> a+b+c < 3b => 7(a+b+c) < 21b , and 15b < 15(a+c)

Hence, The number of days required for b must be in between 15 and 21 (both exclusive).

Hence, the correct option is D

3. Meenu purchases a pair of sunglasses at Rs.1000 and sells to Kanu at 20% profit. Then, Kanu sells it back to Meenu at a 20% loss. Finally, Meenu sells the same pair of sunglasses to Tanu. If the total profit made by Meenu from all her transactions is Rs.500, then the percentage of profit made by Meenu when she sold the pair of sunglasses to Tanu is

(a) 35.42%
(b) 52%
(c) 31.25%
(d) 26%

Solution

The cost price of the sunglasses for Meenu when he purchased it for the first time was 1000 rupees, and he sold it to Kanu at 20% profit. Hence, the selling price of the sunglasses is 1200 rupees, which Kanu purchased. Hence, the profit made by Meenu is (1200-1000) = 200 rupees.

Hence, the cost price of the same sunglass for Kanu is 1200 rupees, and now he sold it to Meenu at a 20% loss. Hence, the selling price of the sunglass now is (1200*0.8) = 960 rupees.

The cost price of the same sunglasses for Meenu when he purchased it for the second time was 960 rupees. Now Meenu sold it Tanu, at a certain price such that the total profit of Meenu becomes 500 rupees.

Hence, on the second transaction (selling it to Tanu), Meenu made a profit of (500-200) = 300 rupees.

Hence, the profit made by Meenu in the second transaction is (300/960)*100% = 31.25%

The correct option is C

4. Anil mixes cocoa with sugar in the ratio 3 : 2 to prepare mixture A, and coffee with sugar in the ratio 7 : 3 to prepare mixture B. He combines mixtures A and B in the ratio 2:3 to make a new mixture C. If he mixes C with an equal amount of milk to make a drink, then the percentage of sugar in this drink will be

(a) 17
(b) 16
(c) 21
(d) 24

Solution

Let the volume of mixture A be 200 ml, which implies the quantity of cocoa in the mixture is 120 ml, and the quantity of sugar In the mixture 80 ml.

Similarly, let the volume of the mixture be 300 ml, which implies the quantity of coffee, and sugar in the mixture is 210, and 90 ml, respectively.

Now we combine mixture A, and B in the ratio of 2:3 (if 200 ml mixture A, then 300 ml of mixture B).

Hence, the volume of the mixture C is (200+300) = 500 ml, and the quantity of the sugar is (90+80) = 170 ml.

Now he mixes C with an equal amount of milk to make a drink, which implies the quantity of the final mixture is (500+500) = 1000 ml.

The quantity of sugar in the final mixture is 170 ml.

Hence, the percentage is 17%

The correct option is A

5. A merchant purchases a cloth at a rate of Rs.100 per meter and receives 5 cm length of cloth free for every 100 cm length of cloth purchased by him. He sells the same cloth at a rate of Rs.110 per meter but cheats his customers by giving 95 cm length of cloth for every 100 cm length of cloth purchased by the customers. If the merchant provides a 5% discount, the resulting profit earned by him is

(a) 4.2%
(b) 9.7%
(c) 15.5%
(d) 16%

Solution

It is given that a merchant purchases a cloth at a rate of Rs.100 per meter and receives 5 cm length of cloth free for every 100 cm length of cloth purchased by him.

Hence, the cost price of 105 cm clothes is 100 rupees.

It is also known that he marked the price of 100 cm clothes as 110 rupees, and gave a 5% discount, and he cheated his customers by giving 95 cm length of cloth for every 100 cm length of cloth purchased by the customers.

Hence, the selling price of 95 cm clothes is 110*(19/20) rupees.

Therefore, the selling price of 105 cm clothes is 115.5 rupees.

Hence, the profit is 15.5%

The correct option is C

6. In a company, 20% of the employees work in the manufacturing department. If the total salary obtained by all the manufacturing employees is one-sixth of the total salary obtained by all the employees in the company, then the ratio of the average salary obtained by the manufacturing employees to the average salary obtained by the non-manufacturing employees is

(a) 6:5
(b) 4:5
(c) 5:4
(d) 5:6

Solution

Let the number of total employees in the company be 100x, and the total salary of all the employees be 100y.

It is given that 20% of the employees work in the manufacturing department, and the total salary obtained by all the manufacturing employees is one-sixth of the total salary obtained by all the employees in the company.

Hence, the total number of employees in the manufacturing department is 20x, and the total salary received by them is (100y/6)

Average salary in the manufacturing department = (100y/6*20x) = 5y/6x

Similarly, the total number of employees in the nonmanufacturing department is 80x, and the total salary received by them is (500y/6)

Hence, the average salary in the nonmanufacturing department = (500y/6*80x) = 25y/24x

Hence, the ratio is:- (5y/6x): (25y/24x)

=> 120: 150 = 4:5

The correct option is B

7. Ravi is driving at a speed of 40 km/h on a road. Vijay is 54 meters behind Ravi and driving in the same direction as Ravi. Ashok is driving along the same road from the opposite direction at a speed of 50 km/h and is 225 meters away from Ravi. The speed, in km/h, at which Vijay should drive so that all the three cross each other at the same time, is

(a) 58.8
(b) 67.2
(c) 61.6
(d) 64.4

Solution

It is given that the speed of Ravi is 40 kmph, which is equal to 100/9 m/s. It is also known that the speed of Ashok is 50 kmph, which is equal to 125/9 m/s.

It is known that the distance between Ravi and Ashok is 225 meters, and the relative speed of Ravi and Ashok is 100/9 + 125/9 = 25 m/s

Hence, they will meet each other in 225/25=9 seconds. The distance traveled by Ravi in these 9 seconds is 100/9× 9=100 meters.

Since Vijay was already 54 meters behind Ravi when they were starting, Vijay must travel (100+54) = 154 meters in these 9 seconds.

Hence, the speed of Vijay is 154/9 m/s, which is equal to 154/9 x 18/5 = 61.6 kmph.

The correct option is C

8. The average score in an examination of 10 students of a class is 60. If the scores of the top five students are not considered, the average score of the remaining students falls by 5. The pass mark was 40 and the maximum mark was 100. It is also known that none of the students failed. If each of the top five scorers had distinct integral scores, the maximum possible score of the topper is

(a) 99
(b) 100
(c) 87
(d) 95

Solution

10 students have scored 600 marks amongst them, and no one is allowed to score lesser than 40 or higher than 100. The idea now is to maximize what the highest scorer gets. The 5 lowest scores have an average of 55, which means that they have scored 55 x 5 = 275 marks amongst them. This leaves 325 marks to be shared amongst the top 5 students.

Let's call them a, b, c, d and e. Now, to maximize what the top scorer "e" gets, all the others have to get the lowest possible scores (and at the same time, they should also get distinct integers.) The least possible score of the top 5 should be at least equal to the highest of the bottom 5.

Now we want to make sure that the highest of the bottom 5 is the least possible. This can be done by making all scores equal to 55. If some scores are less than 55, some other scores have to be higher than 55 to compensate and make the average 55. Thus the highest score is the least only when the range is 0.

So now, we have the lowest value that the top 5 can score, which is 55. The others have to get distinct integer scores, and as few marks as possible, so that “e" gets the maximum. So, 55 + 56 + 57 + 58 + e = 325 => e = 99 marks.

The correct option is C

9. Pipes A and C are fill pipes while Pipe B is a drain pipe of a tank. Pipe B empties the full tank in one hour less than the time taken by Pipe A to fill the empty tank. When pipes A, B and C are turned on together, the empty tank is filled in two hours. If pipes B and C are turned on together when the tank is empty and Pipe B is turned off after one hour, then Pipe C takes another one hour and 15 minutes to fill the remaining tank. If Pipe A can fill the empty tank in less than five hours, then the time taken, in minutes, by Pipe C to fill the empty tank is

(a) 90
(b) 120
(c) 75
(d) 60

Solution

Let the time taken by A to fill the tank alone be x hours, which implies the time taken by B to empty the tank alone is (x-1) hours (B is the drainage pipe), and the time taken by C to fill the tank is y hours.

It is given that when pipes A, B, and C are turned on together, the empty tank is filled in two hours.

Hence, 1/x - 1/x-1 + 1/y = 1/2 …… Eq(1)

It is given that if pipes B and C are turned on together when the tank is empty and Pipe B is turned off after one hour, then Pipe C takes another one hour and 15 minutes to fill the remaining tank.

Hence, B worked for 1 hour, and C worked for 2 hours 15 minutes, which is equal to 9/4 hours.

In 1 hour, B worked - 1/x-1, and in 9/4 hours, C worked 9/4y units.

Hence, 9/4y - 1/x-1 = 1…… Eq(2)

Solving both equations, we get y=3/2, and x = 3

Hence, the time taken by C is 3/2 hours, which is equal to 90 minutes.

The correct option is A

10. The minor angle between a clock's hour hand and minute hand was observed at 8:48 am. The minimum duration, in minutes, after 8.48 am when this angle increases by 50% is

(a) 36/11
(b) 2
(c) 4
(d) 24/11

Solution

The given time is 8:48 AM.

Angle made by hours hand w.r.t 12 is 8 * 30 (30 degrees in 1 hour) + 0.5 * 48 (0.5 degree in 1 minute) = 240 + 24 = 264 degrees.

Angle made by minutes hands w.r.t 12 is 48 * 6 = 288 degrees.

=> The angle between them is 288 - 264 = 24 degrees.

This should further increase by 12 degrees (50% of 24)

After m minutes, the further increase in angle = (6 - 0.5)*m = 11/2 m =12 => m = 24/11

The correct option is D

11. Brishti went on an 8-hour trip in a car. Before the trip, the car had travelled a total of x km till then, where x is a whole number and is palindromic, i.e., x remains unchanged when its digits are reversed. At the end of the trip, the car had travelled a total of 26862 km till then, this number again being palindromic. If Brishti never drove at more than 110 km/h, then the greatest possible average speed at which she drove during the trip, in km/h, was

(a) 110
(b) 90
(c) 100
(d) 80

Solution

Given the total number of kilometres travelled, including the trip = is 26862 Km, and the duration of the trip is 8 hrs.

If the average speed of the car during the trip is 's' => the km travelled till just before the trip is 26862 - 8s, which should also be a palindrome.

=> From the options if s = 110 => The reading will be 26862 - 110*8 = 25982 (Not a palindrome)

=> If s = 100 => The reading will be 26862 - 100*8 = 26062 => It is a palindrome.

=> s = 100

So, C is the correct option.

12. Arvind travels from town A to town B, and Surbhi from town B to town A, both starting at the same time along the same route. After meeting each other, Arvind takes 6 hours to reach town B while Surbhi takes 24 hours to reach town A. If Arvind travelled at a speed of 54 km/h, then the distance, in km, between town A and town B is

Solution

Let us assume the speeds of Arvind and Surbhi are 'a' and 's', respectively.

Let us say they meet after 't' hours

=> Arvind travelled s*t distance in 6 hrs and Surbhi travelled a*t in 24 hrs

=> s*t = a*6 and a*t = s*24 => t2 =6× 24 => t = 12

Given a = 54 => s*12 = 54*6 => s = 27.=> Total distance between A and B is (s+a)*t = (54+27)*12 = 81*12 = 972 Kms.

13. Anil invests Rs. 22000 for 6 years in a certain scheme with 4% interest per annum, compounded half-yearly. Sunil invests in the same scheme for 5 years, and then reinvests the entire amount received at the end of 5 years for one year at 10% simple interest. If the amounts received by both at the end of 6 years are same, then the initial investment made by Sunil, in rupees, is

Solution

Anil invested 22000 for 6 years at 4% interest compounded half-yearly

=> Amount = 22000(1.02) ⁶

Let Sunil invest 'S' rupees for 5 years at 4% C.I. half-yearly and 10% S.I. for 1 additional year

=> Amount = S(1.02) ¹⁰ (1.1)

Given that both amounts are equal

=> 22000(1.02) ¹² = S(1.02) ¹⁰ (1.1)

=> S = 22000(1.02) ² /1.1 = 20808

By practising the sample questions provided, candidates can better prepare for the expected difficulty level of this section, ultimately enhancing their chances of scoring well in the exam. Keep practising, stay consistent, and approach each question with a clear strategy to excel in this section.

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