Arithmetic is one of the most important areas in the CAT Quantitative Aptitude section, contributing significantly to the exam both directly through standalone questions and indirectly as part of word problems, DI sets, and mixed questions. If you are preparing for CAT exam, it is essential to understand the kind of questions asked from Arithmetic, the most frequently covered topics, the complete list of subtopics under Arithmetic, and how this section impacts your overall percentile. Being aware of the weightage and the question patterns from previous years is equally crucial, and this article aims to address all these aspects in detail.
A 99 percentile in the CAT exam puts you in strong contention for many IIMs, but doesn’t guarantee a call, especially from top IIMs like A, B, and C, which often require 99.5+ along with a strong overall profile and WAT/PI performance.
8. Questions on Integration of Ratio, Percentage, and Profit-Loss Combined
9. Simple Equations in Arithmetic
10. Problems using ratio, average, time & work, and TSD
Preparation Tips for CAT Arithmetic
Best resources & practice material for CAT Algebra
CAT 2025 Preparation Resources by Careers360
The 10 Most Repeated Arithmetic Concepts in CAT (2020-2024)
Introduction to CAT Arithmetic Questions
The important arithmetic topics for CAT are as listed here
Why Arithmetic Questions Are Crucial in CAT
Arithmetic questions play a significant role in QA score in CAT. It becomes crucial because of the following reasons:
Arithmetic has a healthy weightage in Quantitative Aptitude section every year. (Approximately 30% to 40%)
In data interpretation also, the concepts of percentages, average, and ratios are widely used. Without a good understanding of these topics, you won’t be able to solve the questions of data interpretation.
Arithmetic questions develop fundamental skills, conceptual understanding. Also, plays a significant role in overall score.
Weightage of Arithmetic in CAT Quant Section As seen in previous years, here is a summary (year-wise and slot-wise) of the number of questions asked from Arithmetic in CAT
S. No.
Year
Slot
%age
Profit and Loss
Interest
Average, Ratio, Mixture and Alligations
TSD
Time and Work
Total
1
2020
1
1
1
1
4
4
11
2
2020
2
1
1
1
2
4
1
10
3
2020
3
1
1
1
3
3
1
10
4
2021
1
1
1
1
3
1
3
10
5
2021
2
2
2
2
1
2
9
6
2021
3
1
1
1
3
1
3
10
7
2022
1
2
2
2
1
2
9
8
2022
2
1
1
5
1
1
9
9
2022
3
1
4
3
2
10
10
2023
1
1
1
1
2
3
1
9
11
2023
2
2
1
4
1
1
9
12
2023
3
2
1
2
1
2
8
13
2024
1
1
1
1
3
1
1
8
14
2024
2
2
1
1
2
1
1
8
15
2024
3
2
1
1
3
1
1
9
Data-Driven Insights: Easy vs. Moderate Questions
We have analysed the last 8 - 10 years' question papers. Before 2019, the level of questions in the quantitative aptitude section of CAT was generally moderate to hard.
But since the last 4 – 5 years, the level of questions in the quantitative aptitude section has ranged from easy to moderate. Out of 22 questions, you will find 6 – 8 questions which are not so tough to attempt. 8 – 10 questions are of a moderate level, and 3 – 4 questions can be moderate to hard.
How Repeated Arithmetic Topics Can Boost Your Score
Repeated concepts help students in several ways:
After knowing about the concepts that are repeating, students can give a direction to their preparation.
Students can develop strong fundamentals.
Practising questions on repeated concepts increases accuracy and speed.
Students became familiar with the question type and it will help in saving time during exams.
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So, repeated arithmetic topics help students to boost their score.
1. Percentage and Profit-Loss
Percentages is one of the most important chapters in the syllabus of the CAT exam and the XAT examination. The importance of ‘percentages’ is supported by the fact that there are a lot of questions related to the use of percentages in all chapters, like Profit and Loss, Ratio and Proportion, Time and Work, Time, Speed and Distance, and Data interpretation.
Understanding the concepts of cost price (CP), selling price (SP), marked price (MP), Profit, Loss and discount is required to solve the questions based on profit and loss.
Repeated Question Types in CAT and PYQs
A few concepts from percentage, profit and loss are repeating consistently. Let’s have a look at these concepts: Percentage (Repeated Concepts):
Questions based on a Venn diagram
Questions on successive change
Income and expenses
Percentages in average and mixtures
Election-based questions
CAT 2025 Percentile Predictor
Use the CAT Percentile Predictor to estimate your score, gauge your performance, and know your chances of getting shortlisted by IIMs.
Q.1) In an election, there were four candidates, and 80% of the registered voters cast their votes. One of the candidates received 30% of the cast votes while the other three candidates received the remaining cast votes in the proportion 1 : 2 : 3. If the winner of the election received 2512 votes more than the candidate with the second highest votes, then the number of registered voters was
A) 50240
B) 40192
C) 60288
D) 60288
Solution:-
Let the number of voters be 300X
Then, votes cast = 240X
Number of votes received by the person who received 30% of the cast votes $\mathrm{=\frac{30}{100} \times 240 X=72 X}$
Remaining votes (240X – 72X) = 168X will be distributed as 28X, 56X, and 84X.
Q.2) In a village, the ratio of number of males to females is 5 : 4. The ratio of number of literate males to literate females is 2 : 3. The ratio of the number of illiterate males to illiterate females is 4 : 3. If 3600 males in the village are literate, then the total number of females in the village is:
A) 43200
B) 42300
C) 43000
D) 43000
Solution:-
Let the number of males to females in the village be 5x and 4x, respectively. Let the number of literate males to literate females in the village be 2y and 3y, respectively. Let the number of illiterate males to illiterate females in the village be 4z and 3z, respectively. $ \text { So, } \frac{2 y+4 z}{3 y+3 z}=\frac{5}{4}$ $ \Rightarrow 8 y+16 z=15 y+15 z $ $ \Rightarrow 7 y=z$ According to the question, $⇒ 2y = 3600$ $\therefore y = 1800$ So, the total number of females in the village is: 3y + 3z = 3y + 21y = 24y = 24 × (1800) = 43200
Hence, the correct answer is option (1).
Q.3) A fruit seller has a stock of mangoes, bananas and apples with at least one fruit of each type. At the beginning of a day, the number of mangoes makes up $40 \%$ of his stock. That day, he sells half of the mangoes, 96 bananas and $40 \%$ of the apples. At the end of the day, he ends up selling $50 \%$ of the fruits. The smallest possible total number of fruits in the stock at the beginning of the day is
Solution:-
Let the initial number of fruits be $x$ and apples be $y$.
The stock of mangoes is given to be 40% of x, which shows that the number of mangoes will be $\frac{2x}{5}$.
The total number of sold fruits is given by including the selling of all the fruits- mango, apple and banana.
Now, the total no. of fruits sold
$=\frac{2x}{10}+\frac{4y}{10}+96=\frac{x}{2}$
$⇒\frac{x}{5}+\frac{2y}{5}+96=\frac{x}{2}$
On simplifying, we get,
$2x+4y+960=5x$
$⇒x=\frac{960+4y}{3}$
For $x$ to be a positive integer, let us check for values of $y$.
$4y+960$ has to be divisible by $3$, which means $4y$ will also be divisible by $3$.
For $\frac{4y}{10}$ to be an integer, $y$ has to be divisible by $5$.
We can say, that the smallest value of $y$ has to be a multiple of both $3$ and $5$, i.e. $15$.
Now, put the value of $y=15$.
We get, $x=\frac{960+4 \times 15}{3}$
$⇒x=\frac{1020}{3}$
$⇒x=340$
So, the smallest possible number of fruits in stock at the start of the day will be $340$.
Hence, the correct answer is $340$.
Q.4) In a group of 250 students, the percentage of girls was at least $44 \%$ and at most $60 \%$. The rest of the students were boys. Each student opted for either swimming or running or both. If $50 \%$ of the boys and $80 \%$ of the girls opted for swimming while $70 \%$ of the boys and $60 \%$ of the girls opted for running, then the minimum and maximum possible number of students who opted for both swimming and running are
A) 75 and 96, respectively
B) 72 and 88, respectively
C) 75 and 90 , respectively
D) 75 and 90 , respectively
Solution:-
We are given: Total students = $250$ Minimum number of girls possible = 44% of 250 = 110, then boys = 140 Maximum number of girls possible = 60% of 250 = 150, then boys = 100
It is given that ; Among boys: $50\%$ swim, $70\%$ run and among girls: $80\%$ swim, $60\%$ run
Let number of girls = $g$, then, boys = $b = 250 - g$
Now, total swimming strength = $0.5b + 0.8g$ total running strength = $0.7b + 0.6g$
Let the number of students who opted for both = $x$
Q.1) Ankita buys 4 kg cashews, 14 kg peanuts and 6 kg almonds when the cost of 7 kg cashews is the same as that of 30 kg peanuts or 9 kg almonds. She mixes all the three nuts and marks a price for the mixture in order to make a profit of Rs.1752. She sells 4 kg of the mixture at this marked price and the remaining at a 20% discount on the marked price, thus making a total profit of Rs. 744. Then the amount, in rupees, that she had spent in buying almonds is:
A) 1680
B) 1176
C) 2520
D) 2520
Solution:-
It is given,7C = 30P = 9A and Ankita bought 4C, 14P and 6A. Let 7C = 30P = 9A = 630k C = 90k, P = 21k, and A = 70k Cost price of 4C, 14P and 6A = 4(90k)+14(21k)+6(70k) = 1074k Marked up price = 1074k + 1752 $\begin{aligned} & \text { S.P }=\frac{1}{6}(1074 k+1752)+\left(\frac{4}{5}\right)\left(\frac{5}{6}\right)(1074 k+1752)=\frac{5}{6}(1074 k+1752) \\ & \text { S.P-C.P }=\text { profit } \\ & 1460-\frac{1074 k}{6}=744 \\ & \frac{1074 k}{6}=716\end{aligned}$ k = 4 Money spent on buying almonds = 420k = 420 × 4 = Rs 1680 The correct answer is Rs. 1680.
Q.2) The selling price of a product is fixed to ensure $40 \%$ profit. If the product had cost $40 \%$ less and had been sold for 5 rupees less, then the resulting profit would have been $50 \%$. The original selling price, in rupees, of the product is
A) 15
B) 20
C) 10
D) 10
Solution:-
Let the cost price be $x.$ Then selling price = $x + 40\% \text{ of } x = x + 0.4x = 1.4x$ Now, new cost price = $x - 40\% \text{ of } x = x - 0.4x = 0.6x$ New selling price = $1.4x - 5$
Given: profit in new case = $50\%$ of new cost price = $0.6x \times 0.5 = 0.3x$ So, New selling price = $0.6x + 0.3x=0.9x$ So, $1.4x - 5 = 0.9x$ $⇒1.4x - 0.9x = 5 \Rightarrow 0.5x = 5 \Rightarrow x = 10$ Original selling price = $1.4x = 1.4 \times 10 = 14$
Hence, the correct answer is option 4.
Q.3) Gopi marks a price on a product in order to make $20 \%$ profit. Ravi gets $10 \%$ discount on this marked price, and thus saves Rs 15. Then, the profit, in rupees, made by Gopi by selling the product to Ravi, is
A) 15
B) 25
C) 10
D) 10
Solution:-
Let the cost price of the product be $100x$.
Gopi wants to make a $20\%$ profit, so the marked price (MP) is:
But the question says Ravi saves Rs 15, So, $12x=15$
$⇒x=\frac{15}{12}=\frac54$
The profit made by Gopi = $108x - 100x = 8x= 10$ rupees.
Hence, the correct answer is option 3.
Shortcuts and Tricks for Quick Calculation
Speed and accuracy matter a lot while solving questions in the CAT exam. So, you should work on how to make quick calculations with accuracy.
Let’s discuss how you can improve your calculation speed:
Learn Vedic maths for quick multiplication, addition, or subtractions.
Learn tables up to 30.
Learn squares up to 30.
Learn cubes up to 30.
Apart from this, to find percentages quickly, you should learn conversion of percentage to fraction and vice versa.
Fractions
Percentage
Fractions
Percentage
1
100%
1/9
11.11%
½
50%
1/10
10%
1/3
33.33%
1/11
9.09%
¼
25%
1/12
8.33%
1/5
20%
1/13
7.69%
1/6
16.66%
1/14
7.14%
1/7
14.28%
1/15
6.66%
1/8
12.5%
1/16
6.25%
Some important Tricks:
1. To find the net percentage change (In Successive Increase or Decrease), you can use the formula. Net R% = $R_1 + R_2 + \frac{(R_1 \times R_2)}{100}$ 2. If two articles were sold at equal price with a profit of a% on one article while loss of a% on the other. Then, there will always be a loss of $\frac {a^2}{100}$ %. 3. If n articles were purchased at equal price and sold at different prices. Then, net profit % is average of all profit %. Loss % will be treated as a negative of profit %.
2. Simple and Compound Interest
Principal: The original amount of money borrowed or invested.
Interest Rate: The interest charged on Rs 100 in one year.
Simple Interest:
Simple Interest (SI} is the interest earned or paid on the original principal during a specific period.
Compound Interest:
Definition: Compound Interest (Cl) refers to the interest that accumulates on both the initial principal and the accumulated interest from previous periods.
Q.1) Anil invests Rs 22000 for 6 years in a scheme with $4 \%$ interest per annum, compounded half-yearly. Separately, Sunil invests a certain amount in the same scheme for 5 years, and then reinvests the entire amount he receives 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 equal, then the initial investment, in rupees, made by Sunil is
A) 20808
B) 20860
C) 20480
D) 20480
Solution:-
Anil’s investment: Principal $ = \text{₹ }22000$ Rate per half year$ = 2\%$ Number of half years$ = 6 × 2 = 12$
$⇒22000 \times \left(\frac{102}{100} \right)^2 = x \times \frac{11}{10}$
$⇒22000 \times \frac{10404}{10000} = x \times \frac{11}{10}$
$⇒20000 \times \frac{10404}{10000} = x $
$⇒x = 2 \times 10404 = 20808$
Hence, the correct answer is option 1.
Q.2) Alex invested his savings in two parts. The simple interest earned on the first part at 15% per annum for 4 years is the same as the simple interest earned on the second part at 12% per annum for 3 years. Then, the percentage of his savings invested in the first part is:
A) 37.5%
B) 62.5%
C) 60%
D) 65%
Solution:-
Let the investments of Alex be ‘a’ and ‘b’ in the two schemes. So, interest earned on the first scheme = 0.15 × a × 4 Interest earned in the second scheme = 0.12 × b × 3
According to the question, 0.15 × a × 4 = 0.12 × b × 3 ⇒ 20a = 12b ⇒ a : b = 3 : 5 $\therefore$ The percentage of Alex’s savings invested in the first scheme $=\frac{3}{3+5}×100=37.5\%$
Hence, the correct answer is 37.5%.
Q.3) Mr. Pinto invests one-fifth of his capital at 6%, one-third at 10% and the remaining at 1%, each rate being simple interest per annum. Then, the minimum number of years required for the cumulative interest income from these investments to equal or exceed his initial capital is:
A) 20
B) 18
C) 16
D) 16
Solution:-
Let the total investment be 15x and the number of years required to be T years.
$ \begin{aligned} & \frac{(3 \mathrm{x} \times 6 \times \mathrm{T})}{100}+\frac{(5 \mathrm{x} \times 10 \times \mathrm{T})}{100}+\frac{(7 \mathrm{x} \times 1 \times \mathrm{T})}{100} \geq 15 \mathrm{x} \\ &⇒ \frac{75 \mathrm{x} T}{100} \geq 15 \mathrm{x} \\ & ⇒ \mathrm{T} \geq 20 \end{aligned} $ So minimum value of T is 20 years.
Relative speed (both moving in opposite direction)
Speed of B with respect to A = $S_B+S_A$
Average Speed
(Total Distance Covered)/(Total Time Taken)
If two equal distances are covered at speeds $S_1$ and $S_2$, then average speed
$(2S_1 S_2)/(S_1+S_2 )$
Effective speed of the boat in upstream (SU)
Speed of boat (SB) - Speed of stream (SR)
Effective speed of the boat in downstream (SD)
Speed of boat (SB) + Speed of stream (SR)
Circular Motion: Time of first meeting
(Circumference of circle)/(Relative Speed)
Circular Motion: Number of distinct meeting points
(Ratio of speeds of two runners be a : b)
(i) If both are moving in the same direction, then find the absolute difference between a and b, i.e., |a - b|. This absolute difference will be the number of distinct meeting points on the circular track.
(ii) If both are moving in the opposite direction, then find the sum of a and b. This sum will be the number of distinct meeting points on the circular track.
Circular Motion: Time of first meeting at starting point
LCM of individual times to complete one round
Previous years questions based on Time, speed, and distance
Q.1) Trains A and B start traveling at the same time towards each other with constant speeds from stations X and Y, respectively. Train A reaches station Y in 10 minutes while train B takes 9 minutes to reach station X after meeting train A. Then the total time taken, in minutes, by train B to travel from station Y to station X is:
A) 6
B) 15
C) 10
D) 10
Solution:-
M – First meeting pointLet the speeds of trains A and B be ‘a’ and ‘b’, respectively. $ \frac{x}{a}=\frac{D-x}{b} $ It is given, $ \begin{aligned} & \frac{D}{a}=10 \text { and } \frac{x}{b}=9\\\ & \frac{x}{\frac{D}{10}}=\frac{D-x}{\frac{x}{9}}\ [\text{After putting the values}] \\ & ⇒\frac{10 x}{D}=\frac{9 D-9 x}{x} \\ &⇒ 10 x^2=9 D^2-9 D x \\ & ⇒10 \mathrm{x}^2+9 \mathrm{Dx}-9 \mathrm{D}^2=0 \end{aligned} $ Solving, we get, $ x=\frac{3 D}{5}$ Solving, we get, $ \frac{x}{b}=9 $ $\begin{aligned} &⇒ \frac{3 D}{b \times 5}=9 \\ & ⇒\frac{D}{b}=15\end{aligned}$
The total time taken by train B to travel from station Y to station X is 15 minutes. Hence, the correct answer is option (2).
Q.2) Two ships meet mid-ocean, and then, one ship goes south and the other ship goes west, both traveling at constant speeds. Two hours later, they are 60 km apart. If the speed of one of the ships is 6 km per hour more than the other one, then the speed, in km per hour, of the slower ship is:
A) 20
B) 12
C) 18
D) 18
Solution:-
Let the speeds of two ships be x and (x + 6) km per hour.
Distance covered in 2 hours will be 2x and (2x + 12).
$ \begin{aligned} & (2 x)^2+(2 x+12)^2=60^2 \\ & ⇒(x)^2+(x+6)^2=30^2 \\ & ⇒ 2 x^2+12 x+36=900 \\ & ⇒x^2+6 x+18=450 \\ & ⇒ x^2+6 x-432=0\\ & ⇒ (x+24)(x-18)=0 \end{aligned} $ Solving, we get $\mathrm{x}=18$ or $x = -24$ As speed can't be negative, the speed of the slower ship is 18 km/hr.
Hence, the correct answer is option (3).
Q.3) A straight road connects points A and B. Car 1 travels from A to B and Car 2 travels from B to A, both leaving at the same time. After meeting each other, they take 45 minutes and 20 minutes, respectively, to complete their journeys. If Car 1 travels at the speed of 60 km/hr, then the speed of Car 2, in km/hr, is
A) 100
B) 90
C) 80
D) 80
Solution:-
Let the speed of Car 2 be ' $x$ ' kmph and the time taken by the two cars to meet be ' $t$ ' hours. In 't' hours, Car 1 travels $(60 \times t)$ km while Car 2 travels $(x \times t) \mathrm{km}$
It is given that the time taken by Car 1 to travel $(x \times t)$ km is 45 minutes or ($\frac34$) hours.
Total work refers to the complete task or job that needs to be accomplished. In questions related to time and work, it's crucial to understand the entire scope of the work because all further calculations, be it regarding time or the number of workers required, will revolve around this concept.
The concept of 1 day work refers to the portion of the total work that an individual or a machine completes in one day. It's an essential tool to compare efficiencies and to calculate combined work when multiple entities work together
Concept of Time and Work in Pipes & Cisterns
Definition: Pipe and cistern problems involve filling or emptying a tank or reservoir. The pipes can either fill (inlet pipes) or empty (outlet pipes) the tank. The 'Time and Work' principles can be directly applied to these problems with the 'work' being represented as the volume of the tank.
Note: Work of filling pipe is treated as positive while the work of emptying pipe/hole/drainage pipe/leak is treated as negative.
Important formulas related to Time and Work
Important Concept/Formula Used
1. Individual time of doing a work is given and time taken if working together is asked. (For equal amount of work) $\frac {1}{T}= \frac {1}{A} + \frac {1}{B} + \frac {1}{C} + ⋯$
T is time taken by all when working together
A is time taken by A when working alone
B is time taken by B when working alone
C is time taken by C when working alone
2. A group of few members with same efficiency is working together, then
R1, R2 rate of doing work of M1 and M2 respectively.
W1, W2 amount of work done by M1 and M2 respectively.
3. Concept of Negative work: Work done by emptying tape/Cistern is taken as negative work in the questions of pipes and cistern.
4. Efficiency is equal to Total work divided by time taken.
PYQs & Frequently Repeated Patterns
Q.1) Renu would take 15 days working 4 hours per day to complete a certain task, whereas Seema would take 8 days working 5 hours per day to complete the same task. They decide to work together to complete this task. Seema agrees to work for double the number of hours per day as Renu, while Renu agrees to work for double the number of days as Seema. If Renu works 2 hours per day, then the number of days Seema will work is
Solution:-
Renu takes 15 days × 4 hours/day = 60 hours to complete the task. So, Renu's efficiency = $\frac{1}{60}$ work per hour.
Seema takes 8 days × 5 hours/day = 40 hours to complete the task. So, Seema's efficiency = $\frac{1}{40}$ work per hour.
Renu works 2 hours/day. Let Seema work for $x$ days. Then, Renu works for $2x$ days and Seema works 4 hours/day (double of Renu).
Total work done by Renu = $2x \times 2 \times \frac{1}{60}$
$= \frac{4x}{60} = \frac{x}{15}$ Total work done by Seema = $x \times 4 \times \frac{1}{40}$
$= \frac{4x}{40} = \frac{x}{10}$
Total work = $\frac{x}{15} + \frac{x}{10} = 1$ $\Rightarrow \frac{2x + 3x}{30} = 1 \Rightarrow \frac{5x}{30} = 1$ $\Rightarrow x = 6$
Hence, the correct answer is $6$.
Q.2) Working alone, the times taken by Anu, Tanu, and Manu to complete any job are in the ratio 5 : 8 : 10. They accept a job that they can finish in 4 days if they all work together for 8 hours per day. However, Anu and Tanu work together for the first 6 days, working 6 hours 40 minutes per day. Then, the number of hours that Manu will take to complete the remaining job working alone is:
A) 6
B) 8
C) 4
D) 4
Solution:-
Let the time taken by Anu, Tanu, and Manu be 5x, 8x, and 10x hours. $\therefore$ Total work = LCM of (5x, 8x, and 10x) = 40x Anu can complete 8 units in one hour. Tanu can complete 5 units in one hour. Manu can complete 4 units in one hour. It is given that, three of them together can complete the work in 32 hours. $32(8+5+4)=40x$ $\therefore x=\frac{68}{5}$ It is given, Anu and Tanu work together for the first 6 days, working 6 hours 40 minutes per day, i.e. 36 + 4 = 40 hours $\begin{aligned} & 40(8+5)+y(4)=40 x \\ &⇒ 4 y=24 \\ & \therefore y=6\end{aligned}$ Manu alone will complete the remaining work in 6 hours.
Hence, the correct answer is option (1).
Q.3) style="text-align:justify">John takes twice as much time as Jack to finish a job. Jack and Jim together take one-third of the time to finish the job than John takes working alone. Moreover, in order to finish the job, John takes three days more than the three of them working together. In how many days will Jim finish the job working alone?
A) 4
B) 5
C) 6
D) 6
Solution:-
Given: John takes twice as much time as Jack to finish a job. Jack and Jim together take one-third of the time to finish the job than John takes working alone. Moreover, in order to finish the job, John takes three days more than the three of them working together. Let Jack take $x$ days to finish a job. So, John will take $2x$ days to finish the job. Also, Jack and Jim take $\frac{2x}{3}$ days to finish the job. Now, let the total work be LCF ($x,2x,\frac{2x}{3}$) = $2x$ units So, the efficiency of Jack = $\frac{2x}{x}$ = 2, The efficiency of John = $\frac{2x}{2x}$ = 1, The total efficiency of Jack and Jim = $\frac{2x}{\frac{2x}{3}}$ = 3 So, the efficiency of Jim = (2 – 1) = 1 According to the question, To finish the job, John takes three days more than the three of them working together. ⇒ $2x=\frac{2x}{2+1+1}+3$ ⇒ $8x=2x+12$ ⇒ $x=2$ Therefore, working alone Jim will finish the job in $\frac{2×2}{1}$ = 4 days. Hence, the correct answer is option (1).
Q.4) 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) 60
C) 120
D) 120
Solution:-
If A takes x hours to fill the tank alone, then B needs (x-1) hours to empty the tank alone, and C needs y hours to fill the tank alone. According to the question: $\frac{1}{x}-\frac{1}{(x-1)} + \frac1y = \frac12$ So, $\frac1y =\frac12-\frac{1}{x}+\frac{1}{(x-1)}$ Pipe B and C worked together for 1 hour. Pipe C further worked for 1 hr and 15 minutes i.e. 1.25 hrs to finish the work. So, $\frac{- 1}{(x-1)} + \frac1y + \frac{1.25}y = 1$ ⇒ $- \frac1{(x-1)} + \frac{2.25}y = 1$ After solving both equations, we have x = 3 and y = $\frac32$ hrs = 90 minutes Therefore, Pipe C takes 90 minutes to fill the empty tank.
Hence, the correct answer is option (1).
5. Ratio and Proportion
A ratio is a comparison between two quantities, showing the relative size of one quantity to another. It is often expressed in the form 'a:b', where "a" and "b" are the two quantities.
Proportion:
1. If two ratios a:b and c:d, are equal then they are said to be in equal proportion.
a/b = c/d
2. In the ratio a:b,
Proportion of a = a/(a+b)
Proportion of b = b/(a+b)
Key Concepts Frequently Tested
The following concepts are repeated asked in CAT exam
Key Concept
Approach
Problems on ages
Partnership
$P_A:P_B:P_C=I_A T_A:I_B T_B:I_C T_C$ where P represents share in profit, I represent investment amount and T represents duration of investment.
Direct and Indirect Variation
If y varies directly as x, then $\frac {y_1}{y_2} = \frac {x_1}{x_2}$
If y varies indirectly as x, then $\frac {y_1}{y_2} = \frac {x_2}{x_1}$
6. Mixtures and Alligations
Questions of mixture and alligation are solved by using either the method of allegation or weighted average.
Rule of Alligation
The Rule of Alligation is a method of finding the ratio in which two or more ingredients at given prices must be mixed to produce a mixture of a desired price. This rule is especially useful in solving problems related to mixtures.
Formula:
If two ingredients A and B are mixed in a ratio x : y respectively, then
Then $\frac {x}{y} = \frac {(m-b)}{(a-m)}$
Shortcut for Quick Solving
1. Quantity of dearer/Quantity of Cheaper = [Mean price(m) - C.P. of cheaper(c)]/[C.P of Dearer (d) - Mean price (m) ]
2. If a Container contains ‘x’ units of pure liquid, and we replace the liquid with ‘y’ units of water:
Then after ‘n’ successive replacements, the units of pure liquid left is $x(1-\frac {y}{x})^n$
7. Averages
Definition: The average of a set of numbers is the sum of those numbers divided by the count of the numbers.
Mathematically:
Average = (Sum of all numbers)/(Count of all numbers)
Concept of Weighted Average:
Definition: The weighted average (or weighted mean) is an average in which each quantity in the dataset is assigned a weight. These weights determine the relative importance of each quantity on the average.
Mathematically:
Average = $\frac {(n_1 x_1+n_2 x_2+⋯)}{(n_1+n_2+⋯)}$
Where, x1, x2, x3, … are the averages of n1, n2, n3, … observations respectively.
Repeated Question Types in CAT and PYQs
Questions are repeatedly asked on mainly three concepts
Calculation of ages
Relating arithmetic mean and geometric mean
Questions involving the concept of weighted average
Q.1) Let A, B, and C be three positive integers such that the sum of A and the mean of B and C is 5. In addition, the sum of B and the mean of A and C is 7. Then the sum of A and B is
A) 5
B) 4
C) 6
D) 6
Solution:-
Given
$A + \frac{(B + C)}2 = 5 ⇒ 2A + B + C = 10$ ….(i)
$\frac{(A + C)}2 + B = 7 ⇒ A + 2B + C = 14$ …..(ii)
(i) – (ii) ⇒ B - A = 4 ⇒ B = 4 + A
Given that A, B, and C are positive integers
If A = 1 then B = 5 and C = 3
If A = 2, then B = 6 and C = 0, but this is invalid as C is positive.
Similarly, if A > 2, C will be negative, and the cases are not valid.
Hence, A + B = 6.
Hence, the correct answer is 6.
Q.2) The mean of all 4-digit even natural numbers of the form 'aabb', where a > 0, is
Q.3) Onion is sold for 5 consecutive months at the rate of Rs 10, 20, 25, 25, and 50 per kg, respectively. A family spends a fixed amount of money on onion for each of the first three months, and then spends half that amount on onion for each of the next two months. The average expense for onion, in rupees per kg, for the family over these 5 months is closest to
A) 26
B) 20
C) 18
D) 18
Solution:-
Let the fixed monthly expenditure be $M$ rupees.
First three months: $10, 20, 25$ per kg. Amount spent $= M$ each month.
Quantity bought: $\frac{M}{10},\ \frac{M}{20},\ \frac{M}{25}$ kg respectively.
Next two months, expenditure $= \frac{M}{2}$ each month, prices $25, 50$ per kg.
Quantity bought: $ \frac{M/2}{25} = \frac{M}{50},\ \frac{M/2}{50} = \frac{M}{100}$ kg.
Average price per kg$= \frac{\text{Total expense}}{\text{Total quantity}} = \frac{4M}{11M/50} = \frac{200}{11} \approx 18.18=18$
Hence, the correct answer is option 3
Variations and Advanced Applications
Average is an important concept which is used to solve the problems of weighted average, problems including ages, mixture and alligations etc.
Averages are used in day-to-day life calculations.
Used in making estimation
Important formulas and tricks
1. When a person replaces another person then average age of the group will be increased or decreased.
Difference between the persons added and who left = change in average x number of persons.
2. When a person joins the group of n persons; if average increases then
Age of new person = Old average + (n+1) $\times$ difference between averages
If average decreases then
Age of new person = Old average - (n+1) $\times$ difference between averages
3. Average of an AP of odd terms is always its middle term.
4. In general case, Average of an AP is Average of its first and last term.
5. Average of an AP of even terms is average of its two middle terms.
6. Arithmetic mean is always greater than or equal to geometric mean.
8. Questions on Integration of Ratio, Percentage, and Profit-Loss Combined
In CAT, questions on integration of ratio with percentages, integration of ratio with profit and loss are being repeatedly asked.
Questions type may be:
Ratio of population of male and female are given and then increased or decreased by some percentages, percentage change in total population may be asked. Such kind of questions can be framed.
Number of articles are given in the form of ratios, selling rate and purchase rate are given in another ratio, and you might be asked the net profit or loss percent. There can be other variations of questions.
PYQs
Q.1) A person buys tea of three different qualities at INR 800,INR 500, and INR 300 per kg, respectively, and the amounts bought are in the proportion 2 : 3 : 5. She mixes all the tea and sells one-sixth of the mixture at 700 per kg. The price, in INR per kg, at which she should sell the remaining tea, to make an overall profit of 50%, is:
A) 653
B) 688
C) 692
D) 692
Solution:-
Let the person bought 2x, 3x, and 5x kg of the three different qualities at INR ₹800, INR 500, and INR 300 per kg respectively. Let the selling price of the remaining tea be ‘P’.
$\begin{aligned} & (2 x \times 800+3 x \times 500+5 x \times 300)(1+50 \%)=\left(\frac{1}{6} \times 10 x \times 700\right)+\left(\frac{5}{6} \times 10 x \times P\right) \\ & 4600\left(\frac{3}{2}\right)=\frac{7000}{6}+\frac{50 P}{6} \text { or } \mathrm{P}=688\end{aligned}$ Hence, the correct answer is 688.
Q.2) The salaries of three friends Sita, Gita, and Mita are initially in the ratio 5:6:7, respectively. In the first year, they get salary hikes of 20%, 25%, and 20%, respectively. In the second year, Sita and Mita get salary hikes of 40% and 25%, respectively, and the salary of Gita becomes equal to the mean salary of the three friends. The salary hike of Gita in the second year is:
A) 26%
B) 28%
C) 25%
D) 25%
Solution:-
Let the initial salaries of Sita, Gita and Mita be 500, 600 and 700 respectively. After getting 20%, 25% and 20% salary hikes respectively, their salaries become 600, 750 and 840 respectively. In the second year, Sita and Mita get 40% and 25% hikes respectively. So, after two years the salaries of Sita and Mita are 840 and 1050 respectively. We also know that Gita’s salary is the average of the salaries of the three which is equal to the average of the other two i.e. $\frac{840+1050}{2} = 945$ So, the hike in the salary of Gita during the second year = $\frac{945-750}{750} \times 100 = 26$% Hence, the correct answer is option (1).
9. Simple Equations in Arithmetic
Questions asked from percentages, profit and loss, and other arithmetic topics can be solved by forming linear equations.
Linear Equations for Word Problems
Word problems on ages are generally solved forming simple linear equations. We can understand this by taking a simple example:
“7 years ago, P was half as old as Q. After 5 years, their total age will be 48. Find their present ages.”
Let present age of P = P, Q = Q.
Equation 1: 2(P – 7) = Q - 7
Equation 2: (P + 7) + (Q + 7) = 48, solve these two equations to get the value of P and Q.
In Profit and Loss:
“A bought 10 pencils for Rs. 80 and sold them at Rs. x each. Find x if his profit % is 20.”
Total selling price = 120% of total cost price = 10x Solve this equation to get the value of x.
Similarly, we can solve the questions on mixture and allegation, time and work, TSD, etc by forming simple equations.
Previously asked questions
Q.1) A gentleman decided to treat a few children in the following manner: He gives half of his total stock of toffees and one extra to the first child, then the half of the remaining stock along with one extra to the second, and continues giving away in this fashion. His total stock exhausts after he takes care of 5 children. How many toffees were in his stock initially?
A) 62
B) 60
C) 50
D) 50
Solution:-
Let the initial number of chocolates be $64x$. The first child gets $(32x + 1)$ and $(32x - 1)$ are left. 2nd child gets $(16x + \frac{1}{2})$ and $(16x - \frac{3}{2})$ are left 3rd child gets $(8x + \frac{1}{4})$ and $(8x - \frac{7}{4})$ are left 4th child gets $(4x + \frac{1}{8})$ and $(4x - \frac{15}{8})$ are left 5th child gets $(2x + \frac{1}{16} )$ and $(2x - \frac{31}{16})$ are left. Given, $2x - \frac{31}{16} = 0 ⇒ 2x=\frac{31}{16} ⇒ x=\frac{31}{32}$. So, initially, the Gentleman has $64x$ i.e. $64 × \frac{31}{32} =62$ chocolates. Hence, the correct answer is option (1).
Q.2) Amal purchases some pens at INR 8 each. To sell these, he hires an employee at a fixed wage. He sells 100 of these pens at INR 12 each. If the remaining pens are sold at INR 11 each, then he makes a net profit of INR 300, while he makes a net loss of INR 300 if the remaining pens are sold at INR 9 each. The wage of the employee, in INR, is:
A) 1000
B) 800
C) 1500
D) 1500
Solution:-
Let the number of pens purchased be n. Then the cost price is 8n. The total expenses incurred would be 8n + W, where W refers to the wage of the employee Then selling price in the first case = 12 × 100 + 11 × (n – 100) Given profit is 300 in this case. So, 1200 + 11n – 1100 – 8n – W = 300 ⇒ 3n – W = 200---------(1) In the second case: 1200 + 9n – 900 – 8n – W = –300 (Loss) ⇒ W – n = 600-----------(2) Adding equation 2 and equation 1, we get, 2n = 800 $\therefore$ n = 400 So, W = 600 + 400 = 1000
Hence, the correct answer is 1000.
Q.3) 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) 1296000
B) 1944000
C) 972000
D) 972000
Solution:-
If W is the weight of the stone and P is the price of that stone, then P = k × W2 Price of unbroken stone = 182 × k = 324 k. Total cost is minimum if the weight of broken stones are close to each other, that is, the weights are 3, 4, 5, and 6 units. Total cost in this case = k(32+42+52+62) = 86k Total cost is maximum when the weights of the broken stones are far from each other, that is, the weights are 1, 2, 3, and 12 units. Total cost in this case = (12+22+32+122)k = 158k According to the question 158k – 86k = 72k = 2,88,000 ⇒ k = 4,000 So, the price of the unbroken stone = 182k = 324k = 12,96,000
Hence, the correct answer is option (1).
10. Problems using ratio, average, time & work, and TSD
In CAT, TSD, time and work make good questions with averages and ratios. Questions on these concepts are frequently asked.
Most Asked Concepts in Past CAT Papers
1. Ratio with time, speed and distance: Important concept: Speed is directly proportional to Distance $\frac {S_1}{S_2} = \frac {D_1}{D_2}$ Speed is inversely proportional to time $\frac {S_1}{S_2} = \frac {T_2}{T_1}$
2. Average with speed, distance, and time: Key concept: Average speed = Total distance covered/Total time taken
3. Ratio with time and Work: Key concept: Time is inversely proportional to efficiency
Practice Questions from 2020–2024 Question Paper
Q.1) A person spent Rs 50000 to purchase a desktop computer and a laptop computer. He sold the desktop at a 20% profit and the laptop at a 10% loss. If overall he made a 2% profit then the purchase price, in rupees, of the desktop is:
A) 20000
B) 18000
C) 16000
D) 16000
Solution:-
Using the Alligation Rule, the ratio of cost prices of desktop and laptop will be:
i.e., 12:18 = 2:3
∴ The cost of desktop = $\frac25$ × 50000 = Rs. 20000
Hence, the correct answer is option (1).
Q.2) A and B are two points on a straight line. Ram runs from A to B while Rahim runs from B to A. After crossing each other. Ram and Rahim reach their destination in one minute and four minutes, respectively. if they start at the same time, then the ratio of Ram's speed to Rahim's speed is
A) $\frac{1}{2}$
B) $\sqrt2$
C) 2
D) 2
Solution:-
The required ratio of speeds = Square root of the inverse ratio of times taken after crossing each other
$\mathrm{=\sqrt{4}: \sqrt{1} \: i.e., 2: 1}$ hence, the correct answer is 2 : 1.
Q.3) The amount Neeta and Geeta together earn in a day equals what Sita alone earns in 6 days. The amount Sita and Neeta together earn in a day equals what Geeta alone earns in 2 days. The ratio of the daily earnings of the one who earns the most to that of the one who earns the least is:
A) 3 : 2
B) 11 : 7
C) 11 : 3
D) 11 : 3
Solution:-
Let Neeta, Geeta, and Sita earn N, G, and S amounts per day.
Given that, N + G = 6S ……(1) S + N = 2G …….(2)
Substituting (2) in (1) for N, 2G – S + G = 6S ⇒ 3G = 7S ⇒ $\frac{\text{G}}{\text{S}}=\frac{7}{3}$
Substituting (2) in (1) for S, N + G = 6(2G – N) ⇒ 7N = 11G ⇒ $\frac{\text{N}}{\text{G}}=\frac{11}{7}$
$\therefore$ N : G : S = 11 : 7 : 3
Hence, the correct answer is 11 : 3.
Q.4) Amar, Akbar and Anthony are working on a project. Working together Amar and Akbar can complete the project in 1 year, Akbar and Anthony can complete in 16 months, Anthony and Amar can complete in 2 years. If the person who is neither the fastest nor the slowest works alone, the time in months he will take to complete the project is
Solution:-
Let Amar, Akbar, Anthony's 1 month work be $a,\ b,\ c$ respectively.
Q: How much time should I give to prepare for Arithmetic?
A:
Well, it varies person to person. Arithmetic as a unit is very vast. You should not focus on the time that you want to give. First, find out the important topics which are frequently asked in the CAT quant section and give time according to your capability to understand the topic well.
Q: How many questions from arithmetic are asked in the quantitative aptitude section of CAT?
A:
In the QA section, you will find around 8–10 questions from arithmetic.
Q: Which topics are important in Arithmetic for CAT?
A:
Important topics in Arithmetic for CAT are: Percentages, Profit & Loss, Simple & Compound Interest, Averages, Ratios & Proportion, Mixtures & Alligations, Time & Work, Time-Speed-Distance.
CAT does not restrict eligibility based on stream. Candidates from commerce, arts, science, engineering or any other background can appear for CAT and apply to MBA programmes offered by IIMs and other top B-schools.
The basic eligibility criteria for CAT are:
You must hold a bachelor’s degree in any discipline from a recognised university with at least 50 percent marks (45 percent for SC, ST and PwD categories). Final-year graduation students are also eligible to apply.
Being a commerce graduate can actually be an advantage in areas such as accounting, finance, economics and business studies, especially during MBA coursework in finance, marketing and operations.
After qualifying CAT, admission depends on multiple factors including CAT percentile, academic background, work experience (if any), performance in GD, WAT and personal interview rounds. Many IIMs and B-schools follow a diverse academic background policy, which means non-engineering candidates, including commerce graduates, often receive additional weightage.
Apart from IIMs, several reputed institutes such as FMS Delhi, SPJIMR Mumbai, MDI Gurgaon, IMT Ghaziabad, and many state and private universities also accept CAT scores for MBA admissions.
So, as a commerce graduate, you are fully eligible to appear for CAT and pursue an MBA, provided you meet the minimum academic requirements and perform well in the selection process.
If you want, I can also help you with realistic CAT percentile targets based on your academic profile or suggest suitable MBA colleges.
With a CAT 2025 raw score of 84 and strong sectionals, your expected percentile may fail in the high 98 range , depending on scaling. Your excellent academics and 2 years of work experience strengthen your profile .IIM Lucknow calls are possible but competitive for EWS engineers , while FMS Delhi looks mainly at CAT score and VARC , so you have a reasonable chance there . Final calls depend on cutoffs and composite score.
For MANAGE Hyderabad MBA , the OBC cutoff in CAT 2024 was generally around 85-90 percentile for shortlist consideration , tough exact figures vary by section and profile . For CAT 2025, if fewer candidates appeared , the cutoff might slightly adjust , but percentile based cutoffs dont change drastically year to year. A minor decrease is possible , but you should still aim for 90+ percentile to be competitive. Factors like VARC, QA-LRDI and overall profile influence the final shortlist.
For a raw score of 56 in CAT 2025 Slot 3, your expected overall percentile is likely to be in the range of the 90th-95th percentile. The exact percentile can vary slightly on the final normalization process and the process and performance of all test-takers. In this
article
you'll find more about the CAT result.
Hello,
With a projected CAT percentile of 87% but not clearing sectional cutoffs, your chances at top IIMs are limited because they require both overall percentile and sectional minimums. However, you still have a good shot at other reputed management institutes and non-IIM B-schools. Consider colleges like NMIMS, SPJIMR, IMT, TAPMI, Great Lakes, and other well-ranked private or state-level B-schools that accept CAT scores and weigh your profile holistically. Your academic record, BSc in Animation with 80%, and 5 years of work experience at Ubisoft India are strong points and may help in institutes that value work experience in their selection process. Also, explore institutes that accept XAT, MAT, or CMAT, where your profile can be competitive.
Hope this helps you.
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M – First meeting pointLet the speeds of trains A and B be ‘a’ and ‘b’, respectively.
