CAT 2026 Arithmetic Preparation: Important Topics, Weightage, PYQs, Formulas, and Strategy to Score High

What is Arithmetic in CAT 2026?

Arithmetic is one of the most important topics in the CAT 2026 Quantitative Aptitude section. It focuses on real-life mathematical applications involving percentages, ratios, profit and loss, averages, time and work, and other business-oriented calculations. Since Arithmetic questions are generally concept-based and calculation-driven, they are often considered among the most scoring questions in CAT Quant.

Why Arithmetic is Important for CAT Quant

Arithmetic consistently contributes the highest number of questions in the CAT Quant section. A strong understanding of Arithmetic concepts can help candidates solve a significant portion of the paper while also improving their overall problem-solving skills.

Key Reasons to Focus on Arithmetic:

• Highest weightage in CAT Quantitative Aptitude

• Questions are mostly application-based

• Easier to master compared to advanced Algebra and Geometry topics

• Builds the foundation for several MBA entrance exams

• Helps improve speed and accuracy in the exam

Weightage of Arithmetic in CAT 2026

Based on recent CAT trends, Arithmetic typically accounts for a substantial share of Quant questions. Topics such as Percentages, Ratio and Proportion, Time and Work, and Profit and Loss appear frequently across different CAT slots.

Is Arithmetic Enough to Score Well in CAT Quant?

Arithmetic alone may not be sufficient to maximize your CAT Quant score, but it can provide a strong foundation. Candidates who are comfortable with Arithmetic can potentially solve a large portion of the Quant section and improve their overall percentile.

What Arithmetic Can Do:

• Help secure a strong sectional score

• Improve confidence in Quant

• Cover a large share of easy-to-moderate questions

What Arithmetic Cannot Do:

• Completely replace Algebra, Geometry, or Modern Math preparation

• Guarantee a top percentile without balanced Quant preparation

CAT 2026 Arithmetic Syllabus

The CAT Arithmetic syllabus includes several important topics that are frequently tested in the Quantitative Aptitude section. Candidates should focus on understanding concepts, formulas, and practical applications rather than memorizing shortcuts alone.

CAT 2026 Arithmetic Topics at a Glance

Mastering these Arithmetic topics can significantly improve your CAT 2026 Quantitative Aptitude score and help you tackle a large portion of the Quant section with confidence.

Arithmetic Section Weightage in the Past Year Papers (2025-2020)

Arithmetic has consistently been the highest-weightage topic in CAT Quantitative Aptitude over the past few years. In most CAT slots, Arithmetic alone has contributed a significant share of the Quant questions, making it one of the most important areas for CAT 2026 preparation. Analyzing previous year papers can help aspirants identify recurring topics, understand question trends, and prioritize high-scoring chapters.

CAT 2025 All Slots Arithmetic Weightage

CAT Arithmetic Weightage (2024-2020): All Slots Analysis

Key Trends (2020-2024)

Observation: Arithmetic consistently contributed 8-11 questions per slot between CAT 2020 and CAT 2024, making it the single most important area in Quantitative Aptitude preparation.

1. Percentage and Profit & Loss

Percentage and Profit & Loss are among the most important Arithmetic topics for CAT 2026. Percentage forms the foundation for several Quant topics, while Profit & Loss questions frequently test real-world business calculations involving profit, discount, and marked price concepts.

Important Concepts

• Percentage Increase and Decrease

• Successive Percentage Change

• Income and Expenditure

• Cost Price (CP), Selling Price (SP), and Marked Price (MP)

• Profit Percentage and Loss Percentage

• Discounts and Successive Discounts

Frequently Asked CAT PYQ Concepts

CAT 2026 Tip: A strong grasp of Percentage and Profit & Loss can help you tackle a significant portion of Arithmetic questions, as these concepts frequently appear both directly and indirectly in CAT Quant.

Previous Year Question Types & Frequently Repeated Patterns based on Percentage

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.

Given that, 84X - 72X = 2512

$\Rightarrow \mathrm{X=\frac{628}{3}}$

Thus, $\mathrm{300 X=300 \times \frac{628}{3}=62800}$

Hence, the correct answer is 62800.

Q.2) In a village, the ratio of the number of males to females is 5: 4. The ratio of the 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 = $x$

Using the formula:

$\text{(Swim)} + \text{(Run)} - \text{(Both)} = 250$

So, $(0.5b + 0.8g) + (0.7b + 0.6g) - x = 250$

$⇒(1.2b + 1.4g) - x = 250 \Rightarrow x = 1.2b + 1.4g - 250$

Substitute $b = 250 - g$:
$x = 1.2(250 - g) + 1.4g - 250$
$⇒x = 300 - 1.2g + 1.4g - 250 = 50 + 0.2g$

Now check for minimum and maximum values of $x$ by plugging in $g = 110$ and $g = 150$:

• Minimum (at $g = 110$):

$x = 50 + 0.2 \times 110 = 50 + 22 = 72$

• Maximum (at $g = 150$):

$x = 50 + 0.2 \times 150 = 50 + 30 = 80$

Hence, the correct answer is option 4.

Frequently Asked CAT PYQ Concepts: Profit & Loss

CAT 2026 Tip: Most Profit & Loss questions combine concepts from Percentages, Ratios, and Alligation. Focus on understanding the relationships between CP, SP, MP, profit percentage, and discount percentage rather than memorizing formulas.

Previous Year Question Types & Frequently Repeated Patterns based on Profit & Loss

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:

$
\text{MP} = 100x \times \left(1 + \frac{20}{100}\right) = 120x
$

Ravi gets $10$ % discount on the marked price, so the selling price (SP) is:

$
\text{SP} = 120x \times \left(1 - \frac{10}{100}\right) = 120 \times \frac{90}{100} = 108x
$

Now, the discount Ravi gets = $120x - 108x = 12x$

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.

Important Percentage and Profit & Loss Formulas & Shortcuts for CAT 2026

High-Value CAT Shortcuts

Most Important Percentage-Fraction Conversions

CAT 2026 Exam Tip: Questions based on successive percentage changes, CP-SP-MP relationships, discounts, and percentage-fraction conversions appear regularly in CAT Arithmetic. Memorizing these shortcuts can save valuable time during the exam.

2. Simple Interest and Compound Interest

Simple Interest (SI) and Compound Interest (CI) are important Arithmetic topics for CAT 2026 and are frequently tested through questions involving investments, loans, installments, depreciation, and population growth. While Simple Interest is calculated only on the principal amount, Compound Interest is calculated on both the principal and accumulated interest.

Important Concepts

• Principal (P)

• Rate of Interest (R)

• Time Period (T)

• Simple Interest (SI)

• Compound Interest (CI)

• Depreciation

• Population Growth

• Loans and Installments

Important CAT 2026 Formulas

Frequently Asked CAT PYQ Concepts

CAT 2026 Tip: Most CAT questions from this topic are application-based rather than formula-based. Focus on understanding the relationship between SI, CI, growth, and depreciation instead of memorizing formulas in isolation.

Previous Year Question Types & Frequently Repeated Patterns based on Simple Interest and Compound Interest

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$

So, amount = $22000 \times \left(1 + \frac{2}{100} \right)^{12} = 22000 \times \left(\frac{102}{100} \right)^{12}$

Now, Sunil invests ₹x for 5 years under same scheme:
Number of half years = 10

So, amount after 5 years = $x \times \left(\frac{102}{100} \right)^{10}$

This is reinvested at 10% simple interest for 1 year:

Amount = $x \times \left(\frac{102}{100} \right)^{10} \times \left(1 + \frac{10}{100} \right) = x \times \left(\frac{102}{100} \right)^{10} \times \frac{11}{10}$

Now equate this to Anil's amount:

$22000 \times \left(\frac{102}{100} \right)^{12} = x \times \left(\frac{102}{100} \right)^{10} \times \frac{11}{10}$

$⇒22000 \times \left(\frac{102}{100} \right)^2 = x \times \frac{11}{10}$

$⇒22000 \times \frac{10404}{10000} = x \times \frac{11}{10}$

$⇒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.

Hence, the correct answer is option (1).

3. Time, Speed and Distance

Time, Speed and Distance (TSD) is one of the most frequently tested Arithmetic topics in CAT 2026, with 2-3 questions commonly appearing every year. Questions are generally based on relative speed, average speed, trains, boats and streams, and circular motion. A clear understanding of the relationship between time, speed, and distance is essential for solving these questions quickly.

Important Concepts

• Speed = Distance ÷ Time

• Distance = Speed × Time

• Time = Distance ÷ Speed

• Relative Speed

• Average Speed

• Boats and Streams

• Circular Motion

• Train Problems

Important CAT 2026 Formulas and Shortcuts

Frequently Asked CAT PYQ Concepts

CAT 2026 Tip: Most TSD questions can be solved quickly by identifying the correct concept—relative speed, average speed, or circular motion. Focus on understanding applications rather than memorizing lengthy formulas.

Previous Year Question Types & Frequently Repeated Patterns based on Time, Speed and Distance

Q.1) Trains A and B start travelling 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.

$\therefore \frac{(x \times t)}{60}=\frac{3}{4} ⇒ t=\frac{180}{4 x}$......(i)

Similarly, the time taken by Car 2 to travel $(60 \times t)$ is 20 minutes or $(\frac1 3)$ hours.

$\therefore \frac{(60 \times t)}{x}=\frac{1}{3}$ or

$\therefore t=\frac{x}{180}$ ....(ii)

Equating the values in (i) and (ii), and solving for x:

$\therefore \frac{180}{4 x}=\frac{x}{180} \Rightarrow x=90$ kmph

Hence, the correct answer is 90.

4. Time and Work

Time and Work is one of the most important CAT Arithmetic topics and regularly appears in CAT previous year papers. Questions are based on work efficiency, combined work, work distribution, and pipes & cisterns. The key to solving these questions is understanding the relationship between work, efficiency, and time.

Important Concepts

• Total Work

• One Day's Work

• Work Efficiency

• Combined Work

• Pipes and Cisterns

• Positive and Negative Work

• Work Distribution

Important CAT 2026 Formulas and Shortcuts

Frequently Asked CAT PYQ Concepts

CAT 2026 Tip: Most Time and Work questions become easy once you convert everything into one day's work or efficiency units. Focus on understanding efficiency concepts rather than relying solely on formulas.

Previous Year Question Types & Frequently Repeated Patterns based on Time and Work

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) 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

Ratio and Proportion is a fundamental Arithmetic topic for CAT 2026 and serves as the foundation for several other topics, including Mixtures, Alligation, Partnership, and Percentages. Questions are usually straightforward and test a candidate's ability to compare quantities and establish relationships between them.

Important Concepts

• Ratio Comparison

• Proportion

• Direct Variation

• Inverse Variation

• Partnership

• Age-Based Problems

Important Ratio & Proportion Formulas and Shortcuts for CAT 2026

CAT Shortcuts

Frequently Asked CAT PYQ Concepts

CAT 2026 Tip: Most Ratio and Proportion questions can be solved using simple comparisons and proportional reasoning. Focus on understanding relationships between quantities rather than relying on lengthy calculations.

Question 1: In a company, the ratio of men to women is 5:3. If 40 men leave and 80 women join, the ratio becomes 7:5. Find the original number of employees.

Solution:

Let the original numbers be $5x$ and $3x$.

According to the question,

$\frac{5x-40}{3x+80}=\frac{7}{5}$

$25x-200=21x+560$

$4x=760$

$x=190$

Total employees

$=5x+3x$

$=8x$

$=1520$

Answer: 1520

Question 2: The incomes of A and B are in the ratio 4:5 and their expenditures are in the ratio 3:4. If each saves ₹20,000 and B's income exceeds A's income by ₹25,000, find A's income.

Solution:

Let incomes be $4x$ and $5x$.

$5x-4x=25000$

$x=25000$

Therefore,

A's income

$=4x$

$=₹100000$

Verification:

Expenditures become ₹80,000 and ₹105,000 respectively, satisfying the condition.

Answer: ₹1,00,000

Question 3: Two numbers are in the ratio 3:5. If 9 is added to each number, the ratio becomes 4:6. Find the larger number.

Solution:

Let the numbers be $3x$ and $5x$.

$\frac{3x+9}{5x+9}=\frac{4}{6}$

$18x+54=20x+36$

$2x=18$

$x=9$

Larger number

$=5x$

$=45$

Answer: 45

6. Mixtures and Alligation

Mixtures and Alligation is one of the most scoring Arithmetic topics in CAT 2026. Questions are generally based on mixing two quantities with different values, concentrations, or costs to achieve a desired result. The Rule of Alligation provides a quick shortcut for solving most mixture problems.

Important Concepts

• Rule of Alligation

• Mean Value of Mixture

• Replacement of Liquid

• Concentration Problems

• Cost-Based Mixtures

Important Mixtures & Alligation Formulas and Shortcuts for CAT 2026

CAT Shortcuts

Frequently Asked CAT PYQ Concepts

CAT 2026 Tip: The Rule of Alligation can significantly reduce calculation time. Practice cost-price and concentration-based questions, as they are among the most frequently tested CAT concepts.

Question 1: A merchant mixes two varieties of rice costing ₹60/kg and ₹90/kg. He sells the mixture at ₹84/kg and gains 20%. Find the ratio in which the two varieties were mixed.

Solution:

Cost price of mixture

$=\frac{84}{1.2}$

$=70$

Using alligation:

Ratio

$=(90-70):(70-60)$

$=20:10$

$=2:1$

Answer: 2:1

Question 2: A container contains 80 litres of milk. 20 litres are removed and replaced with water. The process is repeated three times. Find the quantity of milk left.

Solution:

Milk left

$=80\left(1-\frac{20}{80}\right)^3$

$=80\left(\frac34\right)^3$

$=80\times\frac{27}{64}$

$=33.75$

Answer: 33.75 litres

Question 3: A solution contains milk and water in the ratio 7:3. How much water must be added to 40 litres of solution so that the ratio becomes 7:5?

Solution:

Milk

$=\frac{7}{10}\times40$

$=28$

Water

$=12$

Let $x$ litres of water be added.

$\frac{28}{12+x}=\frac75$

$140=84+7x$

$56=7x$

$x=8$

Answer: 8 litres

7. Averages

Averages is a high-frequency CAT Arithmetic topic that is often combined with Percentages, Ratios, and Mixtures. Most questions are easy to moderate in difficulty and can be solved quickly if the underlying concepts are clear.

Important Concepts

• Arithmetic Mean

• Weighted Average

• Average of Groups

• Age-Based Averages

• Combined Average

Important Averages Formulas and Shortcuts for CAT 2026

CAT Shortcuts

Frequently Asked CAT PYQ Concepts

CAT 2026 Tip: Weighted Average is one of the most important concepts in this chapter and frequently appears in CAT previous year papers. Mastering weighted averages can also help in solving Mixtures and Alligation questions more efficiently.

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

A) 4466

B) 5050

C) 4864

D) 4864

Solution:-

The four-digit even numbers will be of form:

$1100, 1122, 1144, ... 1188, 2200, 2222, 2244 ... 9900, 9922, 9944, 9966, 9988$

Their sum 'S' will be $(1100+1100+22+1100+44+1100+66+1100+88) + (2200+2200+22+2200+44+...) .... + (9900+9900+22+9900+44+9900+66+9900+88)$

⇒ $S = 1100×5 + (22+44+66+88) + 2200×5 + (22+44+66+88)....+ 9900×5 + (22+44+66+88)$

⇒ $S = 5×1100(1+2+3+...9) + 9(22+44+66+88)$

⇒ $S=5×1100×9×10/2 + 9×11×20$

The total number of numbers is 9 × 5 = 45

∴ Mean will be $S/45 = 5 × 1100 + 44 = 5544$.

Hence, the correct answer is 5544.

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.

Total expense $ = 3M + \frac{M}{2} + \frac{M}{2} = 4M$

Total quantity $ = \frac{M}{10} + \frac{M}{20} + \frac{M}{25} + \frac{M}{50} + \frac{M}{100} $

$= \frac{10M + 5M + 4M + 2M + M}{100}$

$= \frac{22M}{100} = \frac{11M}{50} $

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 an estimate

Important formulas and tricks

1. When a person replaces another person, the 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 the 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. Integration of Ratio, Percentage, and Profit & Loss

Questions that combine Ratio, Percentage, and Profit & Loss are frequently asked in CAT Quantitative Aptitude. These questions typically require candidates to apply multiple concepts together rather than solve them independently. Such problems are common in population growth, business transactions, investments, and profit-sharing scenarios.

Important Concepts

• Ratio and Percentage Conversion
• Population Growth and Decline
• Profit & Loss Based on Ratios
• Markup and Discount
• Successive Percentage Changes
• Ratio-Based Business Transactions

Important CAT 2026 Formulas and Shortcuts

Frequently Asked CAT PYQ Concepts

High-Value CAT Shortcuts

CAT 2026 Tip: Most CAT questions from this area are multi-concept problems where Ratio, Percentage, and Profit & Loss are tested together. Focus on understanding how ratios change after percentage increases or decreases and how profit calculations can be represented using ratios. These questions often appear as moderate-to-difficult CAT Arithmetic problems and can be highly scoring with sufficient practice.

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

Many CAT Arithmetic questions can be solved efficiently by forming simple linear equations. Topics such as Ages, Profit & Loss, Mixtures, Time & Work, and Time-Speed-Distance often require candidates to translate verbal information into mathematical equations and solve them systematically.

Important Concepts

• Linear Equations in One Variable
• Linear Equations in Two Variables
• Age-Based Problems
• Profit & Loss Equations
• Mixture Equations
• Time and Work Equations
• Time-Speed-Distance Equations

Important Equation-Based Approaches for CAT 2026

Frequently Asked CAT PYQ Concepts

High-Value CAT Shortcuts

CAT 2026 Tip: Many moderate and difficult CAT Arithmetic questions appear complex at first glance but become straightforward once translated into equations. Developing the habit of converting word problems into simple algebraic expressions can significantly improve both speed and accuracy in CAT Quant.

Previously Asked Question Types

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 Time-Speed-Distance

CAT frequently asks multi-concept Arithmetic questions that combine Ratios, Averages, Time & Work, and Time-Speed-Distance (TSD). These questions test a candidate's ability to identify relationships between quantities and apply multiple concepts simultaneously. Such questions are often moderate to difficult in difficulty and appear regularly in CAT previous year papers.

Important Concepts

• Ratio and Speed Relationship

• Ratio and Time Relationship

• Average Speed

• Efficiency and Work

• Combined Work

• Relative Speed

• Weighted Average

Important CAT 2026 Formulas and Shortcuts

Frequently Asked CAT PYQ Concepts

High-Value CAT Shortcuts

CAT 2026 Tip: Most CAT questions from this area are not direct formula-based questions. Instead, they test how well candidates can connect Ratios, Averages, Efficiency, Speed, and Time in a single problem. Focus on understanding proportional relationships rather than memorizing formulas, as this approach can significantly reduce calculation time in the exam.

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.

$a+b = \frac{1}{12},\quad b+c = \frac{1}{16},\quad c+a = \frac{1}{24}$

Adding all: $2(a+b+c) = \frac{1}{12} + \frac{1}{16} + \frac{1}{24}$

$2(a+b+c) = \frac{4+3+2}{48} = \frac{9}{48} = \frac{3}{16}$

$a+b+c = \frac{3}{32}$

$a = \frac{3}{32} - \frac{1}{16} = \frac{1}{32}$

$b = \frac{1}{12} - \frac{1}{32} = \frac{8-3}{96} = \frac{5}{96}$

$c = \frac{1}{16} - \frac{5}{96} = \frac{6-5}{96} = \frac{1}{96}$

Speeds: $a=\frac{1}{32},\ b=\frac{5}{96},\ c=\frac{1}{96}$

Order of speed: fastest $b$, slowest $c$, middle $a$.

Time taken by Amar alone $ = \frac{1}{a} = 32$ months

Preparation Tips for CAT Arithmetic

To prepare CAT arithmetic section, here is a structured approach that will help you boost your CAT percentile.

Common mistakes in arithmetic and their remedies

There are many common mistakes made by a student during preparation and appearing for the CAT exam.

How to Improve Accuracy and Speed

To improve accuracy and speed,

• Learn Vedic maths for the CAT exam for multiplication and finding the square of larger numbers.

• Learn percentage to fraction conversion and vice versa.

• Learn squares, cubes, and tables up to 30.

• Practice questions daily on each topic according to the plan made by you. Also, practice PYQs.

• Do mental maths.

Time Management for Quant Section

During the exam, you should do:

• Allocate time to each question separately.

• Do not waste much time on one question.

• Avoid lengthy calculations.

• Avoid the use of pen when the calculations can be done mentally.

• Always check on time shown on your screen.

Mock Test & Revision Strategy

Practice sets on arithmetic play an important role in boosting your CAT percentile.

What you should do:

• Attempt at least one small practice set in a day.

• Identify the areas where you are making mistakes most and work on them to rectify.

• Make a record of the time taken by you in every test and the score of each test.

• Identify the question patterns you are taking more time on and try to reduce the time in the next test.

• Check if your score improves after the test. In these ways, you can track your performance.

These practices will certainly help you to improve.

CAT 2026 Preparation Resources by Careers360

The candidates can download the various CAT preparation resources designed by Careers360 using the links given below.