CAT 2025 Cover 27% Syllabus to Score Upto 100% | Chapter-wise MCQs | Study Material

Understanding the 27% Rule

“27% syllabus” is not mystical; it is tactical. 27% syllabus consists of the core topics, which are repeated and covers most of the questions.

Last 5 years’ CAT analysis:

Across previous year CAT question papers, a concentrated set of topic clusters generates around 60–80% of solvable, high-confidence scoring questions in QA and DILR. To understand the fact, go through the complete analysis of previous years CAT papers through the given links:

How mastering these topics can lead to 95–99%ile Range

As the previous years’ CAT papers’ analysis suggest, 27% syllabus covers most of the solvable questions that may lead you to maximize your score. For this you need to follow some rules:

• First identify the correct high yield topics in each section through proper analysis of previous years papers.

• Master the basics of each topic

• Balance all the three sections

• Utilize a single and reliable source for each section

• Practice regularly

• Take mock tests and identify your mistakes and rectify them

Section-Wise High-Impact Topics (The 27%)

Identify the 27% high-yield topics in each section and that needs to be prioritised in your preparation. Provided below are the important topics for CAT- QA,VARC,DILR:

Quantitative Aptitude (~27% of QA syllabus)

Why to choose these topics?

Questions from these topics are direct, time-efficient, and being asked frequently. Also, chances of being correct are high. Choosing 5 – 6 topics from these topics form around 27% of the complete QA syllabus but covers 70 to 80% of the total Quantitative Aptitude questions that may help you to get more than 95%ile in QA.

Data Interpretation and Logical Reasoning (4 key set Types)

Here is a list of common Data Interpretation and Logical Reasoning (DILR) sets that are repeatedly asked in CAT previously which covers around 27% of the syllabus but have potential to help you to get more than 95 percentiles.

Key Strategy:

• Focus on 3 to 4 set types that are repeated every year.

• Practice at least 20 sets on each type

Verbal Ability and Reading Comprehension

Here is a list of common VARC sets that are repeatedly asked in CAT previously which covers around 27% of the syllabus but have potential to help you to get more than 95 percentiles.

45-Day Smart Plan Using the 27% Rule

We divide the complete study plan in 45 days in 3 phases as discussed below:

Phase 1 (Day 1–15): Foundation & concept maps

During first 15 days of your preparation, you should focus on to build core concepts for the 27% list. Prepare formula sheets, learn rules to solve questions.

Daily Target: 1 chapter + 15 targeted practice Questions.

Phase 2 (Day 16–30): Consolidation & mixed practice

During the next 15 days, give sectional tests (alternate sections daily).

Practice and begin timed sets.

Start weekly mock tests of half-length on each section.

Phase 3 (Day 31–45): Mock cycle & fine-tune

1st phase of your preparation is very crucial. Your focus should be solely on mock tests.

You should take at least 3 Mocks every week.

Review each thoroughly and prepare an error log.

Last 7 days strategies:

Focus on low volume, high quality revision, formula sheet, and mental conditioning.

Chapter-wise representative MCQs

Below are representative MCQs covering the high-yield topics. Solve them under time pressure to mimic exam conditions.

Quantitative Aptitude — Arithmetic: Percentage & Ratio

Q1. Rajesh and Vimal own 20 hectares and 30 hectares of agricultural land, respectively, which are entirely covered by wheat and mustard crops. The cultivation area of wheat and mustard in the land owned by Vimal are in the ratio of $5: 3$. If the total cultivation area of wheat and mustard are in the ratio $11: 9$, then the ratio of cultivation area of wheat and mustard in the land owned by Rajesh is

1) $4: 3$

2) $3: 7$

3) $1: 1$

4) $7: 9$

Solution: Let the wheat and mustard areas on Vimal's 30 hectares be $5x$ and $3x$ respectively.

So, total area for Vimal = $5x + 3x = 8x = 30 \Rightarrow x = \frac{30}{8} = 3.75$

Therefore, wheat on Vimal’s land = $5x = 18.75$ hectares
Mustard on Vimal’s land = $3x = 11.25$ hectares

Now, total land = $20$ (Rajesh) $+$ $30$ (Vimal) = $50$ hectares

Total land for wheat = $\frac {11}{20} \times 50$ = 27.5 hectares
So, Total land for mustard = $\frac {9}{20} \times 50$ = 22.5 hectares

Now, for Rajesh land for wheat = $27.5-18.75=8.75$ hectares
land for mustard for Rajesh = $22.5-11.25=11.25$ hectares

So, ratio of wheat ($w$) to mustard ($m$) on Rajesh’s land:

$\frac{w}{m} = \frac{8.75}{11.25} = \frac{7}{9}$

Thus, the ratio is $7:9$

Hence, the correct answer is option 4.

Q2. After two successive increments, Gopal's salary became $187.5 \%$ of his initial salary. If the percentage of salary increase in the second increment was twice of that in the first increment, then the percentage of salary increase in the first increment was

1) 25

2) 20

3) 27.5

4) 30

Solution: Let initial salary = $100$

Let the first increment be $x%$
Then salary becomes: $100 \times \left(1 + \frac{x}{100}\right) = 100 + x$

Second increment = $2x%$
Final salary = $(100 + x) \times \left(1 + \frac{2x}{100}\right)$

Given final salary = $187.5%$ of original salary = $187.5$

So: $
(100 + x)\left(1 + \frac{2x}{100}\right) = 187.5
$

$⇒
(100 + x)\left(\frac{100 + 2x}{100}\right) = 187.5
$

$⇒
(100 + x)(100 + 2x) = 18750
$

$⇒
10000 + 200x + 100x + 2x^2 = 18750$

$
\Rightarrow 2x^2 + 300x + 10000 = 18750$

$
\Rightarrow 2x^2 + 300x - 8750 = 0
$

$⇒
x^2 + 150x - 4375 = 0
$

$⇒
x^2 + 175x-25x - 4375 = 0
$

$⇒
(x + 175)(x-25) = 0
$

Take positive root: $
x = 25
$

$25\%$ increase in the first increment.

Hence, the correct answer is option 1.

Quantitative Aptitude — Number Systems / Remainders

Q3. The number of all positive integers up to 500 with non-repeating digits is

Solution: We want to count all positive integers $\leq 500$ with non-repeating digits.

We'll consider $1$-digit, $2$-digit, and $3$-digit numbers, ensuring they are $\leq 500$ and digits do not repeat.

Case 1: 1-digit numbers (1 to 9):

All digits are unique. So, the total possible such numbers = 9

Case 2: 2-digit numbers (10 to 99):

Digit at Tens place can be from 1 to 9, so possible ways to fill tens place = 9
Digit at Units place can be any digit except the digit at tens place ie. 9 choices

So, the total possible such numbers = $9 \times 9 = 81$

Case 3: 3-digit numbers (100 to 499):

We limit to numbers from $100$ to $499$ (since $\leq 500$). So hundreds digit = $1$ to $4$ (4 choices)

For each:

Tens digit: choose from $0$–$9$, excluding hundreds digit $\Rightarrow$ 9 choices
Unit digit: choose from remaining digits $\Rightarrow$ 8 choices

So, the total possible such numbers = $4 \times 9 \times 8 = 288$

Case 4: 500

Digits: $5, 0, 0$ → Repeated digits → Not allowed.

So, exclude $500$

Total count$ = 9 + 81 + 288 = 378$

Hence, the correct answer is $378$.

Quantitative Aptitude — Algebra (Linear & Quadratic)

Q4. The sum of all distinct real values of $x$ that satisfy the equation $10^x+\frac{4}{10^x}=\frac{81}{2}$, is

1) $4 \log _{10} 2$

2) $3 \log _{10} 2$

3) $2 \log _{10} 2$

4) $\log _{10} 2$

Solution: Let $y = 10^x$
Then the equation becomes:

$y + \frac{4}{y} = \frac{81}{2}$

$⇒y^2 + 4 = \frac{81}{2}y$

$⇒2y^2 + 8 = 81y$

$⇒2y^2 - 81y + 8 = 0$

Since $y = 10^x$ must be positive, both roots are valid as long as they are positive real numbers.
Let the roots be $y_1$ and $y_2$. Then:

So, the roots of original equation, $x_1 = \log_{10} y_1$ or $x_2= \log_{10} y_2$

We know: $y_1 y_2 = \frac{8}{2} = 4$

So: $x_1 + x_2 = \log_{10}(y_1y_2)= \log_{10}(4) = \log_{10} 4$

The sum of all distinct real values of $x$ is $2 \log _{10} 2$.

Hence, the correct answer is option 3.

Q5. For some constant real numbers $p, k$ and $a$, consider the following system of linear equations in $x$ and $y$ :

$\begin{aligned}
& p x-4 y=2 \\
& 3 x+k y=a
\end{aligned}$

A necessary condition for the system to have no solution for ( $x, y$ ), is

1) $2 a+k \neq 0$

2) $k p+12 \neq 0$

3) $a p+6=0$

4) $a p-6=0$

Solution: The given equations are:
$px - 4y = 2$
$3x + ky = a$

For a system of two linear equations $a_1x+b_1y=c_1$ and $a_2x+b_2y=c_2$, the condition for no solution is given by
$\frac{a_1}{a_2}=\frac{b_1}{b_2} \neq \frac{c_1}{c_2}$

So, from the given set of equations
$\frac{p}{3}=\frac{-4}{k} \neq \frac{2}{a}$

$⇒\frac{-4}{k} \neq \frac{2}{a}$

$⇒-4a \neq 2k$
$⇒2k+4a \neq 0$
$⇒k+2a \neq 0$

Hence, the correct answer is option 1.

Quantitative Aptitude — Modern Maths

Q6. All the values of $x$ satisfying the inequality $\frac{1}{x+5} \leq \frac{1}{2 x-3}$ are

1) $-5<x<\frac{3}{2}$ or $\frac{3}{2}<x \leq 8$

2) $x<-5$ or $x>\frac{3}{2}$

3) $x<-5$ or $\frac{3}{2}<x \leq 8$

4) $-5<x<\frac{3}{2}$ or $x>\frac{3}{2}$

Solution: Given: $\frac{1}{x+5} \leq \frac{1}{2x - 3}$

$⇒2x-3 \leq x+5$

$⇒x \leq 8$

But $x \neq -5$ and $x \neq \frac 32$ as at these values fractions are not defined.

So, the set of solutions becomes

$x<-5$ or $\frac{3}{2}<x \leq 8$

Hence, the correct answer is option 3.

Q7. For any non-zero real number $x$, let $f(x)+2 f\left(\frac{1}{x}\right)=3 x$. Then, the sum of all possible values of $x$ for which $f(x)=3$, is

1) -2

2) 2

3) -3

4) 3

Solution: We are given:
$
f(x) + 2f\left(\frac{1}{x}\right) = 3x \quad-------- \text{(1)}
$
Also, we are told: $ f(x) = 3 $

Substituting into (1):
$
3 + 2f\left(\frac{1}{x}\right) = 3x \Rightarrow f\left(\frac{1}{x}\right) = \frac{3x - 3}{2} \quad --------\text{(2)}
$

Now use (1) again but switch $ x $ with $ \frac{1}{x} $:
$
f\left(\frac{1}{x}\right) + 2f(x) = \frac{3}{x}
$
Substitute $ f(x) = 3 $:
$
f\left(\frac{1}{x}\right) + 6 = \frac{3}{x} \Rightarrow f\left(\frac{1}{x}\right) = \frac{3}{x} - 6 \quad------ \text{(3)}
$

Equating (2) and (3):
$
\frac{3x - 3}{2} = \frac{3}{x} - 6
$

Multiply both sides by 2:
$
3x - 3 = \frac{6}{x} - 12
$

Bring all terms to one side:
$
3x + 9 - \frac{6}{x} = 0
$

Multiply through by $ x $:
$
3x^2 + 9x - 6 = 0
$

Divide by 3:
$
x^2 + 3x - 2 = 0
$

Solve using quadratic formula:
$
x = \frac{-3 \pm \sqrt{3^2 + 4 \cdot 1 \cdot 2}}{2} = \frac{-3 \pm \sqrt{17}}{2}
$

Sum of all possible values of $ x $ is:
$
\frac{-3 + \sqrt{17}}{2} + \frac{-3 - \sqrt{17}}{2} = -3
$

Hence, the correct answer is option 3.

Quantitative Aptitude — Geometry and Mensuration

Q8. A circular plot of land is divided into two regions by a chord of length $10 \sqrt{3}$ meters such that the chord subtends an angle of $120^{\circ}$ at the center. Then, the area, in square meters, of the smaller region is

1) $20\left(\frac{4 \pi}{3}+\sqrt{3}\right)$

2) $20\left(\frac{4 \pi}{3}-\sqrt{3}\right)$

3) $25\left(\frac{4 \pi}{3}-\sqrt{3}\right)$

4) $25\left(\frac{4 \pi}{3}+\sqrt{3}\right)$

Solution: The chord subtends an angle of $120^\circ$ at the center.
Let the radius of the circle be $r$.

In a triangle formed by two radii and the chord, apply the Cosine Rule:

Using the cosine rule in $\triangle AOB$ (where $AB = 10\sqrt{3}$ and angle $\angle AOB = 120^\circ$):

$
AB^2 = AO^2 + BO^2 - 2 \cdot AO \cdot BO \cdot \cos(120^\circ)
$

$
⇒(10\sqrt{3})^2 = 2r^2 - 2r^2 \cdot \cos(120^\circ)
$

$
⇒300 = 2r^2 - 2r^2 \cdot \left(-\frac{1}{2}\right)
$

$
⇒300 = 2r^2 + r^2 = 3r^2 \Rightarrow r^2 = 100 \Rightarrow r = 10
$

Now, area of the smaller region = area of sector $AOB$ – area of triangle $AOB$

Angle at center = $120^\circ = \frac{2\pi}{3}$ radians
Radius = $10$

Area of sector: $
\frac{1}{2} \cdot r^2 \cdot \theta = \frac{1}{2} \cdot 100 \cdot \frac{2\pi}{3} = \frac{100\pi}{3}
$

Area of triangle (using formula for triangle with angle between two sides): $
\frac{1}{2} \cdot r^2 \cdot \sin(120^\circ) = \frac{1}{2} \cdot 100 \cdot \frac{\sqrt{3}}{2} = 25\sqrt{3}
$

Area of smaller region (segment) $ =
\frac{100\pi}{3} - 25\sqrt{3}
=25\left(\frac{4 \pi}{3}-\sqrt{3}\right)$ sq. meters.

Hence, the correct answer is option 3.

DILR — (Data Interpretation)

Q9. Comprehension:
Out of 10 countries -- Country 1 through Country 10 -- Country 9 has the highest gross domestic product (GDP), and Country 10 has the highest GDP per capita. GDP per capita is the GDP of a country divided by its population. The table below provides the following data about Country 1 through Country 8 for the year 2024.

- Column 1 gives the country's identity.
- Column 2 gives the country's GDP as a fraction of the GDP of Country 9.
- Column 3 gives the country's GDP per capita as a fraction of the GDP per capita of Country 10.
- Column 4 gives the country's annual GDP growth rate.
- Column 5 gives the country's annual population growth rate.

Assume that the GDP growth rates and population growth rates of the countries will remain constant for the next three years.

Question: Which one among the countries 1 through 8 has the smallest population in $2024 ?$

1) Country 8

2) Country 3

3) Country 7

4) Country 5

Solution: When we need to compare two countries in the list, either by GDP, GDP per capita or population, we may simply use the values found in the table, as they all depend on a common reference point.

In this question, we need to compare the populations of the countries.

Population = $\frac{GDP}{\text{GDP per capita}}$

So, let's analyse the options:

Option A: Country 8
GDP = 0.07 and GDP Per Capita = 0.41
So, Population = $\frac{0.07}{0.41} \ or \ \frac{7}{41}$

Option B: Country 3
GDP = 0.13 and GDP Per Capita = 0.02
So, Population = $\frac{0.13}{0.02} \ or \ \frac{13}{2}$

Option C: Country 7
GDP = 0.08 and GDP Per Capita = 0.3

Population=$\frac{0.08}{0.3} \ or \ \frac{8}{30}$

Option D: Country 5
GDP = 0.1 and GDP Per Capita = 0.36

Population=$\frac{0.1}{0.36} \ or \ \frac{10}{36}$

Comparing the four values, we are looking for the smallest value among the four,

$\frac{7}{41}, \frac{13}{2}, \frac{8}{30}, \frac{10}{36}$

From the above four $\frac{7}{41}$ is the smallest.

Hence, the first option is correct.

DILR — (Logical Reasoning)

Q10. Comprehension:
Eight gymnastics players numbered 1 through 8 underwent a training camp where they were coached by three coaches - Xena, Yuki, and Zara. Each coach trained at least two players. Yuki trained only even numbered players, while Zara trained only odd numbered players. After the camp, the coaches evaluated the players and gave integer ratings to the respective players trained by them on a scale of 1 to 7 , with 1 being the lowest rating and 7 the highest.

The following additional information is known.
1. Xena trained more players than Yuki.
2. Player-1 and Player-4 were trained by the same coach, while the coaches who trained Player-2, Player-3 and Player-5 were all different.
3. Player-5 and Player-7 were trained by the same coach and got the same rating. All other players got a unique rating.
4. The average of the ratings of all the players was 4.
5. Player-2 got the highest rating.
6. The average of the ratings of the players trained by Yuki was twice that of the players trained by Xena and two more than that of the players trained by Zara.
7. Player-4's rating was double of Player-8's and less than Player-5's.

Question: What was the rating of Player-6?

Solution: From conditions 3 and 4, only two players got the same ratings. Thus, let $K$ be the repetitive number, then:

$\frac{(1+2+3+4+5+6+7+K)}{8} = 4$

It means $K$ = 4

Thus, players 5 and 7 got the same rating. As Xena trained more players than Yuki. So, there are two possibilities, which are as follows:

Possibility 1, X = $\frac{38}{13}$, (Not possible)

Possibility 2, X = 3, (Possible)

Step 2:

From condition 5, player 2 got the rating of 7. Also, player 4 got the rating of 2, and player 8 got the rating of 1.

So, we get:

Yuki has 2 players, and their total score is 12. The only available combination that works is 7 + 5. Because player-2 got a score of 7, the other score can only be that of player-6, that is 5. This is also consistent with the fact that the only even-numbered player left is player-6, and Yuki trained only even-numbered players.
Therefore, Yuki trains player-2 and player-6.

The overall score of Zara is 8. The correct combinations used in 8 are 2+6 and 4+4.
The point of 2 is player 4, but Zara trained only those players who have an odd number; therefore, it is not possible.
The only valid move is 4 + 4, which can be given to player-5 and player-7, both belong to odd-numbered.
Therefore, Zara has to be above player-5 and player-7.

Thus, the table of the players and their trainers is as follows:

Thus, the rating of player 6 is 5.

Hence, 5 is the correct answer.

VARC — Reading Comprehension

Q11. The passage given below is followed by four alternate summaries. Choose the option that best captures the essence of the passage.

When the tradwife puts on that georgic, pinstriped dress, she is not just admiring the visual cues of a fantastical past. She takes these dreams of storybook bliss literally, tracing them backward in time until she reaches a logical conclusion that satisfies her. And by doing so, she ends up delivering an unhappy reminder of just how much our lives consist of artifice and playacting. The tradwife outrages people because of her deliberately regressive ideals. And yet her behaviour is, on some level, indistinguishable from the nontradwife's. The tradwife's trollish genius is to beat us at our own dress-up game. By insisting that the idyllic cottage daydream should be real, right down to the primitive gender roles, she leaves others feeling hollow, cheated. The hullabaloo and headaches she causes may be the price we pay for taking too many things at face value: our just deserts, served Instagram-perfect by a manicured hand on a gorgeous ceramic dish, with fat, mouthwatering maraschino cherries on top.

1) The tradwife's vintage dress and adherence to traditional roles reveal the artificial nature of modern life and its superficial values.

2) By promoting an idealized past, the tradwife exposes the artifice of contemporary values and mocks societal norms.

3) The tradwife, with her vintage dress and traditional roles, highlights the superficiality of modern life and challenges current societal norms.

4) The tradwife's commitment to outdated gender roles and retro fashion critiques the superficiality of today's societal ideals.

Solution: Option 3 is the most accurate because it captures the two main ideas of the passage:

The tradwife uses old-fashioned clothing and traditional gender roles to create a staged version of the past. In doing so, she ends up exposing how shallow and artificial modern life and social behaviour have become, making people uncomfortable by reflecting their own pretence back at them.

The other options miss either the element of social critique or the idea of exposing superficiality. Option 3 includes both.

How to turn the 27% focus into full paper readiness

With proper and smart planning, you can convert 27% syllabus into almost 100 percentiles in CAT. You can follow some strategies as suggested by experts and 100 percentile achievers are listed here:

1. Template notebook: Keep short templates or CAT formula sheet per topic.

Example:

5 – 6 lines approach to seating arrangement problems.

Solving percentage questions through fractions.

List of algebra formulas etc.

2. Cope up with mock tests: Mock tests are very crucial as it helps you to find your weaker sections, improve your accuracy and efficiency.

You can start with 1 CAT 2025 mock test /week and then escalate to 3/week in the last 15 days.

Review each mock for the types of mistakes you are making.

3. Sectional mastery: To gain sectional mastery, practice 50-100 high-quality problems from each high-yield topic, then move to mixed sets.

Example:

Build foundation in ratio, practice 50 – 80 questions from this topic and then apply the concepts in TSD, Time and Work etc.

4. Time Management: Practice with the approach of selective skipping to manage time during actual CAT exam. In the exam, allot max 35–40 minutes per section; leave tricky DILR sets if they are taking more than 12 minutes per set.

Common mistakes & how to avoid them

Recommended study material & resources

Here, we suggest you some useful CAT 2025 study materials which are available in hard bound, on online platforms and some free practice materials.

Quantitative Aptitude:

1. Arun Sharma: A Quantitative Approach for CAT (7th Edition)
2. Quantitative Aptitude for CAT by Nishit K Sinha
3. NCERT class 11 and class 12

Logical Reasoning and Data Interpretation:

1. Data Interpretation & Logical Reasoning by Gautam Puri

2. Logical Reasoning and Data Interpretation for CAT by Nishit K Sinha (Pearson)

3. Data Interpretation & Logical Reasoning by Arun Sharma (McGraw Hill)

Verbal Aptitude and Reading Comprehension:

1. "Verbal Ability & Reading Comprehension for the CAT" by Arun Sharma and Meenakshi Upadhyay

2. "Word Power Made Easy" by Norman Lewis

3. "Verbal Ability for the CAT" by Thorpe

4. Read editorials (Economist, The Hindu), practice RC passages from previous CATs.

CAT 2025 Preparation Resources by Careers360

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