CAT 2025 Quantitative Aptitude Practice Questions with Solutions

CAT 2025 Quantitative Aptitude Practice Questions - The candidates preparing for the CAT 2025 examination may face the biggest challenge when attempting the CAT quantitative aptitude section. One of the major reasons candidates find the CAT quantitative aptitude section the most challenging is that it has a vast syllabus, and the questions are lengthy and time-consuming. The right approach to ensure a good CAT score under the quantitative aptitude section is by solving as many CAT quantitative aptitude practice problems as possible. Being one of the most difficult sections of the CAT examination, securing a good CAT score under the quantitative aptitude section can significantly enhance the CAT percentile of the candidates and thereby the chances of getting selected to the best B schools in India. Solving as many CAT quantitative aptitude practice questions is the only way through which the candidates can gain the necessary practice required for scoring well in the CAT quantitative aptitude section. In this article of Careers360, we will discuss the various sets of CAT quant practice questions necessary for the candidate’s CAT 2025 preparation.

This Story also Contains

1. CAT Quantitative Aptitude Section – Exam Overview
2. CAT Quantitative Aptitude 2025 Syllabus
3. CAT 2025 Quantitative Aptitude Questions
4. CAT Quantitative Aptitude Important Topics and Practice Questions
5. Best Books for CAT Quantitative Aptitude Preparation
6. CAT Preparation Materials Designed by Careers360

CAT Quantitative Aptitude Section – Exam Overview

For the past years, the CAT exam pattern for the quantitative aptitude section has remained consistent without a lot of changes. The candidates are supposed to answer 22 questions asked from a variety of CAT quant topics. The candidates are given a fixed sectional timing of 40 minutes to answer the questions. The CAT expected quantitative aptitude exam pattern for the CAT 2025 examination conducted by IIM Kozhikode is provided below.

CAT Quantitative Aptitude 2025 Syllabus

When compared to the other sections of the CAT examination, such as the CAT VARC and CAT DILR sections, the CAT Quantitative Aptitude topic is very broad and requires a lot of time to completely solve. This is one of the major issues faced by the candidates during their CAT quantitative aptitude preparation. It is necessary for the candidates to curate their own achievable CAT preparation timetable for the CAT Quantitative Aptitude section and strictly follow it to ensure that they do not miss any CAT Quant topic. The comprehensive CAT Quantitative Aptitude syllabus, after analysing the CAT previous year questions, is provided below for the reference of the candidates.

CAT 2025 Quantitative Aptitude Questions

The efficiency of the candidate’s CAT Quantitative Aptitude preparation is greatly determined by the number of CAT Quantitative Aptitude practice questions attempted by the candidates. It is important for the candidates to solve a minimum of 50 questions of different types under the CAT QA section to enhance their preparation. A few selected CAT 2025 quantitative aptitude practice questions are listed below for the reference of the candidates.

CAT 2025 Quantitative Aptitude Practice Questions: Arithmetic Section

1. A certain amount of water was poured into a 300-litre container, and the remaining portion of the container was filled with milk. Then an amount of this solution was taken out from the container which was twice the volume of water that was earlier poured into it, and water was poured to refill the container again. If the resulting solution contains $72 \%$ milk, then the amount of water, in litres, that was initially poured into the container was

1. 30

2. 40

3. 50

4. 60

Solution:

Let the amount of water initially poured = $x$ litres
Then, amount of milk initially = $300 - x$ litres

An amount of solution equal to $2x$ litres is removed.
This removed solution contains:
Water = $\frac{x}{300} \times 2x = \frac{2x^2}{300}$ litres
Milk = $\frac{300 - x}{300} \times 2x = \frac{2x(300 - x)}{300}$ litres

Now refill the container with $2x$ litres of water, so final quantities are:
Water = $x - \frac{2x^2}{300} + 2x = x\left(1+2 - \frac{2x}{300}\right) = x\left(3 - \frac{2x}{300}\right)$

Milk = $(300 - x) - \frac{2x(300 - x)}{300}$
= $(300 - x)\left(1 - \frac{2x}{300}\right)$

Given: milk is $72\%$ of 300 litres = $216$ litres
So,

$(300 - x)\left(1 - \frac{2x}{300}\right) = 216$
$⇒(300 - x)\left(\frac{300 - 2x}{300}\right) = 216$
$\Rightarrow \frac{(300 - x)(300 - 2x)}{300} = 216$
$⇒(300 - x)(300 - 2x) = 64800$
$⇒90000 - 900x + 2x^2 = 64800$
$\Rightarrow 2x^2 - 900x + 25200 = 0$
$⇒x^2 - 450x + 12600 = 0$
$⇒x^2 - 420x-30x + 12600 = 0$
$⇒(x-420)(x-30) = 0$

So, $x = 420$ or $x = 30$

Reject $x = 420$ as it exceeds 300 litres.

Therefore, the amount of water initially poured is $30$ litres.

Hence, the correct answer is $30$.

2. A train travelled a certain distance at a uniform speed. Had the speed been 6 km per hour more, it would have needed 4 hours less. Had the speed been 6 km per hour less, it would have needed 6 hours more. The distance, in km, travelled by the train is

1. 800

2. 720

3. 780

4. 640

Solution:

Let the original speed of the train be $x$ km/hr and the distance be $d$ km.

Then, time taken = $\frac{d}{x}$

According to the question:

If speed was increased by $6$ km/hr, time taken would be $4$ hours less:

$\frac{d}{x} - \frac{d}{x+6} = 4 \quad----- \text{(1)}$

If speed was decreased by $6$ km/hr, time taken would be $6$ hours more:

$\frac{d}{x-6} - \frac{d}{x} = 6 \quad ------\text{(2)}$

Add equations (1) and (2):

$\left( \frac{d}{x} - \frac{d}{x+6} \right) + \left( \frac{d}{x-6} - \frac{d}{x} \right) = 4 + 6$

$⇒\frac{d}{x-6} - \frac{d}{x+6} = 10$

Take $d$ common:

$d \left( \frac{1}{x-6} - \frac{1}{x+6} \right) = 10$

$\Rightarrow d \cdot \frac{(x+6)-(x-6)}{(x-6)(x+6)} = 10$

$\Rightarrow d \cdot \frac{12}{x^2 - 36} = 10$

$\Rightarrow d = \frac{10(x^2 - 36)}{12} = \frac{5(x^2 - 36)}{6}$

Now, plug this $d$ into equation (1):

$\frac{d}{x} - \frac{d}{x+6} = 4$

$\Rightarrow \frac{5(x^2 - 36)}{6x} - \frac{5(x^2 - 36)}{6(x+6)} = 4$

Take common factor:

$\frac{5(x^2 - 36)}{6} \left( \frac{1}{x} - \frac{1}{x+6} \right) = 4$

$⇒\frac{5(x^2 - 36)}{6} \cdot \frac{6}{x(x+6)} = 4$

$\Rightarrow \frac{5(x^2 - 36)}{x(x+6)} = 4$

Now simplify:

$5(x^2 - 36) = 4x(x+6)$

$⇒5x^2 - 180 = 4x^2 + 24x$

$⇒5x^2 - 4x^2 - 24x - 180 = 0$

$⇒x^2 - 24x - 180 = 0$

Solve the quadratic:

$x = \frac{24 \pm \sqrt{576 + 720}}{2} = \frac{24 \pm \sqrt{1296}}{2} = \frac{24 \pm 36}{2}$

So $x = 30$ (discard $-6$)

The distance:

$d = \frac{5(x^2 - 36)}{6} = \frac{5(900 - 36)}{6} = \frac{5 \cdot 864}{6} = 720$ km

Hence, the correct answer is option 2.

3. 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 $6$.

CAT 2025 Quantitative Aptitude Practice Questions: Algebra Section

1. If $a, b$ and $c$ are positive real numbers such that $a>10 \geq b \geq c$ and $\frac{\log _8(a+b)}{\log _2 c}+\frac{\log _{27}(a-b)}{\log _3 c}=\frac{2}{3}$, then the greatest possible integer value of $a$ is

1. 14

2. 15

3. 16

4. 17

Solution:

$\frac{\log _8(a+b)}{\log _2 c}+\frac{\log _{27}(a-b)}{\log _3 c}=\frac{2}{3}$

$⇒\frac{\log _{2^3}(a+b)}{\log _2 c}+\frac{\log _{3^3}(a-b)}{\log _3 c}=\frac{2}{3}$

$⇒\frac13\frac{\log _2(a+b)}{\log _2 c}+\frac13 \frac{\log _3(a-b)}{\log _3 c}=\frac{2}{3}$

$⇒\frac{\log _2(a+b)}{\log _2 c}+ \frac{\log _3(a-b)}{\log _3 c}=2$

We use change of base formula:

$\log _c(a+b)+\log _c(a-b)=2$

Use log addition:

$\log_c[(a + b)(a - b)] = 2$

So:

$(a + b)(a - b) = c^2$
$⇒a^2 - b^2 = c^2$
Also, it is given that $a > 10 \geq b \geq c > 0$

To maximize $a$, $b$ should be minimized, which is $b=10$ according to the given condition.
Now $a$ is maximum if the value of $c$ is close to 10.

Since $a$ should be an integer satisfying $a^2 = b^2 + c^2$,
$b^2 = 100$, $c^2$ is also close to 100 so as to maximize $a$, the perfect square close to 200 is 196.
So, $a^2 = 196 \implies a =14$

Hence, the correct answer is $14$.

2. If $(a+b \sqrt{3})^2=52+30 \sqrt{3}$, where $a$ and $b$ are natural numbers, then $a+b$ equals

1. 9

2. 10

3. 7

4. 8

Solution:

We are given that:

$(a + b\sqrt{3})^2 = 52 + 30\sqrt{3}$

$⇒ (a^2 + 3b^2) + 2ab\sqrt{3}=52 + 30\sqrt{3}$

Now, compare real and irrational parts:

• Real part: $a^2 + 3b^2 = 52$

• Irrational part: $2ab = 30$

From $2ab = 30$, we get:

$ab = 15 \quad ------\text{(i)}$

Substitute $b = \dfrac{15}{a}$ into the first equation:

$a^2 + 3\left(\dfrac{15}{a}\right)^2 = 52$
$⇒a^2 + \dfrac{3 \cdot 225}{a^2} = 52$
$⇒a^4 + 675 = 52a^2$
$⇒a^4 - 52a^2 + 675 = 0$

Let $x = a^2$, then: $x^2 - 52x + 675 = 0$

Solve the quadratic equation:

$x = \dfrac{52 \pm \sqrt{52^2 - 4 \cdot 675}}{2} = \dfrac{52 \pm \sqrt{2704 - 2700}}{2} = \dfrac{52 \pm 2}{2}$
$⇒x = 27 \Rightarrow a = \sqrt{27} \quad (\text{not natural}), \quad x = 25 \Rightarrow a = 5$

So $a = 5$, from (i) $b = \dfrac{15}{5} = 3$

Therefore: $a+b=5+3=8$

Hence, the correct answer is option 4.

3. The roots $\alpha, \beta$ of the equation $3 x^2+\lambda x-1=0$, satisfy $\frac{1}{\alpha^2}+\frac{1}{\beta^2}=15$. The value of $\left(\alpha^3+\beta^3\right)^2$, is

1. 1

2. 4

3. 9

4. 16

Solution:

Given: $\frac{1}{\alpha^2} + \frac{1}{\beta^2} = 15$

We know: $\frac{1}{\alpha^2} + \frac{1}{\beta^2} = \frac{\alpha^2 + \beta^2}{\alpha^2 \beta^2}$

From the equation $3x^2 + \lambda x - 1 = 0$,
sum of roots $\alpha + \beta = -\frac{\lambda}{3}$, product of roots $\alpha \beta = -\frac{1}{3}$
So, $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = \left(\frac{\lambda^2}{9}\right) + \frac{2}{3}$

Also, $\alpha^2 \beta^2 = (\alpha \beta)^2 = \left(\frac{1}{9}\right)$

Now,
$\frac{\alpha^2 + \beta^2}{\alpha^2 \beta^2} = 15$
$\Rightarrow \left(\frac{\lambda^2}{9} + \frac{2}{3} \right) \div \frac{1}{9} = 15$
$\Rightarrow 9\left( \frac{\lambda^2}{9} + \frac{2}{3} \right) = 15$
$\Rightarrow \lambda^2 + 6 = 15 \Rightarrow \lambda^2 = 9 \Rightarrow \lambda = \pm 3$

Now, $\alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta)$

We have: $\alpha + \beta = -\frac{\lambda}{3} = \pm 1$, $\alpha\beta = -\frac{1}{3}$

Case 1: $\alpha + \beta = - 1$

$\alpha^3 + \beta^3 = (-1)^3 - 3\left(-\frac{1}{3}\right)(-1) = -1 - 1 = -2$

We get, $(\alpha^3 + \beta^3)^2 = (-2)^2 = 4$

Case 2: $\alpha + \beta = 1$

$\alpha^3 + \beta^3 = (1)^3 - 3\left(-\frac{1}{3}\right)(1) = 1 + 1 = 2$

We get, $(\alpha^3 + \beta^3)^2 = (2)^2 = 4$

Hence, the correct answer is option 2.

CAT 2025 Quantitative Aptitude Practice Questions: Geometry Section

1. Let C be the circle $x^2+y^2+4x−6y−3=0$ and L be the locus of the point of intersection of a pair of tangents to C with the angle between the two tangents equal to 60o. Then, the point at which L touches the line $x = 6$ is:

1. (6,4)

2. (6,8)

3. (6,3)

4. (6,6)

Solution:

We need to find the equation of a circle at the point of intersection of two tangents (at the same centre).

Centre of circle x2+ y2+ 2gx + 2fy + c = 0 is (-g, -f) and radius = $\sqrt {g^2+f^2-c}$

So, the centre of x2+y2+4x−6y−3=0 is (-2, 3) and radius = $\sqrt {2^2+3^2+3} = 4$

Now, $\sin 30° = \frac {4}{OL}$

So, OL = 8 units

Equation of circle passing through L: $(x +2)^2 + (y -3)^2 = 8^2$

So, L touches the line $x=6$ at (6, 3). (put x = 6 in the above equation to get the coordinates).

Hence, the correct answer is option (3).

2. In a rectangle ABCD, AB = 9 cm and BC = 6 cm. P and Q are two points on BC such that the areas of the figures ABP, APQ, and AQCD are in geometric progression. If the area of the figure AQCD is four times the area of triangle ABP, then BP: PQ: QC is

1. 1:1:2

2. 1:2:4

3. 2:4:1

4. 1:2:1

Solution:

The areas of the figures ABP, APQ, and AQCD are in geometric progression.Area of ABP = m
Area of APQ = mr
Area of AQCD = mr2
Also Area of AQCD = 4(Area of ABP)
⇒ mr2 = 4m
⇒ r = 2
So, the total area = Area of rectangle = 9 $\times$ 6 = 54
So, m + 2m + 4m = 54
⇒ m = $\frac{54}{7}$

Area of ABP = m = $\frac{54}{7}$
⇒ ½ (AB) (BP) = $\frac{54}{7}$

⇒ BP = $\frac{12}{7}$ (Since AB = 9)

Area of ABQ = 2m + m = $\frac{162}{7}$
⇒ ½ (AB) (BQ) = $\frac{162}{7}$

⇒ BQ = $\frac{36}{7}$ (Since AB = 9)

Now, PQ = BQ – BP =$\frac{36}{7}$ – $\frac{12}{7}$ = $\frac{24}{7}$
QC = BC – BQ = 6 – $\frac{36}{7}$ = $\frac{6}{7}$

Now, BP: PQ: QC = ($\frac{12}{7}$) : ($\frac{24}{7}$) : ($\frac{6}{7}$) = 2:4:1

Hence, the correct answer is option (3).

3. A triangle is drawn with its vertices on the circle C such that one of its sides is a diameter of C and the other two sides have their lengths in the ratio a:b. If the radius of the circle is r, then the area of the triangle is

1. $\frac {abr^2}{2(a^2+b^2)}$

2. $\frac {abr^2}{(a^2+b^2)}$

3. $\frac {4abr^2}{(a^2+b^2)}$

4. $\frac {2abr^2}{(a^2+b^2)}$

Solution:

Let the sides are $ak$, and $bk$
The triangle formed should be right-angled as the angle subtends by the diameter of a circle on its circumference is a right angle.

Using Pythagoras theorem,
$(2r)^2 = (ak)^2 + (bk)^2$

⇒ $k^2 = \frac {4r^2}{a^2 + b^2}$
Area of triangle = $\frac 12 \times ak \times bk$

= $\frac 12 \times abk^2$

= $\frac 12 \times ab \times \frac {4r^2}{a^2 + b^2}$

= $\frac {2abr^2}{a^2 + b^2}$

Hence, the correct answer is option (4).

CAT 2025 Quantitative Aptitude Practice Questions: Modern Maths Section

1. The sum of all four-digit numbers that can be formed with the distinct non-zero digits $a, b, c$, and $d$, with each digit appearing exactly once in every number, is $153310+n$, where $n$ is a single-digit natural number. Then, the value of $(a+b+c+d+n)$ is

Solution:

Total number of 4-digit numbers formed using $a, b, c, d = 4! = 24$
Each digit appears $6$ times in each place (units, tens, hundreds, thousands), so:

Sum of all numbers = $6(a + b + c + d)(1 + 10 + 100 + 1000) = 6(a + b + c + d) \cdot 1111 = 6666(a + b + c + d)$

Given: $6666(a + b + c + d) = 153310 + n$
The sum of a, b, c, and d should be so that $6666(a + b + c + d)$ should lie between 153310 and 153320.

Estimate: $\frac{153310}{6666} \approx 22.99 \Rightarrow a + b + c + d = 23$

Now, $6666 \times 23 = 153318 \Rightarrow n = 153318 - 153310 = 8$
So, $a + b + c + d + n = 23 + 8 = 31$

Hence, the correct answer is $31$.

2. In a box of chocolates, there are 4 different types of chocolates. If you want to select 2 chocolates to eat, how many different pairs of chocolates can you choose?

1. 4 pairs

2. 6 pairs

3. 8 pairs

4. 12 pairs

Solution:

To select 2 chocolates from the 4 different types, we will use combinations.

${ }^4 \mathrm{C}_2=\frac{4!}{2!(4-2)!}=6$

So, you can choose from 6 different pairs of chocolates.

Hence, the correct answer is 6 pairs.

3. How many ways can one choose 6 face cards from a deck of 52 cards such that exactly two kings of the same colour are there?

1. 364

2. 123

3. 234

4. 567

Solution:

There are only 12 face cards.
Let $B_1$ and $B_2$ be two black kings and $R_1$ and $R_2$ be two red kings.
Case 1: $B_1$ and $B_2$ are selected then, $R_1$ and $R_2$ can not be selected together. If $R_1$ is selected and $R_2$ is not selected then the remaining 3 cards must be chosen from 8 cards i.e. $\mathrm{{}^8C_3}$ ways.
If $R_2$ is selected and $R_1$ is not selected then the remaining 3 cards must be chosen from the rest of 8 cards i.e. $\mathrm{{}^8C_3}$ ways.
If $R_1$ and $R_2$ both are not selected then the remaining 4 cards must be chosen from the rest of 8 cards i.e. $\mathrm{{}^8C_4}$ ways.
Total ways $=\mathrm{{}^8C_3}+\mathrm{{}^8C_3}+\mathrm{{}^8C_4}=56 + 56 + 70 = 182$

Case 2: $R_1$ and $R_2$ are selected then, $B_1$ and $B_2$ cannot be selected together. If $B_1$ is selected and $B_2$ is not selected then the remaining 3 cards must be chosen from the rest of 8 cards i.e. $\mathrm{{}^8C_3}$ ways.
If $B_2$ is selected and $B_1$ is not selected then the remaining 3 cards must be chosen from the rest of 8 cards i.e. $\mathrm{{}^8C_3}$ ways.
If $B_1$ and $B_2$ both are not selected then the remaining 4 cards must be chosen from the rest of 8 cards i.e. $\mathrm{{}^8C_4}$ ways.
Total ways $=\mathrm{{}^8C_3}+\mathrm{{}^8C_3}+\mathrm{{}^8C_4}=56 + 56 + 70 = 182$
So, the answer is (182 + 182) = 364 ways.
Hence, the correct answer is option (1).

CAT 2025 Quantitative Aptitude Practice Questions: Number Systems Section

1. When $10^{100}$ is divided by 7, the remainder is

1. 3

2. 4

3. 6

4. 1

Solution:

We know that $10 \div 7$ leaves a remainder of $3$, so $10^{100}$ will leave the same remainder as $3^{100}$ when divided by $7$
Now, $3^3 = 27$ and $27 \div 7$ leaves remainder $-1$ (since $27 = 28 - 1$)
So, $3^{100} = (3^3)^{33} \times 3 = 27^{33} \times 3$
This becomes $(-1)^{33} \times 3 = -3$

So, the remainder is $-3$.

Since we want a positive remainder, add $7$: $-3 + 7 = 4$
The remainder is 4.

Hence, the correct answer is option 2.

2. When $3^{333}$ is divided by 11 , the remainder is

1. 5

2. 6

3. 1

4. 10

Solution:

We want the remainder when $3^{333}$ is divided by 11.

First, observe the powers of 3 modulo 11:

$
\begin{aligned}
3^1 &\equiv 3 \pmod{11} \\
3^2 &\equiv 9 \pmod{11} \\
3^3 &\equiv 5 \pmod{11} \\
3^4 &\equiv 4 \pmod{11} \\
3^5 &\equiv 1 \pmod{11}
\end{aligned}
$

So, the powers of 3 repeat every 5 terms modulo 11.

Now, divide 333 by 5:

$
333 \div 5 = 66 \text{ remainder } 3
$

We can solve, $
3^{333} \equiv 3^3 \pmod{11} \Rightarrow 3^3 = 27 \equiv 5 \pmod{11}
$

We get the remainder as $5$.

Hence, the correct answer is option 1.

CAT Quantitative Aptitude Important Topics and Practice Questions

As mentioned above, the number of topics that the candidates should prepare under the CAT quantitative aptitude section is very broad and time-consuming. To help the candidates in this regard, Careers360 has performed a comprehensive analysis of the CAT previous year question papers to list the most important CAT quantitative aptitude topics. The candidates are encouraged to learn CAT quantitative aptitude questions from these topics as they have a high probability of being asked in the CAT examination. The most important CAT quantitative aptitude topics, along with the CAT quantitative aptitude practice questions, are provided in the table below.

Best Books for CAT Quantitative Aptitude Preparation

If a candidate wishes for more CAT quantitative aptitude practice questions, they can refer to the best CAT preparation books under the CAT quantitative aptitude section. These books are equipped with a variety of CAT quantitative aptitude questions, including the previous year's CAT quant questions and various CAT sample papers and much more, which will enhance the candidate’s CAT preparation and help them stay focused.

CAT Preparation Materials Designed by Careers360

Careers360 designs a set of CAT 2025 preparation resources such as the complete CAT syllabus guides, topic-wise study materials, necessary CAT mock tests, and expert CAT preparation strategy articles. Candidates can download these CAT resources using the links provided below.