CAT 2025 Quantitative Aptitude Practice Questions with Solutions

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.

Common Mistakes to Avoid in CAT Quantitative Aptitude

Many CAT aspirants lose valuable marks in the Quantitative Aptitude section due to avoidable errors. Understanding these common mistakes can help improve accuracy and speed. By avoiding calculation slips, poor time allocation, and weak topic prioritisation, candidates can significantly boost their CAT percentile and maximise their Quant preparation efforts.

1. Not Focusing on High-Weightage Topics – Ignoring arithmetic, algebra, and geometry in favour of low-priority topics can cost valuable marks. Analyse the CAT Quantitative Aptitude syllabus and prioritise chapters with higher question frequency.

2. Over-Reliance on Shortcuts Without Concept Clarity – Many aspirants use tricks blindly. In CAT Quant, shortcuts only work when backed by strong conceptual understanding.

3. Skipping Regular Practice of Previous Year CAT Quant Questions – Previous year CAT papers reveal question trends and difficulty levels. Not solving them reduces familiarity with the real exam pattern.

4. Poor Time Management in CAT Quant Section – Spending too much time on one tough question can lead to leaving multiple easy questions unanswered. Practise timed mock tests to improve speed.

5. Ignoring Accuracy for Speed – Speed without accuracy lowers your CAT Quant percentile. Maintain a balance by solving with precision while keeping track of time.

6. Not Revising CAT Quantitative Aptitude Formulas – Forgetting essential formulas during the exam slows problem-solving. Keep a formula sheet and revise it daily before mock

CAT 2025 Quantitative Aptitude Previous Year Paper Analysis

Analysing CAT quant previous year papers helps aspirants identify trends, difficulty levels, and topic-wise weightage to optimise preparation strategies, improve accuracy, and master time management.

• Overall Difficulty: Easy to moderate for most slots, though Slot 3 was slightly tougher with a few calculation-heavy questions.

• Slot 1: Balanced paper with strong representation from arithmetic and algebra. Fewer geometry questions, and about 8 TITA questions were present.

• Slot 2: Featured repeated concepts like Simple and Compound Interest (SICI) and Time & Work. Arithmetic dominated, but some questions required multi-step calculations.

• Slot 3: Heavily focused on arithmetic topics such as SICI, Speed-Time-Distance, Time & Work, and Ratios. Around 10 good attempts could fetch a solid score, with a target of 35 marks for a high percentile.

CAT 2023 Quant Analysis

• Overall Difficulty: Slightly tougher than 2022, with an overall moderate to difficult level.

• Topic Trends: Algebra overtook arithmetic in terms of question count, although arithmetic remained strong with key topics like Speed-Time-Distance, Mixtures, and Alligations.

• Slot 3: Considered the lengthiest and most demanding. It contained 22 questions, including 14 MCQs and 8 TITA questions. Many algebra questions were equation-based, and the section also included geometry, modern maths, and number theory.

CAT 2022 Quant Analysis

• Overall Difficulty: Moderate and manageable for most well-prepared candidates.

• Topic Distribution: Arithmetic remained the most tested area, especially topics like Time-Speed-Distance, Mixtures, and Alligations. Algebra came second in frequency, while geometry had fewer questions compared to other years.

• Structure: Followed the traditional CAT format with 66 questions overall, a 40-minute sectional time limit, and about 8 TITA questions in Slot 1.

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High-Scoring Topics in CAT Quantitative Aptitude

Focusing on high-scoring topics in CAT Quantitative Aptitude can significantly boost your percentile. These areas often have higher weightage, predictable question patterns, and quicker solving approaches. By mastering topics like arithmetic, algebra, and geometry, aspirants can secure more marks in less time, improving both accuracy and overall performance in the CAT exam.

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 = x300×2x=2x2300 litres
Milk = 300−x300×2x=2x(300−x)300 litres

Now refill the container with 2x litres of water, so final quantities are:
Water = x−2x2300+2x=x(1+2−2x300)=x(3−2x300)

Milk = (300−x)−2x(300−x)300
= (300−x)(1−2x300)

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

(300−x)(1−2x300)=216
⇒(300−x)(300−2x300)=216
⇒(300−x)(300−2x)300=216
⇒(300−x)(300−2x)=64800
⇒90000−900x+2x2=64800
⇒2x2−900x+25200=0
⇒x2−450x+12600=0
⇒x2−420x−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 = dx

According to the question:

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

dx−dx+6=4−−−−−(1)

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

dx−6−dx=6−−−−−−(2)

Add equations (1) and (2):

(dx−dx+6)+(dx−6−dx)=4+6

⇒dx−6−dx+6=10

Take d common:

d(1x−6−1x+6)=10

⇒d⋅(x+6)−(x−6)(x−6)(x+6)=10

⇒d⋅12x2−36=10

⇒d=10(x2−36)12=5(x2−36)6

Now, plug this d into equation (1):

dx−dx+6=4

⇒5(x2−36)6x−5(x2−36)6(x+6)=4

Take common factor:

5(x2−36)6(1x−1x+6)=4

⇒5(x2−36)6⋅6x(x+6)=4

⇒5(x2−36)x(x+6)=4

Now simplify:

5(x2−36)=4x(x+6)

⇒5x2−180=4x2+24x

⇒5x2−4x2−24x−180=0

⇒x2−24x−180=0

Solve the quadratic:

x=24±576+7202=24±12962=24±362

So x=30 (discard −6)

The distance:

d=5(x2−36)6=5(900−36)6=5⋅8646=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 = 160 work per hour.

Seema takes 8 days × 5 hours/day = 40 hours to complete the task.
So, Seema's efficiency = 140 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×2×160

=4x60=x15
Total work done by Seema = x×4×140

=4x40=x10

Total work = x15+x10=1
⇒2x+3x30=1⇒5x30=1
⇒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≥b≥c and log8⁡(a+b)log2⁡c+log27⁡(a−b)log3⁡c=23, then the greatest possible integer value of a is

1. 14

2. 15

3. 16

4. 17

Solution:

log8⁡(a+b)log2⁡c+log27⁡(a−b)log3⁡c=23

⇒log23⁡(a+b)log2⁡c+log33⁡(a−b)log3⁡c=23

⇒13log2⁡(a+b)log2⁡c+13log3⁡(a−b)log3⁡c=23

⇒log2⁡(a+b)log2⁡c+log3⁡(a−b)log3⁡c=2

We use change of base formula:

logc⁡(a+b)+logc⁡(a−b)=2

Use log addition:

logc⁡[(a+b)(a−b)]=2

So:

(a+b)(a−b)=c2
⇒a2−b2=c2
Also, it is given that a>10≥b≥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 a2=b2+c2,
b2=100, c2 is also close to 100 so as to maximize a, the perfect square close to 200 is 196.
So, a2=196⟹a=14

Hence, the correct answer is 14.

2. If (a+b3)2=52+303, 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+b3)2=52+303

⇒(a2+3b2)+2ab3=52+303

Now, compare real and irrational parts:

• Real part: a2+3b2=52

• Irrational part: 2ab=30

From 2ab=30, we get:

ab=15−−−−−−(i)

Substitute b=15a into the first equation:

a2+3(15a)2=52
⇒a2+3⋅225a2=52
⇒a4+675=52a2
⇒a4−52a2+675=0

Let x=a2, then: x2−52x+675=0

Solve the quadratic equation:

x=52±522−4⋅6752=52±2704−27002=52±22
⇒x=27⇒a=27(not natural),x=25⇒a=5

So a=5, from (i) b=155=3

Therefore: a+b=5+3=8

Hence, the correct answer is option 4.

3. The roots α,β of the equation 3x2+λx−1=0, satisfy 1α2+1β2=15. The value of (α3+β3)2, is

1. 1

2. 4

3. 9

4. 16

Solution:

Given: 1α2+1β2=15

We know: 1α2+1β2=α2+β2α2β2

From the equation 3x2+λx−1=0,
sum of roots α+β=−λ3, product of roots αβ=−13
So, α2+β2=(α+β)2−2αβ=(λ29)+23

Also, α2β2=(αβ)2=(19)

Now,
α2+β2α2β2=15
⇒(λ29+23)÷19=15
⇒9(λ29+23)=15
⇒λ2+6=15⇒λ2=9⇒λ=±3

Now, α3+β3=(α+β)3−3αβ(α+β)

We have: α+β=−λ3=±1, αβ=−13

Case 1: α+β=−1

α3+β3=(−1)3−3(−13)(−1)=−1−1=−2

We get, (α3+β3)2=(−2)2=4

Case 2: α+β=1

α3+β3=(1)3−3(−13)(1)=1+1=2

We get, (α3+β3)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 x2+y2+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 = g2+f2−c

So, the centre of x2+y2+4x−6y−3=0 is (-2, 3) and radius = 22+32+3=4

Now, sin⁡30°=4OL

So, OL = 8 units

Equation of circle passing through L: (x+2)2+(y−3)2=82

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 × 6 = 54
So, m + 2m + 4m = 54
⇒ m = 547

Area of ABP = m = 547
⇒ ½ (AB) (BP) = 547

⇒ BP = 127 (Since AB = 9)

Area of ABQ = 2m + m = 1627
⇒ ½ (AB) (BQ) = 1627

⇒ BQ = 367 (Since AB = 9)

Now, PQ = BQ – BP =367 – 127 = 247
QC = BC – BQ = 6 – 367 = 67

Now, BP: PQ: QC = (127) : (247) : (67) = 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. abr22(a2+b2)

2. abr2(a2+b2)

3. 4abr2(a2+b2)

4. 2abr2(a2+b2)

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

⇒ k2=4r2a2+b2
Area of triangle = 12×ak×bk

= 12×abk2

= 12×ab×4r2a2+b2

= 2abr2a2+b2

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)⋅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: 1533106666≈22.99⇒a+b+c+d=23

Now, 6666×23=153318⇒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.

4C2=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 B1 and B2 be two black kings and R1 and R2 be two red kings.
Case 1: B1 and B2 are selected then, R1 and R2 can not be selected together. If R1 is selected and R2 is not selected then the remaining 3 cards must be chosen from 8 cards i.e. 8C3 ways.
If R2 is selected and R1 is not selected then the remaining 3 cards must be chosen from the rest of 8 cards i.e. 8C3 ways.
If R1 and R2 both are not selected then the remaining 4 cards must be chosen from the rest of 8 cards i.e. 8C4 ways.
Total ways =8C3+8C3+8C4=56+56+70=182

Case 2: R1 and R2 are selected then, B1 and B2 cannot be selected together. If B1 is selected and B2 is not selected then the remaining 3 cards must be chosen from the rest of 8 cards i.e. 8C3 ways.
If B2 is selected and B1 is not selected then the remaining 3 cards must be chosen from the rest of 8 cards i.e. 8C3 ways.
If B1 and B2 both are not selected then the remaining 4 cards must be chosen from the rest of 8 cards i.e. 8C4 ways.
Total ways =8C3+8C3+8C4=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 10100 is divided by 7, the remainder is

1. 3

2. 4

3. 6

4. 1

Solution:

We know that 10÷7 leaves a remainder of 3, so 10100 will leave the same remainder as 3100 when divided by 7
Now, 33=27 and 27÷7 leaves remainder −1 (since 27=28−1)
So, 3100=(33)33×3=2733×3
This becomes (−1)33×3=−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 3333 is divided by 11 , the remainder is

1. 5

2. 6

3. 1

4. 10

Solution:

We want the remainder when 3333 is divided by 11.

First, observe the powers of 3 modulo 11:

31≡3(mod11)32≡9(mod11)33≡5(mod11)34≡4(mod11)35≡1(mod11)

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

Now, divide 333 by 5:

333÷5=66 remainder 3

We can solve, 3333≡33(mod11)⇒33=27≡5(mod11)

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.