MAH MBA CET Quantitative Aptitude Sample Questions 2025: MAH MBA CET is a state-level exam conducted for admission into the MBA and MMS programs of many institutes in Maharashtra. The test evaluates various sections, such as Logical Reasoning, Abstract Reasoning, Verbal Ability, and Quantitative Aptitude. Quantitative Aptitude is one of the most important sections of the exam, where the students are questioned on mathematics and analytical awareness. Understanding concepts is the secret to success and reaching the desired score.
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Here is the complete list of MAH MBA CET Quant Questions Topics and question types appearing in the exam. It has sample test questions, practice sets, and expert advice on improving problem-solving speed and accuracy.
MAH MBA CET 2025 is a state-level entrance test for MBA and MMS admission in different Maharashtra institutes. The authority that conducts and organizes the test is the State Common Entrance Test Cell, Maharashtra. Candidates must be informed about MAH MBA CET 2025 aspects, such as the MAH MBA CET exam pattern, eligibility, syllabus and exam dates, etc., which will help them to study properly for the test. Also practicing using MAH MBA CET mock tests can also be extremely helpful.
Particulars | Details | ||
Exam duration | 150 minutes (2 hours 30 minutes) | ||
Medium of Exam | English | ||
Mode of exam | Online | ||
Type of question | Multiple Choice Objective Questions (5 options) | ||
Sections | Logical Reasoning Abstract Reasoning Quantitative Aptitude Verbal Ability/Reading Comprehension | ||
No of Questions | 200 | ||
Marking Scheme | 200 Marks (+1 for each correct answer) No negative marking | ||
Section | No. of Questions | Max. Marks |
Logical Reasoning | 75 | 75 |
Abstract Reasoning | 25 | 25 |
Quantitative Aptitude | 50 | 50 |
Verbal Ability/Reading Comprehension | 50 | 50 |
The candidates preparing for the MAH MBA CET must be in the final stage of their preparation. To help them excel during this phase, Careers360 has introduced a new online MAH MBA CET mock test. Candidates can refer to this article to gain a detailed understanding of the MAH MBA CET online mock test designed by Careers360. The link to attend the MAH MBA CET mock test is provided below.
| Title | Mock Test Link |
| Free MAH MBA CET online mock test by Careers360 | Attempt Now |
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Solving many of the problems of the MAH MBA CET Quantitative Section will not just improve the fluency but also make the habit of quick and correct solving of the problems. This section of the test needs knowledge of the fundamental Arithmetic, Algebra, Geometry, and reading graphical data, and therefore, the candidate must have a well-executed plan to achieve a good score bracket. Here, you will find a few MAH MBA CET Quantitative Aptitude questions of practice to help you understand the format of the test, improve your mathematical skills, and raise the discussion to the test.
1- Suppose f (x, y) is a real-valued function such that f (3x + 2y,2x −5y) =19x, for all real numbers x and y. The value of x for which f (x, 2x) = 27, is:
1) 3
2) 2
3) 4
4) 5
Solution:
Let m = 3x + 2y; and n = 2x - 5y
Multiply 1st equation by 5 and the second by 2 and then add both.
We get, 5m + 2n = 19x
So, we write: f (m, n) = 5m + 2n
Put m = x and n = 2x
So, f (x, 2x) = 5x + 4x = 9x = 27
So, x = 3
Hence, the correct answer is option (1).
2- The value of $1+\left(1+\frac{1}{3}\right) \frac{1}{4}+\left(1+\frac{1}{3}+\frac{1}{9}\right) \frac{1}{16}+\left(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}\right) \frac{1}{64}+\cdots$ is:
1) $\frac{15}{13}$
2) $\frac{27}{12}$
3) $\frac{16}{11}$
4) $\frac{15}{8}$
Solution:
First, find the general term of the series (Series in the bracket is the sum of n terms with first term 1 and common ratio $\frac13$, and the terms in the multiplication form a GP with common ratio $\frac14$)
Since the general term is the sum of $n$ terms,
$A=\frac{a(1- r^n)}{1-r}$
where $a=1$
$r=\frac{1}{3}$
$T_n=A\times R^{n-1}$
$R=\frac{1}{4}$
$T_n=\left( \frac{1- \frac {1}{3^n}}{1-\frac{1}{3}}\right)\times (\frac{1}{4})^{n-1}$
$=\left( \frac{(3^n- 1) \times 3}{2 \times 3^n}\right)\times (\frac{1}{4})^{n-1}$
$= \frac32[1 - \frac{1}{3^n}] \times (\frac{1}{4})^{n-1}$
$= [\frac32 - \frac{1}{2 \times 3^{n-1}}] \times (\frac{1}{4})^{n-1}$
$= [\frac3{2 \times 4^{n-1}} - \frac{1}{2 \times 12^{n-1}}] $
This is an infinite GP.
So, Sum = $\frac{\frac32}{1- \frac14}-\frac{\frac12}{1-\frac12} =2 - \frac6{11} = \frac{16}{11}$
Hence, the correct answer is option (3).
3- Let an = 46 + 8n and bn = 98 + 4n be two sequences for natural numbers n ≤ 100. Then, the sum of all terms common to both the sequences is:
1) 14900
2) 15000
3) 14798
4) 14602
Solution:
For better understanding, take $n = m$ in the second series.
Find the terms where $a_{n}=b_{m}$
So, $46 + 8n = 98 + 4m$
⇒ $8n – 4m = 52$
⇒ $2n – m = 13$
If $m = 1, n = 7$
So, the first common term = 98 + 4 = 102
Common difference for this series of common terms = LCM of 8 and 4 = 8
So, the series is like $102, 110, 118,$ ……….
Now, find the largest such value of $m$,
$2n – m = 13$
⇒ $m = 2n – 13$
If $n = 56$, we get, $m = 99$.
Last term of common series $= 98 + 4(99) = 494$
Number of common terms = $\frac{(494 – 102)}{8}+ 1 = 50$
$\therefore$ Required sum = $\frac{50(102 + 494)}{2} = 14900$
Hence, the correct answer is option (1).
4- In a regular polygon, any interior angle exceeds the exterior angle by 120 degrees. Then, the number of diagonals of this polygon is:
1) 54
2) 34
3) 44
4) 24
Solution:
Interior angle of a $n$ sided regular polygon $=\frac{180(n-2)}{n}$
Exterior angle of a $n$ sided regular polygon $=\frac{360}{n}$
According to the question:
$
\frac{180(n-2)}{n}-\frac{360}{n}=120
$
$
\begin{aligned}
& \Rightarrow \frac{180 n-360-360}{n}=120 \\
& \Rightarrow 180 n-720=120 n \\
& \Rightarrow n=12
\end{aligned}
$
$\therefore$ Number of diagonals $=\frac {n(n-3)}{2}=\frac{12(12-3)}{2}=54$
Hence, the correct answer is option (1).
5- A rectangle with the largest possible area is drawn inside a semicircle of radius 2 cm. Then, the ratio of the lengths of the largest to the smallest side of this rectangle is:
1) $1:1$
2) $\sqrt{5}: 1$
3) $\sqrt{2}: 1$
4) $2:1$
Solution:
A rectangle is symmetric about the center.
If the length is $2L$ and the breadth is $B$.
Using Pythagoras theorem, we get,$L^2+B^2 = 2^2$
So, $B^2 = 2^2-L^2$
⇒ $B = \sqrt {2^2-L^2}$
Area = $2L × B$
= $2L × \sqrt {4-L^2}$
Squaring both sides,
Area2 = $4L^2 × (4-L^2)$
= $4(4L^2-L^4)$
= $4(4L^2-L^4 + 4 -4)$
= $4(4-(2 - L^2)^2)$
So, for the area to be maximum,
$2 = L^2$
$⇒L=\sqrt2$
So, $B = \sqrt {2^2-(\sqrt2)^2}= \sqrt2$
$\therefore$ Required ratio = $\frac{2L}{B} = 2\sqrt2: \sqrt2 = 2: 1$
Hence, the correct answer is option (4).
6- Let ABC be an isosceles triangle such that AB and AC are of equal length. AD is the altitude from A on BC and BE is the altitude from B on AC. If AD and BE intersect at O such that $\angle AOB = 105º$, then $\frac{AD}{BE}$ =?
1) sin15°
2) cos15°
3) 2cos15°
4) 2sin15°
Solution:
$\angle \mathrm{AOB}=105^{\circ}$So, $\angle \mathrm{DOE}=105^{\circ}$ (vertically opposite angles are equal)
$\angle \mathrm{C}=75^{\circ}$ (Sum of all angles of a quadrilateral OECD is $360^{\circ}$ and the other two angles are $90^{\circ}$ each)
$\angle B=75^{\circ}$ (Angles opposite to equal sides are equal; $A B=A C$ )
So, $\angle \mathrm{OBD}=15^{\circ}$ (Using sum of all angles in a triangle is $180^{\circ}$ )
$\angle \mathrm{A}=30^{\circ}$ (Using sum of all angles in a triangle is $180^{\circ}$ )
Now, Area of triangle $A B C=\frac12(A D)(B C)=\frac12(A C)(B E)$
So, $\frac{A D}{B E}=\frac{A C}{B C}$
In triangle $A B E, \sin 30^{\circ}=\frac{B E}{A B}=\frac{B E}{A C}$ (Since $A B=A C$ )
$\Rightarrow \frac12=\frac{B E}{A C}$
$\Rightarrow \mathrm{AC}=2 \mathrm{BE}$
In triangle $B E C, \operatorname{cos} 15^{\circ}=\frac{B E}{B C}$
From equations 1, 2, and 3,
$
\frac{A D}{B E}=\frac{2 B E}{\frac{B E }{\cos 15^{\circ}}}=2 \cos 15^{\circ}
$
Hence, the correct answer is option (3).
7- A fruit seller has a stock of mangoes, bananas, and apples with at least one fruit of each type. At the beginning of the 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:
1) 340
2) 280
3) 320
4) 360
Solution:
Let the total fruits at the beginning be 100x.
Mangoes = 40x
Banana + apple = 60x
If Banana = y, then Apple = 60x – y
Total fruits sold = 50% = 50x
According to the question,
50x = 20x + 96 + 40% of (60x -y)
⇒ 30x = 96 + 24x – 40% of y
⇒ 6x = 96 – 40% of y = 96 – 0.4y
⇒ x = 16 – $\frac{2y}{30}$ = 16 – $\frac{y}{15}$
⇒ 100x = 1600 – $\frac{100y}{15}$
⇒ 100x = 1600 – $\frac{20y}{3}$
To get a minimum value of 100x, y should be maximum and a multiple of 3 so that 100x remains positive and integer. Also, y ≥ 96.
Also 60x – y should be a non-zero integer.
i.e. 60 (16 – $\frac{y}{15}$) – y is non-zero.
If we put this value equal to zero, we get,
⇒ 960 – 5y = 0
⇒ y = 192
So, y should be less than 192.
So, the maximum multiple of 3 between 96 and 192 is 189.
So, y = 189
Hence, total fruits = 100x = 1600 – 20($\frac{189}{3}$) = 1600 – 1260 = 340
Hence, the correct answer is option (1).
8- The number of coins collected per week by two coin collectors A and B are in the ratio 3:4. If the total number of coins collected by A in 5 weeks is a multiple of 7, and the total number of coins collected by B in 3 weeks is a multiple of 24, then the minimum possible number of coins collected by A in one week is:
1) 42
2) 52
3) 62
4) 72
Solution:
Let the total coins collected by A in 5 weeks = 7x
Let the total coins collected by B in 3 weeks = 24y
According to the question,
$\frac{7x}{5}:\frac{24y}{3} = 3: 4$
So, $x : y = 30 : 7$
For the minimum number of coins collected by A, take x = 30
So, coins collected by A in a week = $\frac75$ of 30 = 42
Hence, the correct answer is option (1).
9- There are three people, A B, and C in a room. If a person D joins the room, the average weight of the persons in the room reduces by x kg. Instead of D, if person E joins the room, the average weight of the persons in the room increases by 2x kg. If the weight of E is 12 kg more than that of D, then the value of x is:
1) 1.5
2) 0.5
3) 1
4) 2
Solution:
According to the question,
$\frac{A+B+C}{3} - \frac{A+B+C+D}{4} = x$
And $\frac{A+B+C+E}{4} - \frac{A+B+C}{3} = 2x$
Adding both equations, we get,
$\frac{E-D}{4} = 3x$
⇒ $E-D = 12x$
Also, $E-D=12$
So, x = 1
Hence, the correct answer is option (3).
10- If is a positive real number such that $x^8 + (\frac {1}{x})^8 = 47$ , then the value of $x^9 + (\frac {1}{x})^9$ is:
1) $34 \sqrt{5}$
2) $40 \sqrt{5}$
3) $36 \sqrt{5}$
4) $30 \sqrt{5}$
Solution:
$x^8 + (\frac {1}{x})^8 = 47$
⇒ $(x^4 + \frac {1}{x^4})^2 - 2 = 47$
⇒ $(x^4 + \frac {1}{x^4})^2 = 49$
⇒ $x^4 + \frac {1}{x^4} = 7$
⇒ $(x^2 + \frac {1}{x^2})^2 - 2 = 7$
⇒ $(x^2 + \frac {1}{x^2})^2 = 9$
⇒ $x^2 + \frac {1}{x^2} = 3$
⇒ $(x + \frac {1}{x})^2 -2 = 3$
⇒ $x + \frac {1}{x} = \sqrt 5$
Cubing both sides, we get,
⇒ $(x + \frac {1}{x})^3 = 5 \sqrt 5$
⇒ $x^3 + \frac {1}{x^3} +3 \times \sqrt 5 = 5 \sqrt 5$
⇒ $x^3 + \frac {1}{x^3} = 2 \sqrt 5$
Cubing both sides, we get,
⇒ $x^9 + \frac {1}{x^9} + 3 \times 2 \sqrt 5 = 40 \sqrt 5$
$\therefore x^9 + \frac {1}{x^9} = 34 \sqrt 5$
Hence, the correct answer is option (1).
11- Let an and bn be two sequences such that an= 13 + 6(n−1) and bn = 15 + 7(n−1) for all natural numbers n. Then, the largest three-digit integer that is common to both these sequences, is:
1) 967
2) 850
3) 758
4) 980
Solution:
Here, $a_n=13+6(n-1)=7+6n$ and $b_{n}=8+7n$
The common difference of terms that exists in both series is LCM of $(6, 7) = 42$
Now, by writing the series up to a few terms, the first common term of the first two series = 43
Now, $t_m=a+(m-1)d<1000$
⇒ $43 + (m - 1)42 < 1000$
On solving, we get $m < 23.8$
So, take $m = 23$
Now, the largest three-digit common term $= 43 + (22 × 42) = 967$
Hence, the correct answer is option (1).
12- 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:
1) 26%
2) 28%
3) 25%
4) 30%
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).
13- Let $n$ be the least positive integer such that $168$ is a factor of $1134^n$. If $m$ is the least positive integer such that $1134^n$ is a factor of $168^m$, then $(m+n)$ equals:
1) 12
2) 9
3) 15
4) 24
Solution:
The prime factorizations of 168 and 1134 are as follows:
$\begin{aligned}
& 168=2^3 \times 3 \times 7 \\
& 1134=2 \times 3^4 \times 7
\end{aligned}$
As we can see the power of $2$ in the factor of $168$ is $3$.
So, the smallest positive integral value of $n$, such that 168 is a factor of $1134^{\text {n }}$ is 3.
Now, $1134^n=1134^3=2^3 \times 3^{12} \times 7^3$
Since the power of 3 in the factor of $1134^3$ is $12$, and the smallest positive integral value of $m$, such that $1134^3$ is a factor of $168^{\mathrm{m}}$ is $12$.
Therefore, $m+n=12+3=15$
Hence, the correct answer is option (3).
14- Let α and β be the two distinct roots of the equation $2x^2−6x+k=0$, such that (α + β) and αβ are the distinct roots of the equation $x^2+px+p=0$. Then, the value of $8(k−p)$ is:
1) 6
2) 4
3) 3
4) 2
Solution:
For, $2x^2−6x+k=0$
$α+β = \frac{6}{2}=3$
and, $αβ = \frac{k}{2}$
For, $x^2+px+p=0$
$α+β +αβ = \frac{-p}{1}=-p$
and,$(α+β)αβ = \frac{p}{1}=p$
Adding these two equations,
$α+β +αβ +(α+β)αβ = 0$
Put the values of ($α+β$) and $αβ$, we get
$3 +\frac{k}{2}+ \frac{3k}{2}= 0$
So, $k = \frac{-3}{2}$
Also, we get using the above equations, $p= \frac{3k}{2} = \frac{-9}{4}$
So, $8(k−p)= 8[\frac{-3}{2}-(\frac{-9}{4})]=6$
Hence, the correct answer is option (1).
15- The sum of all possible values of $x$ satisfying the equation $2^{4x^2}−2^{2x^2+x+16}+2^{2x+30} = 0$, is:
1) $\frac{3}{2}$
2) $\frac{5}{2}$
3) $\frac{1}{2}$
4) $3$
Solution:
Given: $2^{4x^2}−2^{2x^2+x+16}+2^{2x+30} = 0$,
⇒ $(2^{2x^2})^2−2^{2x^2} \times 2^{x+15} \times 2 + (2^{x+15})^2 = 0$
⇒ $(2^{2x^2} - 2^{x+15})^2 = 0$
⇒ $2^{2x^2} = 2^{x+15}$
⇒ $2x^2 = x+15$
⇒ $2x^2- x-15=0$
⇒ $(2x+5)(x-3)=0$
⇒ $x = - \frac 52, 3$
So, the sum of all values of $x$ = $\frac 12$
Hence, the correct answer is option (3).
MAH MBA CET 2025 preparation demands a strategic approach to perform well in sections such as Logical Reasoning, Abstract Reasoning, Quantitative Aptitude, and Verbal Ability & Reading Comprehension. Effective study planning, consistent practice through MAH MBA CET mock tests, and good time management are necessary for success. The candidates must follow expert-suggested MAH MBA CET preparation strategies as well as the subject-wise tips, learn using the best MAH MBA CET study materials, and practice intelligent revision methods to achieve a great MAH MBA CET score.
Understand the Syllabus & Exam Pattern – Know the question distribution section-wise and important topics to concentrate on high-scoring sections.
Practice Mock Tests & Previous Year Papers – Keep practicing mock tests and reviewing MAH MBA CET previous years' papers to enhance speed, accuracy, and time management.
Enhance Logical & Abstract Reasoning – Spend sufficient time on reasoning questions, as they carry considerable weight in the test.
Enhance Speed & Accuracy in Quantitative Aptitude – Emphasize mental calculation, shortcuts, and practice assorted numerical problems.
Improve Verbal Ability & Reading Comprehension – Read newspapers, enhance vocabulary, and practice RC passages to do well in the Verbal Section.
During the last lap of the candidate’s MAH MBA CET 2025 preparation, they should ensure that they do not miss out on going through the expert-suggested books on the exam. The list of best books for making the candidate’s MAH MBA CET exam fruitful is given below.
Book Name | Author |
50+ Solved Papers MBA | Arihant Publications |
Maharashtra MBA MH-CET | Arihant Publications |
How to Prepare for Logical Reasoning for CAT | Arun Sharma |
MH-CET (MBA/ MMS) Entrance Guide | Disha Publications |
How to Prepare for Quantitative Aptitude for CAT | Arun Sharma |
How to Prepare for Verbal Ability and Reading Comprehension for CAT | Arun Sharma |
The Pearson Guide to CET: MBA Maharashtra | Vandana Thorpe |
Target MHCET (MBA/MMS) – Past papers + 6 Mock Tests | Disha Publications |
Maharashtra MBA CET Guide by RPH Editorial Board | Priyanka Prakashan |
Study material on all sections of MBA CET Maharashtra | Chandresh Agarwal |
With the MAH MBA CET 2025 examination set to take place within a month, the candidates are expected to be in their last round of preparation. To assist them in their preparation journey, Careers360 have designed a set of ebooks to enhance the overall preparation of the candidates. The links to the ebook are given below.
TITLE | DOWNLOAD LINK |
MAH MBA CET 2024 Official Sample Paper | |
MAH MBA CET 2025 Preparation Tips | |
MAH CET MBA Syllabus | |
MAH MBA CET Question Paper 2018 | |
MAH MBA CET Question Paper 2017 | |
MAH CET MBA Question Paper 2016 | |
MAH MBA CET Question Paper 2015 | |
MAH MBA CET Question Paper 2014 |
Frequently Asked Questions (FAQs)
The level of difficulty is usually moderate, but Logical Reasoning is usually the most challenging section. Speed and accuracy are very important in order to score well.
There are 200 multiple-choice questions (MCQs), with one mark awarded per question. The test lasts for a total of 150 minutes.
No, there is no negative marking in the MAH MBA CET exam. Students can answer all questions without fear of losing marks for wrong answers.
The test contains four sections: Logical Reasoning, Abstract Reasoning, Verbal Ability & Reading Comprehension, and Quantitative Aptitude. Logical Reasoning has the maximum weightage.
MAH MBA CET exam is a state-level entrance test held for entry into MBA/MMS courses of Maharashtra. It is an internet-based test testing the candidate’s logical, verbal, and quantitative aptitude.
On Question asked by student community
Hello,
MHCET is mainly for engineering, pharmacy, law, and agriculture courses.
For business courses like BBA, BMS, or MBA, MHCET is not used.
Business courses usually have their own entrance exams like:
MAH BBA CET
MAH MBA CET
CUET (for some colleges)
Hope it helps!
Hi
The MHA MBA CET is available on our official website. The MBA CET 2025 exam for MBA admission takes place on April 1, 2 and 3, 2025. Candidates check the previous year's questions using the previous year's question paper and understand the structure of the exam pattern. Using the
With a 91.16 percentile in MAH MBA CET 2025, you have a good chance of securing admission in decent MBA colleges in Maharashtra. While top institutes like JBIMS or Sydenham may be out of reach, colleges such as PUMBA, SIES, Chetana’s, MET, and DY Patil are realistic options. Among these,
With 91.16 percentile, you can target top-tier MBA colleges like PUMBA (Pune University), MET Mumbai, SIESCOMS Navi Mumbai, and NL Dalmia. Welingkar and K J Somaiya may be slightly out of reach but possible under lower cutoffs or spot round.
With 42 percentile in MAH MBA CET, it is very difficult to get admission in good or top MBA colleges through CAP rounds. Most of the well-known colleges in Maharashtra like JBIMS, SIMSREE, Welingkar, and PUMBA usually take students who score more than 90 or 99 percentile. Even average colleges
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