MAH MBA CET Previous Paper 2025 Slot-1

Let Correct Answers, Wrong Answers and Unattempted be C, W and N respectively. 

Total questions: C+W+N=50

Score: $$C-\dfrac{W}{3}-\dfrac{N}{6}=32$$

Multiply the score equation by 6 to eliminate the fraction: 6C - 2W - N = 192

Substituting N with N=50−C−W from equation 1:

6C − 2W − (50 − C − W) = 192

6C − 2W − 50 + C + W = 192

7C − W − 50 = 192

7C − W = 242

W = 7C − 242

We want minimum W, so minimise C, but ensure: $$W\ge\ 0,\ N\ge\ 0$$

Condition 1: W ≥ 0

7C − 242 ≥ 0 ⇒ C ≥ 34.57

So C ≥ 35

Condition 2: N ≥ 0 

N = 50 − C − W

Substitute W: N = 50 − C − (7C − 242) = 292 − 8C ≥ 0

C ≤ 36.5 ⇒ C ≤ 36

So, C = 35 or 36

If C = 35:  W = 7(35) − 242 = 245 − 242 = 3

If C = 36: W = 252 − 242 = 10

Hence, the minimum possible wrong answers = 3

All options seem to be arrangements of the same 5 letters:  F, G, H, J, S

But if we notice, the sequence is basically the rotations. 

Option B: F G H J S → perfectly alphabetical order

Option C: J S F G H → rotation of B

Option D: H J S F G → rotation

Option E: S F G H J → rotation

So B, C, D, E are all cyclic rotations of the same sequence: F → G → H → J → S

Option A: A: G S F H J - This is not a cyclic rotation of F G H J S

The pattern:

a appears 1 time
b appears 2 times
c appears 3 times
d appears 4 times
e appears 5 times, and so on. 

Hence, we need cumulative sums. The sum of the first n numbers can be given by the formula: $$\dfrac{n\left(n+1\right)}{2}$$

We want to find the 288th term. 

Hence, $$\dfrac{n\left(n+1\right)}{2}\ge\ 288$$

Checking:

When n = 23, $$\dfrac{\left(23\times24\right)}{2}=276$$

When n = 24, $$\dfrac{\left(24\times25\right)}{2}=300$$

So, the 288th term lies in the 24th group

24th alphabet is x. 

Since both move in the same direction: 

Relative Speed = 20 - 15 = 5 Km/hr

Convert to m/s: $$5\times\dfrac{1000}{3600}=\dfrac{25}{18}$$ m/sec

Time taken to catch up = $$\dfrac{200}{\dfrac{25}{18}}=200\times\dfrac{18}{25}=144\ $$ seconds

Friend’s speed = 15 km/h 

So the distance walked by that friend = $$15\times\dfrac{1000}{3600}=\dfrac{25}{6}$$ m/sec

Distance = $$\dfrac{25}{6}\times144=600\ m$$

Solutions: lets look at all the statements one by one.

A. All teachers are doctors
This statement is false as the triangular area that represents Teachers is not completely under the circular region that represents doctors.

B. Some doctors are teachers.

This statement is True as some part of the triangular region that represents teachers is under the circular region that represents doctors.

C. All surgeons are doctors

This statement is also true as the entire square region that represents surgeons is under the circular region that represents doctors.

Therefore option c is true.

Solution: Based on data given we can identify the following pattern :

              Dot: Moves 1 step right in next image 

              Star: Moves 1 Step up in next image 

              whenever the position of star and dot is same then only the dot appears in the image

              All these things are happening together.

Hence the solution would be option E    

Solution: Based on data provided we can identify the following pattern here 

              The resultant image is the image that remains after removing the common things in both images and showing only the distinct                    things in both images and Based on this pattern Option C is correct     

Solution: Based on the images given we can see that images A, B, D and E are symmetric about Y axis and only the image C is not following this pattern. Correct option that is odd out of all is option C

The example case follows the pattern that the dotted area disappears in next image and the shaded area appears Blank in next image.

Therefore as per the pattern the answer should be option A here.