MAH MBA CET Previous Paper 2025 Slot-1

The total marks of every student is follows:
Maya-440
Arya-478
Riya-465
Sreya-430
Roopesh-455
Chandesh-396
Hanish-387

So, the topper of the class is Arya. Option A is the answer.

The total marks of every student is follows:
Maya-440
Arya-478
Riya-465
Sreya-430
Roopesh-455
Chandesh-396
Hanish-387

Total marks of Riya - Total marks of Roopesh = 465-455 = 10.
Option E is the answer.

Chandesh scored 75 out of 120 marks in Hindi. So, out of 200 marks, his score = $$\ \frac{\ 75}{120}\times\ 200$$ = 125
Hanish scored 35 out of 120 marks in Hindi. So, out of 200 marks, his score = $$\ \frac{35}{120}\times\ 200$$ = 58.33
Difference in their marks = 125-58.33 = 66.67
The nearest option is option E, which is the answer.

Sunil checks 1000 items in 5 hours, so Sunil's rate = 1000/5 = 200 items/hour

Nilesh completes 75% of the same job (1000 items) in 3 hours. 75% of 1000 is 750 items. = 750/3 = 250 items/hour

In the first 2 hours, both Sunil and Nilesh are working together.  Combined Rate: 200 + 250 = 450 items / hour

Items checked in 2 hours: 450*2 = 900 

The total target is 1300 items. Remaining Items = 1300 - 900 = 400 items. 

Now, Nilesh stops checking after the first 2 hours, so Sunil must finish the remaining 400 items alone.

Time required by Sunil = 400 items / 200 items/hour = 2 hours

Total Time = 2 hours + 2 hours = 4 hours.

Let the initial weights of Aashiqui, Aarohi and Aamrohi be 4x,9x and 7x respectively.
We are given total initial weight is 100kg.
So,4x+9x+7x=100, which gives x=5.
Initial Weights of Aashiqui, Aarohi and Aamrohi are 20kg,45kg and 35kg respectively.
Let the weights of Aashiqui, Aarohi and Aamrohi be 4y,7y and 9y respectively after Aamrohi weight gain.
So, we are given 4y+7y+9y=140.
We get y=7kg
Aamrohi's weight after weight gain=9y=63kg. 
Aamrohi's weight gain = Final weight - Initial weight = 63kg-35kg = 28kg.
Option A is the answer.

We can use the Angle Bisector Theorem:

In triangle ABC, if AD bisects ∠A, then $$\dfrac{AB}{AC}=\dfrac{BD}{DC}$$

We know: BC=10 and DC=6

So, BD = BC − DC = 10 − 6 = 4

Applying the theorem

$$\dfrac{AB}{AC}=\dfrac{BD}{DC}$$ 

Given
AC=4.2, so:$$AB=\dfrac{2}{3}\times4.2=2.8\ cm$$

Assign Task 2 first: 2 choices (Person 3 or 4).

Total ways to assign the remaining 5 tasks: 5! = 120.

Total without Task 1 restrictions: 2*120 = 240.

Subtract restricted cases for Task 1:

If Task 1 is given to Person 1: 2*1*4! = 48.

If Task 1 is given to Person 2: 2*1*4! = 48.

Final Result: 240 - 48 - 48 = 144.

Let the speed of stream be 's' km/hr.
$$\ \frac{\ 20}{5-s}=\ 3\times\frac{20}{5+s}$$
5+s=15-3s
s=2.5km/hr
Option B is the answer.

Umla: 24 sheep * 3 months = 72 sheep-months

Vimla: 10 sheep * 5 months = 50 sheep-months

Wilma: 35 sheep * 4 months = 140 sheep-months

Xema: 21 sheep * 3 months = 63 sheep-months

Total sheep-months = $72 + 50 + 140 + 63 = 325 sheep-months

We are given that Umla’s share of the rent is Rs. 720. Since her contribution was 72 sheep-months, we can find the cost per sheep-month:

Rate = $$\dfrac{720}{72}$$ = Rs. 10 per sheep-month

Total Rent = 325 sheep-months * Rs. 10 per sheep-month = Rs. 3,250

Volume remains conserved during melting.
Volume before melting = Volume after melting
Before melting, the wire was in cylindrical shape and after melting, it became sphere.
So, Vol. before = $$\pi r^2l$$ = Vol. after = $$\ \frac{\ 4}{3}\pi\ R^3$$, where R is the radius of sphere formed.
$$\left(2\times\ 10^{-3}\right)^2\times\ 243\ =\ \frac{\ 4}{3}R^3$$ 
$$R^3=729\times\ 10^{-6}$$
R=9cm.
So, diameter of the sphere = 18cm.
Option C is the answer.