MAH MBA CET Previous Paper 2025 Slot-1

Let the no. be x
LCM(x,990) = 6930
6930=$$2\times3^2\times5\times7\times11$$
990=$$2\times3^2\times5\times11$$
Factor of 7 is missing in 990, so x has to be multiple of 7. 
x=7a
GCD(x,550)=110
x=110b
x=7a=110b. So, x=770k
Least value of x is 770.
Sum of digits of 770=7+7+0=14
Option B is the answer.    

Let the sides of the isosceles triangle be (a, a, b), where a and b are integers.

Case 1: The sum of the two equal sides is 12

a + a = 12 ⇒ a = 6

a + a = 12 ⇒ a = 6

Triangle inequality: 6 + 6 > b ⇒ b < 12

Also b > 0 and integer.

So, b = 1, 2, 3, …, 11

So, Total = 11 triangles

Case 2: One equal side + base = 12 

a + b = 12 ⇒ b = 12 − a

Triangle inequality: a + a > b ⇒ 2a > 12 − a ⇒ 3a > 12 ⇒ a > 4

Also: b > 0 ⇒ 12 − a > 0 ⇒ a < 12

So, a = 5,6,7,8,9,10,11

Hence, Total = 7 triangles

But when a=6, the triangle is (6,6,6), which was already counted in Case 1. So subtract 1 duplicate.

So final count: 11 + 7 - 1 = 17

Let the base of building be 'x' m and height of student be 'k' m
tan 45 = $$\ \frac{\ 100}{x}$$
x=100m
tan 60 = $$\frac{100+k}{100}$$
k=100($$\sqrt{3}$$ - 1) = 73.2m
Option D is the answer.

Let the total portion of tank to be filled be '3x' units.
Let the amount of units filled per minute by A,B and C be a,b and c respectively.
We are given that, at+10b+(t-12)c = 3x
Also, each pipe fills equal share of tank i.e. x units
at=x, 10b=x, (t-12)c=x. 
$$\therefore\ $$ b=$$\ \frac{\ x}{10}$$
Also, at+bt=3x
bt=2x, substitute b=$$\ \frac{\ x}{10}$$, t=20 min.
(t-12)c=x
c=$$\ \frac{\ x}{8}$$
Let k be time taken by c to fill the tank alone,
ck=3x
k=$$\ \ \frac{\ 3x}{c}$$=24 minutes.
Option B is the answer.

a, b, c, d and e are distinct.
Given that c is the arithmetic mean of a and b, there can be 2 cases. 
Case 1 : The sequence is a,c,b
Case 2: The sequence is b,c,a
Given that d is the arithmetic mean of b anc c, there can be 2 cases.
Case 1 : The sequence is b,d,c
Case 2: The sequence is c,d,b
Only 2 final cases are possible here.
Sequence 1 : a,e,c,d,b
Sequence 2 : b,d,c,e,a

Statement (i) is true as in both sequence 1 and sequence 2, the average of all terms is c.
Statement (ii) is true as in both sequence 1 and sequence 2, the average of d and e is c and average of a and b is c. c is not greater than c.
Statement (iii) is false as average of b and c, which is d is not greater than average of a and d in both sequences simultaneously.
Option A is the answer.

Let the eggs bought by Vijay be N.
Vijay throws 2 eggs, so remaining=N-2
Eggs in Vijay's fridge=$$\frac{5}{9}\left(N-2\right)$$
Remaining eggs=$$\frac{4}{9}\left(N-2\right)$$
Rashmi cooks 2 eggs, so Eggs in her fridge = $$\frac{4}{9}\left(N-2\right)\ -\ 2$$
$$\frac{4}{9}\left(N-2\ \right)-\ 2\ \le\ 5$$
N$$\le$$17.75
Also, for $$\ \frac{5}{9}\left(N-2\right)$$ and $$\ \frac{4}{9}\left(N-2\right)$$ to be an integer N-2=9k (k is an integer greater than 0)
N=9k+2=11,20,....
Maximum possible value of N is 11.
Option A is the answer. 

We want to minimise the total number of coins under these conditions:

7 pockets, each has at least 1 coin

At most 3 pockets can have the same number of coins

The remaining pockets must all have distinct numbers

To minimise total coins: we can use the smallest possible positive integers and allow exactly 3 pockets to have the same number (since “at most 3” → we use 3 to minimise total). The other 4 pockets must all be distinct and different from the repeated number.

Let the repeated number be 1 (smallest possible): so 3 pockets: 1, 1, 1

Now choose 4 distinct numbers, all different from 1: Next smallest distinct numbers: 2, 3, 4, 5

Hence, Total coins: 1 + 1 + 1 + 2 + 3 + 4 + 5 = 17

Let the monthly rent per person for the apartment be $P.
Statement 1 - P=$375 and Total rent=4P=$1500. So, statement 1 alone is sufficient to answer the question.
Statement 2 - 4P=3(P+125). P=$375, so total rent=4P=$1500. So, statement 2 alone is sufficient to answer the question.
So, Each statement alone is sufficient.
Option D is answer.

Given: 

Rate = x revolutions per minute and Time = k seconds

Convert time into minutes: $$\dfrac{k}{60}$$ minutes

Number of revolutions: $$x\times\dfrac{k}{60}$$ = $$\dfrac{xk}{60}$$

Let the amount invested in Scheme A be Rs. X.

Then, the amount invested in Scheme B = 13900 - X

Interest from Scheme A: $$\dfrac{\left(X\times14\times2\right)}{100}=0.28X$$

Interest from Scheme B: $$\dfrac{\left(\left(13,00-X\right)\times11\times2\right)}{100}=0.22\left(13900-X\right)$$

Total interest: 0.28X + 0.22(13900 - X) = 3508

0.28X + 3058  - 0.22X = 3508

0.06X = 450

X = 7500

Amount in Scheme B: 13900 - 7500 = 6400