MAH MBA CET Previous Paper 2025 Slot-1

Boxes have 2 toys which are of different colors to each other and to the color of box.

so Green box has 1 red toy and 1 blue toy

Blue box has 1 red toy and 1 green toy

Red box has 1 blue toy and 1 green toy

Also

so Green Box should have maximum possible chocolates and the Red box has the least possible chocolates in it. Such that each box has at least one chocolate and no two boxes have equal number of chocolates.

so to minimize chocolates minimum 1 chocolate needs to be there in it. In blue box, in order to maximize chocolates in green box, needs to have only 2 chocolates ( as no two boxes have equal number of chocolates and minimum we can put now here is 2 chocolates ) and finally the green box will have 7 chocolates left which is the maximum that can be done.

Therefore option c : 'The box ( blue colored box )  which has the toys of Green and Red colors has 2 chocolates in it' is correct

A bought R and T. 

It is given that if A bought R, then B will neither buy P nor S. 

So, B can only buy Q and U and C can buy P and S.

Option D.  

Solution: So as per the given data Ambatta's watch skips 35 seconds in total in the week ( 0+1+3+4+7+13+7 ) .

Now total number of seconds in 1 week is : 7 (days) * 24 (hours in 1 day) * 60(minutes in 1 hour) *60 (seconds in 1 minute)

which gives us 6,04,800 seconds in total 

Now in order to arrive at our answer we need to subtract the seconds Ambatta's watch skipped.

Therefore How many Seconds did Ambatta's watch work during the given duration = 6,04,800 - 35 = 604 765 Seconds

Therefore option c is correct

First, we will count odd days from 15 Aug 1947 to 26 Jan 1957

(A) From 15 Aug 1947 to 15 Aug 1956 → 9 years

Leap years in between: 1948, 1952, 1956 → 3 leap years

• Normal years = 6 → 6 × 1 = 6 odd days

• Leap years = 3 → 3 × 2 = 6 odd days

Total = 6 + 6 = 12 odd days → 12 ≡ 5 (mod 7)

Now, (B) From 15 Aug 1956 to 26 Jan 1957

Aug (remaining 16 days)
Sep = 30
Oct = 31
Nov = 30
Dec = 31
Jan = 26

Total: 16 + 30 + 31 + 30 + 31 + 26 = 164 days

164mod7 = 3 odd days

Total odd days: 5 + 3 = 8 ≡ 1 (mod 7)

Hence, the final day: Friday + 1 = Saturday

Let: 

Cost of one orange = O
Cost of one banana = B
Cost of one apple = A

We are given that:

2O + 3B + 4A = 15...(1)

3O + 2B + A = 10...(2)

Now, we need 3O + 3B + 3A = 3(O + B + A)

Now, we will aim to eliminate the variables. 

Multiply (2) by 4: 12O + 8B + 4A = 40...(3)

Subtract (1) from (3): (12O - 2O) + (8B - 3B) + (4A - 4A) = 40 - 15 

We get: 10O + 5B = 25

Or, 2O + B = 5...(4)

Now, From (4): B = 5 - 2O

Substitute into (2): 3O + 2(5  - 2O) + A = 10
3O + 10 - 4O + A = 10
-A + O + 10 = 10
A = O

We need to find O + B + A.

O + B + A = O + (5 - 2O) + O = 5

Final value 3(O + B + A) = 3 × 5 = 15

We are given that:

Type I → Rs. 32/ltr

Type II → Rs. 40/ltr

Ratio = 3 : 5

We can simply solve this using the weighted average method.

Cost = $$\dfrac{\left(32\times3\right)+\left(40\times5\right)}{3+5}=\dfrac{96+200}{8}=\dfrac{296}{8}=37$$

Hence, the answer is Rs. 37 per litre. 

We have to find the probability that the umpire picked up at least 1 sweater. 

It means that we can subtract the probability in which no sweater was picked (i.e. picking up 3 caps) from 1 to get the desired probability. 

Number of ways in which 3 caps were picked = $$\large{6_{\large{C_3}}}$$

Number of ways in which 3 articles were picked = $$\large{10_{\large{C_3}}}$$

Probability of picking up 3 articles with no sweaters = $$\dfrac{\large{6_{\large{C_3}}}}{\large{10_{\large{C_3}}}}$$ = $$\dfrac{1}{6}$$

Probability of picking up at least 1 sweater = $$1-\dfrac{1}{6}=\dfrac{5}{6}$$

Hence, the probability of an umpire picking up 3 articles with at least 1 sweater is $$\dfrac{5}{6}$$.

Option D.   

Let the number be X. 

Forming the equation:

$$\dfrac{1}{3}\times\dfrac{1}{4}\times X=15$$

$$\dfrac{1}{12}\times X=15$$

$$X=12\times15=180$$

Now, three-tenths of the number: $$\dfrac{3}{10}\times180=54$$

Let the number of males = M and females = F

We can now use the weighted average formula:

$$\dfrac{70M+83F}{M+F}=76$$

Or, $$70M+83F=76M+76F$$

$$70M+83F=76M+76F$$

$$7F=6M$$

Hence, ratio: $$\dfrac{M}{F}=\dfrac{7}{6}$$