MAH MBA CET Previous Paper 2025 Slot-1

Let the initial point of Krishna be A.

He travelled for 20 m in the North. Then he turned right and travelled for 30 m. So, now he is 20 m north and 30 m east of A. 

Then again turned right and travelled for 35 m. Now, he is 30 m east and 15 m south of A. 

Then, he turned left and walked for 15 m. Now, he is 45 m east and 15 m south of A. 

Finally, he travelled 15 m north. Now, he is 45 m to the east of A. 

Hence, Krishna is 45 m east of A after travelling the entire distance. 

Option E.

In the given question, we have to find the distance between K and P. 

Now, it is given that, J is 1 km north of K and Q is 2 km south of J. It means that, Q is 1 km south of K. 

It is also given that, P is 1 km west of Q. 

So, P, Q and K forms a right angled triangle at Q. 

So, we can apply the Pythagoras theorem to find the distance between P and K. 

$$\left(PK\right)^2=\left(PQ\right)^2+\left(KQ\right)^2$$

$$\left(PK\right)^2=\left(1\right)^2+\left(1\right)^2$$

$$PK=\sqrt{2}\ km=1.414\ km$$

Hence, the distance between P and K is 1.41 km. 

Option C. 

6 : 222 : : 7 : $$x$$

To get to 222, we do the following operation: $$\left(6\right)^3+6=216+6=222\ $$

Hence, $$\left(7\right)^3+7=x$$

$$343+7=x$$

$$x=350$$ 

Option A.

PATTERN : Options A , B and C follow the rule ( FIRST TERM / SECOND TERM ) = ( FOURTH TERM / THIRD TERM  )  which is equal to (2/3)

Option A : ( 6.40/9.60) =(12.80/19.20) = (2/3)

Option B : ( 1.0/1.50) =(2/3) = (2/3)

Option c : ( 0.40/0.60) =(0.80/1.20) = (2/3)

NOW  

In  Option D : left hand side ( 0.50/1.50) = (1/3)

                      Right hand side (0.25/3.0) = (1/8)

This option breaks the pattern here as neither LHS = RHS nor the resultant on either side equal to (2/3).

Let the number of girls in between Rimmi and Mimmi be $$x$$. 

It is given that Rimmi and Mimmi occupies ninth and tenth place from the right end and left end respectively. 

Number of girls to the right of Rimmi = 8 (as she occupies 9th place from the right end)

Number of girls to the left of Mimmi = 9 (as she occupies 10th place from the left end)

The initial arrangement : $$9,\ M,\ x,\ R,\ 8$$ 

Now, both Rimmi and Mimmi interchanged their positions. 

Now, the new arrangement: $$9,\ R,\ x,\ M,\ 8$$

Now, Rimmi is 17th place from the right. It means that Rimmi has 16 girls to her right. 

$$8+1+x=16$$

$$x=16-9=7$$

We have 7 girls in between Rimmi and Mimmi. 

Total number of girls in the arrangement = $$8+1+7+1+9=26$$

Hence, we have a total of 26 girls in the given row. 

Option C. 

It is given that if a vowel is placed at the beginning or the end of the word, then it will be coded as 9. 

The word that we have to code contains A and I at the beginning and at the end. So, the required code will start and end with 9. 

Now, F, D, Q and N are coded as 7, 2, 8 and 4 (given).

It is also given that if a vowel is not at the start or at the end, then it will be coded as 6. Hence, E will be coded as 6. 

The required code for AFDQENI is 9728649.

Option C.  

It is given that if a vowel is placed at the beginning or the end of the word, then it will be coded as 9.

The word that we have to code contains O at the beginning. So, the required code will start with 9.

Now, L, S, W, Y, C, Z and T are coded as 3, 3, 2, 8, 1, 2 and 1 (given).

It is also given that if a vowel is not at the start or at the end, then it will be coded as 6. Hence, I will be coded as 6.

The required code for OLSWYCZIT is 933281261.

Option D.

It is given that if a vowel is placed at the beginning or the end of the word, then it will be coded as 9.

The word that we have to code contains U at the end. So, the required code will end with 9.

Now, T, J, V, R, Q and X are coded as 1, 5, 4, 5, 8 and 7 (given).

It is also given that if a vowel is not at the start or at the end, then it will be coded as 6. Hence, I and A will be coded as 6.

The required code for TJITAVRQXU is 1561645879.

Option E.

Since, there are no vowels in the given word, so we have to find the code for each digit from the given values. 

So, X, L, P, V, R, Y, T, Q, D and C are coded as 7, 3, 7, 4, 5, 8, 1, 8, 2 and 1 (given).

The required code for XLPVRYTQDC is 7374581821.

Option C.

The clock gains a total of 10 min in 24 hours. 

So, in 3 hours, the clock will gain a total of $$\dfrac{10\times3}{24}=\dfrac{10}{8}=1\dfrac{1}{4}$$ min 

When the clock indicates a time of 11 am, after it started at 8 am, it has gained a total of $$1\dfrac{1}{4}$$ min. 

The true time can be found by subtracting the time gained from the actual time indicating by the clock. 

So, we have to subtract 1 min and 15 seconds from 11 am to find the true time. 

Hence, the true time is 10:58:45 am. 

Option D.