MAH MBA CET Previous Paper 2025 Slot-1

CP = Rs. 225

MP = 2*225 = Rs. 450

Now, to get a profit of 56%, the toy must be sold for 156% of its cost price:

Hence, SP = 225*1.56 = Rs. 351

Discount = MP - SP

Discount = Rs. 450 - Rs. 351 = Rs. 99

Discount % = $$\dfrac{99}{450}\times100=22\%$$

Prime numbers from 0 to 100 are 2,3,5,7,11,........
The multiplication will be very complicated. So, the question requires you to find number of zeros at the end.
Product = 2*3*5*7.....
No. of zeros at end = highest power of 10, which is 1 in this case(since 2,5 occur only once each and no higher powers of 2 and 5 occur).
So, the answer is option D as it has only 1 zero at the end.

17*17*18*18*19*19 = $$2^2\times\ 3^4\times\ 17^2\times\ 19^2$$
971 is not divisible by 2 or 3 or 17 or 19
173,23,5813 are also not divisible by 2 or 3 or 17 or 19
646=2*19*17. So, it can divide the given no.
Option C is the answer.

R($$\ \frac{\ 1719*1721*1723*1725*1727}{18}$$) = R($$\frac{9\times\ 11\times\ 13\times\ 15\times\ 17\ }{18}$$) = 9

Let the cost price be 100x.
Let the part sold at 8% profit be K kg. So, part sold at 18% profit would be (1000-K) kg
Total profit = $$\ \frac{\ SP\ -\ CP}{CP}\times\ 100$$ = $$\ \frac{\ 108x\times K\ +\ 118x\left(1000-K\right)\ \ -\ 1000\times\ 100x}{1000\times\ 100x}\times\ 100$$ = 14
On solving it, we get K=400kg.
Quantity sold at 18% profit = 1000-K = 600 Kg.
So, option E is the answer.

Let the breadth of room be 'b' meters. So, the length will be 'b+2' meters. So, Area = b(b+2)
If the length is increased by 4 meters, then length='b+6' meters and the breadth decreased by 2 meters, then breadth='b-2' meters
So, initial area = new area. 
b(b+2) = (b+6)(b-2)
$$b^2+2b\ =\ b^2+4b-12$$
On solving, b = 6 meters. So, Breadth = 6 meters and length=8 meters.
Surface area of walls = 2*h(l+b) = 2*3(6+8) = 84m^2

Average speed = $$\ \frac{\ Total\ Distance}{Total\ time}$$
Total distance = 50+50+30 = 130km
Total time = $$\ \frac{\ 40}{60}$$+$$\ \ \ \frac{\ 50km}{120kmph}$$+$$\ \ \ \frac{\ 30km}{60kmph}$$ = $$\ \frac{\ 19}{12}hr$$
Avg speed = $$\ \frac{\ 130}{\frac{19}{12}}=82.1\ kmph$$
So, option E is the answer.

Volume of cone = $$\ \frac{\ 1}{3}\pi\ r^2h$$ 
Let initial height be 'h' and radius be 'r' of the cone.
So, Initial volume =  $$\ \frac{\ 1}{3}\pi\ r^2h$$
New height H = h(1+200%) = h(1+$$\ \frac{\ 200}{100}$$) = 3h
New radius R = r(1-50%) = r(1-0.5) = $$\ \frac{\ r}{2}$$
New volume = $$\ \ \frac{\ 1}{3}\pi\ \times\ \left(\frac{r}{2}\right)^2\times\ 3h$$ = $$\ \frac{\ \pi\ r^2h}{4}$$
Change in volume = $$\frac{\left(\frac{\ \pi\ r^2h}{4}-\frac{\ \pi\ r^2h}{3}\right)}{\ \frac{\ \pi\ r^2h}{3}}\times\ 100$$ = -25%
So, volume decreased by 25%.
Option B is the answer.

When tossing 3 coins, each coin has 2 possibilities (H or T). The total number of possible outcomes is 2^3 = 8.

The sample space (S) is:

HHH
HHT
HTH
THH
HTT
THT
TTH
TTT

The condition is "at least two" tails, which means we are looking for outcomes with exactly 2 tails or exactly 3 tails. 

Exactly 2 tails: {HTT, THT, TTH} — 3 outcomes 

Exactly 3 tails: {TTT} — 1 outcome

Total favorable outcomes = 3 + 1 = 4

Hence:

P(Atleast 2 tails) = 4/8 = 1/2 = 50%