MAH MBA CET Previous Paper 2025 Slot-1

We are noticing that each term is obtained by dividing the previous term by 2.

So, the next term: $$\dfrac{1}{4}\ \times\dfrac{1}{2}=\dfrac{1}{8}$$

Let 

P(Alka selected) = 1/7 so P(Alka not selected) = 6/7

P(Alkit selected) = 1/5 sp P (Alkit not selected) = 4/5

So, the probability of only one being selected: 

Case 1: Alka selected, Alkit not selected

$$\dfrac{1}{7}\times\dfrac{4}{5}=\dfrac{4}{35}$$

Case 2: Alka not selected, Alkit selected

$$\dfrac{6}{7}\times\dfrac{1}{5}=\dfrac{6}{35}$$

Total probability: $$\dfrac{4}{35}+\dfrac{6}{35}=\dfrac{10}{35}=\dfrac{2}{7}$$

20 X 8 ÷ 8 - 4 + 2 = 20+8-8÷4*2(Since + sign meaning is not given, we are taking it as multiplication)
Applying BODMAS rule,
20+8-2*2=20+8-4=24.
Option C is the answer.

Total cars sold(SUV+MUV)=56000
Number of SUV cars sold=32000
So, Number of MUV Cars sold=24000
Total cars sold by D=$$\ \frac{\ 15}{100}\times\ 56000$$=8400
Total SUV cars sold by C,F and G=(7%+8%+6%) of 32000=$$\ \frac{\ 21}{100}\times\ 32000$$=6720
Required answer=$$\ \frac{\ \left(8400-6720\right)}{6720}\times\ 100=\ \frac{\ 1680}{6720}\times\ 100=25\ \%\ $$
Option E is the answer.

Total cars sold(SUV+MUV)=56000
Number of SUV cars sold=32000
So, Number of MUV Cars sold=24000
X=Total SUV cars distributed by A and B=(16%+16%) of 32000 = 32*320 = 10240
Y=Total(SUV+MUV) cars distributed by C and F together=(14%+8%) of 56000 = 12320
Required answer=$$\ \frac{\ X}{Y}=\ \frac{\ 10240}{12320}=\frac{\ 64}{77}$$
Option C is the answer.

Total cars sold(SUV+MUV)=56000
Number of SUV cars sold=32000
So, Number of MUV Cars sold=24000
Average number of MUV cars distributed by dealers A, D, E, F and H = $$\ \frac{\ TotalMUVcarsdistributedbydealersA,D,E,FandH}{5}$$
Total MUV cars distributed by dealers A, D, E, F and H = Total(SUV+MUV) cars distributed by dealers A, D, E, F and H - Total SUV cars distributed by dealers A, D, E, F and H
=(18%+15%+20%+8%+9%) of 56000 - (16%+13%+22%+8%+12%) of 32000 = $$\ \frac{\ 70}{100}\times\ 56000\ -\ \ \frac{\ 71}{100}\times\ 32000\ =\ 39200\ -\ 22720\ =\ 16480$$
Average number of MUV cars distributed by dealers A, D, E, F and H = $$\ \frac{\ 16480}{5}$$ = 3296
Option D is the answer.

Unit digit of $$6374^{1793}\times625^{317}\times341^{491}\ $$ = Unit digit of $$4^{1793}\times5^{317}\times1^{491}\ $$ = unit digit of 4*5*1 = 0
Option A is the answer.

Total possible combinations when 2 dice are thrown=6*6=36
Product of 2 numbers is even when either both of them are even or one is even, other is odd.
There are 3 even no.(2,4,6) and 3 odd no.(1,3,5) possible in a dice.
Total combinations when both are even = 3*3 = 9
Total combinations when one is even and other is odd = 3*3(first no. is odd, second even) + 3*3(first no. is even, second odd) = 18
Required probability = $$\frac{\ 9+18}{36}=\ \frac{\ 27}{36}=\ \frac{\ 3}{4}\ $$
Option E is the answer.
 

Perfect square of a quadratic equation is of form $$\left(pa\ +\ q\right)^2$$
Let 'k' be the no. added to make the required equation as a perfect square
$$\left(pa\ +\ q\right)^2=p^2a^2\ +\ 2apq\ +\ q^2=16a^2-12a\ +\ k$$
Comparing the coefficients, we get $$p^2=16,\ p=\pm\ 4.$$
$$2apq=-12a.\ So,\ pq=-6,\ \ $$ which gives q=$$\ \pm\frac{\ 3}{2}$$
k=$$\ q^2$$=$$\ \frac{\ 9}{4}$$
Option A is the answer.

There is 1 ball on top, 3 balls in second layer, 6 balls in third layer,10 balls in fourth layer .......... and so on 
Total balls=8436
Sum of balls across all layers = S = 1+3+6+10+.........
S = 1+3+6+10+.......+$$t_n$$ 
S =     1+3+6+........+$$t_{n-1}\ +\ t_n$$
Subtracting both equations,
0 = 1+2+3+4+......... - $$t_n$$
$$t_n$$ = 1+2+3+4+.......... = $$\ \frac{\ n\left(n+1\right)}{2}$$
$$S_n=\Sigma\ t_n=\Sigma\ \ \frac{\ n\left(n+1\right)}{2}=\ \frac{\ 1}{2}\Sigma\ \left(n^2+n\right)$$
$$S_n=\ \frac{\ 1}{2}\left(\ \frac{\ n\left(n+1\right)\left(2n+1\right)}{6}+\ \frac{\ n\left(n+1\right)}{2}\right)=\ \frac{\ n\left(n+1\right)\left(n+2\right)}{6}=8436$$
n(n+1)(n+2)=50616
On solving, we get n=36.
Option B is the answer.