MAH MBA CET Previous Paper 2025 Slot-1

Total production of all companies = 1500 units
Production of A - $$\ \frac{\ 16}{100}\times\ 1500$$ = 240 units
Production of B -  $$\ \frac{\ 11}{100}\times\ 1500$$ = 165 units
Production of C -  $$\ \frac{\ 21}{100}\times\ 1500$$ = 315 units
Production of D -  $$\ \frac{\ 9}{100}\times\ 1500$$ = 135 units
Production of E -  $$\ \frac{\ 26}{100}\times\ 1500$$ = 390 units
Production of F -  $$\ \frac{\ 5}{100}\times\ 1500$$ = 75 units
Production of G -  $$\ \frac{\ 12}{100}\times\ 1500$$ = 180 units

Statement A - A+G=240+180=420 units, which is less than 500 units. So, A is false.
Statement B - B+F=165+75=240 units, which is less than 250 units. So, B is true.
Statement C - C+E=315+390=705 units, which is equal to 705 units. So, C is true.
Statement D - D+G=135+180=315 units, which is more than 200 units. So, D is false.
Statement E - E+F+G=390+75+180=645 units, which is greater than 545 units. So, E is false.
Only B and C are true.
Option D is the answer.

Total votes polled = 15,200

Invalid votes = 15% of 15,200 = 0.15 × 15,200 = 2,280

So, valid votes = 15,200 − 2,280 = 12,920

Winner got 55% of valid votes ⇒ Winner’s votes = 0.55 × 12,920 = 7,106

So, loser’s valid votes = 12,920 − 7,106 = 5,814

Male workforce in SA sector = $$\ \frac{\ 5}{8}\times\ $$ Total workforce in SA sector = $$\ \frac{\ 5}{8}\times\ \frac{\ 12}{100}\times\ 6000000\ =\ 450000$$
Male workforce in MA sector = $$\ \frac{\ 3}{7}\times\ $$ Total workforce in MA sector = $$\ \frac{\ 3}{7}\times\ \ \frac{\ 21}{100}\times\ 6000000\ =\ 540000$$
Number of male workforce in SA and MA sectors put together = 450000+540000 = 990000
Total workforce in PT sector =  $$\frac{\ 19}{100}\times\ 6000000\ =\ 1140000$$
Required percent = $$\ \frac{\ 990000}{1140000}\times\ 100\ =\ 86.84\%\ \approx\ 87\%$$
Option B is the answer.

Female workforce in SA sector = $$\ \frac{\ 3}{8}\times\ $$ Total workforce in SA sector = $$\ \frac{\ 3}{8}\times\ \frac{\ 12}{100}\times\ 6000000\ =\ 270000$$
Female workforce in MA sector = $$\ \frac{\ 4}{7}\times\ $$ Total workforce in MA sector = $$\ \frac{\ 4}{7}\times\ \frac{\ 21}{100}\times6000000\ =\ 720000\ $$
Number of female workforce in SA and MA sectors put together = 270000+720000 = 990000
Number of female workforce in PT sector = $$\ \frac{3}{8}\times\ $$ Total workforce in PT sector = $$\ \frac{3}{8}\times\ \frac{19\ }{100}\times\ 6000000\ =$$ 427500
Required difference = 990000-427500 = 56250 which is close to 560000
Option A is the answer.

Let the value of the gift coupon be Rs. x.

Total Yearly Payment: Rs. 90 + x

Payment for 9 months = $$\dfrac{3}{4}\left(90+x\right)$$

According to the question, this equals what the servant actually received: $$\dfrac{3}{4}\left(90+x\right)=65+x$$

$$67.5+\dfrac{3x}{4}=65+x$$

$$67.5-65=x-\dfrac{3x}{4}$$

$$2.5=\dfrac{x}{4}$$

$$x=10$$

11x + 15y = -1
Solution of this equation will be of the form x=a+15k , y=b-11k, where (a,b) is any solution of the equation
We can clearly see that x=4,y=-3 is one solution of the equation.
x=4+15k, y=-3-11k
Given x<100,y<100
4+15k<100, k$$\le\ $$6 and -3-11k<100, k$$\ge-9\ $$
$$-9\le\ k\ \le\ 6$$, k can take 16 values.
There exists 16 solutions for the equation.
Option D is the answer.
 

Let the radius of sphere A be 'r' and sphere B be 'R'
Surface area of A = $$4\pi r^2$$
Surface area of B = $$4\pi R^2$$ = (1+300%) of A = 4 * $$4\pi r^2$$
R=2r
Volume of A = $$\frac{4}{3}\pi r^3$$ 
Volume of B = $$\frac{4}{3}\pi R^3\ =\ \ \frac{\ 4}{3}\pi\ \left(2r\right)^3\ =\ \ \frac{\ 32}{3}\pi\ r^3$$
Required percent = $$\ \frac{\ \left(\frac{32}{3}\pi\ r^3\ -\ \frac{4}{3}\pi\ r^3\right)}{\ \frac{32}{3}\pi\ r^3}\times\ 100\ =87.5\ \%\ $$
Option C is the answer.

The ship admits 2.5 metric tons of water in 6 minutes and 30 seconds.
Water Influx rate = $$\frac{2.5}{\ \frac{\ 6.5}{60}}$$ metric tons/hr = 23.07 metric tons/hr
Water outflux rate = 15 metric tons/hr
Net influx rate = 23.07-15 = 8.07 metric tons/hr
Time after which ship starts sinking = $$\ \frac{\ 68}{8.07}$$ = 8.42 hours
Average rate of sailing=$$\frac{156}{8.42}$$=18.53kmph
Closest option is option C.

Total even integers between 100 and 200 = $$\frac{200-100}{2}\ \ +\ \ 1\ =\ 51$$
Even integers divisible by 7 = Number of multiple of 14.
There are 7 integers which are of form 14k(112,126,...) between 100-200
Even integers divisible by 9 = Number of multiple of 18.
There are 6 integers which are of form 18k(108,126,...) between 100-200
Even integers divisible by both 7 and 9 = No. of multiple of LCM(14,18) = No. of multiple of 126k = 1
So, Required answer = 51-(7+6)+1 = 39.
Option C is the answer.

The total cost consists of a fixed daily fee plus a per-show fee: Daily fixed cost: Rs. 10,000 

Cost for 3 shows: 3*5,000 = Rs. 15,000

Total Cost: 10,000 + 15,000 = Rs. 25,000

The theatre has 200 seats and runs 3 shows, meaning there are 600 total available seats.

Total occupancy (80%): 0.80*600 = 480

To maximise profit, we want the most expensive tickets to sell out first.

Show 1 (Rs. 250): Max capacity = 200 viewers.

Show 2 (Rs. 250): Max capacity = 200 viewers.

Show 3 (Rs. 200): Remaining viewers = 480 - (200 + 200) = 80  viewers.

First two shows: 400 × 250 = Rs. 100,000

Late-night show: 80*200 = Rs. 16,000

Total Revenue: 1,16,000

Profit = Revenue - Cost

Profit = 1,16,000 - 25,000 = Rs. 91,000