MAH CET 9th Mar 2024 Slot 1

Aman invested in the business for a complete 3 years = 36 months

Rakhi had joined 6 months after Aman $$\Rightarrow$$ Rakhi invested for total of 30 months

Sagar had joined 6 months after Rakhi $$\Rightarrow$$ Sagar invested for total of 24 months

Ratio of profits = Ratio of (investment $$\times$$ time period)

$$\therefore$$ Aman : Rakhi : Sagar = (70,000 $$\times$$ 36) : (105,000 $$\times$$ 30) : (140,000 $$\times$$ 24)

$$\Rightarrow$$ Aman : Rakhi : Sagar = (70 $$\times$$ 6) : (105 $$\times$$ 5) : (140 $$\times$$ 4)

$$\Rightarrow$$ Aman : Rakhi : Sagar = (2 $$\times$$ 6) : (3 $$\times$$ 5) : (4 $$\times$$ 4)

$$\Rightarrow$$ Aman : Rakhi : Sagar = 12 : 15 : 16

Hence, option A is the correct choice.

Let the man bought 'b' dozens of oranges at Rs 10 per dozen.

CP$$_1$$ = 10a

He also bought the same number, i.e., 'b' dozens, of oranges at Rs 8 per dozen.

CP$$_2$$ = 8a

Total CP = 10a + 8a = 18a

Now he sold all the oranges at Rs 11 per dozen.

Total SP = 11 $$\times$$ 2a = 22a

Profit = Total SP - Total CP = 22a - 18a = 4a

$$\therefore$$ 4a = 120 $$\Rightarrow$$ a = 30

Total oranges bought = 2a dozens = 60 dozens

Hence, option D is the correct choice

Let the number of filling pipes = f

Let the number of waste pipes = w

f + w = 8 ----(1)

Each fill pipe completely fills the empty tank in 12 hours

Each waste pipe empties the completely filled tank in 36 hours

Let total volume of tank = LCM (12, 360) = 36 units

$$\therefore$$ Efficiency of fill pipe = 36/12 = 3 units/hour

$$\therefore$$ Efficiency of waste pipe = 36/36 = 1 unit/hour

Volume of tank filled by 'f' fill pipes and 'w' waste pipes = (3f - w) units/hour

It is given that the tank gets filled in 3 hours

$$\therefore$$ 3 $$\times$$ (3f - w) = 36

$$\therefore$$ 3f - w = 12 ----(2)

Adding (1) and (2), we get,

4f = 20 $$\Rightarrow$$ f = 5

$$\therefore$$ w = 8 - f = 3

Hence, option A is the correct choice.

The company produces on average 4000 items per month for the first 3 months

$$\therefore$$ Total items produced in the first 3 months = 4000 $$\times$$ 3 = 12000

Let the company produce on average x items per month for the first 9 months

$$\therefore$$ Total items produced in the next 9 months = x $$\times$$ 9 = 9x

It needs to average 4375 items per month over the whole.

$$\therefore$$ Total items prouduced in 12 months = 4375 $$\times$$ 12 = 52500

$$\Rightarrow$$ 12000 + 9x = 52500

$$\Rightarrow$$ 9x = 40500

$$\Rightarrow$$ x = 4500

Hence, option C is the correct choice.

Let the number = N

It is doubled $$\Rightarrow$$ 2N

9 is added $$\Rightarrow$$ 2N + 9

It is trebled $$\Rightarrow$$ (2N + 9) $$\times$$ 3 = 6N + 27

$$\therefore$$ 6N + 27 = 75

$$\Rightarrow$$ 6N = 48 $$\Rightarrow$$ N = 8

Hence, option B is the correct choice.

148 = 4 $$\times$$ 37

185 = 5 $$\times$$ 37

LCM (148, 185) = 4 $$\times$$ 5 $$\times$$ 37

$$\Rightarrow$$ LCM(148, 185) = 740

Hence, option A is the correct choice.

A number is divisible by 3 if the sum of its digits is divisible by 3.

Option A: 2345678 $$\Rightarrow$$ Sum of digits = 35 $$\Rightarrow$$ Not divisible by 3

Option B: 2876423 $$\Rightarrow$$ Sum of digits = 32 $$\Rightarrow$$ Not divisible by 3

Option C: 4006020 $$\Rightarrow$$ Sum of digits = 12 $$\Rightarrow$$ Divisible by 3

Option D: 9566003 $$\Rightarrow$$ Sum of digits = 29 $$\Rightarrow$$ Not divisible by 3

Option E: 3456785 $$\Rightarrow$$ Sum of digits = 38 $$\Rightarrow$$ Not divisible by 3

Hence, option C is the correct choice.

Three numbers are in the ratio 3 : 4 : 5

Let the numbers be 3k, 4k and 5k

LCM(3k, 4k , 5k) = 60k

60k = 2400 $$\Rightarrow$$ k = 40

The numbers are 120, 160, 200

$$\therefore$$ HCF(120, 160, 200) = 40

Hence, option C is the correct choice

We have to select at least $$3$$ men.

So we can either select $$3$$, $$4$$ or $$5$$ men

Case 1: $$3$$ Men, $$2$$ Women

Total number of ways = $$^7C_3 \times ^6C_2 = 35 \times 15 = 525$$

Case 2: $$4$$ Men, $$1$$ Woman

Total number of ways = $$^7C_4 \times ^6C_1 = 35 \times 6 = 210$$

Case 3: $$5$$ Men, $$0$$ Woman

Total number of ways = $$^7C_5 \times ^6C_0 = 21 \times 1 = 21$$

Total number of ways = $$525 + 210 + 21 = 756$$

Hence, option D is the correct choice.

Let A and B be two men, and let A be older than B

15 years ago, A's age was twice that of B.

Let 15 years ago, B's age = x

$$\therefore$$ 15 years ago, A's age = 2x

Difference between ages = 2x - x = x = 10 (given)

Present age of A = 2x + 15 = 35 yrs

Hence, option A is the correct choice.