AP ICET 2018 Question Paper Shift-2 (2nd May)

According to the problem,

$$\frac{25}{100}$$m + n = $$\frac{110}{100}$$n

$$\Rightarrow$$  $$\frac{25}{100}$$m = $$\frac{110}{100}$$n - n

$$\Rightarrow$$  $$\frac{25}{100}$$m = $$\frac{10}{100}$$n

$$\Rightarrow$$   m : n = 2 : 5

Hence, the correct answer is Option A

By selling at Rs. 5700, if the loss is 5%, it means that 5700 = 0.95*CP
Hence, CP = 5700/0.95

Now, to make a gain of 5%, the new SP = 1.05*CP = 1.05 * (5700/0.95) = Rs. 6300

After the discount, the loss incurred is = 10% i.e. 10% of 800 = 80 rupees
Hence, the product is sold for 800-80 = 720 rupees

The marked price of article was Rs. 1000 on which the discount D% was given
The actual selling price Rs. 720 is = (720/1000) = 72% of the marked price
This means, the discount percentage D = 100-72 = 28%

Given, R’s investment is Rs. 5630

Q invested twice as much as R

$$\Rightarrow$$  Q’s investment = 2 x 5630 = 11260

P invested half as much as Q

$$\Rightarrow$$  P's investment = $$\frac{11260}{2}$$ = 5630

$$\therefore\ $$Investment of P and Q together = 5630 + 11260 = 16890

Let the actual shares of A,B,C be 5x, 6x, 11x respectively
We are given that 11x-5x = 3 lakhs ===> x = 0.5 lakhs

Hence, share of B = 3 lakhs and 50% of his share = 1.5 lakhs

Let the capacity of the cistern be 600 liters
Hence, capacity of pipe P = 30 liters/min and capacity of Q = 20 liters/min 
Let the draining capacity of R be -r liters per minute. 

Now, when all three are opened together, the cistern gets filled in 24 minutes.
Effective capacity of all three together = 600/24 = 25 liters/min

Hence, 30+20-r = 25 ===> r = 25 liters/min
Hence, time required by R to empty complete cistern =  600/25 = 24 minutes

Let capacity of the tank = 60 liters.
Hence, capacities of the taps respectively are:
A = 6 liters/min
B = 4 liters/min
C = -3 liters/min (minus because it drains the tank)

When all three are opened for a minute, tank filled = 6+4-3 = 7 liters
Tank remaining to be filled = 53 liters

Time needed now to fill the tank = 53/10 = 5.3 minutes

Speed of the person = $$\frac{1200}{12}=100\ \frac{meters}{\min}$$

Multiplying both numerator and denominator by 60 (because we want speed in kmph), 
We get Speed = $$6000\ \frac{meters}{hour}=6\ kmph$$

Let the distance between P and Q be D km.

Total time = $$3\frac{1}{2}$$ hours = 3.5 hours

According to the problem, $$\frac{D}{60}+\frac{D}{45}=3.5$$

$$\Rightarrow$$ $$\frac{1}{15}\left(\frac{D}{4}+\frac{D}{3}\right)=3.5$$

$$\Rightarrow$$ $$\frac{1}{15}\left(\frac{7D}{12}\right)=3.5$$

$$\Rightarrow$$ D = 90 km

$$\therefore\ $$Distance between P and Q = 90 km

It is known that the share of any single worker in the complete pay of the workers is inversely proportional to the number of days each worker takes to complete the task alone. 
Here, P and Q take 50 and 40 days respectively. Hence, the pay off for P and Q will be in the ratio of 40:50 i.e. 4:5 respectively

Hence, for a total pay of Rs. 900, P's share will be = (4/9)*900 = 400 rupees.