AP ICET 2016 Question Paper (16th May)

Original Volume V = $$\pi\ \cdot r^2\cdot h$$

New Volume = $$\pi\ \cdot\left(1.1r\right)^2\cdot\left(1.1\right)h$$ = $$1.331\cdot\pi\ \cdot\left(r\right)^2\cdot h$$ = $$1.331V$$ 

Hence, the volume increases by 33.1% 

Let the marked price be MP. After giving a discount of 30% on it, loss incurred is 2% 
hence, we can say that 0.7MP = 0.98CP 
Thus MP = (0.98/0.7)CP = 1.4*CP 

Hence, marked price must be 40% above the cost price of the article 

Let cost price be CP

$$CP\left(1+\frac{CP}{100}\right)=SP=75$$

$$100CP+CP^2=7500$$

This gives CP = -150 or CP = 50 

Investment*Time product of P = 90,000*12 = 10,80,000 rupees-months
Investment*Time product of Q = 1,25,000*9 = 11,25,000 rupees-months

Total = 22,05,000 rupees-months

Share of P in the year-end profit = (1080/2205)*32,340 = 15,840 rupees 

The year end profits are shared in the ratio made by the individual's sum total of (investment*time period) product for all the 12 months. 
Let the individual investments made by A,B,C be 3x, 4x, 5x respectively. 

A's investment remains unchanged for all 12 months. 
Hence, his Investment*Time product = 36x units

B's investment is 4x for 3 months and then 5x for 9 months
Hence, his Investment*Time product = 12x + 45x = 57x units

C's investment is 5x for 3 months and 4x for 9 months
Hence, his Investment*Time product = 51x units

Total of Investment*Time products = 144x units

Hence, share of B in annual profit of 2,40,000 = (57/144)*2,40,000 = 95,000

Given, a : b = 2 : 5

Let a = 2k and b = 5k

$$a^2-ab+b^2=\left(2k\right)^2-\left(2k\right)\left(5k\right)+\left(5k\right)^2=4k^2-10k^2+25k^2=19k^2$$

$$a^2+ab+b^2=\left(2k\right)^2+\left(2k\right)\left(5k\right)+\left(5k\right)^2=4k^2+10k^2+25k^2=39k^2$$

$$\therefore\ $$ $$a^2 - ab + b^2 : a^2 + ab + b^2 = 19k^2 : 39k^2 = 19 : 39$$

Hence, the correct answer is Option A

Let the capacity of the tank be 72 liters
Hence, A = 9 liters/hour, B = 6 liters/hour, C = -8 liters/hour.

Now, tank filled by A and B in the first 2 hours = 30 liters. 
Tank capacity remaining = 42 liters
After this C is opened, net filling capacity drops to 7 liters/hour
Hence, the time taken to fill remaining tank = 42/7 = 6 hours
Hence, total time taken = 2+6 = 8 hours

Let the capacity of the tank be 210 liters. 
Hence, the capacity of the pump = 70 liters/hour

But because of the leak, it takes 3.5 hours.
Hence, effective filling rate of the tank with the pump and the leak = 210/3.5 = 60 liters/hour

Hence, the leak is occurring at the rate of 10 liters/hour. 
Hence, time taken by the leak to drain a full tank = 210/10 = 21 hours. 

Let the total work be 18 units and the capacities of A,B,C be a,b,c units/day respectively. 
Hence, a+b+c = 6 and b+c = 3  ==> a = 3 units/day 
Also, a + c = 4 hence, c = 1 unit/day. 
Hence, time required by C alone to complete all the work = 18 days

Let the total work be 60 units. 
Hence, capacity of B = 3 units/day and thus, capacity of A = 9 units/day
A and B work together for 4 days, hence, work completed = 48 units
Remaining work = 12 units completed by B in 12/3 = 4 days