**Directions(1-5):** The profit percentage of a company over different months have been given below. Refer the following graph & answer the following questions.

- If the expenditures in February and May are equal, then the approximate ratio of the income in February and May respectively ?
31:3527:3117:1931:3411:20Option D

For February 55 =( I1-x)/x*100

I1 = 155x/100

For May 70 =( I2-x)/x*100

I2 = 170x/100

Required ratio = I1/I2 = 31:34 - What is the average profit earned for the given months?
55(4/5)57(2/5)41(5/6)50(3/4)55(5/6)Option E

Average percent profit earned for the given month

= 1/6*(40+55+45+65+70+60)

= 55(5/6) - If the profit in May was Rs.4 crores, what was the profit in June?
11.12%33.33%50%25%Cannot be determined.Option E

From the obtain information about the percentage profit only.

To find the profit in June we must have the data for the income or expenditure in June.

Therefore, the profit for June cannot be determined. - If the income in April was Rs.264 crores, what was the expenditure in April?
168 crores152 crores135 crores171 crores160 croresOption E

For April 65 = (264-x)/x*100

x = 160 crores - In which month is the expenditure minimum?
AprilMarchJanuaryJuneCannot be determined.Option E

The comparison of percent profit for different months the comparison of the expenditures is not possible without more data. Therefore, the year with minimum expenditure cannot be determined. - Two pipes, P3 and P4 are opened simultaneously and it is found that due to the leakage in the bottom, 17/7 minutes are taken extra to fill the tank. If the tank is full, in what approximate time would the leak empty it?
29 min.35 min.22 min.40 min.39 min.Option E

Total time taken by both pipes before the leak was developed = 60/7 minutes

Now, leaks is developed which will take T time to empty the tank.

So, (1/15 +1/20 – 1/T) = 60/7 + 17/7

=> (1/15 +1/20 – 1/T) = 1/11

solve for T 660/17 minutes == 39 minutes - Two pipes P7 and P8 can fill a cistern. If they are opened on alternate minutes and if pipe P7 is opened first, in how many minutes will the tank be full?
7 min.5 min.6 min.4 min.3 min.Option C

Pipe P7 can fill = 1/12

Pipe P8 can fill = ¼

For every two minutes, 1/12 + 1/4 = 1/3 Part filled

Total = 6 minutes - A large cistern can be filled by two pipes P1 and P2. How many minutes will it take to fill the Cistern from an empty state if P2 is used for half the time and P1 and P2 fill it together for the other half?
4.8 min.6.0 min.7.5 min.6.2 min.5.5 min.Option C

Part filled by P1 and P2 = 1/15 + 1/10 = 1/6 Part filled by P2 = 1/10

Required time = x/2(1/6 + 1/10) = 2/15

= 15/2 = 7.5 minutes - Two pipes P5 and P6 can fill a tank. If both the pipes are opened simultaneously, after how much time should P6 be closed so that the tank is full in 8 minutes?
18 min.12 min.20 min.15 min.32 min.Option A

Required time = y (1-(t/x)) = 27(1-(8/24))

= 18 minutes - Three pipes P9, P10, and P18 can fill a tank. If Pipe P18 alone can fill a tank in 24 minutes then the pipe P18 is closed 12 minutes before the tank is filled. In what time the tank is full?
6 (3/10)3 (4/15)5 (2/11)8 (4/13)7 (2/13)Option D

Let T is the time taken by the pipes to fill the tank

=> (1/12 + 1/18 + 1/24)*(T – 12) + (1/12 + 1/18)*12 = 1

=> Time = 108/13 = 8(4/13)

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Directions(6-10):** Study the following graph carefully to answer the given questions

Time taken by the pipes to fill a tank/cistern (hours/minutes).