Robert Bosch Pipes and Cisterns Quiz-2

Answer: option b

Explanation: 1/10+1/15-1/x=1/18
3/30+2/30-1/18=1/x
1/6 -1/18=1/x
1/9=1/x
x=9

Answer: option b

Explanation:Part of the filled by all the three pipes in one minute
= 1/18 + 1/15 - 1/45 = (5 + 6 - 2)/90 = 9/90 = 1/10
So, the tank becomes full in 10 minutes

Answer: option c

Explanation: 1/20 + 1/30 - 1/40 = 7/120
=>120/7 = 17 1/7

Answer:Option b

Explanation: Net part filled in 1 hour(1/5 +1/10 -1/12)=13/60

The tank will be full in 60/13hours.

Answer: Option B
Explanation:

Part filled by (A + B + C) in 3 minutes = 3(1/30+1/20+1/10)
=(3*11/60)
=11/20
Part filled by C in 3 minutes=3/10

Required ratio=(3/10*20/11)=6/11

Answer : Option C

Explanation: As cistern is filled in 20 hours, therefore in 1 hour, filled part → 1/20th
Now, due to outlet pipe, filled part in 1 hour = 1/30th
Part of the cistern emptied, due to leakage in 1 hour = 1/20 - 1/30 = 1/60th
Therefore the leak will empty the full cistern in 60 hrs.

Correct answer (b)

Assume the total time required = x + 2 min
Part filled by pipe A in 1min = 1/10 & part filled by B in 1 min = 1/15
After 2 min, part filed by A & B together = 2 [ 1/10 + 1/15] = 1/6

Remaining part = 1 – 1/6 = 5/6
Part filled by pipe B in 1 min = 1/15

As pipe A is turned off after 2 minutes, we get the relation as,
Part filled by (A + B) in 1 hr /Remaining part :: Total part filled/ Time taken

1/15 : 1/6 :: 1: x

x = 15/6 min

Tank will be full in ( 2 min + 15/6 min) = 9/2 min = 4.5 min

Correct answer: (c)

Consider a pipe fills the tank in 'x' hrs. If there is a leakage in the bottom, the tank is filled in 'y' hrs.
The time taken by the leak to empty the full tank = xy/y-x hrs

But, direct values of x & y are not given. So, we need to find the work done by the two pipes in 1 hr = (1/20) + (1/25) = 9/100

Hence, the time taken by these pipes to fill the tank = 100/9 hrs = 11.11 hrs = 11 hrs 6 min

Due to leakage, time taken = 11 hrs 6 min + 54 min = 12 hrs

Thus, work done by (two pipes + leak) in 1 hr = 1/12
Hence, work done by leak in 1 hr = work done by two pipes – 1/9
= 9 /100 – 1/12 = 1/12

Therefore, leak will empty the full cistern in 12 hours.

Correct answer: b

Work done by waste pipe in 1 min = (Part filled by total pipes together) -(part filled by first pipe + part filled by second pipe)
= 1/8 – [(1/8) + (1/12)] = 1/8 – (5/24) = -2/24
Here, negative (-) sign indicates emptying of tank.

To find the capacity, we need to determine the volume of 1/210 part.
Therefore, volume of 2/24 part = 5 gallons ---------(given condition)
Hence, the capacity of tank = volume of whole = (5 x 24)/2 = 60 gallons.

Correct answer:(b)

If a pipe requires 'x' hrs to fill up the tank, then part filled in 1 hr = 1/x

If pipe A requires 15 hrs to fill the tank, then part filled by pipe A in 1 hr = 1/15

If pipe B requires 30 hrs to fill the tank, then part filled by pipe B 1 hr = 1/30

Hence, part filled by (A + B) together in 1 hr = 1/15 + 1/30

= 3 / 30 = 1/10

In 1 hr both pipes together fill 1/10th part of the tank. This means, together they fill the tank in 10 hrs.