Test: Pipes & Cisterns - 1 - UPSC MCQ

Net part filled in 1 hour
The tank will be full in 60/17 hours i.e., hours.

Concept:

Hole in this case do negative work.

Calculation:

LCM of 4 and 5 = 20

Let the total quantity of the bucket is 20 units

Efficiency of the tap = 20 units / 4 min = 5 units per minute

But due to hole tap took 5 min to fill the bucket.

Now, new efficiency of the tap = 20 units/ 5 min = 4 units/min

Tap old efficiency – tap new efficiency = efficiency of the hole

5 – 4 = 1 units/min

Hole can empty the tank in = 20 / 1 = 20 min

Work done by the waste pipe in 1 minute = 

=  [-ve sign means emptying]
∴ 40 Volume of 1/40 part = 3 gallons.
Volume of whole = (3 x 40) gallons = 120 gallons.

Given:

Time taken for A and B to fill tank = 20 hours.

Time taken for B and C to fill tank = 30 hours.

Time taken for A and C to fill tank = 15 hours.

Concepts used:

Basic time and work concept.

Calculation:

(1 / A + 1 / B) + (1 / B + 1 / C) + (1 / C + 1 / A) = 1 / 20 + 1 / 30 + 1 / 15

2 (1 / A + 1 / B + 1 / C) = 9 / 60

Time for A, B and C = 120/9 hours = 40/3 hours

Reservoir filled in 1 hour by flood gates A, B and C together = 1/6

⇒ Reservoir filled in 2 hours by all gates = 1/3

⇒ Reservoir left to be filled = 1 – 1/3 = 2/3

⇒ Reservoir filled in 1 hour by flood gates A and B = 2/21

⇒ Reservoir filled in 1 hour by gate C = 1/6 – 2/21 = 3/42

∴ Flood gate C alone filled the reservoir in 14 hours.

Given:
Pipe A and B can fill the tank = 6 hours
Formula used:
Work done = Time × Efficiency

Calculation:
Let pipe B can fill the tank in x hours
Pipe A will take x - 5 hours
Now
1/x + 1/(x - 5) = 1/6
⇒ 6(x - 5 + x) = x² - 5x
⇒ 12x - 30 = x² - 5x
⇒ x² - 17x + 30 = 0
⇒ x² - 15x - 2x + 30 = 0
⇒ x(x - 15) - 2(x - 15) = 0
⇒ (x -15)(x - 2) = 0
x = 15 and x = 2
If x = 2, A = - 3 and time cannot be negative
So x = 15 hours
∴ B will fill the tank in 15 hours.

Given:

A tap can fill the tank in 8 hours

Concept:

Work = 1/Efficiency

Calculation:

⇒ One hour work of one tap = 1/8

⇒ One hour work of 4 taps = 4/8 = 1/2

⇒ Total time to fill the tank = 4 hours to first half + 1 hour to second half

∴ The required result will be 5 hours.

Given:

Pipe P filled = 15 min

Pipe Q filled = 25 min

Pipe R can empty tank = 11 gallon/min

If Pipe P, Q, and R opened together tank filled = 30 min

Concept used:

Total capacity = LCM

Efficiency of inlet tap always positive and efficiency of outlet tap is always negative

Formula used:

Efficiency = Total capacity/Time

Calculation

Total capacity = LCM of 15, 25, 30 = 150

Efficiency R = Efficiency of P + Q + R – Efficiency of P + Q

⇒ 5 – (10 + 6) = -11 (Here – ive sign represents that R is outlet pipe)

Time taken by R to empty the full tank = 150/11

According to the question

Capacity of tank = (150/11) × 11 = 150 gallons

∴ Capacity of the tank is 150 gallons

Given:

The first pipe can fill the tank in 10 min

the second pipe can fill the tank in 12 min

the waste pipe can empty 2 gallons per minute

All the three pipes working together can fill the tank in 6 minutes

Formula:

(A + B + C)’s 1 hour work = 1/A + 1/B – 1/C

Calculation:

Let us consider filling pipes as A, B, and waste pipe as  C

Let the time taken by the waste pipe to empty a full tank alone be x minutes

Hence, we get:

(1/10) + (1/12) - (1/x) = 1/6

⇒ 1/x = 1/60

⇒ x = 60 minutes

Hence, the tank capacity is obtained as:

60 × 2 = 120 gallons

∴ The capacity of the tank is 120 gallons

Given

Pipe X filled a tank in 18 hours

Pipe Z filled a tank in 4 hours

Calculation

Total capacity of the tank = 36 units (i.e L.C.M of 18 and 4)

Tank filled by pipe X in 1 hour = 36/18

⇒ 2 units

Tank filled by pipe Z in 1 hour = 36/4

⇒ 9 units

Tank filled by pipe X in 7 hours = 2 × 7

⇒ 14 units

Remaining tank is filled by pipe X and Z = 36 - 14

⇒ 22 units

Tank filled in 1 hour by X and Z from 4:00 pm to 5:00 pm = 2 + 9

⇒ 11 units

Tank filled by X and Z from 5:00 pm to 6:00 pm = 2 + 9

⇒ 11 units

∴ Tank will be filled by 6:00 p.m