Math, asked by dommarajukusuma, 6 months ago

a city water supply tank has two inlet pipes x and y which can fill it in 20 and 30 hours respectively. and an outlet pipe z which can empty a full tank in 40 hours if the tank is empty and the taps are opened in succession for 1 hour each, and the process continues, in how many hours the tank get filled?​

Answers

Answered by Nani8523
1

Answer:

Let the capacity of the tank be 60 units (LCM of 15, 20 & 30).

Pipe A fills =

15

60

=4 units / hour

Pipe B fills =

20

60

=3 units / hour

Pipe A fills =

30

60

=2 units / hour

(A + B + C)'s together per hour work = (4+3−2)=5 units/hour

Time taken by all three to fill an empty tank =

5

60

=12 hours.

Answered by kavinsk158
9

Answer:

51 3/7

Step-by-step explanation:

Let pipe X and Y can fill the tank in 20 hours and 30 hours.

In 1 hour, X and Y can fill the 1/20 and 1/30 of tank's capacity.

Let pipe Z can empty the tank in 40 hours, In 1 hour Z can empty 1/40 of tank's capacity.

while X and Y filling the tank from empty in successive hours but Z emptying the tank in next hour.

In 2 hours , Both X and Y can fill -

1/20 + 1/30 = (20+30)/(20*30) = 50/600 =1/12 (of tank's capacity)

In 3rd hour, Z empties

1/12 - 1/40 = (40-12)/(12*40) = 28/480 =7/120

Each 3 hours the 7/120 of tank's capacity is filled.

Time required for completing Full tank , (120/7)*3 =360/7 =51(3/7).

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