Question

The tank at a water supply station is filled with water by several pumps. At first, three pumps of the same capacity are turned on 2.5 h later, two more
pumps (both the same) of a different capacity are set into operation. After 1 h the additional pumps were set into operation; the tank was almost filled
to its capacity (15 m3 were still lacking); in another hour, the tank was full. One of the two additional pumps could have filled the tank in 40 h.
6. What is the volume of the tank?
(a) 60 m3
(b) 80 m3
(c) 75 m3
(d) 90 m^3

Answer 1

Given :- The tank at a water supply station is filled with water by several pumps. At first, three pumps of the same capacity are turned on 2.5 h later, two more pumps (both the same) of a different capacity are set into operation. After 1 h the additional pumps were set into operation; the tank was almost filled to its capacity (15 m3 were still lacking); in another hour, the tank was full. One of the two additional pumps could have filled the tank in 40 h. What is the volume of the tank?

(a) 60 m³

(b) 80 m³

(c) 75 m³

(d) 90 m³

Answer :-

Let us assume that, first three pumps fills the tank in x hours .

so,

→ Efficiency of each pump = (1/x) m³ / hour .

then,

→ Efficiency of three pump = (3/x) m³ / hour .

now,

→ First three pumps works for = 2.5h + 1h + 1h = 4.5 hours.

so,

→ Water filled by 3 pumps in 4.5 hours = 4.5 * (3/x) = (13.5/x) m³ .

now, given that,

→ Time taken by additional pump to fill the tank = 40 hours.

so,

→ Efficiency of 2 additional tanks = 2 * (1/40) = (1/20) m³ / h .

and,

→ Additional pumps work for = 1 + 1 = 2 hours.

so,

→ Water filled by additional pumps in 2 hours = 2 * (1/20) = (1/10) m³ .

therefore,

→ (13.5/x) + (1/10) = 1

→ (13.5/x) = 1 - (1/10)

→ (13.5/x) = (9/10)

→ x = 135/9 = 15 hours.

since given that, in last 1 hour they filled 15 m³ .

hence,

→ 3 * (1/15) + (1/20) = 15 m³

→ (1/5) + (1/20) = 15

→ (4 + 1)/20 = 15

→ (5/20) = 15

→ (1/4) = 15

→ 1 = 60 m³ (Ans.) (Option A)

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