Question

It can take 12 minutes to fill a water tank using two pipes. If the pipe of smaller diameter is used for 9 minutes
and the pipe of larger diameter for 4 minutes, only, half the tank can be filled.

The time taken by the pipe of smaller diameter to fill the tank is

The time taken by the pipe of larger diameter to fill the tank is

If both the
pipes are working together for 10 minutes, then the tank will filled by

(iv) The ratio
of time taken by the pipes can
be​

Answer 1

Given :- It can take 12 minutes to fill a water tank using two pipes. If the pipe of smaller diameter is used for 9 minutes and the pipe of larger diameter for 4 minutes, only, half the tank can be filled. Based on given information.

Answer the following questions :-

(i) The time taken by the pipe of smaller diameter to fill the tank is ?

(ii) The time taken by the pipe of larger diameter to fill the tank is ?

(iii) If both the pipes are working together for 10 minutes, then the tank will filled by ?

(iv) The ratio of time taken by the pipes can be ?

Answer :-

Let us assume that,

• Pipe of smaller diameter can fill the tank alone in = x min. .
• Pipe of larger tank can fill the tank alone in = y min.
• Here, smaller diameter pipe takes more time , so x > y .

now,

→ Efficiency of smaller diameter pipe = (1/x) / min. .

→ Efficiency of larger diameter pipe = (1/y) / min. .

given that, both fills the water tank in 12 minutes .

so,

→ 12(1/x + 1/y) = 1

→ (1/x + 1/y) = (1/12) ---------- Eqn.(1)

and,

→ 9(1/x) + 4(1/y) = (1/2)

→ (9/x) + (4/y) = (1/2) --------- Eqn.(2)

Let us assume that,

• (1/x) = m
• (1/y) = n .

then,

→ m + n = (1/12) ------ Eqn.(3)

→ 9m + 4n = (1/2) ------- Eqn.(4)

multiply Eqn.(3) by 4 and subtracting from Eqn.(4),

→ (9m + 4n) - 4(m + n) = (1/2) - 4(1/12)

→ 9m - 4m + 4n - 4n = (1/2) - (1/3)

→ 5m = (1/6)

→ m = (1/30)

putting value of m in Eqn.(3)

→ (1/30) + n = (1/12)

→ n = (1/12) - (1/30)

→ n = (5 - 2)/60

→ n = (3/60)

→ n = (1/20)

therefore,

→ m = (1/x) => (1/30) = (1/x) => x = 30 minutes.

→ n = (1/y) => (1/20) = (1/y) => y = 20 minutes.

hence,

→ The time taken by the pipe of smaller diameter to fill the tank is = 30 minutes .

→ The time taken by the pipe of larger diameter to fill the tank is = 20 minutes .

now,

→ in 10 minutes both fill = 10(1/30 + 1/20) = 10(5/60) = (5/6) of the tank.

and,

→ The ratio of time taken by the pipes = 30 : 20 = 3 : 2 .

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