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. 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 ?
only answer r required​

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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