Two pipes A and B can fill a tank in 12 minutes and 16 minutes respectively. If both the pipes are opened together, then after how much time, B should be closed so that the tank is full in 9 minutes ?
3 ½ min
4 min
4 ½ min
4 ¾ min
Answer is Right!
Answer is Wrong!
This question was previously asked in
UPSC CAPF – 2017
– Pipe B fills the tank in 16 minutes, so its filling rate is $1/16$ of the tank per minute.
– Let $t$ be the time (in minutes) for which both pipes are open together.
– After time $t$, pipe B is closed, and only pipe A continues to fill the tank for the remaining time.
– The total time to fill the tank is 9 minutes. So, pipe A works for the entire 9 minutes. Pipe B works only for the first $t$ minutes.
– Amount filled by A in 9 minutes = Rate of A $\times$ Time A worked = $(1/12) \times 9 = 9/12 = 3/4$ of the tank.
– Amount filled by B in $t$ minutes = Rate of B $\times$ Time B worked = $(1/16) \times t = t/16$ of the tank.
– The total amount filled is the sum of the amounts filled by A and B, which is 1 full tank.
– So, $(3/4) + (t/16) = 1$.
– To solve for $t$, multiply the entire equation by the least common multiple of 4 and 16, which is 16:
$16 \times (3/4) + 16 \times (t/16) = 16 \times 1$
$4 \times 3 + t = 16$
$12 + t = 16$
$t = 16 – 12 = 4$.
– Therefore, pipe B should be closed after 4 minutes.
Work done by A and B together in $t$ minutes = $(1/12 + 1/16) \times t = (4/48 + 3/48) \times t = (7/48)t$.
Work done by A alone in $(9-t)$ minutes = $(1/12) \times (9-t)$.
Total work = $(7/48)t + (1/12)(9-t) = 1$.
Multiply by 48: $7t + 4(9-t) = 48 \Rightarrow 7t + 36 – 4t = 48 \Rightarrow 3t + 36 = 48 \Rightarrow 3t = 12 \Rightarrow t = 4$.
Both methods yield the same result.