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Question Detail
Taps A and B can fill a bucket in 12 minutes and 15 minutes respectively. If both are opened and A is closed after 3 minutes, how much further time would it take for B to fill the bucket?
- 8 min 15 sec
- 7 min 15 sec
- 6 min 15 sec
- 5 min 15 sec
Answer: Option A
Explanation:
Part filled in 3 minutes =
\begin{aligned}
3*\left(\frac{1}{12} + \frac{1}{15}\right) \\
= 3*\frac{9}{60} = \frac{9}{20}\\
\text{Remaining part }= 1-\frac{9}{20} \\
= \frac{11}{20} \\
=> \frac{1}{15}:\frac{11}{20}=1:X \\
=> X = \frac{11}{20}*\frac{15}{1} \\
=> X = 8.25 mins
\end{aligned}
So it will take further 8 mins 15 seconds to fill the bucket.
1. A tap can fill a tank in 6 hours. After half the tank is filled then 3 more similar taps are opened. What will be total time taken to fill the tank completely.
- 2 hours 30 mins
- 2 hours 45 mins
- 3 hours 30 mins
- 3 hours 45 mins
Answer: Option D
Explanation:
Half tank will be filled in 3 hours
Lets calculate remaining half,
Part filled by the four taps in 1 hour = 4*(1/6) = 2/3
Remaining part after 1/2 filled = 1-1/2 = 1/2
\begin{aligned}
\frac{2}{3}:\frac{1}{2}::1:X \\
=> X = \left( \frac{1}{2}*1*{3}{2} \right) \\
=> X = \frac{3}{4} hrs = 45 \text{ mins} \\
\end{aligned}
Total time = 3 hours + 45 mins = 3 hours 45 mins
2. Two pipes A and B can fill a tank in 6 hours and 4 hours respectively. If they are opened on alternate hours and if pipe A is opened first, in how many hours, the tank shall be full ?
- 3 hours
- 5 hours
- 7 hours
- 10 hours
Answer: Option B
Explanation:
(A+B)'s 2 hour's work when opened =
\begin{aligned}
\frac{1}{6}+\frac{1}{4} = \frac{5}{12} \\
(A+B)'s \text{ 4 hour's work} = \frac{5}{12}*2 \\
= \frac{5}{6}
\text{Remaining work = } 1-\frac{5}{6} \\
= \frac{1}{6} \\
\text{Now, its A turn in 5 th hour} \\
\frac{1}{6} \text{ work will be done by A in 1 hour}\\
\text{Total time = }4+1 = 5 hours
\end{aligned}
3. A tank is filled by three pipes with uniform flow. The first two pipes operating simultaneously fill the tank in the same time during which the tank is filled by the third pipe alone. The second pipe fills the tank 5 hours faster than the first pipe and 4 hours slower than the third pipe. Find the time required by the first pipe to fill the tank ?
- 10 hours
- 15 hours
- 17 hours
- 18 hours
Answer: Option B
Explanation:
Suppose, first pipe alone takes x hours to fill the tank .
Then, second and third pipes will take (x -5) and (x - 9) hours respectively to fill the tank.
As per question, we get
\begin{aligned}
\frac{1}{x} + \frac{1}{x-5} = \frac{1}{x-9} \\
=> \frac{x-5+x}{x(x-5)} = \frac{1}{x-9}\\
=> (2x - 5)(x - 9) = x(x - 5)\\
=> x^2 - 18x + 45 = 0
\end{aligned}
After solving this euation, we get
(x-15)(x+3) = 0,
As value can not be negative, so x = 15
4. There are two pipes which are functioning simultaneouly to fill a tank in 12 hours, if one pipe fills the tank 10 hours faster than other then how many hours second pipe will take to fill the tank ?
- 30 hours
- 35 hours
- 40 hours
- 42 hours
Answer: Option A
Explanation:
Lets suppose tank got filled by first pipe in X hours,
So, second pipe will fill the tank in X + 10 hours.
\begin{aligned}
=> \frac{1}{X} + \frac{1}{X} + 10 = \frac{1}{12} \\
=> X = 20
\end{aligned}
5. A tank can be filled by two pipes A and B in 60 minutes and 40 minutes respectively. How many minutes will it take to fill the tank from empty state if B is used for the first half time and then A and B fill it together for the other half.
- 15 mins
- 20 mins
- 25 mins
- 30 mins
Answer: Option D
Explanation:
Let the total time be x mins.
Part filled in first half means in x/2 = 1/40
Part filled in second half means in x/2 = \begin{aligned}
\frac{1}{60}+\frac{1}{40} \\
= \frac{1}{24} \\
\text{ Total = } \\
\frac{x}{2}*\frac{1}{40} + \frac{x}{2}*\frac{1}{24} = 1 \\
=> \frac{x}{2} \left(\frac{1}{40}+\frac{1}{24} \right) = 1 \\
=> \frac{x}{2}*\frac{1}{15} = 1 \\
=> x = 30 mins
\end{aligned}
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