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 ?
Answer: B
(A+B)'s 2 hour's work when opened =
Now, its A turn in 5th hour
1/6 work will be done by A in 1 hour
Total time = 4 + 1 = 5 hours
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Two taps can separately fill a cistern 10 minutes and 15 minutes respectively and when the waste pipe is open, they can together fill it in 18 minutes. The waste pipe can empty the full cistern in?
Answer: D
Work in rates: a pipe that fills a tank in \( n \) units of time fills \( \tfrac{1}{n} \) of it per unit, and rates add (an emptying pipe counting negative).
\[ \frac{1}{10} + \frac{1}{15} - \frac{1}{w} = \frac{1}{18} \]
\[ \frac{3 + 2}{30} - \frac{1}{w} = \frac{1}{18} \quad\Longrightarrow\quad \frac{1}{6} - \frac{1}{w} = \frac{1}{18} \]
\[ \frac{1}{w} = \frac{3 - 1}{18} = \frac{1}{9} \quad\Longrightarrow\quad w = 9 \text{ minutes} \]
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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. The time required by the first pipe is:
Answer: C
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.
(2x - 5)(x - 9) = x(x - 5)
x2 - 18x + 45 = 0
(x - 15)(x - 3) = 0
x = 15. [neglecting x = 3]
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A water tank is two-fifth full. Pipe A can fill a tank in 10 minutes and pipe B can empty in 6 minutes. If both the pipes are open, how long will it take to empty or fill the tank completely ?
Answer: A
There are two important points to learn in this type of question,
First, if both will open then tank will be empty first.
Second most important thing is,
If we are calculating filling of tank then we will subtract as (filling-empting)
If we are calculating empting of thank then we will subtract as (empting-filling)
So lets come on the question now,
Part to emptied 2/5
Part emptied in 1 minute =
1/6 - 1/10 = 1/15 => 1/15 : 25 :: 1: x =>25 × 15 = 6mins
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A cistern is filled by Pipe A and Pipe B together in 2.4 hours. Pipe A alone can fill the cistern at the rate of 100 litres per hour. Pipe B alone can fill the cistern in 4 hours. What is the capacity of the cistern?
Answer: B
Work in rates: a pipe that fills a tank in \( n \) units of time fills \( \tfrac{1}{n} \) of it per unit, and rates add (an emptying pipe counting negative).
\[ \text{A's rate} = \frac{1}{2.4} - \frac{1}{4} = \frac{5 - 3}{12} = \frac{1}{6} \]
So A alone would fill the cistern in 6 hours. At 100 litres an hour:
\[ \text{capacity} = 6 \times 100 = 600 \text{ litres} \]
Pipe B's own time is only needed to isolate A's rate; the capacity then follows from A's known flow.
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An electric pump can fill a tank in 3 hours. Because of a leak in the tank, it took 3 hours 30 min to fill the tank. In what time the leak can drain out all the water of the tank and will make tank empty ?
Answer: D
We can get the answer by subtracting work done by leak in one hour by subtraction of
filling for 1 hour without leak and with leak, as
Work done for 1 hour without leak = 1/3
Work done with leak =
So tank will be empty by the leak in 21 hours.
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A class eats 2/5 of chocolates on 1st day. On the 2nd day they eat 3/4 of the remainder. How many chocolates were there initially if still 75 chocolates are left?
Answer: A
Let there be 100 chocolates
215th of 100 40 chocolates eaten on first day
314th of the remaining 314th of (100-40) 45 chocolates
Chocolates left 100-40-45 15 chocolates
15 chocolates are left when 100 are there
75 chocolates are left when ? chocolates are there? = 500 Chocolates
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Pipes A and B can fill a tank in 20 hrs and 30 hrs respectively and pipe c can epmty the full tank in 40 hrs. If all the pipes are opened together, how much time will it take for the Tank to fill completely .. ?
Answer: C
Two pipes fill and one empties, so the outflow is subtracted:
\[ \frac{1}{20} + \frac{1}{30} - \frac{1}{40} \]
Over a common denominator of 120:
\[ \frac{6 + 4 - 3}{120} = \frac{7}{120} \]
\[ \therefore\; \text{time} = \frac{120}{7} = 17\tfrac{1}{7} \text{ hours} \]
The net rate is positive, so the tank does fill; had pipe C been faster than A and B combined the tank would never fill at all.
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There is a leak at the bottom of a cistern. Due to this it takes 8 hours to fill the cistern. Had there not been a leak, it would take one hour less to fill the cistern. How much time does it take for the leak to completely empty the cistern?
Answer: C
Work in rates: a pipe that fills a tank in \( n \) units of time fills \( \tfrac{1}{n} \) of it per unit, and rates add (an emptying pipe counting negative).
Without the leak the cistern fills in \( 8 - 1 = 7 \) hours.
\[ \frac{1}{7} - \frac{1}{L} = \frac{1}{8} \]
\[ \frac{1}{L} = \frac{1}{7} - \frac{1}{8} = \frac{8 - 7}{56} = \frac{1}{56} \]
\[ L = 56 \text{ hours} \]
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Pipe A can fill a tank 5 times faster than Pipe B and takes 32 minutes less than Pipe B to fill the tank. If both the pipes are opened together, then in how much time the tank would be full?
Answer: C
Work in rates: a pipe that fills a tank in \( n \) units of time fills \( \tfrac{1}{n} \) of it per unit, and rates add (an emptying pipe counting negative).
A is five times as fast, so A's time is a fifth of B's.
\[ B - \frac{B}{5} = 32 \quad\Longrightarrow\quad \frac{4B}{5} = 32 \]
\[ B = 40, \qquad A = 8 \text{ minutes} \]
\[ \frac{1}{8} + \frac{1}{40} = \frac{5 + 1}{40} = \frac{3}{20} \]
\[ t = \frac{20}{3} \text{ minutes} \]
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