Question Details

The data shows volume and height of 4 right circular tanks P, Q, R and S. The given data also depicts time taken to fill or empty the tank by inlet pipe A, B and outlet pipe C.
1. Ratio between time taken by pipe B and that by pipe C is different because pipes were used by the operator in different ways for different tanks.
2. Radius of tank S is 10.5 m.
Tank Volume Height Time taken by pipe A to fill tank (in hours) Ratio of time taken by pipe B to fill and pipe C to empty the tank
P 38808 28 10 2: 3
Q -------- 10 12 3: 2
R 5390 35 18 1: 3
S 19250 -------- --------- 2: 5

When all 3 pipes were opened simultaneously, volume of tank S filled in 1 hour is 3575 cubic meter. Pipe B alone can fill 25% of tank S in 3.5 hours. In how much time can pipe A fill 40% of tank S alone.

Tank	Volume	Height	Time taken by pipe A to fill tank (in hours)	Ratio of time taken by pipe B to fill and pipe C to empty the tank
P	38808	28	10	2: 3
Q	--------	10	12	3: 2
R	5390	35	18	1: 3
S	19250	--------	---------	2: 5

Options

6 hours

4.8 hours

5.4 hours

2.8 hours

1.6 hours

Show Answer

Correct Answer :

Option D

2.8 hours

Solution :

To find the time taken by pipe A to fill 40% of tank S alone, let us analyze the data given for tank S:
- Volume of tank S = 19250 m³
- Radius of tank S ( $r$ ) = 10.5 m
- Ratio of time taken by pipe B to fill and pipe C to empty tank S = 2 : 5

First, let us calculate the rate and time taken by pipe B to fill tank S:
Pipe B alone can fill 25% of tank S in 3.5 hours.
Therefore, the time taken by pipe B to fill 100% of tank S is:
$T_{B} = \frac{3.5}{0.25} = 14 hours$

Next, let us use the ratio of the time taken by pipe B to fill and pipe C to empty tank S:
The ratio is given as 2 : 5.
Since the time taken by pipe B ( $T_{B}$ ) is 14 hours:
$\frac{T_{B}}{T_{C}} = \frac{2}{5} \Rightarrow \frac{14}{T_{C}} = \frac{2}{5} \Rightarrow T_{C} = 35 hours$

Now, let us determine the work done per hour (in cubic meters) by each pipe:
- Rate of pipe B (filling) = $\frac{19250}{14} = 1375$ m³/hour
- Rate of pipe C (emptying) = $\frac{19250}{35} = 550$ m³/hour

Let the time taken by pipe A to fill tank S be $T_{A}$ hours.
- Rate of pipe A (filling) = $\frac{19250}{T_{A}}$ m³/hour

When all three pipes are opened simultaneously, the net volume of tank S filled in 1 hour is 3575 m³:
$Rate of A + Rate of B - Rate of C = 3575$
Substituting the known rates:
$\frac{19250}{T_{A}} + 1375 - 550 = 3575$
$\frac{19250}{T_{A}} + 825 = 3575$
$\frac{19250}{T_{A}} = 2750$
$T_{A} = \frac{19250}{2750} = 7 hours$

Thus, pipe A alone can fill the entire tank S in 7 hours.
The time required for pipe A to fill 40% of tank S is:
$40 % of T_{A} = 0.4 \times 7 = 2.8 hours$