Question:
The rate of water flow through three pipes A, B and C are in the ratio 4 : 9 : 36. An empty tank can be filled up completely by pipe A in 15 hours. If all the three pipes are used simultaneously to fill up this empty tank, the time, in minutes, required to fill up the entire tank completely is nearest to
The rate of water flow through three pipes A, B and C are in the ratio 4 : 9 : 36. An empty tank can be filled up completely by pipe A in 15 hours. If all the three pipes are used simultaneously to fill up this empty tank, the time, in minutes, required to fill up the entire tank completely is nearest to
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In pipes and cisterns problems:
Convert ratios into actual rates using any one known filling time.
Add rates (not times) when pipes work together.
Do all calculations in hours first, and convert to minutes only at the end to avoid mistakes.
Convert ratios into actual rates using any one known filling time.
Add rates (not times) when pipes work together.
Do all calculations in hours first, and convert to minutes only at the end to avoid mistakes.
Updated On: Jul 29, 2026
- \(76\)
- \(78\)
- \(73\)
- \(71\)
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The Correct Option is C
Approach Solution - 1
Approach: Convert one known fill-time into a rate, scale up to all three pipes using the ratio, then invert to get the combined time.
Step 1: Rates are in ratio \(4:9:36,\) so write them as \(4u,9u,36u\) tanks/hour.
Step 2: Pipe A alone fills in \(15\) hours, so A's rate \(=\dfrac1{15}.\) Hence \(4u=\dfrac1{15}\Rightarrow u=\dfrac1{60}.\)
Step 3: Combined rate \(=(4+9+36)u=49u=\dfrac{49}{60}\) tank/hour.
Step 4: Time to fill one tank \(=\dfrac1{49/60}=\dfrac{60}{49}\) hours \(=\dfrac{60}{49}\times60=\dfrac{3600}{49}\approx73.47\) minutes.
Final answer: Nearest to \(73\) minutes.
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Approach Solution -2
Step 1: Convert the rate ratio into actual rates.
Let the rates of flow (in tank/hour) of pipes A, B and C be:
\[
\text{A : B : C} = 4 : 9 : 36.
\]
Suppose the common factor is \(k\). Then:
\[
\text{Rate of A} = 4k.
\]
Given that pipe A alone can fill the empty tank in 15 hours, its rate is:
\[
\text{Rate of A} = \frac{1}{15} \text{ tank/hour}.
\]
So:
\[
4k = \frac{1}{15} \quad \Rightarrow \quad k = \frac{1}{60}.
\]
Therefore:
\[
\text{Rate of A} = 4k = \frac{4}{60} = \frac{1}{15},
\]
\[
\text{Rate of B} = 9k = \frac{9}{60} = \frac{3}{20},
\]
\[
\text{Rate of C} = 36k = \frac{36}{60} = \frac{3}{5}.
\]
Step 2: Find the combined rate of all three pipes. \[ \text{Combined rate} = \frac{1}{15} + \frac{3}{20} + \frac{3}{5}. \] Take LCM of 15, 20 and 5, which is 60: \[ \frac{1}{15} = \frac{4}{60}, \quad \frac{3}{20} = \frac{9}{60}, \quad \frac{3}{5} = \frac{36}{60}. \] So: \[ \text{Combined rate} = \frac{4}{60} + \frac{9}{60} + \frac{36}{60} = \frac{49}{60} \text{ tank/hour}. \]
Step 3: Calculate the total time to fill the tank. \[ \text{Time taken} = \frac{1}{\text{Combined rate}} = \frac{1}{\frac{49}{60}} = \frac{60}{49} \text{ hours}. \] Convert this into minutes: \[ \frac{60}{49} \times 60 = \frac{3600}{49} \text{ minutes}. \] Approximate: \[ 49 \times 73 = 3577,\quad 49 \times 74 = 3626. \] So: \[ \frac{3600}{49} \approx 73.47 \text{ minutes}, \] which is closest to \(73\) minutes. \[ \boxed{73} \]
Step 2: Find the combined rate of all three pipes. \[ \text{Combined rate} = \frac{1}{15} + \frac{3}{20} + \frac{3}{5}. \] Take LCM of 15, 20 and 5, which is 60: \[ \frac{1}{15} = \frac{4}{60}, \quad \frac{3}{20} = \frac{9}{60}, \quad \frac{3}{5} = \frac{36}{60}. \] So: \[ \text{Combined rate} = \frac{4}{60} + \frac{9}{60} + \frac{36}{60} = \frac{49}{60} \text{ tank/hour}. \]
Step 3: Calculate the total time to fill the tank. \[ \text{Time taken} = \frac{1}{\text{Combined rate}} = \frac{1}{\frac{49}{60}} = \frac{60}{49} \text{ hours}. \] Convert this into minutes: \[ \frac{60}{49} \times 60 = \frac{3600}{49} \text{ minutes}. \] Approximate: \[ 49 \times 73 = 3577,\quad 49 \times 74 = 3626. \] So: \[ \frac{3600}{49} \approx 73.47 \text{ minutes}, \] which is closest to \(73\) minutes. \[ \boxed{73} \]
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