A tanker is provided with two pipes X and Y. X can fill it in 20 minut

1.	A tanker is provided with two pipes X and Y. X can fill it in 20 minutes and Y can empty it in 30 minutes. If X and Y be kept open ultimately from one minute each, then how soon will the tanker be filled?1. 121 minutes2. 110 minutes3. 115 minutes4. 120 minutes
Answer» Correct Answer - Option 3 : 115 minutes Given: X can fill the tanker in 20 minutes and Y can empty it in 30 minutes. Concept: If a tap can fill a tank in x hours, then the tank filled by the tap in 1 hour = 1/x of the total tank. Calculation: Part of the tanker filled in first 2 minutes ⇒ \(\frac{1}{{20}} - \frac{1}{{30}}{\rm{}} = {\rm{}}\frac{{3 - 2}}{{60}}{\rm{}} = {\rm{}}\frac{1}{{60}}\) Part of tanker filled in 114 minutes ⇒ \(\frac{{57}}{{60}}\) ⇒ \(\frac{{19}}{{20}}\) ∴ Remaining part of tanker will be filled in 115^th minutes.

A tanker is provided with two pipes X and Y. X can fill it in 20 minutes and Y can empty it in 30 minutes. If X and Y be kept open ultimately from one minute each, then how soon will the tanker be filled?1. 121 minutes2. 110 minutes3. 115 minutes4. 120 minutes

Answer» Correct Answer - Option 3 : 115 minutes

Given:

X can fill the tanker in 20 minutes and Y can empty it in 30 minutes.

Concept:

If a tap can fill a tank in x hours, then the tank filled by the tap in 1 hour = 1/x of the total tank.

Calculation:

Part of the tanker filled in first 2 minutes

⇒ \(\frac{1}{{20}} - \frac{1}{{30}}{\rm{}} = {\rm{}}\frac{{3 - 2}}{{60}}{\rm{}} = {\rm{}}\frac{1}{{60}}\)

Part of tanker filled in 114 minutes

⇒ \(\frac{{57}}{{60}}\)

⇒ \(\frac{{19}}{{20}}\)

∴ Remaining part of tanker will be filled in 115^th minutes.

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