A tank containing 270 L of water starts leaking at 1:00 PM. If the leak drains 30 L in 5 minutes, the volume of water in the tank at 1:01 PM will be ______

262 L

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268 L

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264 L

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266 L

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Solution

The correct option is C 264 L
$Volume of water leaked in 5 minutes = 30 L$

$Ratio of water leaked to time taken = 30 : 5$

$The fraction=30 L5$

$To find the unit rate, divide both the numerator and the denominator by 5$

$Water leaked in 1 minute = \frac{30}{5} = \frac{30 \div 5}{5 \div 5} = \frac{6}{1} = 6 L /min$

$1:01 PM is 1 minute past 1:00 PM$

$∴ The water left in the tank at 1:01 PM$ $will be 6 L less than 270 L$

$⟹ Volume of water in the tank at$ $1:01 PM = (270 - 6) L$
$= 264 L$

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