A train increases its normal speed by $12.5 %$ and reaches its destination $20$ minutes earlier. What is the actual time taken by the train in the journey?

220

min

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180

min

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145

min

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160

min

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Solution

The correct option is B $180$ min
Let the normal speed of the train be $x$ metres/minute and the actual time taken by the train in the journey be $t$ minutes. Then
$x \times t = \frac{112.5 x}{100} \times (t - 20) \Rightarrow t = \frac{112.5 t}{100} - 22.5 \Rightarrow \frac{112.5 t}{100} - t = 22.5$
$\Rightarrow \frac{12.5 t}{100} = 22.5 \Rightarrow t = \frac{22.5 \times 100}{12.5}$ minutes = $180$ min

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A train increases its normal speed by 12.5% and reaches its destination 20 minutes earlier. What is the actual time taken by the train in the journey?

The correct option is B 180 minLet the normal speed of the train be x metres/minute and the actual time taken by the train in the journey be t minutes. Thenx×t=112.5x100×(t−20)⇒t=112.5t100−22.5⇒112.5t100−t=22.5⇒12.5t100=22.5⇒t=22.5×10012.5 minutes = 180 min

A train increases its normal speed by $12.5 %$ and reaches its destination $20$ minutes earlier. What is the actual time taken by the train in the journey?