Question

Walking at 60% of his usual speed, a man reaches his destination 1hour 40minutes late. His usual time (In hours) to reach the destination is:
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Answer 1

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

GIVEN

Walking at 60% of his usual speed, a man reaches his destination 1 hour 40 minutes late

TO DETERMINE

The usual time (In hours) to reach the destination

CALCULATION

Let total distance = y km and

usual speed = x km / min

$\displaystyle \sf{Time \: taken \: to \: reach \: the \: destination \: = \frac{y}{x} \: \: \: min }$

Now if the man is walking at 60% of his usual speed

Then the reduced speed of the man is

$= \displaystyle \sf{x \: \times \frac{60}{100} = \frac{3x}{5} \: \: \: }km / min$

Now

$\displaystyle \sf{Time \: taken \: to \: reach \: the \: destination \:}$

$\displaystyle \sf{ = \frac{y}{ \frac{3x}{5} } \: min }$

$\displaystyle \sf{= \frac{5y}{3x} \: \: \: min }$

1 hour 40 minutes = 100 minutes

So by the given condition

$\displaystyle \sf{ \frac{5y}{3x} \: - \frac{y}{x} = 100 }$

$\implies \: \displaystyle \sf{ \frac{5y - 3y}{3x} \: = 100 }$

$\implies \: \displaystyle \sf{ \frac{2y }{3x} \: = 100 }$

$\implies \: \displaystyle \sf{ \frac{y }{x} \: = 100 \times \frac{3}{2} }$

$\implies \: \displaystyle \sf{ \frac{y }{x} \: = 150 }$

RESULT

Hence The usual time (In hours) to reach the destination

= 150 minutes

= 2 hour 30 minutes

= 2.5 hours

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