A man driving his moped at $24$ kmph reaches his destination $5$ minutes late to an appointment. If he had driven at $30$ kmph he would have reached his destination $4$ minutes before time. How far is his destination?

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Solution

Let the distance be ' $x$ ' km
Therefore time taken to cover ' $x$ ' km at $24$ kmph $= \frac{x}{24}$ hr
Time taken to cover ' $x$ 'km at $30$ kmph $= \frac{x}{30}$ hr
But it is given that the difference between two timinings $= 9 m i n = \frac{9}{60}$ hr
$∴$ $\frac{x}{24} - \frac{x}{30} = \frac{9}{60}$
$∴$ $\frac{5 x - 4 x}{120} = \frac{9}{60}$
$\Rightarrow$ $\frac{x}{120} = \frac{9}{60}$
$\Rightarrow$ $x = \frac{9}{60} \times 120 = 18$
Therefore the distance is $18$ km.

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A man driving his moped at 24 kmph reaches his destination 5 minutes late to an appointment. If he had driven at 30 kmph he would have reached his destination 4 minutes before time. How far is his destination?

Let the distance be 'x' kmTherefore time taken to cover 'x' km at 24 kmph =x24 hrTime taken to cover 'x'km at 30 kmph =x30 hrBut it is given that the difference between two timinings =9min=960 hr∴ x24−x30=960∴ 5x−4x120=960⇒ x120=960⇒ x=960×120=18Therefore the distance is 18 km.

A man driving his moped at $24$ kmph reaches his destination $5$ minutes late to an appointment. If he had driven at $30$ kmph he would have reached his destination $4$ minutes before time. How far is his destination?