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Standard XII

Mathematics

Average Rate of Change

A person reac...

Question

A person reaches his destination $40$ minutes late if his speed is $3$ km/hr and reaches $30$ minutes before time if his speeed is $4$ km/hr. Then the distance of his destination from his starting point is

14

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Solution

The correct option is A $14$ km
Let the required distance be $x$ km.
Difference in the time taken at two speeds $= 70 min = \frac{7}{6} hr$
$∴ \frac{x}{3} - \frac{x}{4} = \frac{7}{6}$
$\Rightarrow 4 x - 3 x = 14$
$\Rightarrow x = 14$ km

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A person reaches his destination 40 minutes late if his speed is 3 km/hr and reaches 30 minutes before time if his speeed is 4 km/hr. Then the distance of his destination from his starting point is

The correct option is A 14 kmLet the required distance be x km. Difference in the time taken at two speeds =70 min=76 hr ∴x3−x4=76 ⇒4x−3x=14 ⇒x=14 km

A person reaches his destination $40$ minutes late if his speed is $3$ km/hr and reaches $30$ minutes before time if his speeed is $4$ km/hr. Then the distance of his destination from his starting point is

The correct option is A $14$ km
Let the required distance be $x$ km.
Difference in the time taken at two speeds $= 70 min = \frac{7}{6} hr$
$∴ \frac{x}{3} - \frac{x}{4} = \frac{7}{6}$
$\Rightarrow 4 x - 3 x = 14$
$\Rightarrow x = 14$ km