Question

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

Answer 1

let the distance be D m
and time required in reaching be T hr

Answer 2

Answer: The distance of his destination from his starting point is 14 km.

Explanation:

Given that,

Speed $v_{1} = 3 km/h$

Speed $v_{2} = 4 km/h$

Time $t_{1} = 40 min$

Time$t_{2}= 30 min$

Now, using formula

v = \dfrac{d}{t}

If his speed is 3 km/h then he reaches 40 min late

$\dfrac{d}{3km/h}= T+\dfrac{40}{60}h$....(I)

If his speed is 4 km/h then he reaches 30 min before

$\dfrac{d}{4km/h}= T+\dfrac{30}{60}h$....(II)

On solving equation(I) and (II)

$\dfrac{d}{3} -\dfrac{d}{4} = \dfrac{40}{60} + \dfrac{30}{60}$

$\dfrac{d}{12km/h} = \dfrac{7}{6}h$

$d = 14 km$

Hence, the distance of his destination from his starting point is 14 km.