Question

Akash when going slower by by 15 km/hr, reaches late by 45 hours. if he goes faster by 10 km/hr from his original speed, he reaches early by by 20 hours than the original time. find the distance he covers.

Answer 1

Let the actual speed of Akash is "v"

now if Akash is going slow by 15 km/h then he will reach late by 45 hours

Let the total distance of travel is "d"

so the actual time to reach while he was travelling with actual speed is given by

$T = \frac{d}{v}$

now if his speed is lesser by 15 km/h the the time taken will be more by 45 hours

$T + 45 = \frac{d}{v - 15}$

now if he increase his speed by 10 km/h then his time take is less by 20 hours and he will reach early

so we can say

$T - 20 = \frac{d}{v + 10}$

now by above 3 equations we can say

$\frac{d}{v} + 45 = \frac{d}{v - 15}$

$\frac{d}{v} - 20 = \frac{d}{v + 10}$

now we will divide the two equations above

$\frac{\frac{d}{v-15} - \frac{d}{v}} {\frac{d}{v} - \frac{d}{v + 10}}= \frac{45}{20}$

$\frac{15* (v)(v + 10)}{10 * (v)(v -15)} = \frac{9}{4}$

$2* (v + 10)  = 3* (v - 15)$

$v = 65 km/h$

now from above equation

$\frac{d}{v} + 45 = \frac{d}{v - 15}$

put v = 65

$\frac{d}{65} + 45 = \frac{d}{50}$

$d= 9750 km$

so the total distance will be 9750 km

Answer 2

9750

Explanation:

concept used

distance covered in just extra time with original speed = distance covered in net time with extra speed

u×45 = 15(t-45)....(1)

u×20 = 10(t-20).....(2)

from (1) and (2)

u = 65kmph

t = 150 hrs

distance = 150×65 = 9750km