If the seed of boat in still water on Tuesday was 15 k/ hr and the speed of boat in still water on Wednesday was `66(2)/(3)%` more than that of Tuesday and time taken to travel upstream on Wednesday is `(9)/(10)` times than time taken by it to trae downstrem on Tuesday, then find the speed of stream (in kmph) on Wednesday ?

Options

A. `1.5`
B. `2.5`
C. 2
D. 1

Answer

D

Explanation

Speed o boat in still water on Tuesday =15 km/hr
`66(2)/(3)E=(2)/(3)`
Speed of boat in still water on Wednesday
`=15+(2)/(3)xx15`
=25 km/hr
According to question,
`(14xx18)/(25-x)=(14xx15)/((15+3))xx(9)/(10)`
`(6)/(25-x)=(5)/(18)xx(9)/(10)`
`(6)/(25-x)=(5)/(18)xx(9)/(10)`
`(6)/(0.2)=25-x`
`25-x=24`
x=1 km/hr
`:.` Speed of boat in still waer on Wednesday
= km/hr