Question

A cyclist after riding a certain distance ,stopped for half an hour to repair his bicycle, after which he completes the whole 30km at half speed in 5hours.if break down has occurec 10km farther off he would have done the whole journey in 4hr Find where the breakdown occured and orignal speed.
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Answer 1

$\huge\mathfrak{ANSWER}$

$\sf\huge{Question!!?}$

A cyclist after riding a certain distance ,stopped for half an hour to repair his bicycle, after which he completes the whole 30km at half speed in 5hours.if break down has occurec 10km farther off he would have done the whole journey in 4hr . To Find,, where the breakdown occured i.e, at what distance he stopped his bicycle and orignal speed...

【1】case=>

Let the distance be 'x' at which the cyclist with velocity v km\h stopped . Let t1 be the time taken to cover the distance 'x'

SIMILARLY, after repairing out his bicycle he moves with velocity of $\frac{v}{2}$km\h the distance of 30-x, since total distance covered is 30km. Let t2 be the time taken to cover the distance (30-x)m

.

from the question...

t1 + t2 = 5hrs..==(1)

where,

t1=$\frac{x}{v}$ ,t2= $\frac{30-x}{v/2}$

from 1..

$\frac{x}{v}$ + $\frac{30-x}{v/2}$ = 5

$\frac{x}{v}$ + $\frac{(30-x)2}{v}$= 5

$\frac{x}{v}$ + $\frac{(60-2x}{v}$ =5

$\frac{x+60-2x}{v}$ = 5

$-x+60$ = 5v

x + 5v =60==(2)..

【2】case==>

If the cyclist stopped after covering 10 km more , then, distance covered in time t1 with same velocity 'v' will be (x+10)km

After repairing cycle, he travels a distance 30-(10+x)km in time t2, with half velocity $\frac{v}{2}$km\h..

given that==>

t1 + t2 = 4hrs...

where

t1 = $\frac{x+10}{v}$ ,t2= $\frac{30-x-10}{v/2}$= $\frac{(20-x)2}{v}$

substituting....

$\frac{x+10}{v}$ + $\frac{(20-x)2}{v}$ =4

$\frac{x+10}{v}$ + $\frac{40-2x}{v}$ =4

$\frac{x+10+40-2x}{v}$ = 4

$\frac{50-x}{v}$ = 4

50 - x = 4v

x + 4v = 50===(3)

from equation (2)&(3)

x + 5v = 60

x + 4v = 50

___________

v = 10..

from value of 'v' in equation 3, we get

x + 4(10) = 50

x + 40 = 50

x = 10km..

hence the distance at which cyclist stopped is at $\sf\huge{\boxed{\boxed{\underline{10km}}}}$.. and the original velocity (or) speed of the cyclist is $\bold{\boxed{\underline{10km/h}}}$

Answer 2

let s = original speed

then

.5s = half speed

:

let d = that "certain distance" where the breakdown occurred

:

He rode d at original speed and (30-d) at half speed

:

Write a time equation, time = dist/speed

:

Time at full speed + repair time + time at half speed = 5 hrs

d%2Fs + .5 + %28%2830-d%29%29%2F%28.5s%29 = 5

Multiply equation by s

d + .5s + 2(30-d) = 5s

d + .5s + 60 - 2d = 5s

combine like terms

d - 2d + .5s - 5s + 60 = 0

-d - 4.5s + 60 = 0

multiply by -1

d + 4.5s - 60 = 0

d + 4.5s = 60