### Ifeanyi walks a distance of 1.5km to his school everyday, his rate of walking is...

Ifeanyi walks a distance of 1.5km to his school everyday, his rate of walking is always constant.
On a certain day,owing to I'll health, he had to reduce his walking rate by 0.5 km/hr and as a result he took extra 6 minutes to reach the school than usually. Find the normal average rate which ifeanyi uses to reach the school.
(Pls show workings)?

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Distance to school=1.5km
let his walking rate be= x km/hr
let the time taken to reach the school be y
On a certain day,owing to I'll health, he had to reduce his walking rate by 0.5 km/hr
his walking rate due to illness=( x - 0.5)km/hr -------- 1
time taken to school due to illness=(y + 6)min = [(y + 6)/60]hr -------2

Normal average rate to school ( x )= 1.5/(y/60)km/hr
x = (1.5 * 60)/y
x = 90/y
his walking rate due to illness= 1.5/[(y + 6)/60]km/hr
=(1.5 *60)/(y +6)
= 90/(y + 6)km/hr
his walking rate due to illness=( x - 0.5)km/hr
90/(y + 6) =(90/y - 0.5)
90/(y + 6)= (90 - 0.5y)/y cross multiply
90 * y = (90 - 0.5y) *(y + 6)
90y=90y + 540 - 0.5y^2 - 3y
90y -90y - 540 + 0.5y^2 + 3y = 0
0.5y^2 + 3y -540=0 divide all through by 0.5
y^2 + 6y - 1080
y^2 +(36 - 30)y - 1080
y^2 +36y - 30y - 1080
(y^2 +36y) ( - 30y - 1080)
y (y +36)-30 ( y + 36)
(y + 36)(y - 30)
y + 36 = 0 or y - 30 = 0
y = -36 or y = 30
the normal average rate which ifeanyi uses to reach the school
x = 90/30
x = 3km/hr
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Let the speed be x
time taken=1.5km/x
timetaken=1.5/x-0.5km
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• Eddymontana: 1.5/x-0.5=1.5/x=6
1.5 - 1.5m + 0.75=6x-3
0.75=6x-3
3.75=6x
x=6/3.75=1.6
1 month ago
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