A motorist leaves his home at 8 a.m. to a place 90 km away. If he drives at an average speed of (x-5)km/h, he will be 15 minutes late for his appointment. If driving at (x+10) km/h, he will be 15 minutes early. Find the value of x and the time fixed for the appointment.
Answers
let his actual time be y
case-i
distance(s)=90 km
average speed(v)=(x-5)kmph
time(t)=(y+15)÷60hr
case-ii
distance(s)=90 km
average speed(v)=(x+10)kmph
time(t)=(y-15)÷60hr
case-i equation
distance(s)=speed(v)×time(t)
90km=((x-5)kmph)((y+15)÷60hr)
90km=(x-5)×((y+15)÷60)km
90km=((xy+15x-5y-75)÷60)km --------------i
case-ii equation
distance(s)=speed(v)×time(t)
90km=((x+10)kmph)((y-15)÷60hr)
90km=(x+10)((y-15)÷60)km
90km=((xy-15x+10y-150)÷60)km--------------ii
substitute equation i and ii
90km=((xy+15x-5y-75)÷60)km --------------i
90km=((xy-15x+10y-150)÷60)km--------------ii
we get,
((xy+15x-5y-75)÷60)km=((xy-15x+10y-150)÷60)km
((xy+15x-5y-75)÷60)=((xy-15x+10y-150)÷60)
xy+15x-5y-75=xy-15x+10y-150
15x-5y-75= -15x+10y-150
30x-15y= -75
15(2x-y)= -75
2x-y= -5
y-2x=5
☆y=5+2x
90km=((xy-15x+10y-150)÷60)km--------------ii
90=((x(5+2x)-15x+10(5+2x)-150)÷60)
5400=2x^2+10x-100
0=2x^2+10x-5500
☆x=50 (or)-55
☆y=5+2x=105 (or)-105
☆The value of x is 50 because as distance and time are always positive speed is also positive.
☆time fixed for the appointment is 105 min and not -105 min because time is never negative.
Hope it helps you.
Mark it has brainlist answer.