A car covers a distance of 90 km at a uniform speed. Had the speed been 15 km/hr more, car would have taken 30 minutes less for the journey. Find the original speed of the car.
Answers
Answered by
13
let original speed of car is x km/h and time is t hour
now
according to question
90/x=t -----------------(1)
90/(x+15)=t-30/60 ------------(2)
solve both equation
subtracting (1) to (2)
e.g.
90 {1/x-1/(x+15)}=1/2
90{15/(x^2+15x)=1/2
2700=x^2+15x
x^2+15x-2700=0
now use quadratic equation formula
x=45km/h
hence original speed of car is 45km/h
now
according to question
90/x=t -----------------(1)
90/(x+15)=t-30/60 ------------(2)
solve both equation
subtracting (1) to (2)
e.g.
90 {1/x-1/(x+15)}=1/2
90{15/(x^2+15x)=1/2
2700=x^2+15x
x^2+15x-2700=0
now use quadratic equation formula
x=45km/h
hence original speed of car is 45km/h
Answered by
11
distance =90km
speed =x
time = t
x=90÷t
x+15=90\t-1÷2
so ,
x^2-30x-675=0
by using quadratic formula ,
b^2-4ac=3600
so ,
30+-(plus minus)60\2(whole divided by 2)
=45kmph
ie the speed of car is 45 kmph
speed =x
time = t
x=90÷t
x+15=90\t-1÷2
so ,
x^2-30x-675=0
by using quadratic formula ,
b^2-4ac=3600
so ,
30+-(plus minus)60\2(whole divided by 2)
=45kmph
ie the speed of car is 45 kmph
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