Question

By increasing the speed of car by 10km/hr , the time of journey for a distance of 72km is reduced by 36 min . Find the original speed of the car

Answer 1

Answer:

the original speed of the car = 30 km/Hr

Step-by-step explanation:

By increasing the speed of car by 10km/hr , the time of journey for a distance of 72km is reduced by 36 min .

Let say the original speed of the car =  S km/Hr

Distance = 72 km

Time Taken = Distance / Speed =  72/S  hr

Increased speed = S + 10 km/hr

Time taken with increased speed = 72/(S + 10)   hr

72/S  - 72/(S + 10)  = 36/60

=> 1/S  - 1/(S + 10)  =  1/120

=> (S + 10) - S = S(S+10)/120

=> 1200 = S² + 10S

=> S² + 10S - 1200 = 0

=> S² + 40S - 30S -1200 = 0

=> S(S + 40) - 30(S + 40) = 0

=> (S-30)(S+40) = 0

=> S = 30   S = -40 is neglected as speed can not be negative

Original Speed of the car = 30 km/hr

Answer 2

Answer:

The speed of car is 30km/hr

Step-by-step explanation:

$let \: the \: speed \: be \: x \: kmph \\ time = \frac{distance}{speed} \\ t1 = \frac{72}{x} \\ time \: after \: increasing \: the \: speed \: \\ t2 = \frac{72}{x + 10} \\ \\ now. \: as \: per \: condition \\ t1 - t2 = \frac{36}{60} \\ \frac{72}{x} - \frac{72}{x + 10} = \frac{36}{60} \\ \frac{72(x + 10) - 72(x)}{ (x)(x + 10} = \frac{3}{5} \\ \frac{72x + 720 - 72x}{ {x}^{2} + 10x} = \frac{3}{5} \\ cross \: multiplication \\ 72(5) = 3( {x}^{2} + 10x) \\ 3600 = 3 {x}^{2} +30x \\ 0 = {x}^{2} + 10x - 1200 \\ 0 = {x}^{2} + 40x - 30x - 1200 \\ 0 = x(x - 30) - 30( x + 4) \\ ( x - 30)or(x + 40) \\ x = 30km h \\ because \: negative \: speed \: is \: not \: possible.$