A car travels at a constant speed from Q to R, a distance of 300 km apart.?

Question

If the driver increase the speed by 5 km/h, the journey will take 2 hours less. Find the original speed of the car.

Answer 1

Distance = Rate * Time

Time = Distance / Rate

T1 = 300 / R1
T2 = 300 / R2

T2 = T1 - 2

R2 = R1 + 5

T1 = 300 / R1
T1 - 2 = 300 / (R1 + 5)
T1 = 300 / (R1 + 5) + 2

300 / R1 = 300 / (R1 + 5) + 2

Multiply both sides by (R1)(R1 + 5):

300(R1 + 5) = 300(R1) + 2(R1)(R1 + 5)
300(R1) + 1500 = 300(R1) + 2(R1)² + 10(R1)

2(R1)² + 10(R1) - 1500 = 0
2(R1² + 5(R1) - 750) = 0
2(R1 + 30)(R1 - 25) = 0
R1 = 25 or -30

Since the first rate can't be negative, the first rate must be 25 km/h.

To prove this, calculate the time taken for a 300km trip at 25km/h:

300 = 25 * T
T = 300 / 25 = 12 hrs

Now, calculate the time for a 300km trip at 30km/h:

300 = 30 * T
T = 300 / 30 = 10 hrs

The difference is two hours. QED

Answer 2

speed=dist/time

for the first case
s1=300/t1 or t1*s1=300...................(1)
for the second case
s2=300/t2 or (s1+5)=300/(t1-2)...................(2)
from this eq. (t1-2)*(s1+5)=300
t1*s1+5*t1-2*s1-10=300 substitude (1) in this eq.
300+5*t1-600/t1-10=300
5*t1-600/t1-10=0
5*t1^2-10*t1-600=0 divide both sides by 5

t1^2-2*t1-120=0
from this (t1-12)*(t1+10)=0
this results t1=12h.

put this in (1) s1=300/12=25km/h