Jason drives to his parents' house and back home the same way. On the

Question

Jason drives to his parents' house and back home the same way. On the way to his parents' house, his speed is 10 kilometers per hour slower than his speed on his way back, and therefore it takes him 20 minutes longer. If Jason drives to his parents' house at 50 kilometers per hour, how many minutes does he spend on his way back from his parents' house?

A. 60
B. 80
C. 100
D. 120
E. 150

A. 60
B. 80
C. 100
D. 120
E. 150

JerryAtDreamScore · Answer

Breaking Down the Info:

We know Jason drives at 50 km/h to his parents' house and 40 km/h on the way back. Set the amount of time it took as t in minutes, and we would have:

50 * t = 40 * (t + 20)

Then  t = 80 , so it took 80 + 20 = 100 min to get back.

Note that the units on either side do not cooperate in the equation, but this doesn't create any problems as we are using the same units on both sides.

Answer: C

Fdambro294 · Answer

Speed to: 50 mph

Therefore his speed coming home is: 60 mph

Since the distance is constant, the ratio of the speeds driven at will be inversely proportional to the time taken

Speed TO : Speed BACK = 5 : 6

Time TO : Time BACK = 6g : 5g

And we are told that it takes him 20 minutes longer to drive to his parents house

(Time TO) — (Time BACK) = 6g - 5g = g = 20

And the Time back from his parent’s house is = 5g

5(20) = 100 minutes

C 100 minutes

Posted from my mobile device

AnandDoss2004 · Answer

it seems that there are so many issues in your approach. 
1. the speed back is 10 kmph more = 60 kmph
2. if you add 20, which is in minutes, then you should convert the speed in kmph to kmpm
3. when you take t as the time to go parents home, then finally 20 minutes should be reduced as the return is faster.

thkkrpratik · Answer

The answer will be C itself but yes, as you have pointed out, the calculations are given wrong in the solution by  JerryAtDreamScore

azamatboden · Answer

50 * (t+1/3) = 60 t

t = 100 mins

ThatDudeKnows · Answer

He drives to his parents house at 50kph, which is  \frac{5}{6} kpminute. That is 10kph slower than the speed on the way back, so the speed on the way back is 60kph, which is 1kpminute.

Time spent driving back is m minutes. Times spent driving there is m+20 minutes.

Distance going is  \frac{5}{6}(m+20) . Distance returning is  1m

Distance going = distance returning.

\frac{5}{6}(m+20) = m

\frac{5}{6}m+\frac{100}{6} = m

\frac{100}{6} = \frac{m}{6}

m=100

Answer choice C.

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