M04-21

Question

A soccer coach takes  x  hours to cycle from his house to his office. When riding at 24 km/h, he arrives 5 minutes late, but at 30 km/h, he arrives 4 minutes early. What is the distance between his house and the office?

A. 18 km
B. 24 km
C. 36 km
D. 40 km
E. 72 km

Bunuel · Accepted Answer

Official Solution:

A soccer coach takes  x  hours to cycle from his house to his office. When riding at 24 km/h, he arrives 5 minutes late, but at 30 km/h, he arrives 4 minutes early. What is the distance between his house and the office?

A. 18 km
B. 24 km
C. 36 km
D. 40 km
E. 72 km

Denote the distance as  d .
 
Keep in mind that the rates are given in kilometers per hour, while the time is in minutes. Convert the minutes to hours: 5 minutes is  \frac{1}{12}  hours, and 4 minutes is  \frac{1}{15}  hours.
 
The equation for the first scenario is  \frac{d}{24} = x + \frac{1}{12} . Multiplying by 24, we get:  d = 24x + 2 .
 
The equation for the second scenario is  \frac{d}{30} = x - \frac{1}{15} . Multiplying by 30, we get:  d = 30x - 2 .
 
Subtract the second equation from the first one to obtain:  6x = 4 .
 
Solving for  x , we find that  x = \frac{2}{3} .
 
Now, substitute this value of  x  into either equation to find the distance  d . For example, using the first equation:  d = 24*\frac{2}{3} + 2 = 18 .
 
Therefore, the distance between the coach's house and his office is 18 kilometers.

Answer: A

Raihanuddin · Answer

D/24 -1/12 = D/30 +1/15

By solving the equation we get D =18

How do we get the equation?

Here we are making equation for usual time.

In the first case 5 minutes late from usual time. So we have to subtract the extra time,  which is 5 min or 5/60 or 1/12 hr,  to get the usual time

Similarly we have to add  4/60 or 1/15 hr  in the 2nd case to get the usual time.

gmatser1 · Answer

I solved this problem in a very prevaricate way. I got the right answer but I'm just concerned I'm not solving these the fastest way. It took me almost three minutes and involved some calculation.

The overall approach was to create two equations, set them equal to each other and after finding t, and plugging t back into the original equation to find d.

Let d equal distance to office.
Let t equal scheduled time to office.

d=rt

d=24 km/h*(t + 5) 
 d=30 km/h*(t - 4)

30(t - 4) = 24(t+5) 
 30t - 120 = 24t + 120 
 6t=240 
 t=40

Plugging t into the first equation:
 d=24(40+5) = 1080 km/h

But that's per hour. We need to find the distance, so we must divide that by 60.

\frac{1080}{60} = 18 km

Answer choice A.

Any tips on how I can see different ways to solve this? I'm getting closer to solving these harder problems, but I'm not quite there yet to doing it flawlessly.

mbadude2017 · Answer

I did this similar to gmatser1, but a bit different (in hours).

24 km/hr -> 1/12 hrs late
30 km/hr -> 1/15 hrs early

D=24*(t+1/12)=30*(t-1/15)
solve for t=2/3 hours (how long it should take at the "typical" speed - not given)

go back to the equation above to find D:
D=24*(2/3+1/12)=24*(9/12)= 18km

Spartan85 · Answer

Answer: A

Initial speed = 24 km/hr
Later speed = (5/4) *24 km/hr = 30 km/hr

Now, Difference of time in hrs, D/ 24 - D/30 = 9/60

Therefore D = 18

omullick · Answer

Can someone explain why this reasoning is wrong:

I assumed for path 1, time = t

for path 2, time = t - 9/60

Using this method doesn't yield the same results, what is incorrect in my underlying assumption?

Thanks,

Spartan85 · Answer

Path 1 : speed : 24km/hr ; time: t hrs
Path 2 : speed : 30km/hr ; time: t -9/60 hrs

therefore, Distance= 24*t = 30* (t-9/60)
t = 3/4 hrs
d = 24*t = 18

mpmbtr · Answer

I think this is a  high-quality  question and I agree with explanation. Its dev. test

Bunuel · Answer

I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.

BottomJee · Answer

I think this is a  high-quality  question and I agree with explanation.

vn049636 · Answer

I think this is a  high-quality  question and I agree with explanation.

KarishmaB · Answer

When his speed is 30 kmph, he reaches at say 12:00 noon (which is 4 mins early)
So normally he reaches at 12:04.
When his speed is 24 kmph, he reaches at 12:09 (5 mins late)

Hence the diff between time taken with speed 24 to speed 30 is 9 mins.

Ratio of speeds = 24:30
Ratio of time taken = 30:24 = 5:4
The diff of 1 on ratio scale is equal to 9 mins which means time taken at speed of 30 kmph is 4*9 = 36 mins
Distance travelled at speed 30 kmph in 36 mins =  30 * \frac{36}{60} = 18  km

Answer (A)

Nikhil17bhatt · Answer

My solution is a little different and more on logic.

so Speed and Time are inversely proportional if speed increases by 25% then time will decrease by 20%.

So speed changes from 24 to 30, an, an increase of 25%, and hence time decreases by 20%. Now, 20% of something is 9 mins, so the time taken when speed was 24kmph is 45 mins.

hence distance is 18km

Manvi01 · Answer