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Question of Average

Problem Solving — algebra and arithmetic (GMAT Focus Edition)
Expert replies
by zerotoinfinite2006 » Sat Jul 31, 2010 8:15 am
So the question is:

With an average speed of 40km/h, a train reaches to the destination in time. If it goes with an speed of 25km/h, it is late by 15 min. The length of the total journey is:


[1]40km
[2]70km
[3]30km
[4]80km
[5]60km


Now let see the shortest possible you can track out to find out the answer of this question.
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Source: — Quantitative Reasoning |

by outreach » Sat Jul 31, 2010 10:11 am
t-time
40t=25*(t+1/4)
t=5/12

D=50/3

2d=100/3

bu there is no option with this ans
where am i wrong?
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by Stuart@KaplanGMAT » Sat Jul 31, 2010 10:34 am
zerotoinfinite2006 wrote:So the question is:

With an average speed of 40km/h, a train reaches to the destination in time. If it goes with an speed of 25km/h, it is late by 15 min. The length of the total journey is:


[1]40km
[2]70km
[3]30km
[4]80km
[5]60km


Now let see the shortest possible you can track out to find out the answer of this question.
In future:

1) please post the source of your questions;
2) post your questions in the proper forum (this should be in problem solving, not general math); and
3) post questions in GMAT format (on the GMAT, answer choices are always arranged in ascending or descending order).

Point (3) leads me to believe this is a made-up question, but let's fix it and solve anyway.

Rearranging the choices:

[1]30km
[2]40km
[3]60km
[4]70km
[5]80km

Let's use logic, our bestest GMAT friend!

If the trip were 40km, then it would take 1h at 40kph and almost 2h at 25kph - that's way more than a 15 minute difference.

So, the answer must be less than 40km - choose (1) 30km.

However, small problem:

At 40kph, 30km takes:

t = d/r = 30/40 = 3/4h = 45 minutes

At 25kph, 30km takes:

t = d/r = 30/25 = 1h 12minutes

That's not a 15 minute difference, so the correct answer is actually:

"this question is inherently flawed and its source should not be trusted".

(Which is why, again, you should ALWAYS post the source.)
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by zerotoinfinite2006 » Sat Jul 31, 2010 12:04 pm
Ok, I got the rules to post the questions I apologize to made this mistake.

let me finish this question;

The train needs to travel 15 min extra @35 kmph. Hence, it is behind by 8.5 kms. The rate of losing distance is 5kmph. Hence, the train must have travelled for 8.75/5 = 1 hour 45 min @40kmph -> 70 km

Alternatively,

you can also see that 12.5% drop in speed results in 14.28% increase in time. Hence, total time required is 105 min @40kmph -> 70 km

Do you still think that the question is incorrect ?

for the next time I make sure to post the question in the right section, please share your thought if you still think I answered it wrong.

Thanks
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by Stuart@KaplanGMAT » Sat Jul 31, 2010 12:13 pm
zerotoinfinite2006 wrote:Ok, I got the rules to post the questions I apologize to made this mistake.

let me finish this question;

The train needs to travel 15 min extra @35 kmph. Hence, it is behind by 8.5 kms. The rate of losing distance is 5kmph. Hence, the train must have travelled for 8.75/5 = 1 hour 45 min @40kmph -> 70 km

Alternatively,

you can also see that 12.5% drop in speed results in 14.28% increase in time. Hence, total time required is 105 min @40kmph -> 70 km

Do you still think that the question is incorrect ?

for the next time I make sure to post the question in the right section, please share your thought if you still think I answered it wrong.

Thanks
Well, since the original question said 25kph and your explanation said 35kph, that's the problem.
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by shashank.ism » Wed Aug 04, 2010 12:26 am
this is a simple problem...
let total distance =d and time taken with 40 km/h be t1 hr. & with speed 25km/h is (t1-0.25)hr.
so 40 t1 = d , 25 (t1+0.25)=d --> 25(d/40 + 0.25) = d --> 25d-250=40d --> d=250/15 km = 50/3 Km .no such option..

If you take it as 35 km/h then,
40 t1 = d , 35 (t1+0.25)=d --> 35(d/40 + 0.25) = d --> 25d - 350 = 30d --> d = 350/5 = 70 km Ans 2..
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