average of two averages approach

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dla5h expert: Newbie | Next Rank: 10 Posts; Posts: 8; Joined: Thu Dec 25, 2008 12:20 am

average of two averages approach

by dla5h expert » Fri Oct 02, 2009 4:58 pm

Hi guys.. in my last practice test I got this Q .. and since the distance is equal, I tried to average the two averages ,,

please can someone tell me why I am Wrong ?? and whats better approach for this kind of questions

thanks

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Source: Beat The GMAT — Problem Solving | Fri Oct 02, 2009 4:58 pm

xcusemeplz2009: Master | Next Rank: 500 Posts; Posts: 399; Joined: Wed Apr 15, 2009 3:48 am; Location: india; Thanked: 39 times

by xcusemeplz2009 » Fri Oct 02, 2009 7:34 pm

IMO C

we need to remember that
avg speed=total distance/total time

1.e
total time=total dist/avg speed

foe m distance avg speed is V mph
total time=m/V hours..........T1

for 2m miles avg speed is 5V/3
toatal time=6m/5V...........T

T=T1+T2
T2=T-T1=6M/5V-m/V=m/5V hrs

sp avg speed for rest of m miles = m/T2=m/m/5V=5V

It does not matter how many times you get knocked down , but how many times you get up

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dla5h expert: Newbie | Next Rank: 10 Posts; Posts: 8; Joined: Thu Dec 25, 2008 12:20 am

by dla5h expert » Fri Oct 02, 2009 10:23 pm

xcusemeplz2009 wrote:IMO C

we need to remember that
avg speed=total distance/total time

1.e
total time=total dist/avg speed

foe m distance avg speed is V mph
total time=m/V hours..........T1

for 2m miles avg speed is 5V/3
toatal time=6m/5V...........T

T=T1+T2
T2=T-T1=6M/5V-m/V=m/5V hrs

sp avg speed for rest of m miles = m/T2=m/m/5V=5V

Thank you xcusemeplz2009 .. that is an easy and correct approach

but my question is ..

since we know that both distance for T1 and T2 are equal = m

why cant we say ..

Total avg = avg1 + avg2 / 2 ??

I will give you an example :

distance from A to B is 200 Mile ,, for the first 100 m my avg was 100m/h and the total trip avg speed is 200 m/hours ? can you tell me what is my avg speed for the second 100 mile ?

notice here that we know only the total avg and the avg for the first 100 mile ,, and we need the avg for the secon 100 mile !!

my approach is :
total avg = (avg1 + avg2)/2
200 = (100 + avg2)/2
avg2 = 300 m/h

is that correct ? if NO Why ?

thank you.

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xcusemeplz2009: Master | Next Rank: 500 Posts; Posts: 399; Joined: Wed Apr 15, 2009 3:48 am; Location: india; Thanked: 39 times

by xcusemeplz2009 » Tue Oct 06, 2009 6:55 am

dla5h expert wrote:
xcusemeplz2009 wrote:IMO C

we need to remember that
avg speed=total distance/total time

1.e
total time=total dist/avg speed

foe m distance avg speed is V mph
total time=m/V hours..........T1

for 2m miles avg speed is 5V/3
toatal time=6m/5V...........T

T=T1+T2
T2=T-T1=6M/5V-m/V=m/5V hrs

sp avg speed for rest of m miles = m/T2=m/m/5V=5V
Thank you xcusemeplz2009 .. that is an easy and correct approach

but my question is ..

since we know that both distance for T1 and T2 are equal = m

why cant we say ..

Total avg = avg1 + avg2 / 2 ??

I will give you an example :

distance from A to B is 200 Mile ,, for the first 100 m my avg was 100m/h and the total trip avg speed is 200 m/hours ? can you tell me what is my avg speed for the second 100 mile ?

notice here that we know only the total avg and the avg for the first 100 mile ,, and we need the avg for the secon 100 mile !!

my approach is :
total avg = (avg1 + avg2)/2
200 = (100 + avg2)/2
avg2 = 300 m/h

is that correct ? if NO Why ?

thank you.

i think u are getting confused b/n simple avg and weighted avg..... in simple avg we can simply do the avg of averages but in weighted average we cant take average of averages .

coming back to your approach...

from the answer u got if u do a back calculation then u will find that your value for time is not matching.

for 1st 100 milles speed is 100mph therefore time is 1 hr.........(1)

for next 100 miles from your value 300mph time taken is 100/300=1/3 hr........(2)

total time is 200/100=2 hr
adding 1 and 2 u get 1hr20 min which is not correct

HTH

It does not matter how many times you get knocked down , but how many times you get up

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Stuart@KaplanGMAT: GMAT Instructor; Posts: 3225; Joined: Tue Jan 08, 2008 2:40 pm; Location: Toronto; Thanked: 1710 times; Followed by:614 members; GMAT Score:800

by Stuart@KaplanGMAT » Tue Oct 06, 2009 2:21 pm

xcusemeplz2009 does a good job explaining this particular question; let's address the more general issue.

In multiple-part distance/rate/time questions, each part of the trip is weighted based on the time spent at a particular speed.

For example, if you spend 10 hours at 50kph and 5 hours at 100kph, your average speed will be closer to 50 than to 100, since you spent more time at 50. The only case in which the average speed will be the dead average of the individual speeds is if you traveled the same length of time at each speed.

For example, if you spend 10 hours and 50kph and 10 hours at 100kph, the average speed for the 20 hours will be 75kph.

Understanding this often gives us the opportunity to make a quick guess on round trip questions. If the distance is the same for both legs of the journey, the average speed will always be closer to the slower of the two speeds (since, if we're covering the same distance, it will take longer to do so if we're moving more slowly).

Here's an example:

Carolyn bikes uphill to work at 10kph and downhill from work at 30kph. If she takes the same route to and from work, what's her average speed in kph for the round trip?

a) 12
b) 15
c) 20
d) 22
e) 25

We know it's a round trip question, so her average speed is going to be closer to 10 than to 30: eliminate c, d and e. Worse case, we have a 50/50 shot at the question.