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by bpgen » Fri Jun 18, 2010 6:36 pm
A riverboat leaves Mildura and travels upstream to Renmark at an average speed of 6 miles/hr. It returns by the same route at an average speed of 9 miles/hr. What is the average speed of the round trip, in miles/hr?

A- 7.0
B- 7.2
C- 7.8
D- 7.8
E- 8.2
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Source: — Problem Solving |

by bpgen » Fri Jun 18, 2010 6:38 pm
OA- B
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by Stuart@KaplanGMAT » Fri Jun 18, 2010 8:29 pm
bpgen wrote:A riverboat leaves Mildura and travels upstream to Renmark at an average speed of 6 miles/hr. It returns by the same route at an average speed of 9 miles/hr. What is the average speed of the round trip, in miles/hr?

A- 7.0
B- 7.2
C- 7.8
D- 7.8
E- 8.2
There are a number of ways we could approach this question, including solving equations, using weighted averages and a matrix. The third option works for all multiple part distance/rate/time questions and will almost certainly be applicable on test day.

Across the top of the matrix we have the applicable formula: distance = rate * time.
Down the side of the matrix we have the parts of the trip and the entire trip: upstream, downstream, total.

To start, it looks like this:

..............Distance....=......Rate............*..............Time

Upstm........D......................6.................................t(up)

Dnstm........D......................9.................................t(dn)

Total.........2D.....................r...............................t(up) + t(dn)

Now, we could solve as is, or we could make our lives easier by picking a distance - no matter what number we choose, we'll always get the same average rate. Since we're going to be dividing distance by rate, we want a distance divisible by both 6 and 9 - let's choose 18.

So, back to our chart:

..............Distance....=......Rate............*..............Time

Upstm........18......................6.................................t(up)

Dnstm........18......................9.................................t(dn)

Total...........36.....................r...............................t(up) + t(dn)

Now we can solve for t(up) and t(dn):

..............Distance....=......Rate............*..............Time

Upstm........18......................6.................................3

Dnstm........18......................9.................................2

Total...........36......................r.................................5

Finally, we solve the total equation on the bottom line:

36 = r*5
36/5 = r
7.2 = r

One of the primary advantages of the matrix is that it allows you to see what information you have and what pieces are missing, so you can plan your attack to find whatever that particular question is asking about.
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by selango » Fri Jun 18, 2010 8:59 pm
If object travels from point A to B with speed p and point B to A with speed q,then

average speed=2pq/(p+q)

=2*6*9/15=7.2
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by saurabhmahajan » Sat Jun 19, 2010 12:49 am
Other way to avoid formula :-)

Take any value that is multiple of 6 and 9...i took 18 (18 miles is the distance)

so 18/6=3 hrs upstream
18/9=2 hrs downstream
total =5hrs
total distance=18+18=36
avg = 36/5 =7.2
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