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by sparkle6 » Sun Sep 25, 2011 6:41 am
A car runs 5 successive 6 km stretches at speeds of 5 km/hr, 12 km/hr, 15 km/hr, 18 km/hr and 20 km/hr. Find the Average speed of the car

a. 8 km/hr
b. 9 km/hr
c. 11 km/hr
d. 13 km/hr
e. 18 km/hr
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by amit2k9 » Sun Sep 25, 2011 7:08 am
speed = distance/ time

distance = 5*(6/5)+12*(6/12)+ 15* (6/15)+ 18 * (6/18) + 20 * (6/20) = 30km

time = 6/5 + 6/12+ 6/15+ 6/18 + 6/20= 139/30

thus avg speed = 30*30/139 = 9km/hr approx
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by GMATGuruNY » Sun Sep 25, 2011 10:33 am
sparkle6 wrote:A car runs 5 successive 6 km stretches at speeds of 5 km/hr, 12 km/hr, 15 km/hr, 18 km/hr and 20 km/hr. Find the Average speed of the car

a. 8 km/hr
b. 9 km/hr
c. 11 km/hr
d. 13 km/hr
e. 18 km/hr
The question should ask for the APPROXIMATE average speed.

The distance can be any value.
Let distance = 180km.

The 5 rates will yield the following times:
180/5 = 36 hours.
180/12 = 15 hours.
180/15 = 12 hours.
180/18 = 10 hours.
180/20 = 9 hours.

Total distance = 5*180 = 900km.
Total time = 36+15+12+10+9 = 82 hours.
Average speed = 900/82 ≈ 11.

The correct answer is C.
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by sparkle6 » Sun Sep 25, 2011 11:48 pm
Don't we need to use 6 km somehow for the distance?
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by GMATGuruNY » Mon Sep 26, 2011 2:58 am
sparkle6 wrote:Don't we need to use 6 km somehow for the distance?
Nope. Since the question is asking for the AVERAGE speed, and the SAME DISTANCE is being traveled at each speed, we can plug in ANY VALUE for the distance.
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by seema19 » Mon Sep 26, 2011 4:01 am
I didn't get the above reply. Can someone plz elaborate..
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by seema19 » Mon Sep 26, 2011 4:02 am
I didn't get the above reply. Can someone plz elaborate..
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by GMATGuruNY » Mon Sep 26, 2011 4:48 am
seema19 wrote:I didn't get the above reply. Can someone plz elaborate..
Let d = the distance traveled at each speed.

The 5 rates will yield the following times:
d/5, d/12, d/15, d/18, and d/20.

Total time = d/5 + d/12 + d/15 + d/18 + d/20 = 36d/180 + 15d/180 + 12d/180 + 10d/180 + 9d/180 = 82d/180.
Total distance = 5d.
Average rate = (total distance)/(total time) = 5d/(82d/180) = (5d*180)/82d = 900d/82d = 900/82 ≈ 11.

The value of d is irrelevant: notice how it cancels out in the fraction above.
Whatever the value of d, the average speed will be approximately 11kph.
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