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avg speed qs

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by rijul007 » Thu Nov 10, 2011 10:48 am
A cyclist rides to the top of a hill and then coasts back down by the same route. The entire trip takes 4 hours. What is his average (arithmetic mean) speed when climbing the hill?

(1) The cyclist's average speed over the entire trip was 6 miles per hour.

(2) The distance to the top of the hill was 12 miles, and the cyclist's average speed on the downward trip was 18 miles per hour.
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Source: — Data Sufficiency |

by amit2k9 » Thu Nov 10, 2011 10:59 am
a 2D/ 4 = 6
however, t|up and t|down is not known.Not sufficient.

b D=12 S|down = 18

thus t|down = 12/18 = 2/3 hrs

t|up = 4-2/3
hence, D/t|up = avg speed|up. sufficient.

B it is.
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by shankar.ashwin » Thu Nov 10, 2011 11:07 am
Time = Distance / speed

Statement 1:

4 = 2D / 6 - D = 12 miles.

His decrease in speed going uphill would be compensated when coming downhill.

Let T1 be uphill time and T2 be downhill time,

T1 + T2 = 4hrs

12/(S1) + 12/(S2) = 4 - InSufficient


Statement 2:

D = 12 given and downward speed = 18, so time would be 40 mins. Upward time can be found.(3 hrs 20 mins)

B IMO
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by vaibhavgupta » Thu Nov 10, 2011 5:31 pm
rijul007 wrote:A cyclist rides to the top of a hill and then coasts back down by the same route. The entire trip takes 4 hours. What is his average (arithmetic mean) speed when climbing the hill?

(1) The cyclist's average speed over the entire trip was 6 miles per hour.

(2) The distance to the top of the hill was 12 miles, and the cyclist's average speed on the downward trip was 18 miles per hour.
+1 for B! :)
If OA is A, IMO B
If OA is B, IMO C
If OA is C, IMO D
If OA is D, IMO E
If OA is E, IMO A

FML!! :/
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