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A cyclist rides his bicycle over a route which is 1/3 uphill [#permalink]

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21 Mar 2007, 18:20

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A cyclist rides his bicycle over a route which is 1/3 uphill, 1/3 level, and 1/3 downhill. If he covers the uphill part of the route at the rate of 16 miles per hour and the level part at the rate of 24 miles per hour, what rate in miles per hour would he have to travel the downhill part of the route in order to average 24 miles per hour for the entire route?

(A) 32 (B) 36 (C) 40 (D) 44 (E) 48

Please solve and explain why this doesn't this work: 1/3(16) + 1/3(24) + 1/3(x) = 24

And don't just give me your solution. Please let me know what's wrong w/the above. This equation was the first thing that came to my mind. I need to understand what I'm missing.

"Please solve and explain why this doesn't this work:
1/3(16) + 1/3(24) + 1/3(x) = 24 "

When you calculate average speed it is equal to the time traveled times the speed or if you travel half an hour at speed 60 mph and half an hour at 30 mph then the average speed would be 45 mph
If you travel 1/2h at 60mph and 1/6h at 120mph then the aver speed would be the distance traveled( 50 miles) for 40 min or av speed is 3000/40=300/4=75 mph
Hope it helps

Take total distance = 24 miles. so each section uphill,level,downhill is 8 miles.

Now he covered uphill at 16 miles/hour so it will take him 30 minutes for 8 miles
He covered level at 24 miles/hour so he would take 20 minutes for that part

Now, overall, if he has to average 24 miles/hours for all the trip, then he has to cover 24 miles in 1 hour. So that means he has only 10 minutes to cover remaning 8 miles.
So his speed must be 8 miles in 10 minutes which is same as 48 miles/hour

As you see we dont need to write any equations and can simply do mental calculations to get to answer

Thanks everyone. Kyatin, I like your method. With time, hopefully I'll use more reason and less math to solve these. But for now, I'm quick to pull out some sort of equation. This is how I reason it. If you're bored you can let me know if my reasoning holds up.

rt=d, r=d/t

The total distance is unknown. We'll let that equal d.

We're given the average rate. The avg. rate, 24 mph., is the aggregate of the 3 legs of our cyclist's journey.

So, since r = d/t, let's have d/24 = d/16(1/3) + d/24(1/3) + d/x(1/3) =
d/24 = d/48 + d/72+ d/3x =
3d/24 = d/16 + d/24 + d/x =
6d/48 = 3d/48 + 2d/48 + d/48

Sorry. I thought I had something here.
ok, here it is.
432 is the LCM of 3, 48 and 72.
d/24 = d/48 = 9d/432 + d/72 = 6d/432 + d/3x = dx/432
432/3 = 144
3x = 144
x = 48

"Please solve and explain why this doesn't this work: 1/3(16) + 1/3(24) + 1/3(x) = 24 "

When you calculate average speed it is equal to the time traveled times the speed or if you travel half an hour at speed 60 mph and half an hour at 30 mph then the average speed would be 45 mph If you travel 1/2h at 60mph and 1/6h at 120mph then the aver speed would be the distance traveled( 50 miles) for 40 min or av speed is 3000/40=300/4=75 mph Hope it helps

You can only take the average of the three speeds if they involve the same amount of time. Clearly, they do not in this case, the leg uphill is the longest in terms of time and thus carries the greatest weight

Re: A cyclist rides his bicycle over a route which is 1/3 uphill [#permalink]

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05 Mar 2013, 20:42

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Expert's post

ggarr wrote:

A cyclist rides his bicycle over a route which is 1/3 uphill, 1/3 level, and 1/3 downhill. If he covers the uphill part of the route at the rate of 16 miles per hour and the level part at the rate of 24 miles per hour, what rate in miles per hour would he have to travel the downhill part of the route in order to average 24 miles per hour for the entire route?

(A) 32 (B) 36 (C) 40 (D) 44 (E) 48

Please solve and explain why this doesn't this work: 1/3(16) + 1/3(24) + 1/3(x) = 24

And don't just give me your solution. Please let me know what's wrong w/the above. This equation was the first thing that came to my mind. I need to understand what I'm missing.

Please Show ALL work

Responding to a pm:

There are two problems with using the formula given on my blog on this question:

In case of three speeds, you can simply use the formula: Cavg = (C1*W1 + C2*W2 + C3*W3)/(W1 + W2 + W3)

Weight in case of speed is 'time taken'

Hence, Avg Speed = Total distance / Total time (which we know)

In this question, I would like to assume that the total distance is 48*3 such that the distance traveled in each leg of the journey uphill, level and downhill is 48 miles (you can assume it to be something else or x)

Time taken to go uphill = 48/16 = 3 hrs Time taken on level = 48/24 = 2 hrs Time taken to go downhill = 48/d

You can also look at it in another way - A Shortcut:

Speed on level plain is 24 miles/hr and average speed is also 24 miles/hr. So we can just ignore the level plain since it is at the average. We need to average out the rest of the journey to 24 miles/hr. Again, assuming that distance of each leg is 48 miles,

24 = 48*2/(3 + 48/d) d = 48 miles/hr _________________

Re: A cyclist rides his bicycle over a route which is 1/3 uphill [#permalink]

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12 Aug 2013, 15:02

ggarr wrote:

A cyclist rides his bicycle over a route which is 1/3 uphill, 1/3 level, and 1/3 downhill. If he covers the uphill part of the route at the rate of 16 miles per hour and the level part at the rate of 24 miles per hour, what rate in miles per hour would he have to travel the downhill part of the route in order to average 24 miles per hour for the entire route?

(A) 32 (B) 36 (C) 40 (D) 44 (E) 48

Please solve and explain why this doesn't this work: 1/3(16) + 1/3(24) + 1/3(x) = 24

And don't just give me your solution. Please let me know what's wrong w/the above. This equation was the first thing that came to my mind. I need to understand what I'm missing.

Please Show ALL work

Let, each of three equal part of distance = 48 (LCM of 16 and 24)

24 = (total distance)/(total time) or, 24 = 3 × 48 / (3+2+t) or, t = 1 hour (for downhill part, time=t) so, 48 miles/1 hours = 48m/h have to travel. _________________

Re: A cyclist rides his bicycle over a route which is 1/3 uphill [#permalink]

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13 Aug 2013, 12:17

A cyclist rides his bicycle over a route which is 1/3 uphill, 1/3 level, and 1/3 downhill. If he covers the uphill part of the route at the rate of 16 miles per hour and the level part at the rate of 24 miles per hour, what rate in miles per hour would he have to travel the downhill part of the route in order to average 24 miles per hour for the entire route?

Rate = distance/time Time = distance/rate

In this problem, we are looking for the rate for a series of separate events (i.e. different segments of the distance traveled at different speeds). We know that each leg of the journey is the same distance and because we are looking for rate, not distance, we can choose a number to represent d.

To solve for the problem we also need the time taken for each leg of the journey.

t1 = d/16 t2 = d/24 t3 = d/x

let d = 48 (which is the lowest common multiple of 16 and 24)

Re: A cyclist rides his bicycle over a route which is 1/3 uphill [#permalink]

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17 Jan 2015, 04:41

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Re: A cyclist rides his bicycle over a route which is 1/3 uphill [#permalink]

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18 Oct 2015, 04:54

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Expert's post

Let us assume that distance for each of the three parts is 48 km( Since 48 is multiple for 16 and 24 , hence it will be convenient )

Time taken to go uphill = 48/16 = 3 hours Time taken on level = 48/24 = 2 hours Time taken to go downhill = 48/Sd , where Sd= Speed of cyclist in downhill journey

Total time for the entire journey = Distance/ Speed = 48x3/24 = 6 hours

Time taken to go downhill = 1 hour Hence, Sd = 48 miles/hour Answer E _________________

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Re: A cyclist rides his bicycle over a route which is 1/3 uphill
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