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

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

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New post 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.

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

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New post 05 Mar 2013, 20:42
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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:

1. There are three different speeds which you need to average out.
2. The weights in case of speed is 'time taken' not distance traveled. (explained here: bill-travels-first-40-of-the-distance-to-his-destination-at-137000.html#p1172411)

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

Avg Speed = 24 = 48*3/(3 + 2 + 48/d)
48/d = 1
d = 48 miles/hr

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
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New post 21 Mar 2007, 18:58
here is a hint.

Please check what two quantities you are multiplying when you do 16 * 1/3 etc..

are u multiplying distance into speed?? Is that correct??
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New post 21 Mar 2007, 20:40
I tried this method and for some reason I am not coming out with a good answer.

Here is my line of reasoning:

16(u)=1/3x, 24(L)=1/3x, r(d)=1/3x


we also know that we want the avg speed to be 24.


total distance/total time= 24


x/(x/32+x/72+x/3r)=24

and you want to solve for r

x=24(x/32+x/72+x/3r)

1=24(1/32+1/72+1/3r)



and then solve for r

This doesn't yield a good answer, perhaps I made a miscalculation.
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New post 21 Mar 2007, 20:51
Tuneman

Why use X when you can take a value and eliminate one variable.

I have a solution that does not involve writing lot of equations. Give it a shot you will get to it.

Hint : Take some distance which will ease calculations.
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New post 21 Mar 2007, 20:51
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Average Speed= total distance/total time

24=x/((x/3*16)+(x/3*24)+(x/3*Y)
where, Y is the speed for the downhill slope.

Hence, 24=1/1/(1/16+1/24+1/3Y)

which gives, 1/24=1/16+1/72+1/3Y (not 24=1/16+1/72+1/3Y)

Solving for Y, you will get Y=48
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New post 21 Mar 2007, 23:12
ahh yes I see, dumb miscalculation by me, but thank you for the explanation. Well done.

that is also true you could take the LCM of those denominators, would probably save you 30-40 seconds
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New post 22 Mar 2007, 08:50
"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
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New post 22 Mar 2007, 10:35
This is how I did it.

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
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New post 22 Mar 2007, 23:53
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

check: rt=d => (48mph.)(d/48mph.) = d
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New post 23 Mar 2007, 07:23
Quote:
d/24 = d/48 + d/72+ d/3x =
d/24 = d/16 + d/24 + d/x =
d/24 = 3d/48 + 2d/24 + d/48


Not sure what you are trying to do here.
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New post 23 Mar 2007, 11:16
kyatin wrote:
Quote:
d/24 = d/48 + d/72+ d/3x =
d/24 = d/16 + d/24 + d/x =
d/24 = 3d/48 + 2d/24 + d/48


Not sure what you are trying to do here.
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

A bit messy.
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New post 23 Mar 2007, 18:00
BG wrote:
"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
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Re: A cyclist rides his bicycle over a route which is 1/3 uphill  [#permalink]

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New post 12 Aug 2013, 03:41
Hi Guys,

when distance intervals are equal then, the average speed is given by the harmonic mean of the speeds.

thus,

Avg Speed=24= 3/[1/16+1/24+1/x]
thus, x=48.


Thanks,

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

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

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New post 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)

24 = (48+48+48) / [(48/16) + (48/24) + (48/x)
24 = (144) / [2+3+(48/x)]
24 = (144) / [5 + (48/x)]
120 + (1152/x) = 144
120x + 1152 = 144x
1152 = 24x
x = 48

The speed he would have to travel at for the third leg of the journey (48/x) is 48 miles/hour.

ANSWER: (E) 48
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Re: A cyclist rides his bicycle over a route which is 1/3 uphill  [#permalink]

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New post 31 Aug 2015, 21:50
Harmonic mean formula is good for equal subdistances to find mean rate:

3/(1/16+1/24+1/x)=24

can calculate but we can see that in this case denominator should equal 1/8 to get 24. The only option is x=48

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

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New post 18 Oct 2015, 04:54
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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  [#permalink]

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New post 14 Jul 2017, 02:44
the clue to this question is that you need total distance and total time..let say the total distance is 36 and each uphill, level and downhill cover an equal distance of 12 miles
time taken for uphill = distance/speed = 12/16 = 3/4 T (uphill)
time taken for level = 12/24 = 1/2 (level)
time taken for downhill = 12/x as we dont know its speed
Average speed of entire distance is 24
total time taken is 3/4 + 1/2 + 12/x
average speed = total distance /total time
24 = 36/(3/4 +1/2 +12/x)
solve this and you will get 48 as your answer
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Re: A cyclist rides his bicycle over a route which is 1/3 uphill  [#permalink]

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New post 14 Jul 2017, 06:01
Let the total distance travelled be \(D\) miles, and \(S\) be the rate at which he covered the downhill.

\(\frac{D}{3*16} + \frac{D}{3*24} + \frac{D}{3*S} = \frac{D}{24}\)

\(\frac{D}{48} + \frac{D}{72} + \frac{D}{3S} = \frac{D}{24}\)

\(\frac{D}{3} (\frac{1}{16} + \frac{1}{24} + \frac{1}{S}) = \frac{D}{3}(\frac{1}{8})\)

\(\frac{1}{16} + \frac{1}{24} + \frac{1}{S} = \frac{1}{8}\)

\(\frac{1}{S} = \frac{1}{8} - (\frac{5}{48}) = \frac{6}{48} - \frac{5}{48} = \frac{1}{48}\)

So, \(S = 48\). Ans - E.
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Re: A cyclist rides his bicycle over a route which is 1/3 uphill   [#permalink] 14 Jul 2017, 06:01

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