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Bicycle rate problem

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Bicycle rate problem

by sdotcruz » Tue Nov 23, 2010 5:43 am
I was using the "beat the Gmat practice" problem question bank and came across this problem:

It took Ellen 6 hours to ride her bike a total distance of 120 miles. For the first part of the trip, her speed was constantly 25 miles per hour. For the second part of her trip, her speed was constantly 15 miles per hour. For how many miles did Ellen travel at 25 miles per hour?

a. 60
b. 62
c. 66 2/3
d. 75
e. 90

OA:D

My question is not how the correct answer was calculated, but if my alternate method is valid or dumb luck? The test bank official explanation demonstrated the algebraic approach. Where D was used for the unknown distance traveled at 25 mph and d-120 was used for distance traveled at 15 mph. These values were plugged into D/R = T equation.

Since i was timing myself i did not want to spend too much time on this problem using algebra. Instead I calculated the average speed of the rider (total distance/ total time ;120mi / 6 h = 20 mph). Then i used the average balancing technique to calculate how many hours were required for 15 mph to balance 25 mph. After drawing the average balancing figure, I noticed that 25 mph was +5 above the average and 15 mph was -5 below the average. I knew that the time traveled at the different speeds had to be equal and deduced 3 hrs was spent at both since the total hours had to add to 6. Then I quickly multiplied 25mph * 3hrs = 75. I have not worked out my approach algebraically yet, but i will later on. Was my approach valid? It seems like alot of work when i write it out, but i only spent 1:45 mins on the problem and was able to do most of the work in my head. I wrote down a few notes and drew the average balance figure. What does everyone think?
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by Geva@EconomistGMAT » Tue Nov 23, 2010 6:08 am
Your approach works, and relies on the following general principle:

In weighted average, the "split ratio" is equal to the weights ratio.

The split ratio is the way the range between the two groups is split by the final average. In your question the two groups were 15 mph and 25 mph, each with a weight composed of the time Ellen drove at each speed. The final average was 20, splitting the range of 15-25 at a 1:1 ratio - straight down the middle. According to the principle above, the weight ratio will also be 1:1 - meaning that ellen drove the same number of hours at each speed.

Note that in average speed problems, the weight is measured in hours, not miles - the time ratio is 1:1, but the distance is different, as 3 hours at 15 mph will cover 45 miles, while the same time at 25 mph covers 75 miles. Just remember that in average distance questions (which is weighted average question, when you come down to it), the time units are the weight of each group, and then use the principle above.

Let's mix things up a bit: for the same problem, the average speed is still 20 mph, but the speeds are 10 mph and 25 mph. What's the distance traveled at 25 mph?

[spoiler]Answer: 100 miles.
split ratio: average of 20 splits the 10-25 range at 1:2 ratio. Therefore -
weight ratio (time ratio) is also 1:2, so the 6 hours now split at 2 hours to 4 hours.
One last consideration - which is the 2 and which is the 4? since the average (20 mph) is closer to 25 mph, Ellen must have spent more time at the faster 25 mph speed (it has a bigger weight, so it pulls the average towards it). Thus, Ellen drove 4 hours at 25 mph (100 miles total) and 2 hours at the slower speed of 10 mph (20 miles total). [/spoiler]
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by Night reader » Tue Nov 23, 2010 6:24 am
sdotcruz wrote:I was using the "beat the Gmat practice" problem question bank and came across this problem:

It took Ellen 6 hours to ride her bike a total distance of 120 miles. For the first part of the trip, her speed was constantly 25 miles per hour. For the second part of her trip, her speed was constantly 15 miles per hour. For how many miles did Ellen travel at 25 miles per hour?

a. 60
b. 62
c. 66 2/3
d. 75
e. 90

OA:D

My question is not how the correct answer was calculated, but if my alternate method is valid or dumb luck? The test bank official explanation demonstrated the algebraic approach. Where D was used for the unknown distance traveled at 25 mph and d-120 was used for distance traveled at 15 mph. These values were plugged into D/R = T equation.

Since i was timing myself i did not want to spend too much time on this problem using algebra. Instead I calculated the average speed of the rider (total distance/ total time ;120mi / 6 h = 20 mph). Then i used the average balancing technique to calculate how many hours were required for 15 mph to balance 25 mph. After drawing the average balancing figure, I noticed that 25 mph was +5 above the average and 15 mph was -5 below the average. I knew that the time traveled at the different speeds had to be equal and deduced 3 hrs was spent at both since the total hours had to add to 6. Then I quickly multiplied 25mph * 3hrs = 75. I have not worked out my approach algebraically yet, but i will later on. Was my approach valid? It seems like alot of work when i write it out, but i only spent 1:45 mins on the problem and was able to do most of the work in my head. I wrote down a few notes and drew the average balance figure. What does everyone think?
sdoctruz, this problem assigns coefficients for time (1) and (2) with the speeds (1) and (2). The time variables are dependent on the speed values. Time varies with the speeds, if we have less speed, then time is much; if the speed is higher, the traveling time is less.

Your calculation is cumbersome, because you find the average speed, then calculate weights per each speed and assign these new values to time (1) and (2) variables. This seems a long way.

While you read the problem, try to note its wording in math.

We have a total distance (S), distance (S1) and distance (S2); S=S1+S2
clear with traveled distance, move onto speeds while you read--
We got V(1) for one speed and V(2) for the other speed, there should be time too (T1) and (T2)

Now grid these in 5 seconds

T(1) + T(2) = 6
? ?
V(1) + V(2) = n/a
25 __ 15

S=120

build the obvious equations:
25*T(1)+15*T(2)=120
T(1)+T(2)=6, T(1)=6-T(2); 25*(6-T[2]) + 15*T(2) = 120 because>>>> S=S1+S2, S1=T(1)*V(1) and S(2)=T(2)*V(2) it easy to see these from our grid, therefore sketch some grid on a note pad - I do this always when I read and prepare to solve a problem.

So, 150-T(1)=120, T(2)=3, we need T(1) and it's 6-3 or 3 hours, S(1)=3*25=75

This should take you maximum 2 minutes 40 seconds. Timing in quant is saved on DS questions, about 1.5 minutes per DS question and expended on word problems. I usually allow maximum 2 minutes and 40 seconds per non-DS question.
My knowledge frontiers came to evolve the GMATPill's methods - the credited study means to boost the Verbal competence. I really like their videos, especially for RC, CR and SC. You do check their study methods at https://www.gmatpill.com
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by rishab1988 » Tue Nov 23, 2010 6:51 am
Here is how I approached the problem:

Let the first part of the trip @ 25mph be x.
Then the time taken to travel this distance (t1) =distance/speed= x/25.

Then the second part of the trip @ 15 mph is 120-x
The time taken to travel this distance (t2) = distance/speed=(120-x)/15

We are given total time =6 hrs

t1+t2=6
x/25 + (120-x)/15 =6
3x+600-5x = 450 or 3x + 6*100 -5x = 75*6 [lcm of denominators 25 and 15 is 75]
-2x=-150 or -2x = 6 (75-100)
x=75 or x = 3*25=75


You can use either brute multiplication (if you are good at it) or cute factoring (if you find multiplication difficult).
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by GMATGuruNY » Tue Nov 23, 2010 8:41 am
sdotcruz wrote:I was using the "beat the Gmat practice" problem question bank and came across this problem:

It took Ellen 6 hours to ride her bike a total distance of 120 miles. For the first part of the trip, her speed was constantly 25 miles per hour. For the second part of her trip, her speed was constantly 15 miles per hour. For how many miles did Ellen travel at 25 miles per hour?

a. 60
b. 62
c. 66 2/3
d. 75
e. 90

OA:D
Always look at the answer choices! Since the total time was an integer, the distance travelled at 25 mph is likely to be a multiple of 25. Only answer choice D is a multiple of 25. Let's try it:

Answer choice D: 75 miles at 25mph, so 120-75 = 45 miles at 15mph
Time spent at 25mph = 75/25 = 3.
Time spent at 15mph = 45/15 = 3.
Total time = 3+3 = 6.

The correct answer is D. No algebra and very little arithmetic needed.
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