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Two dogsled teams raced across a 300 mile course in Wyoming. Team A fi

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Two dogsled teams raced across a 300 mile course in Wyoming. Team A fi [#permalink] New post   18 Feb 2011, 14:57
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Two dogsled teams raced across a 300 mile course in Wyoming. Team A finished the course in 3 fewer hours than team B. If team A's average speed was 5 mph greater than team B's, what was team B's average mph?

A. 12
B. 15
C. 18
D. 20
E. 25
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Re: Two dogsled teams raced across a 300 mile course in Wyoming. Team A fi [#permalink] New post   18 Feb 2011, 15:54
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gc92677 wrote:
Two dogsled teams raced across a 300 mile course in Wyoming. Team A finished the course in 3 fewer hours than team B. If team A's average speed was 5 mph greater than team B's, what was team B's average mph?

A. 12
B. 15
C. 18
D. 20
E. 25


This is a very specific format that has appeared in a handful of real GMAT questions, and you may wish to learn to recognize it: here we have a *fixed* distance, and we are given the difference between the times and speeds of two things that have traveled that distance. This is one of the very small number of question formats where backsolving is typically easier than solving directly, since the direct approach normally produces a quadratic equation.

Say Team B's speed was s. Then Team B's time is 300/s.

Team A's speed was then s+5, and Team A's time was then 300/(s+5).

We need to find an answer choice for s so that the time of Team A is 3 less than the time of Team B. That is, we need an answer choice so that 300/(s+5) = (300/s) - 3. You can now immediately use number properties to zero in on promising answer choices: the times in these questions will always work out to be integers, and we need to divide 300 by s, and by s+5. So we want an answer choice s which is a factor of 300, and for which s+5 is also a factor of 300. So you can rule out answers A and C immediately, since s+5 won't be a divisor of 300 in those cases (sometimes using number properties you get to the correct answer without doing any other work, but unfortunately that's not the case here). Testing the other answer choices, if you try answer D, you find the time for Team B is 15 hours, and for Team A is 12 hours, and since these differ by 3, as desired, D is correct.
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Re: Two dogsled teams raced across a 300 mile course in Wyoming. Team A fi [#permalink] New post   01 Mar 2011, 22:58
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In fact, you don't need to go till factors of 300.
If the speed of B be s, then the equation becomes: 300/s - 300/s+5 = 3, thta is, 100/s - 100/s+5 = 1.
12, 15 and 18 are not factors of 100, 20 and 25 are, 30 isn't, that directly leaves us with one option, which is D. 20

gmat1220 wrote:
Friend ! Use the format of the exam against the exam. After all we are aspiring mbas :wink: See Ian's post.

A. 12 --> 12 + 5 = 17 OUT 17 is not a factor of 300
B. 15
C. 18 ---> 18 + 5 = 23 OUT 23 is not a factor of 300
D. 20
E. 25

Likely answer will be B or D.

B - 20 * 15 = 300 problem is 15 + 3 = 18 is not a factor of 300.
D - 25 * 12 = 300. 12 + 3 = 15 is a factor of 300. Bingo !
E - 30 * 10 = 300. problem is 10 + 3 = 13 is not a factor of 300.

beyondgmatscore wrote:
From your statement B, t = 300/s
Put this in A, (s+5)*(300/s-3)=300
or, (s+5)*(300-3s)=300s
0r, 300s + 1500 - 3s^2-15s = 300s
or, 3s^2+15s-1500 = 0
or, 3s^2+75s-60s-1500=0
0r, (3s-60)(3s+75)=0
or s = 20
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Re: Two dogsled teams raced across a 300 mile course in Wyoming. Team A fi [#permalink] New post   19 Feb 2011, 14:22
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From your statement B, t = 300/s
Put this in A, (s+5)*(300/s-3)=300
or, (s+5)*(300-3s)=300s
0r, 300s + 1500 - 3s^2-15s = 300s
or, 3s^2+15s-1500 = 0
or, 3s^2+75s-60s-1500=0
0r, (3s-60)(3s+75)=0
or s = 20
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Re: Two dogsled teams raced across a 300 mile course in Wyoming. Team A fi [#permalink] New post   01 Mar 2011, 19:55
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Friend ! Use the format of the exam against the exam. After all we are aspiring mbas :wink: See Ian's post.

A. 12 --> 12 + 5 = 17 OUT 17 is not a factor of 300
B. 15
C. 18 ---> 18 + 5 = 23 OUT 23 is not a factor of 300
D. 20
E. 25

Likely answer will be B or D.

B - 20 * 15 = 300 problem is 15 + 3 = 18 is not a factor of 300.
D - 25 * 12 = 300. 12 + 3 = 15 is a factor of 300. Bingo !
E - 30 * 10 = 300. problem is 10 + 3 = 13 is not a factor of 300.

beyondgmatscore wrote:
From your statement B, t = 300/s
Put this in A, (s+5)*(300/s-3)=300
or, (s+5)*(300-3s)=300s
0r, 300s + 1500 - 3s^2-15s = 300s
or, 3s^2+15s-1500 = 0
or, 3s^2+75s-60s-1500=0
0r, (3s-60)(3s+75)=0
or s = 20
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Re: Two dogsled teams raced across a 300 mile course in Wyoming. Team A fi [#permalink] New post   01 Mar 2011, 22:34
Nice work on this one. An example of a case where jumping head first into algebra is counterproductive.
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Re: Two dogsled teams raced across a 300 mile course in Wyoming. Team A fi [#permalink] New post   02 Mar 2011, 02:50
The plugging-in works only because the options are of a certain nature. What if all the options satisfied the condition involving being multiples, then we would end up doing the plugging in first and then come back to solve the algebraic equation. Just because plugging in works sometime doesn't mean it is always the optimal and fastest way. If the algebraic method just involves solving a simple quadratic equation, then it is GUARANTEED to give an answer in under one minute, whereas plugging-in may take less than that or greater than that depending on the kind of options that are there. So, as a rule, if I can quickly get the answer following a general algebraic principle, I would always prefer that over plugging in the numbers.





abmyers wrote:
Nice work on this one. An example of a case where jumping head first into algebra is counterproductive.
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Re: Two dogsled teams raced across a 300 mile course in Wyoming. Team A fi [#permalink] New post   02 Mar 2011, 09:04
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beyondgmatscore wrote:
The plugging-in works only because the options are of a certain nature. What if all the options satisfied the condition involving being multiples, then we would end up doing the plugging in first and then come back to solve the algebraic equation. Just because plugging in works sometime doesn't mean it is always the optimal and fastest way. If the algebraic method just involves solving a simple quadratic equation, then it is GUARANTEED to give an answer in under one minute, whereas plugging-in may take less than that or greater than that depending on the kind of options that are there. So, as a rule, if I can quickly get the answer following a general algebraic principle, I would always prefer that over plugging in the numbers.


If you look at this question:

difficult-ps-problem-help-100767.html

which is in essentially the same format as the one in the original post above, the algebra is considerably more time consuming than using the answer choices and divisibility properties. Using divisibility (we want to get an integer when we divide 90 by the right answer, and by the right answer plus 0.5) you can get the answer within a few seconds.

I've mentioned quite a few times that I think the usefulness of backsolving is vastly overstated in many prep books. For most GMAT questions, you end up solving the same problem several times, rather than once, when you backsolve, so it's typically a very inefficient approach. However, for questions in the very specific format of the one in this post, I find backsolving to be much faster than any direct algebraic approach, provided you use number theory to zero in on the correct answer. In most questions of this type, only one answer is even plausible when you apply divisibility principles, so you can often pick the right answer without doing any calculation. The algebra is never quite that fast.
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Re: Two dogsled teams raced across a 300 mile course in Wyoming. Team A fi [#permalink] New post   02 Mar 2011, 09:31
IanStewart wrote:
beyondgmatscore wrote:
The plugging-in works only because the options are of a certain nature. What if all the options satisfied the condition involving being multiples, then we would end up doing the plugging in first and then come back to solve the algebraic equation. Just because plugging in works sometime doesn't mean it is always the optimal and fastest way. If the algebraic method just involves solving a simple quadratic equation, then it is GUARANTEED to give an answer in under one minute, whereas plugging-in may take less than that or greater than that depending on the kind of options that are there. So, as a rule, if I can quickly get the answer following a general algebraic principle, I would always prefer that over plugging in the numbers.


If you look at this question:

difficult-ps-problem-help-100767.html

which is in essentially the same format as the one in the original post above, the algebra is considerably more time consuming than using the answer choices and divisibility properties. Using divisibility (we want to get an integer when we divide 90 by the right answer, and by the right answer plus 0.5) you can get the answer within a few seconds.

I've mentioned quite a few times that I think the usefulness of backsolving is vastly overstated in many prep books. For most GMAT questions, you end up solving the same problem several times, rather than once, when you backsolve, so it's typically a very inefficient approach. However, for questions in the very specific format of the one in this post, I find backsolving to be much faster than any direct algebraic approach, provided you use number theory to zero in on the correct answer. In most questions of this type, only one answer is even plausible when you apply divisibility principles, so you can often pick the right answer without doing any calculation. The algebra is never quite that fast.


Ian - I think both of us are saying the same thing - Having backsolving as your primary technique would lead to more situations where we would end up solving the problem multiple times. I am all for being smart in reducing the steps in algebra, but a little hesitant on relying on backsolving as a primary method. for.e.g. in the above problem that you linked to, if my answer choices were such that 2.5 was at the end and if 4.5 was listed as one of the choices (which also gives integer when 90 is divided by it and 4.5+0.5 = 5.0 also yields an integer) then we would need to resort to algebra anyway AFTER spending considerable time in plugging in - and we have no way of knowing before we start plugging in that divisibility condition would be met by just one of the choices.

As per algebra there, we quickly get the equation as v^2-9 = 1080 or (v+3)*(v-3) = 36*30 and hence v = 33. Personally, I find this more intuitive and predictable way of solving it then to try and check various choices.

But then again, one should stick to what works best for them - its just that I prefer a more predictable method than the one where examiner has the ability to fox me by giving options that rule out back solving or plugging in and discover the same after spending time trying to back solve.
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Re: Two dogsled teams raced across a 300 mile course in Wyoming. Team A fi [#permalink] New post   02 Mar 2011, 09:39
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IanStewart wrote:
beyondgmatscore wrote:
The plugging-in works only because the options are of a certain nature. What if all the options satisfied the condition involving being multiples, then we would end up doing the plugging in first and then come back to solve the algebraic equation. Just because plugging in works sometime doesn't mean it is always the optimal and fastest way. If the algebraic method just involves solving a simple quadratic equation, then it is GUARANTEED to give an answer in under one minute, whereas plugging-in may take less than that or greater than that depending on the kind of options that are there. So, as a rule, if I can quickly get the answer following a general algebraic principle, I would always prefer that over plugging in the numbers.


If you look at this question:

difficult-ps-problem-help-100767.html

which is in essentially the same format as the one in the original post above, the algebra is considerably more time consuming than using the answer choices and divisibility properties. Using divisibility (we want to get an integer when we divide 90 by the right answer, and by the right answer plus 0.5) you can get the answer within a few seconds.

I've mentioned quite a few times that I think the usefulness of backsolving is vastly overstated in many prep books. For most GMAT questions, you end up solving the same problem several times, rather than once, when you backsolve, so it's typically a very inefficient approach. However, for questions in the very specific format of the one in this post, I find backsolving to be much faster than any direct algebraic approach, provided you use number theory to zero in on the correct answer. In most questions of this type, only one answer is even plausible when you apply divisibility principles, so you can often pick the right answer without doing any calculation. The algebra is never quite that fast.


Similarly, the algebra for current question in this thread is 300/s - 300/(s+5) = 3 or 100/s - 100/(s+5) = 1 or s*(s+5) = 500
So we need to factorize 500 in two factors that are 5 apart, so 500 = 20*25 and s = 20 , its really not that tough and probably more predictable :)
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Two dogsled teams raced across a 300 mile course in Wyoming. Team A fi [#permalink] New post   Updated on: 30 Dec 2020, 06:54
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\(\frac{300}{s}-\frac{300}{s+5}=3\)

\(=\frac{(300s+1500-300s)}{s(s+5)}=3\)

\(=s^2+5s-500=0\)

\(=s^2+25s-20s-500=0\)

\(s=20\)

Ans. D

Originally posted by MHIKER on 25 Mar 2011, 11:04.
Last edited by MHIKER on 30 Dec 2020, 06:54, edited 1 time in total.
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Re: Two dogsled teams raced across a 300 mile course in Wyoming. Team A fi [#permalink] New post   Updated on: 21 Jul 2018, 20:35
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Two dogsled teams raced across a 300 mile course in Wyoming. Team A finished the course in 3 fewer hours than team B. If team A's average speed was 5 mph greater than team B's, what was team B's average mph?

A. 12
B. 15
C. 18
D. 20
E. 25

let t=team A's time
(300/t)-5=300/(t+3)
t=12 hours
t+3=15 hours for team B
300/15=20 average mph for team B
20
D

Originally posted by gracie on 26 Nov 2015, 16:54.
Last edited by gracie on 21 Jul 2018, 20:35, edited 1 time in total.
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Re: Two dogsled teams raced across a 300 mile course in Wyoming. Team A fi [#permalink] New post   29 Aug 2016, 10:50
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let t=B's time
300/(t-3)-300/t=5
t=15
300/15=20 mph
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Re: Two dogsled teams raced across a 300 mile course in Wyoming. Team A fi [#permalink] New post   29 Aug 2016, 13:24
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Bunuel wrote:
Two dogsled teams raced across a 300-mile course in Wyoming. Team A finished the course in 3 fewer hours than did team B. If team A’s average speed was 5 miles per hour greater than that of team B, what was team B’s average speed, in miles per hour?

A. 12
B. 15
C. 18
D. 20
E. 25



300/T- 300/T+3 = 5

300( 1/T - 1/T+3) = 5

Can form the quadratic T^2 +3T -180 = 0

or can go for option choice substitution fro here.

Answer D
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Re: Two dogsled teams raced across a 300 mile course in Wyoming. Team A fi [#permalink] New post   29 May 2017, 02:22
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gc92677 wrote:
Two dogsled teams raced across a 300 mile course in Wyoming. Team A finished the course in 3 fewer hours than team B. If team A's average speed was 5 mph greater than team B's, what was team B's average mph?

A. 12
B. 15
C. 18
D. 20
E. 25

1. Let s1 be speed of A and s2 be speed of B.
2. s1 =300/t1 and
3. s2=300/t2 or
s1-5 = 300/(t1+3)
4. Solving the equations we get t1=12 and so s2=20
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Re: Two dogsled teams raced across a 300 mile course in Wyoming. Team A fi [#permalink] New post   20 Jan 2019, 16:54
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gc92677 wrote:
Two dogsled teams raced across a 300 mile course in Wyoming. Team A finished the course in 3 fewer hours than team B. If team A's average speed was 5 mph greater than team B's, what was team B's average mph?

A. 12
B. 15
C. 18
D. 20
E. 25


Here's my reasoning if it helps anyone:

Distance is the same here so all we need to do is set A = B

Let B's time be represented by variable t and speed by variable s.

Let A's time be t-3 and speed be s+5

st = (t-3)(s+5)
\(t = \frac{(3s+15)}{5}\)

s*t = 300 hence:

\(s * \frac{(3s+15)}{5}\) = 300

(s+25)(s-20) = 0

s = 20
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Two dogsled teams raced across a 300 mile course in Wyoming. Team A fi [#permalink] New post   08 Mar 2020, 20:36
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gc92677 wrote:
Two dogsled teams raced across a 300 mile course in Wyoming. Team A finished the course in 3 fewer hours than team B. If team A's average speed was 5 mph greater than team B's, what was team B's average mph?

A. 12
B. 15
C. 18
D. 20
E. 25



300/x - 300/(x+5) = 3

(300x + 1500 - 300x)/(x^2 + 5x) = 3
500 = x^2 + 5x

Method 1:
x(x+5) = 500
Working backwards: 20 * 25 = 500 => x = 20 => D

Method 2:
x^2 + 5x - 50 = 0
(x + 25) (x - 20) = 0
x = 20 => D
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Re: Two dogsled teams raced across a 300 mile course in Wyoming. Team A fi [#permalink] New post   30 Dec 2020, 07:57
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gc92677 wrote:
Two dogsled teams raced across a 300 mile course in Wyoming. Team A finished the course in 3 fewer hours than team B. If team A's average speed was 5 mph greater than team B's, what was team B's average mph?

A. 12
B. 15
C. 18
D. 20
E. 25


Some aspirants solved it as t+3 but it should be t+5 because A's speed is 5 miles per hour more than B. So if B's speed is t miles per hour then A's speed will be t+5 miles per hour. The equation will be:

\(\frac{300}{t}-\frac{300}{t+5}=3\)

We can avoid cumbersome calculations with back solving from answer choices.


A. If B =12, A=12+5=17; Not possible, because 300/17 will be a fraction and the time of B-A will not be 3. Eliminated

B. If B =15, A=20; Not possible, because the time of B-A (20-15=5) is not 3. Eliminated

C. If B =18, A= 18+5=23; Not possible, because the result will be a fraction and the time of B-A will not be 3. Eliminated

D. If B =20, A= 25; B will take 15 hours and A will take 12 hours; B- (15-12) 3 hours.

The answer is D.
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Two dogsled teams raced across a 300 mile course in Wyoming. Team A fi [#permalink] New post   19 Sep 2022, 10:45
gc92677 wrote:
Two dogsled teams raced across a 300 mile course in Wyoming. Team A finished the course in 3 fewer hours than team B. If team A's average speed was 5 mph greater than team B's, what was team B's average mph?

A. 12
B. 15
C. 18
D. 20
E. 25


Time taken by A = \(\frac{300}{A}\)
Time Take By B = \(\frac{300}{B}\)
so \(\frac{300}{A}\) + 3 = \(\frac{300}{B}\)
And A = B + 5
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Re: Two dogsled teams raced across a 300 mile course in Wyoming. Team A fi [#permalink] New post   18 Nov 2023, 08:41
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