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Avg. Speed - Solve Algebraically

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Avg. Speed - Solve Algebraically

by MBALA2009 » Wed Jun 04, 2008 7:55 pm
Please solve the following algebraically:

2 dogsleds raced across a 300 mile course. 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

Answer: D
Last edited by MBALA2009 on Wed Jun 04, 2008 8:26 pm, edited 1 time in total.
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Re: Avg. Speed - Solve Algebraically

by Stuart@KaplanGMAT » Wed Jun 04, 2008 8:20 pm
MBALA2009 wrote:Please solve the following algebraically:

2 dogsleds raced across a 300 mile course. Team A finished the course in 3 fewer hours than did Team B. If Team A's average speed was 5 miles oer 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

Answer: D
Algebraic approach (although not sure why anyone would want to solve this algebraically):

Let time for team A = t
Let rate for team A = r

300 = rt
300 = (r-5)(t+3)

So:

rt = (r-5)(t+3)
rt = rt -5t + 3r - 15
5t = 3r - 15
t = (3/5)r - 3

Subbing into equation 1:

300 = r * ((3/5)r - 3)

300 = (3/5)r^2 - 3r
1500 = 3(r^2) - 15r
3(r^2) - 15r - 1500 = 0
r^2 - 5r - 500 = 0
(r-25)(r+20) = 0
r=25 or r=-20

Since we can't have a negative rate, r=25

HOWEVER - Step 4 of the Kaplan method for problem solving: double check the question. We just solved for r, which is A's rate; the question asks for B's rate, which is 5 mph slower. So, final step:

25-5 = 20: choose (D).

Now, the above method is nightmarishly long. This is a perfect example of a question on which backsolving is MUCH quicker, even if you're great at math.
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by MBALA2009 » Wed Jun 04, 2008 8:29 pm
Thanks Stuart. I solved it by backsolving but wanted to see how if could be solved algebraically for my own edification.
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by ksh » Wed Jun 04, 2008 11:01 pm
Hi Stuart,

I think you have established the equations in respect of A. It would be simpler if we establish in terms of B. Such as:

let avg speed of B=s
and time taken byB =t
300=s*t--- eqn1

then in respect of A
300=(s+5)*(t-3)

So, s*t=(s+5)*(t-3)
5t-3s-15=0
t=(3/5)*s+3

from eqn1
300=s*{(3/5)*s+3}
s^2+5s-500=0
s=20,-25
So speed of B is 20 mph.
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by Ian Stewart » Thu Jun 05, 2008 4:13 pm
No need for long algebra here. Use t = d/s. Let s be the speed of the slower sled. The times are 300/s and 300/(s+5) for teams B and A respectively. These times differ by 3 hours:

300/s - 300/(s+5) = 3
1/s - 1/(s+5) = 1/100

Lots of ways to finish this, including the conventional common denominator approach. I prefer number theory: from the answer choices, s is a whole number. From the above equation, it's clear s and s+5 are divisors of 100 (indeed 100 must be their LCM), so the only possible answer is s=20.
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by gmataspirant » Sat Jun 07, 2008 5:59 pm
I am asking a basic question
300 = (r-5)(t+3)
Time should be t-3 and Rate should be r+5 right ?

The problem says " 3 fewer hours than Team B" and Team's A speed 5 miles greater than "Team B".

I just trying to fix my basics ? Your help is appreciated.[/quote]
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by Ian Stewart » Sun Jun 08, 2008 5:42 am
gmataspirant wrote:I am asking a basic question
300 = (r-5)(t+3)
Time should be t-3 and Rate should be r+5 right ?

The problem says " 3 fewer hours than Team B" and Team's A speed 5 miles greater than "Team B".

I just trying to fix my basics ? Your help is appreciated.

It all depends on what t and r mean in your solution:

If t and r represent the time and speed for Team A, then the time and speed for Team B would be t+3 (B takes three hours longer than A) and r-5 (B travels 5 mph slower than Team A). Now, if you solve for t, that will tell you how long it takes A to finish the race, and r will represent the speed at which A traveled.

You could instead make t represent: 'the time it takes Team B to finish', and you could let r mean: 'the speed at which Team B travels'. Then A's time will be t-3, and A's speed with be r+5. Now, if you solve for t, that will tell you how long it takes B to finish the race. You'd need to subtract 3 to find how long it takes A to finish the race. Similarly, here r represents B's speed, and you'd need to add 5 to find A's speed.

That's a long way of saying that you're right, but so is the text you've quoted: it just depends on what the letters represent.
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