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Difficult Math Question #6

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Difficult Math Question #6

by 800guy » Thu Aug 31, 2006 6:19 pm
OA coming when a few people have answered...

A and B ran a race of 480 m. In the first heat, A gives B a head start of 48 m and beats him by 1/10th of a minute. In the second heat, A gives B a head start of 144 m and is beaten by 1/30th of a minute. What is B’s speed in m/s?
(A) 12
(B) 14
(C) 16
(D) 18
(E) 20
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Not good at these

by bensanderson » Fri Sep 01, 2006 7:32 am
I'm glad you put this question up because I'm really not very good at rate/time/distance questions especially when they involve different variables, times, etc.

My guess is that the answer is A, 12 m/s.

At first I was trying to figure out what runner A's time (T) was but I couldn't get it to work. But since I knew that runner B's time was T-2 for 336 meters and T+6 for 432 meters, I thought that the difference between the heats for runner B would be 8 seconds. I know this a long way of doing this but I checked the answers to find out which rate would result in a difference of 8 seconds for the two different times. That's how I got A. The 336m run would take B 28 seconds and the 432m run would take B 36 seconds for a difference of 8 seconds.

Is that correct? If not, please let me know. If it is correct, then is there a faster way to get the answer? Thanks for posting these questions.

cheers.
good luck!
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800guy

by marco » Sun Sep 03, 2006 1:53 am
First i do agree that the answer is 12

but the only way i can think of speeding it up is as follows

if for A B

S=? S= ?
T= t then T= t+6
D= d D= 9d/10

Now

S=? S= ?
T= t still but now T= t-2
D= d still D= 7d/10 ( so if d =480 as we know
then it is 7/10 x 480)

now you have 2 equations for b which is what we are concerned with, Ai is only relative in this question

so speed of B in firts heat will equal that in the second heat

s=d/t 9d/10(t+6) = 7d/10(t-2)

now solve for t which will = 30 sec

and substitute back in the equation giving you an answer of 12m/s

hope that helps
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Re: Difficult Math Question #6

by piren » Tue Sep 05, 2006 12:23 am
800guy wrote:OA coming when a few people have answered...

A and B ran a race of 480 m. In the first heat, A gives B a head start of 48 m and beats him by 1/10th of a minute. In the second heat, A gives B a head start of 144 m and is beaten by 1/30th of a minute. What is B’s speed in m/s?
(A) 12
(B) 14
(C) 16
(D) 18
(E) 20
Here's my solving method. Pay attention to units -- I suggest not to solve the 48-which is 10% of 480 and 144 (30%) until the very end.

First race:
A runs 480 m in t seconds
B runs 0.9 x 480 m in t+6 seconds

Second race:
A runs 480 m in t seconds
B runs 0.7 x 480 m in t-2 seconds

Speeds of runners are presupposed constant between the two races
which means that Sb1 = Sb2, where Sb1= speed of B in the first race and Sb2 = speed of B in the second race

Speed= Distance / Time, hence

Sb1 = 0.9 * 480 / (t+6)
Sb2 = 0.7 * 480 / (t-2)

Sb1 = Sb2 <=> 0.9*480 / (t+6) = 0.7*480 / (t-2)
<=> 0.9*480*t - 1.8*480 = 0.7*480*t + 4,2*480
<=> 0.9 t - 1.8 = 0.7 t + 4.2
<=> 0.2 t = 6 <=> t = 30

Plug in the value of t found in any of the formulas for Sb above: that leads us to

Sb1 = 0.9 * 480 / (36) = 12 m / s

Answer is A.
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by 800guy » Tue Sep 05, 2006 7:43 pm
OA:

Ans: race 1 :- ta = tb-6 ( because A beats B by 6 sec)
race 2 :- Ta = tb+2 ( because A looses to B by 2 sec)

By the formula D= S * T
we get two equations
480/Sa = 432/Sb -6 ------------1)
480/Sa = 336/Sb +2------------2)
Equating these two equations we get Sb = 12
ta,Sa stand for time taken by A and speed of A resp.
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by naveen1003 » Thu Jun 09, 2011 3:00 am
A and B ran a race of 480 m. In the first heat, A gives B a head start of 48 m and beats him by 1/10th of a minute. In the second heat, A gives B a head start of 144 m and is beaten by 1/30th of a minute. What is B's speed in m/s?
(A) 12
(B) 14
(C) 16
(D) 18
(E) 20
distance = 480m = d
B is given head start of 48 meters (which is 10% of 480) or we can say that B travels .9d
time taken by A = t sec and thus time taken by B = t+6 (=1/10 of a minute = 6 seconds)
B is given head start of 144 meters (30% of 480) or B travels .7d
time taken by A = t sec and time taken by B = t-2 (B beats A by 1/30 of a minute which is 2 seconds)
speed of B = distance/time = (.9d)/(t+6) = (.7d)/(t-2)
9t-18 = 7t + 42
=> t=30
Thus speed of B = .9d/(t+6) = .9*480/36 = [spoiler]12 m/sec[/spoiler]
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by GMATGuruNY » Thu Jun 09, 2011 5:44 am
800guy wrote:OA coming when a few people have answered...

A and B ran a race of 480 m. In the first heat, A gives B a head start of 48 m and beats him by 1/10th of a minute. In the second heat, A gives B a head start of 144 m and is beaten by 1/30th of a minute. What is B�s speed in m/s?
(A) 12
(B) 14
(C) 16
(D) 18
(E) 20
We can plug in the answers, which represent B's rate.
The correct answer must be a factor of 480.
Eliminate B and D.

In the first heat, A beats B by (1/10)*60 = 6 seconds.
In the second heat, B beats A by (1/30)*60 = 2 seconds.

Answer choice C: B = 16 meters per second.
First heat:
Time for B = d/r = 480/16 = 30 seconds.
Headstart given to A = 48/16 = 3 seconds.
Since A beats B by 6 seconds, total time for A = 30-(3+6) = 21 seconds.
21 is not a factor of 480.
Eliminate C.

Answer choice A: B = 12 meters per second.
First heat:
Time for B = d/r = 480/12 = 40 seconds.
Headstart given to A = 48/12 = 4 seconds.
Since A beats B by 6 seconds, total time for A = 40-(4+6) = 30 seconds.
Rate for A = d/t = 480/30 = 16 meters per second.

Second heat:
Headstart given to B = 144/12 = 12 seconds.
From the first heat: Time for B - Time for A = 40-30 = 10 seconds.
Since B is given a 12-second headstart, and the time for A is only 10 seconds less than the time for B, B beats A by 12-10 = 2 seconds.
Success!

The correct answer is A.
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by Vish_ » Thu Jun 09, 2011 6:52 am
IMO A
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by sunilramu » Thu Jun 09, 2011 4:46 pm
Mitch,

Can you please elaborate on how you arrived at the premise that speeds of both A and B should to be factors of 480.

Thanks
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by GMATGuruNY » Thu Jun 09, 2011 5:00 pm
sunilramu wrote:Mitch,

Can you please elaborate on how you arrived at the premise that speeds of both A and B should to be factors of 480.

Thanks
When the given times are converted into seconds, the results are integers:
1/10 of a minute = 6 seconds.
1/30 of a minute = 2 seconds.

Thus, the correct rate for B almost certainly will be a factor of the distance; otherwise, the resulting times will not be integer values.
Since the distance is 480, the correct answer should be a factor of 480.
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