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distance problem

by sudi760mba » Fri Mar 13, 2009 7:48 pm
For the past x laps around the track, Steven's average time per lap was 51 seconds. If a lap of 39 seconds would reduce his average time per lap to 49 seconds, what is the value of x?

(A)2
(B)5
(C)6
(D)10
(E)12

I could probably solve this by numbers but I wanted to try algebra:

R * T = D
__________

51secs/lap * x laps = 51x
39 secs /lap * x+1 laps = 39(x+1)

51x + 39(x+1)/(2x+1) = 49
x=-5/4

What am I doing wrong?

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by cramya » Fri Mar 13, 2009 7:54 pm
I am thinking

(x-1)51 + 39 = 49x

x = 6
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by sudi760mba » Fri Mar 13, 2009 8:04 pm
cramya wrote:I am thinking

(x-1)51 + 39 = 49x

x = 6
Actually the OA is B. In any case, could you explain your thinking process?

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by cramya » Fri Mar 13, 2009 8:11 pm
I must be missing something in interpretting the question or the OA must be wrong.


Here's how I see it.

The runner runs x number of laps each with a timing of 51 secs

Total time 51x

If he runs one lap for 39 sec instead of 51 then his total time becomes 49x.

(x-1)51 + 39*1(lap) = 49x

x=6



Plug in numbers 5 doesnt work whereas 6 does

x=6

Original time: 6*51 = 306

51*5 + 39*1 = 255+39 = 294

New time : 294/6 = 49


x=5

Original time: 5*51 = 255

51*4 + 39*1= 243

New time : 243/5 -> definitely not 49


I could be mistaken since my answer 6 and the OA seems have a difference of 1.... I still think x should be 6 since thats the toal number of laps ran.

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CR
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by sureshbala » Fri Mar 13, 2009 8:29 pm
In the final lap also had he averaged 51 seconds, the average of the x+1 laps would have been 51. Since he averaged 39 seconds instead of 51 seconds, i.e there is a decrease of 12 seconds and this decrease of 12 seconds resulted in a decrease of 2 seconds in the average. So 12 seconds will give an average of 2 seconds, if its divided among 6 laps. So x = 5.

To understand this concept more, follow the Averages link in my signature.
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by cramya » Fri Mar 13, 2009 8:32 pm
Sudi,
I did misinterpret the question. Missed the past x laps meaning x+1 is the total number of laps.

The algebric equation would be


51x + 39*1 = 49(x+1)


x=5

I stand corrected. Nice question; learnt from the mistake I made with the x part instead of x+1

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CR
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by anniev2 » Fri Mar 13, 2009 9:06 pm
In general, I believe this is a simple Averages problem and the rates were introduced to throw off the reader to use the d = rt formula which is not useful in solving this problem.

(x1 + x2)/2 = AVG rate

knowing that the Sum of Values = avg value x # of values we determine from the question stem that:

- x laps (x1) were done at 51 seconds
- 1 lap (x2) was done at 39 seconds and doing this lap reduced the average time to 49 seconds
- we need the total of x1 + total of x2; finding the total of x1 is the avg * sum of values (Initial average was 51 seconds at x laps so 51x)


(51x + 39)/2 = 49 seconds
x = 5

Edited to add:

For example, Average MPH = TOTAL MILES / TOTAL HOURS

In this case the question asks for: Average Seconds per Lap = TOTAL SECONDS / TOTAL LAPS
Last edited by anniev2 on Fri Mar 13, 2009 9:13 pm, edited 2 times in total.
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by cramya » Fri Mar 13, 2009 9:10 pm
Anniev,
Loosk like the answer is 5 and not 6.

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by anniev2 » Fri Mar 13, 2009 9:14 pm
cramya wrote:Anniev,
Loosk like the answer is 5 and not 6.

Regards,
CR
Thanks! I made that correction!
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Re: distance problem

by logitech » Sat Mar 14, 2009 8:32 am
The difference between the averages: 51-49 = 2 seconds

Lap Number x Difference = 2 * x is actually equal to what is added to the average pool, which is 49-39

So actually problem comes down to:

2x=10, and x=5
LGTCH
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