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Weekly Math Quest - Nov19th,2006

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gmat_enthus Senior | Next Rank: 100 Posts Default Avatar
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Weekly Math Quest - Nov19th,2006

Post Tue Dec 05, 2006 3:46 am
Q. A coin is tossed from the top of the cliff of height c, at time t = 0 secs. The coin follows the parabolic path h=at^2+bt+c, where h is the height from the ground at time t . What is the height of the stone at t = 5 sec?

1) 0.25a+0.5b+c=24
2) 0.04a+0.008b+0.0016c=1.5

The answer will be posted along with the solution on Sunday, Dcember 10th, 2006. Till then you can disucss your solutions here.

reference: www.scorechase.com




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Post Sat Jan 20, 2007 4:53 pm
Technically, the question should state that the coin is tossed UPWARD from the top of the cliff.

h = at^2 + bt + C
find h for t = 5

rephrase: h = 25a + 5b + c

(1) 24 = 0.25a + 0.5b + c This is a different equation from the rephrased equation above, since the coeffient for c is 1 in both equations, but the coefficients for b are 5 and 0.5 respectively. Since there are 3 variables and only two different questions, we can't solve.

(2) 1.5 = 0.04a + 0.008b + 0.0016c This is the SAME equation as the rephrased equation above. The coefficients for c are 1 and 0.0016 respectively. .0016/.0016 = 1. Divide the other coefficients in this equation by .0016 also to see what happens.
.008/.0016 = 5
.04/.0016 = 25
So if I divide the entire equation given by statement 2 by .0016, I get:
937.5 = 25a + 5b + c. When I compare with my rephrased equation, I see that h must be 937.5.

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abhizin Newbie | Next Rank: 10 Posts Default Avatar
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Post Fri Jan 19, 2007 2:45 pm
Muliply option B with 625 and it becomes

25a+5b+c= 937.5

which is what is supposed to be calculated Smile

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gmat_enthus Senior | Next Rank: 100 Posts Default Avatar
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Post Tue Dec 12, 2006 5:03 am
The OA is B.

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tenpercenter76 Newbie | Next Rank: 10 Posts Default Avatar
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Post Wed Dec 13, 2006 8:04 pm
could someone please post an explanation to this? i'm not seeing it.

thanks!

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