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In a field day at a school, each child who competed in n events and sc

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In a field day at a school, each child who competed in n events and sc [#permalink]

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In a field day at a school, each child who competed in \(n\) events and scored a total of \(p\) points was given an overall score of \(\frac{p}{n} + n\). Andrew competed in 1 event and scored 9 points. Jason competed in 3 events and scored 5, 6, and 7 points, respectively. What was the ratio of Andrew's overall score to Jason's overall score?


A. \(\frac{10}{23}\)

B. \(\frac{7}{10}\)

C. \(\frac{4}{5}\)

D. \(\frac{10}{9}\)

E. \(\frac{12}{7}\)
[Reveal] Spoiler: OA

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Last edited by Bunuel on 24 Jul 2017, 00:57, edited 1 time in total.
Edited the question.

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Re: In a field day at a school, each child who competed in n events and sc [#permalink]

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New post 23 Jul 2017, 18:01
carcass wrote:
In a field day at a school, each child who competed in n events and scored a total of p points was given an overall score of P/n + n. n Andrew competed in 1 event and scored 9 points. Jason competed in 3 events and scored 5, 6, and 7 points, respectively. What was the ratio of Andrew's overall score to Jason's overall score?

A. \(\frac{10}{23}\)

B.\(\frac{7}{10}\)

C. \(\frac{4}{5}\)

D.\(\frac{10}{9}\)

E. \(\frac{12}{7}\)


It should be D

For readers, denominator for P is just 1 n. The fraction then adds into so you can rewrite it as \(n + \frac{p}{n}\) for a clearer understanding.

Andrew:

\(1 + \frac{9}{1}\) \(= 1 + 9 = 10\)

Jason:
\(3 + \frac{(5+6+7)}{3}\) \(= 3 + \frac{18}{3} = 9\)

Andrew : Jason \(= 10 : 9\)
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Re: In a field day at a school, each child who competed in n events and sc [#permalink]

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New post 24 Jul 2017, 07:37
Answer D

Andrew: n=1, p=9 So: 9/1+1=10
Jason: n=3, p=5+6+7=18 So: 18/3+3=9
Ratio: 10/9

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Re: In a field day at a school, each child who competed in n events and sc [#permalink]

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New post 24 Jul 2017, 09:00
Andrew's overall score = \(\frac{9}{1} + 1 = 10\).

Jason's overall score \(= \frac{5+6+7}{3} + 3 = 6+3 = 9\).

Ratio = \(10:9\). Ans - D.
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Re: In a field day at a school, each child who competed in n events and sc [#permalink]

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New post 26 Jul 2017, 16:38
carcass wrote:
In a field day at a school, each child who competed in \(n\) events and scored a total of \(p\) points was given an overall score of \(\frac{p}{n} + n\). Andrew competed in 1 event and scored 9 points. Jason competed in 3 events and scored 5, 6, and 7 points, respectively. What was the ratio of Andrew's overall score to Jason's overall score?


A. \(\frac{10}{23}\)

B. \(\frac{7}{10}\)

C. \(\frac{4}{5}\)

D. \(\frac{10}{9}\)

E. \(\frac{12}{7}\)


Since Andrew competed in 1 event and received 9 points, his overall score was:

9/1 + 1 = 10

Since Jason completed in 3 events and scored 5, 6, and 7 points, he scored a total of 18 points and his overall score was:

18/3 + 3 = 9

Thus, the ratio of Andrew’s score to Jason’s score is 10/9.

Answer: D
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Re: In a field day at a school, each child who competed in n events and sc [#permalink]

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New post 05 Sep 2017, 06:00
Andrew / Jason = X = Ratio of andrews overall over Jason's overall

Andrew = P/n + n = 9/1 + 1 = 1 9/1 = 10/1 = 10 = Andrews overall score
Jason = 18/3 + 3 = 3 18/3 = 27/3 = 9 = Jason's overall score
X = 10/9

Therefore, the correct answer is (D) 10/9

Good warm up question. I wonder what level it is.

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Re: In a field day at a school, each child who competed in n events and sc   [#permalink] 05 Sep 2017, 06:00
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