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For a bake sale, Simon baked 2n more pies than Theresa. Theresa baked

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Bunuel
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For a bake sale, Simon baked 2n more pies than Theresa. Theresa baked [#permalink] New post   04 Mar 2022, 02:30
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For a bake sale, Simon baked 2n more pies than Theresa. Theresa baked half as many pies as Roger, who baked \(\frac{1}{3}n\) pies. No other pies were baked for the sale. What fraction of the total pies for sale did Roger bake?


A. \(\frac{1}{16}\)

B. \(\frac{1}{8}\)

C. \(\frac{3}{16}\)

D. \(\frac{3}{8}\)

E. \(\frac{13}{16}\)




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kungfury42
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Re: For a bake sale, Simon baked 2n more pies than Theresa. Theresa baked [#permalink] New post   04 Mar 2022, 03:52
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Roger bakes = n/3
Theresa bakes = (1/2)*(n/3) = n/6
Simon bakes = n/6 + 2n = 13n/6

Total bakes = n/3 + n/6 + 13n/6 = 16n/6 = 8n/3

Roger's share = (n/3)÷(8n/3) = 1/8

Hence, B.

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Aham56
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For a bake sale, Simon baked 2n more pies than Theresa. Theresa baked [#permalink] New post   04 Mar 2022, 16:00
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Pick a smart number, n = 30, Roger bakes 10 pies, Theresa bakes 15 pies and Simon 65 pies. Roger bakes 10 out of 80 pies so 1/8

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Re: For a bake sale, Simon baked 2n more pies than Theresa. Theresa baked [#permalink] New post   14 Sep 2023, 10:22
For a bake sale, Simon baked 2n more pies than Theresa. Theresa baked half as many pies as Roger, who baked (1/3)n pies. No other pies were baked for the sale. What fraction of the total pies for sale did Roger bake?

Let the no. of pies baked by Theresa = x
Therefore the no. of pies baked by Rogers = 2x = n/3 => n = 6x
Therefore the no. of pies baked by Simon = 2n+x = 13x

The fraction of the total pies for sale did Roger bake = 2x/(x+2x+13x) = 1/8 = 0.125

Hence B
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Re: For a bake sale, Simon baked 2n more pies than Theresa. Theresa baked [#permalink] New post   14 Sep 2023, 10:37
Bunuel wrote:
For a bake sale, Simon baked 2n more pies than Theresa. Theresa baked half as many pies as Roger, who baked \(\frac{1}{3}n\) pies. No other pies were baked for the sale. What fraction of the total pies for sale did Roger bake?


A. \(\frac{1}{16}\)

B. \(\frac{1}{8}\)

C. \(\frac{3}{16}\)

D. \(\frac{3}{8}\)

E. \(\frac{13}{16}\)
\(R = \frac{n}{3}\)

\(T = \frac{n}{3*2}=\frac{n}{6}\)

\(S = \frac{n}{6}+2n\)

Thus, \(R + S+ T = \frac{n}{3}+\frac{n}{6}+\frac{n}{6}+2n\)

Let \(n = 30\)

So, \(R + S+ T = \frac{30}{3}+\frac{30}{6}+\frac{30}{6}+2*30\)

Or, \(R + S+ T = 10 + 5 +5 + 60 = 80\) , where \(R = \frac{30}{3} = 10\)

Hence, \(\frac{R}{(R+S+T) }= \frac{10}{80} = \frac{1}{8}\), Answer must hence be (B)
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Re: For a bake sale, Simon baked 2n more pies than Theresa. Theresa baked [#permalink]
14 Sep 2023, 10:37
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