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13 Jun 2024, 01:30
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Difficulty:
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Question Stats:
83% (02:03) correct
17%
(02:48)
wrong
based on 121
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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. \(\frac{13}{16}\)
A. \(\frac{1}{16}\)
B. \(\frac{1}{8}\)
C. \(\frac{3}{16}\)
D. \(\frac{3}{8}\)
E. \(\frac{13}{16}\)
13 Jun 2024, 04:09
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Given - For a bake sale, Simon baked 2n more pies than Theresa. Theresa baked half as many pies as Roger, who baked n/3 pies. No other pies were baked for the sale.
From this lets form table
Roger Theresa Simon Total
n/3 n/6 n/6 + 2n 8n/3
So, fraction of the total pies for sale did Roger bake = (n/3) / (8n/3) = 1/8.
Hence B.
From this lets form table
Roger Theresa Simon Total
n/3 n/6 n/6 + 2n 8n/3
So, fraction of the total pies for sale did Roger bake = (n/3) / (8n/3) = 1/8.
Hence B.
Shwarma
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13 Jun 2024, 08:04
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Option B
R= 1/3 n
T= 1/6 n
S= 2n+(1/6)n
Total =2n+(2/3)n
Roger fraction = (1/3)n/(2n+(2/3)n) = 1/8n
R= 1/3 n
T= 1/6 n
S= 2n+(1/6)n
Total =2n+(2/3)n
Roger fraction = (1/3)n/(2n+(2/3)n) = 1/8n
