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29 Mar 2024, 05:22
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
39% (02:49) correct
61%
(03:23)
wrong
based on 95
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A certain café sells bagels for $1.10 and muffins for $1.50. Last Wednesday, muffins accounted for M percent of all revenue from baked goods, and b percent of the baked goods sold were bagels. If the café sold only muffins and bagels last Wednesday, which of the following expresses the value of M in terms of b ?
A. \(\frac{25b}{375-b}\)
B. \(\frac{300-3b}{200}\)
C. \(\frac{1500-15b}{1500-4b}\)
D. \(\frac{20b}{300-3b}\)
E. \(\frac{375(100-b)}{375-b}\)
A. \(\frac{25b}{375-b}\)
B. \(\frac{300-3b}{200}\)
C. \(\frac{1500-15b}{1500-4b}\)
D. \(\frac{20b}{300-3b}\)
E. \(\frac{375(100-b)}{375-b}\)
Updated on: 30 Mar 2024, 09:45
Originally posted by HarshaBujji on 29 Mar 2024, 22:32.
Last edited by HarshaBujji on 30 Mar 2024, 09:45, edited 1 time in total.
Last edited by HarshaBujji on 30 Mar 2024, 09:45, edited 1 time in total.
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devashish2407
There might be a traditiona way to slove this problem by determining the % & costs etc. But I feel this might be time consuming.
We know that m + b =100% as the café sold only muffins and bagels last Wednesday.
So let's assume that the no bagels are sold, means b=0;
Hence m = 100%.
Lets check the options, by substituting m=100, b=0; Only E full fills the equation. Hence E is the answer.
General Discussion
30 Mar 2024, 05:31
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devashish2407
Given: A certain café sells bagels for $1.10 and muffins for $1.50. Last Wednesday, muffins accounted for M percent of all revenue from baked goods, and b percent of the baked goods sold were bagels.
Asked: If the café sold only muffins and bagels last Wednesday, which of the following expresses the value of M in terms of b ?
Let the number of bagels sold be x & number of muffins sold be y.
Revenue from baked goods = 1.1x + 1.5y
Revenue from muffins = 1.5y
M% = 1.5y/(1.1x + 1.5y) * 100%
(1.1x + 1.5y)*M = 150y
1.1Mx + 1.5My = 150y
1.1Mx = (150-1.5M)y
\(\frac{x}{y} = \frac{(150-1.5M)}{1.1M} = \frac{15}{11} * \frac{(100-M)}{M}\)
Baked goods sold = x+ y
b% = x/(x+y) *100%
(x + y)b = 100x
(100 - b)x = by
\(\frac{x}{y} =\frac{ b}{(100-b)} = \frac{(150-1.5M)}{1.1M} = \frac{1.5}{1.1} *\frac{(100-M)}{M}= \frac{15}{11} * \frac{(100-M)}{M} \)
\(\frac{(100-M)}{M} = \frac{11b}{15(100-b)}\)
\(\frac{M}{100} = \frac{15(100-b)}{(1500-4b)}\)
\(M = \frac{1500(100-b)}{(1500-4b)} = \frac{375(100-b)}{(375-b)}\)
IMO E
Given: A certain café sells bagels for $1.10 and muffins for $1.50. Last Wednesday, muffins accounted for M percent of all revenue from baked goods, and b percent of the baked goods sold were bagels.
Asked: If the café sold only muffins and bagels last Wednesday, which of the following expresses the value of M in terms of b ?
Let the number of bagels sold be x & number of muffins sold be y.
Revenue from baked goods = 1.1x + 1.5y
Revenue from muffins = 1.5y
M% = 1.5y/(1.1x + 1.5y) * 100%
(1.1x + 1.5y)*M = 150y
1.1Mx + 1.5My = 150y
1.1Mx = (150-1.5M)y
\(\frac{x}{y} = \frac{(150-1.5M)}{1.1M} = \frac{15}{11} * \frac{(100-M)}{M}\)
Baked goods sold = x+ y
b% = x/(x+y) *100%
(x + y)b = 100x
(100 - b)x = by
\(\frac{x}{y} =\frac{ b}{(100-b)} = \frac{(150-1.5M)}{1.1M} = \frac{1.5}{1.1} *\frac{(100-M)}{M}= \frac{15}{11} * \frac{(100-M)}{M} \)
\(\frac{(100-M)}{M} = \frac{11b}{15(100-b)}\)
\(\frac{M}{100} = \frac{15(100-b)}{(1500-4b)}\)
\(M = \frac{1500(100-b)}{(1500-4b)} = \frac{375(100-b)}{(375-b)}\)
IMO E
