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06 Aug 2026, 06:14
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A bakery display case contains only croissants, muffins, and scones. If the ratio of croissants to muffins is \(4:9\) and the display case contains 120 pastries in total, how many more muffins than scones are in the display case?
(1) The ratio of croissants to scones is \(4:7\).
(2) The number of scones is 18 more than the number of croissants.
(1) The ratio of croissants to scones is \(4:7\).
(2) The number of scones is 18 more than the number of croissants.
06 Aug 2026, 06:14
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Official Solution:
A bakery display case contains only croissants, muffins, and scones. If the ratio of croissants to muffins is \(4:9\) and the display case contains 120 pastries in total, how many more muffins than scones are in the display case?
Let \(C\), \(M\), and \(S\) be the numbers of croissants, muffins, and scones, respectively.
Given:
\(C = 4x\) and \(M = 9x\)
Also:
\(C + M + S = 120\)
So:
\(4x + 9x + S = 120\)
\(13x + S = 120\)
The question asks for:
\(M - S\)
(1) The ratio of croissants to scones is \(4:7\).
Since \(C = 4x\), this means \(S = 7x\).
Then:
\(13x + 7x = 120\)
\(20x = 120\)
\(x = 6\)
So \(M\) and \(S\) are fixed, and \(M - S\) can be determined.
Sufficient.
(2) The number of scones is 18 more than the number of croissants.
So:
\(S = C + 18\)
Since \(C = 4x\):
\(S = 4x + 18\)
Substitute into \(13x + S = 120\):
\(13x + 4x + 18 = 120\)
\(17x = 102\)
\(x = 6\)
So \(M\) and \(S\) are fixed, and \(M - S\) can be determined.
Sufficient.
Answer: D
A bakery display case contains only croissants, muffins, and scones. If the ratio of croissants to muffins is \(4:9\) and the display case contains 120 pastries in total, how many more muffins than scones are in the display case?
Let \(C\), \(M\), and \(S\) be the numbers of croissants, muffins, and scones, respectively.
Given:
\(C = 4x\) and \(M = 9x\)
Also:
\(C + M + S = 120\)
So:
\(4x + 9x + S = 120\)
\(13x + S = 120\)
The question asks for:
\(M - S\)
(1) The ratio of croissants to scones is \(4:7\).
Since \(C = 4x\), this means \(S = 7x\).
Then:
\(13x + 7x = 120\)
\(20x = 120\)
\(x = 6\)
So \(M\) and \(S\) are fixed, and \(M - S\) can be determined.
Sufficient.
(2) The number of scones is 18 more than the number of croissants.
So:
\(S = C + 18\)
Since \(C = 4x\):
\(S = 4x + 18\)
Substitute into \(13x + S = 120\):
\(13x + 4x + 18 = 120\)
\(17x = 102\)
\(x = 6\)
So \(M\) and \(S\) are fixed, and \(M - S\) can be determined.
Sufficient.
Answer: D
