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05 Aug 2026, 12:16
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At a food bank, volunteers packed delivery boxes in only two sizes: standard boxes and family boxes. Each standard box weighed 6 kilograms, and each family box weighed \(w\) kilograms. If the average (arithmetic mean) weight of all the delivery boxes was 12 kilograms, what is the value of \(w\)?
(1) The number of family boxes was 3 times the number of standard boxes.
(2) There were 12 more family boxes than standard boxes.
(1) The number of family boxes was 3 times the number of standard boxes.
(2) There were 12 more family boxes than standard boxes.
05 Aug 2026, 12:16
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Official Solution:
At a food bank, volunteers packed delivery boxes in only two sizes: standard boxes and family boxes. Each standard box weighed 6 kilograms, and each family box weighed \(w\) kilograms. If the average (arithmetic mean) weight of all the delivery boxes was 12 kilograms, what is the value of \(w\)?
Let \(S\) be the number of standard boxes and \(F\) be the number of family boxes.
The average weight is 12 kilograms, so:
\(\frac{6S + wF}{S + F} = 12\)
\(6S + wF = 12S + 12F\)
\(wF = 6S + 12F\)
\(w = 12 + \frac{6S}{F}\)
So, to find \(w\), we need the ratio \(\frac{S}{F}\).
(1) The number of family boxes was 3 times the number of standard boxes.
This tells us that \(F = 3S\), so \(\frac{S}{F} = \frac{1}{3}\).
Thus:
\(w = 12 + 6 * \frac{1}{3}\)
\(w = 14\)
Sufficient.
(2) There were 12 more family boxes than standard boxes.
This tells us that \(F = S + 12\). This gives the difference between the two numbers of boxes, but not the ratio \(\frac{S}{F}\).
For example, if \(S = 12\) and \(F = 24\), then:
\(w = 12 + 6 * \frac{12}{24} = 15\)
But if \(S = 24\) and \(F = 36\), then:
\(w = 12 + 6 * \frac{24}{36} = 16\)
So \(w\) can have different values.
Not sufficient.
Answer: A
At a food bank, volunteers packed delivery boxes in only two sizes: standard boxes and family boxes. Each standard box weighed 6 kilograms, and each family box weighed \(w\) kilograms. If the average (arithmetic mean) weight of all the delivery boxes was 12 kilograms, what is the value of \(w\)?
Let \(S\) be the number of standard boxes and \(F\) be the number of family boxes.
The average weight is 12 kilograms, so:
\(\frac{6S + wF}{S + F} = 12\)
\(6S + wF = 12S + 12F\)
\(wF = 6S + 12F\)
\(w = 12 + \frac{6S}{F}\)
So, to find \(w\), we need the ratio \(\frac{S}{F}\).
(1) The number of family boxes was 3 times the number of standard boxes.
This tells us that \(F = 3S\), so \(\frac{S}{F} = \frac{1}{3}\).
Thus:
\(w = 12 + 6 * \frac{1}{3}\)
\(w = 14\)
Sufficient.
(2) There were 12 more family boxes than standard boxes.
This tells us that \(F = S + 12\). This gives the difference between the two numbers of boxes, but not the ratio \(\frac{S}{F}\).
For example, if \(S = 12\) and \(F = 24\), then:
\(w = 12 + 6 * \frac{12}{24} = 15\)
But if \(S = 24\) and \(F = 36\), then:
\(w = 12 + 6 * \frac{24}{36} = 16\)
So \(w\) can have different values.
Not sufficient.
Answer: A
