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07 May 2026, 19:00
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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 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.
The OA will be automatically revealed on Monday 11th of May 2026 07:00:44 PM Pacific Time Zone
(1) The number of family boxes was 3 times the number of standard boxes.
(2) There were 12 more family boxes than standard boxes.
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The OA will be automatically revealed on Monday 11th of May 2026 07:00:44 PM Pacific Time Zone
10 May 2026, 04:15
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GMAT Club 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 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:
(6S + wF)/(S + F) = 12
6S + wF = 12S + 12F
wF = 6S + 12F
w = 12 + 6S/F
So, to find w, we need the ratio S/F.
(1) The number of family boxes was 3 times the number of standard boxes.
This tells us that F = 3S, so S/F = 1/3.
Thus:
w = 12 + 6 * 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 S/F.
For example, if S = 12 and F = 24, then:
w = 12 + 6 * 12/24 = 15
But if S = 24 and F = 36, then:
w = 12 + 6 * 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 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:
(6S + wF)/(S + F) = 12
6S + wF = 12S + 12F
wF = 6S + 12F
w = 12 + 6S/F
So, to find w, we need the ratio S/F.
(1) The number of family boxes was 3 times the number of standard boxes.
This tells us that F = 3S, so S/F = 1/3.
Thus:
w = 12 + 6 * 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 S/F.
For example, if S = 12 and F = 24, then:
w = 12 + 6 * 12/24 = 15
But if S = 24 and F = 36, then:
w = 12 + 6 * 24/36 = 16
So w can have different values.
Not sufficient.
Answer: A.
General Discussion
07 May 2026, 20:38
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Always simplify DS question before solving:
Let standard boxes = s and family boxes = f.
Average weight equation:
(6s + wf) / (s + f) = 12
Simplify:
6s + wf = 12s + 12f
wf = 6s + 12f
w = 6s/f + 12
Statement (1): f = 3s
w = 6(s/3s) + 12
w = 2 + 12 = 14
Unique value → sufficient.
Statement (2): f = s + 12
w = 6s/(s+12) + 12
Depends on s, so not unique → insufficient.
Answer: A
Let standard boxes = s and family boxes = f.
Average weight equation:
(6s + wf) / (s + f) = 12
Simplify:
6s + wf = 12s + 12f
wf = 6s + 12f
w = 6s/f + 12
Statement (1): f = 3s
w = 6(s/3s) + 12
w = 2 + 12 = 14
Unique value → sufficient.
Statement (2): f = s + 12
w = 6s/(s+12) + 12
Depends on s, so not unique → insufficient.
Answer: A
