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E
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Difficulty:
55%
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Question Stats:
64% (02:01) correct
36%
(02:21)
wrong
based on 151
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The average (arithmetic mean) of four packages is 4 pounds. If the weight of the first package is x, and each package weighs two more pounds than the previous package, which of the following CANNOT be the total weight, in pounds, of any combination of the packages?
A. 9
B. 10
C. 12
D. 13
E. 14
A. 9
B. 10
C. 12
D. 13
E. 14
GraemeGmatPanda
Joined: 13 Apr 2020
Last visit: 21 Apr 2026
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04 Sep 2025, 03:45
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We represent the weights of the four packages, which increase by 2 pounds each time, as consecutive even numbers starting from x
\(x, x+2, x+4, x+6\)
We translate "The average (arithmetic mean) of four packages is 4 pounds" using the formula for average
\(\frac{x + (x+2) + (x+4) + (x+6)}{4} = 4\)
We multiply both sides by 4 to find the sum of the weights
\(x + (x+2) + (x+4) + (x+6) = 16\)
We combine like terms to simplify the equation
\(4x + 12 = 16\)
We isolate x by subtracting 12 from both sides and then dividing by 4
\(4x = 4\quad\Rightarrow\quad x = 1\)
We list the actual weights by substituting \(x=1\)
\(1, 3, 5, 7\)
We test option A by checking if 9 can be formed as a sum of a combination
\(1 + 3 + 5 = 9\)
We test option B by checking if 10 can be formed
\(3 + 7 = 10\)
We test option C by checking if 12 can be formed
\(5 + 7 = 12\)
We test option D by checking if 13 can be formed
\(1 + 5 + 7 = 13\)
We test option E by checking if 14 can be formed; no two or three weights sum to 14
\(\text{No combination of these weights sums to }14\)
Answer 14 cannot be formed from any combination of the packages, so the answer is E.
Hope this helps!
ʕ•ᴥ•ʔ
\(x, x+2, x+4, x+6\)
We translate "The average (arithmetic mean) of four packages is 4 pounds" using the formula for average
\(\frac{x + (x+2) + (x+4) + (x+6)}{4} = 4\)
We multiply both sides by 4 to find the sum of the weights
\(x + (x+2) + (x+4) + (x+6) = 16\)
We combine like terms to simplify the equation
\(4x + 12 = 16\)
We isolate x by subtracting 12 from both sides and then dividing by 4
\(4x = 4\quad\Rightarrow\quad x = 1\)
We list the actual weights by substituting \(x=1\)
\(1, 3, 5, 7\)
We test option A by checking if 9 can be formed as a sum of a combination
\(1 + 3 + 5 = 9\)
We test option B by checking if 10 can be formed
\(3 + 7 = 10\)
We test option C by checking if 12 can be formed
\(5 + 7 = 12\)
We test option D by checking if 13 can be formed
\(1 + 5 + 7 = 13\)
We test option E by checking if 14 can be formed; no two or three weights sum to 14
\(\text{No combination of these weights sums to }14\)
Answer 14 cannot be formed from any combination of the packages, so the answer is E.
Hope this helps!
ʕ•ᴥ•ʔ
04 Sep 2025, 06:37
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Bunuel
Given that the weight of the first package is x, and each package weighs more than 2 from the previous.
So the weights can be x , (x+2), (x+4), (x+6)
The sum of the all the four packages is 4x+12 = 16
This, implies x =1.
So, the package weights can be 1,3,5,7 respectively.
1+3+5 = 9 (option A)
7+3 =10 (Option B)
5 + 7 = 12 ( Option C )
5 + 7 + 1 = 13 ( Option D)
The sum of 14 is not possible in any combination. So option E is the answer
