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The average (arithmetic mean) of four packages is 15 pounds

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by VJesus12 » Thu Apr 12, 2018 9:19 am
The average (arithmetic mean) of four packages is 15 pounds. If the weight of the first package is x, and each package weighs twice as much as the previous package, which of the following CANNOT be the total weight, in pounds, of any combination of the packages?

A) 12
B) 24
C) 44
D) 46
E) 52

The OA is the option D.

How can I determine the weight of the packages? May anyone helps me? Please. I'd be thankful.
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by Vincen » Thu Apr 12, 2018 10:13 am
Hello.

We have FOUR packages whose average is 15 pounds.

- "If the weight of the first package is x, and each package weighs twice as much as the previous package"

Let be x the weight of the first package, then the other packages are 2x, 4x and 8x of weight.

Therefore, we can set the equation $$\frac{x+2x+4x+8x}{4}=15$$ $$15x=60$$ $$x=4$$ Therefore, the weights of the four packages are: 4, 8, 16, and 32. The possible weight combinations are: $$4+8=12$$ $$8+16=24$$ $$4+8+32=44$$ $$4+16+32=52$$ The unique option that is not possible is [spoiler]D=46[/spoiler].

Therefore, this is the correct answer.
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by Scott@TargetTestPrep » Mon Apr 16, 2018 3:43 pm
VJesus12 wrote:The average (arithmetic mean) of four packages is 15 pounds. If the weight of the first package is x, and each package weighs twice as much as the previous package, which of the following CANNOT be the total weight, in pounds, of any combination of the packages?

A) 12
B) 24
C) 44
D) 46
E) 52
Using the average formula: average = sum/number, we we see that the total weight of all 4 packages is 60 pounds.

We can let the weights of the 4 packages be x, 2x, 4x, 8x; thus:

15x = 60

x = 4

So the total weight of any combination of the packages must be a multiple of 4, so 46 can't be that weight.

Answer: D

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