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A total of 180 passengers on a certain flight registered all of the

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Bunuel
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A total of 180 passengers on a certain flight registered all of the [#permalink] New post   08 May 2022, 02:11
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A total of 180 passengers on a certain flight registered all of the pieces of luggage they had. If 1/10 of the passengers had 4 pieces of luggage each, ​1/4 of the passengers had 2 pieces of luggage each, 1/5 of the passengers had 3 pieces of luggage each, and the remaining passengers had no luggage, what was the average (arithmetic mean) number of pieces of luggage registered per passenger?

A. 1.5
B. 2.0
C. 2.5
D. 3.0
E. 3.5


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Abhishek009
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A total of 180 passengers on a certain flight registered all of the [#permalink] New post   08 May 2022, 02:16
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Bunuel wrote:
A total of 180 passengers on a certain flight registered all of the pieces of luggage they had. If 1/10 of the passengers had 4 pieces of luggage each, ​1/4 of the passengers had 2 pieces of luggage each, 1/5 of the passengers had 3 pieces of luggage each, and the remaining passengers had no luggage, what was the average (arithmetic mean) number of pieces of luggage registered per passenger?

A. 1.5
B. 2.0
C. 2.5
D. 3.0
E. 3.5

Total luggage is \(180*\frac{1}{10}*4 + 180*\frac{1}{4}*2 + 180*\frac{1}{5}*3 = 72 + 90 + 108 => 270\)

So, Avg luggage per passenger will be \(\frac{270}{180} = 1.50\), Answer must be (A)
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A total of 180 passengers on a certain flight registered all of the [#permalink] New post   22 Sep 2022, 06:30
Hello,

I noticed I solved this problem without multiplying the fractions into the total number of passengers:
So \((\frac{1}{10}∗4)+ (\frac{1}{4}∗2)+(\frac{1}{5}∗3)=\frac{4}{10} + \frac{2}{4 }+ \frac{3}{5 }= 2/5 + 1/2 + 3/5 = 5/5 + 1/2 = 1.5\)

So, Avg luggage per passenger will be 1.5 answer (A) still. Is there a downside to this approach?
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Sa800
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Re: A total of 180 passengers on a certain flight registered all of the [#permalink] New post   24 Mar 2024, 13:22
sebastian111 wrote:
Hello,

I noticed I solved this problem without multiplying the fractions into the total number of passengers:
So \((\frac{1}{10}∗4)+ (\frac{1}{4}∗2)+(\frac{1}{5}∗3)=\frac{4}{10} + \frac{2}{4 }+ \frac{3}{5 }= 2/5 + 1/2 + 3/5 = 5/5 + 1/2 = 1.5\)

So, Avg luggage per passenger will be 1.5 answer (A) still. Is there a downside to this approach?

­

can someone say if this is the right approach and if so, why it works?
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Bunuel
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Re: A total of 180 passengers on a certain flight registered all of the [#permalink] New post   25 Mar 2024, 02:25
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Sa800 wrote:
sebastian111 wrote:
Hello,

I noticed I solved this problem without multiplying the fractions into the total number of passengers:
So \((\frac{1}{10}∗4)+ (\frac{1}{4}∗2)+(\frac{1}{5}∗3)=\frac{4}{10} + \frac{2}{4 }+ \frac{3}{5 }= 2/5 + 1/2 + 3/5 = 5/5 + 1/2 = 1.5\)

So, Avg luggage per passenger will be 1.5 answer (A) still. Is there a downside to this approach?

­

can someone say if this is the right approach and if so, why it works?

­
180 is simply factored out, thus there's no need to divide by that number again at the end. So, factoring out and not dividing at the end balance each other out and give a correct answer.
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Re: A total of 180 passengers on a certain flight registered all of the [#permalink]
25 Mar 2024, 02:25
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