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Four students contributed to a charity drive, and the averag
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29 Aug 2018, 16:39
29 Aug 2018, 16:39
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Question Stats:
72% (00:48) correct
27% (01:01) wrong based on 47 sessions
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Four students contributed to a charity drive, and the average (arithmetic mean) amount contributed by each student was $20. If no student gave more than $25, what is the minimum amount that any student could have contributed?
Show: :: OA
$5
Re: Four students contributed to a charity drive, and the averag
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30 Aug 2018, 01:12
30 Aug 2018, 01:12
the avg for the 4 students = 20. while no student contributed more than 25. To get the least amount from a student, 3 students must contribute the maximum amount possible.
Therefore, \(\frac{25*3 +x}{4} = 20\)
or, \(\frac{75+x}{4} = 20\)
or, \(75 +x = 80\)
Therefore, \(x =5\)
Therefore, \(\frac{25*3 +x}{4} = 20\)
or, \(\frac{75+x}{4} = 20\)
or, \(75 +x = 80\)
Therefore, \(x =5\)
Re: Four students contributed to a charity drive, and the averag
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31 Aug 2018, 20:49
31 Aug 2018, 20:49
the answer is 5$
(25*3+x)/4=20
on solving
x=5
(25*3+x)/4=20
on solving
x=5
