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26 Dec 2017, 00:17
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
15%
(low)
Question Stats:
79% (01:16) correct
21%
(01:44)
wrong
based on 73
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The average (arithmetic mean) of five consecutive numbers is 48. What is the average of the three smallest of these numbers?
A. 45
B. 45.5
C. 46
D. 46.5
E. 47
A. 45
B. 45.5
C. 46
D. 46.5
E. 47
26 Dec 2017, 01:28
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Bunuel
Anytime we're give a question involving consecutive integers, we'll look for a solution involving number properties.
This is a Logical approach.
In any sequence containing an odd number of consecutive integers, the average equals the middle number.
Therefore our middle number is 48 so our sequence is 46,47,48,49,50.
Then the average of the 3 smallest numbers is 47 - option (E).
27 Dec 2017, 04:30
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Bunuel
\(Average=\frac{Sum}{Quantity}\), so \(48=\frac{Sum}{5}\), \(Sum=240\).
\(Sum=(x)+(x+1)+(x+2)+(x+3)+(x+4)=5x+10=240\), so \(x=230/5=46\).
\(Sum.Lowest=(x)+(x+1)+(x+2)=3x+3=3(46)+3=141\)
\(Average.Lowest=\frac{141}{3}=47\)
(E) is the answer.
