kushal.adhia wrote:The average (arithmetic mean) of the 43 numbers in list L is a positive number. The average of all 48 numbers in both lists L and M is 50 percent greater than the average of the 43 numbers in list L. What percent greater than the average of the numbers in list L is the average of the numbers in list M?
A. 85%
B. 240%
C. 350%
D. 480%
E. 580%
OA is D
Say the average of all the 43 numbers in list L is x.
=> Sum of this 43 numbers = 43x.
Average of all the 48 numbers in list L and M is = x + (50/100)x = 1.5x (As the average is 50% greater)
=> Sum of this 48 numbers = 48*(1.5x) = 72x.
Therefore sum of all the (48 - 43) = 5 numbers in the list M = Sum of all the numbers in list L and M - Sum of all the numbers in list L = 72x - 43x = 29x
Therefore, average of all the 5 numbers in list M = 29x/5 = 5.8x
5.8x is greater than x by (5.8x - x) = 4.8x
Percentage = (4.8x/x)*100 = 480%
The correct answer is D.
Rahul Lakhani
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