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Bunuel
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Bunuel
List M consists of the numbers 10, 20, 30, 40, 50.

Which of the following lists of numbers have an average (arithmetic mean) that is equal to the average of the numbers in list M?

I. 10, 20, 30, 35, 50
II. 10, 22, 30, 38, 50
III. 0, 0, 0, 0, 150

A. I only
B. II only
C. III only
D. II and III only
E. I, II, and III

Given: List M consists of the numbers 10, 20, 30, 40, 50. Average = 30
Question: Which of the following lists of numbers have an average (arithmetic mean) that is equal to the average of the numbers in list M?

I. 10, 20, 30, 35, 50
\(Average=\frac{10+20+30+35+50}{5}\) = 29
II. 10, 22, 30, 38, 50
\(Average = \frac{10+22+30+38+50}{5}\) = 30
III. 0, 0, 0, 0, 150
Average = \(\frac{150}{3}\) = 30

IMO D
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The thing to notice here is that this is an evenly spaced set and hence median shall be equal to mean. Clearly median=mean=30

For 2- it is the same as original set, take 2 from 22 and give it to 38.
For 3- 150/5- 30

Both 2 and 3 match

Answer D
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Average of list M =30
Only 2nd and 3rd list's average is 30
Hence Answer is D
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