Carcass wrote:
A set has exactly five consecutive positive integers.
Quantity A |
Quantity B |
The percentage decrease in the average of the numbers when one of the numbers is dropped from the set |
20% |
A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
Let the 5 numbers be: \(x, x+1, x+2, x+3, x+4\)
Since the numbers are in Arithmetic Progression (i.e. consecutive numbers have a constant gap), we have:
Mean = Median = 3rd term = \(x+2\)
(Note: You can always add up the terms and check the average)
One of the 5 numbers will be dropped
Maximum change in mean will occur if either of the 2 extreme terms, i.e. \(x\) or \(x+4\) is dropped. The average will decrease if \(x+4\) is dropped (Note: if \(x\) is dropped, the average would actually increase. Also, if the middle number, i.e. \(x+2\) is dropped, there will be no change in the mean)
If \(x+4\) is dropped: The 4 terms are: \(x, x+1, x+2, x+3\)
=>
New Mean = Median = \([(x+1)+(x+2)]/2 = x+1.5\)
=> Percent decrease in mean = \([{(x+2)-(x+1.5)}/(x+2)] * 100\) = \([50/(x+2)]%\)
The above percent will be
maximum if the value of \(x\) is minimum, i.e. \(x=1\)
=> Maximum percent decrease = \([50/(1+2)]% = 16.67%\)
Thus, Quantity B is greater than Quantity A
Answer BNote: Some important results that come up here: In an Arithmetic Progression i.e. consecutive terms having a constant difference of \(d\):
#1. Mean = Median
#2. The average remains unchanged if the middle term (or both middle terms) are removed
#3. The maximum change in mean occurs when one of the extreme terms is removed
#4. The
maximum change in mean (when one of the extreme terms is removed) = \(d/2\), where \(d\) is the constant difference between consecutive terms
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Sujoy Kumar Datta | GMAT - Q51 & CAT (MBA @ IIM) 99.98 Overall with 99.99 QA
IIT Kharagpur, TUD Germany
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