Which of the following series of numbers, if added to the

Question

Which of the following series of numbers, if added to the set {1, 6, 11, 16, 21}, will not change the set’s mean?

I. 1.5, 7.11 and 16.89 
II. 5.36, 10.7 and 13.24 
III. -21.52, 23.3, 31.22

(A) I only 
(B) II only 
(C) III only 
(D) I and III only 
(E) None

Bunuel · Accepted Answer

Mean of the given set is (1+6+11+16+21)/5=11.

Now, in order the mean not to change, the mean of the new set we add to the old one should also be equal to 11 (or as in all 3 new sets there are 3 numbers, then their sum must be 3*11=33). Let's check:

I. 1.5, 7.11 and 16.89 --> will end with 0.5 son not 33. Discard.
II. 5.36, 10.7 and 13.24 --> will end with 0.3 son not 33. Discard.
III. -21.52, 23.3, 31.22 --> -21.52+23.3+31.22=-21.52+54.52=33. Correct.

Answer: C (III only).

Konstantin1983 · Answer

Straightforward. The mean of the current set is (1+6+11+16+21)/5=11. Set has 5 numbers. So keep the mean the same after new numbers are added we need to find the sum that will be added. We can set up equation (55+x)/8=11. 8 is because we add 3 numbers. Hence x=33. Only set III has numbers that sum up to 33. Hope it is clear

EMPOWERgmatRichC · Answer

Hi All,

This question has some great Number Property shortcuts built into it (which you can take advantage of to save some time and avoid some of the math "work").

We're given the set {1, 6, 11, 16, 21}. We're asked which set of additional numbers, when added into this set, will NOT change the set’s mean.

In the original set of numbers, notice how the 5 terms are 'evenly spaced'; this means that the average MUST be the 'middle term' --> the average is 11.

Looking at the three options, notice how each has 3 terms. To add 3 terms to the given set and NOT change the average, we need the average of the 3 terms to be 11. By extension, we need the SUM to = 33.

A quick estimate of Roman Numerals 1 and 2 proves that neither has a sum of 33 (the sums are both TOO SMALL). Eliminate Answers A, B and D.

Adding up the terms in Roman Numeral 3 gives us a sum of 33, so this set 'fits' what we're looking for.

Final Answer:  C

GMAT assassins aren't born, they're made,
Rich

ramblers · Answer

Frustratingly good question. Went through it trying to figure out how much would have been added/subtracted to each side of 11. Was running out of time and estimated incorrectly on III, so picked E.

Rich's point on summing to 33 is a fantastic shortcut. Will try to remember that for future average questions.

KarishmaB · Answer

Use the method of deviations to arrive at the answer quickly. Discussed here: https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2012/0 ... eviations/ {1, 6, 11, 16, 21} is an arithmetic progression with a common difference of 5 between each two consecutive elements. So arithmetic mean is the middle element 11. If you add a bunch of numbers to this set and want that the mean should remain the same, the mean of these numbers should be 11. I. 1.5, 7.11 and 16.89 - Deviation on the left is far more than on the right. Not possible. II. 5.36, 10.7 and 13.24 - Deviation on the left is far more than on the right. Not possible. III. -21.52, 23.3, 31.22 Deviations on left = 11 - (-21.52) = 32.52 Deviations on right = 23.3 - 11 + 31.22 - 11 = 12.3 + 20.22 = 32.52 Since deviations on left and right are same, the mean must be 11. Answer (C)

Cellchat · Answer

For the mean not to change, if we sum the difference of each term in the sequence away from the mean (11), the sum should equal 0.

III. (-21.52-11) + (23.3-11) + (31.22-11) = 0 Therefore no mean change. Only III.

Note this is the same as adding each term and finding the sequence that equals 33 (rearrange the above

)

bumpbot · Answer

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