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# Set J consists of the terms {a,b,c,d,e}, where e>d>c>b>a>1. Which of t

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Re: Set J consists of the terms {a,b,c,d,e}, where e>d>c>b>a>1. Which of t [#permalink]
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Forget conventional ways of solving math questions. In PS, IVY approach is the easiest and quickest way to find the answer.

Set J consists of the terms {a, b, c, d, e}, where e > d > c > b > a > 1. Which of the following operations would decrease the standard deviation of Set J?

A. Multiply each term by e/d
B. Divide each term by b/c
C. Multiply each term by −1
D. Divide each term by d/e
E. Multiply each term by c/e

First of all e/d, b/c, -1, d/e, c/e are all constant value. So we transfer the choices as follows to check whether the constants are less than 1 :
A. Multiply each term by e/d
B. Multiply each term by c/b
C. Multiply each term by −1
D. Multiply each term by e/d
E. Multiply each term by c/e

Since 1 < a < b < c < d < e, e/d, c/b are greater than 1, while c/e is less than 1. Only E is different. So the answer is E.

FACT : If each term is multiplied by a constant k then the standard deviation is also multiplied by k.

So the standard deviations of choices A, B, D are greater than the original SD, while the SD of choices E is less than the original SD and the SD of choices C is equal to the original SD.

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Re: Set J consists of the terms {a,b,c,d,e}, where e>d>c>b>a>1. Which of t [#permalink]
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Bunuel wrote:
Set J consists of the terms {a, b, c, d, e}, where e > d > c > b > a > 1. Which of the following operations would decrease the standard deviation of Set J?

A. Multiply each term by e/d
B. Divide each term by b/c
C. Multiply each term by −1
D. Divide each term by d/e
E. Multiply each term by c/e

Kudos for a correct solution.

VERITAS PREP OFFICIAL SOLUTION:

Standard deviation essentially measures how far the terms in a set deviate from the mean. Because of that, if you can tell whether the new terms are more spread out or more condensed, you can tell what will happen to the standard deviation in comparison with the "old" set. If the numbers are more spread out - for example, if the numbers are multiplied by a number with an absolute value greater than 1 - the standard deviation will increase. And if the numbers are closer together - for example, if they're divided by a number with a standard deviation greater than 1 - the standard deviation will increase.

For quick demonstration, consider the set 1, 2, 3, 4, 5. If you multiply all those numbers by 2, the set becomes farther spread at 2, 4, 6, 8, 10. But if you divide them each by 2 they get closer together: 0.5, 1, 1.5, 2, 2.5.

Multiply each term by e/d - since this fraction is greater than 1, this will increase the dispersion and the standard deviation.

Divide each term by b/c - since this fraction is less than 1 (but greater than 0), dividing by it will again expand the set.

Multiply each term b/y −1 - changing all the terms from positive to negative doesn't change the dispersion, so there will be no change.

Divide each term by d/e - like answer choice B, this will lead to an expanded set.

Multiply each term by c/e - because c is less than e, this fraction is between 0 and 1, meaning that multiplying it has the effect of bringing the terms closer together. This reduces the standard deviation.
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Re: Set J consists of the terms {a,b,c,d,e}, where e>d>c>b>a>1. Which of t [#permalink]
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Bunuel wrote:
Set J consists of the terms {a, b, c, d, e}, where e > d > c > b > a > 1. Which of the following operations would decrease the standard deviation of Set J?

A. Multiply each term by e/d
B. Divide each term by b/c
C. Multiply each term by −1
D. Divide each term by d/e
E. Multiply each term by c/e

Kudos for a correct solution.

Here is what one needs to understand to solve this question:

If you multiply each number of a set by x, the SD also gets multiplied by x.
New standard deviation = Old standard deviation * Multiplied number
When you divide a term by x, it is equal to multiplying by 1/x so the same concept is used for division too.

This and other such concepts are discussed here: https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2015/08 ... -the-gmat/

Back to the question:

If x is greater than 1, SD will increase. If it is less than 1, SD will decrease.

Only in case of option (E), we are multiplying by a number less than 1 so the SD will decrease.
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Re: Set J consists of the terms {a,b,c,d,e}, where e>d>c>b>a>1. Which of t [#permalink]
Multiplying any N>1 will increase SD in N times. So, A is out

Dividing any N<1 will also increase SD. So, B and D are out

Multiplying by -1 only change the direction of inequality but does not change SD. C is out

Multiplying any N<1 will decrease SD

E
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Re: Set J consists of the terms {a,b,c,d,e}, where e>d>c>b>a>1. Which of t [#permalink]
Bunuel wrote:
Set J consists of the terms {a, b, c, d, e}, where e > d > c > b > a > 1. Which of the following operations would decrease the standard deviation of Set J?

A. Multiply each term by e/d
B. Divide each term by b/c
C. Multiply each term by −1
D. Divide each term by d/e
E. Multiply each term by c/e

Kudos for a correct solution.

E as we're essentially multiplying with a value lesser than 1.
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Re: Set J consists of the terms {a,b,c,d,e}, where e>d>c>b>a>1. Which of t [#permalink]
Why is C not the answer? As you said multiplication will change the SD i.e multiply by -1 --> SD *(-1). This makes the SD negative. which is a decrease?
GMATinsight wrote:
Bunuel wrote:
Set J consists of the terms {a, b, c, d, e}, where e > d > c > b > a > 1. Which of the following operations would decrease the standard deviation of Set J?

A. Multiply each term by e/d
B. Divide each term by b/c
C. Multiply each term by −1
D. Divide each term by d/e
E. Multiply each term by c/e

Kudos for a correct solution.

CONCEPT: Standard Deviation is Defined as Average Deviation of Terms in the set from the Mean value of the set. i.e.
1) It depends on the separation between the successive terms of the set
2) If a Constant Value is Added/Subtracted in every terms of set then the Separation between successive terms does NOT change Hence S.D. remains Constant
3) If a Constant Value is Multiplied in every terms then the Separation between succesive terms gets multiplied by the constant Hence S.D. remains gets multiplied by same Number

A. Multiply each term by e/d is equivalent to multiplying each term with some value>1 which will Increase the SD
B. Divide each term by b/c is equivalent to Dividing each term with some value<1 which will Increase the SD because b/c<1
C. Multiply each term by −1 will keep the deviation of the terms same only the signs of the terms in set will change hence, No change in SD
D. Divide each term by d/e is equivalent to Dividing each term with some value<1 which will Increase the SD because d/e<1
E. Multiply each term by c/e is equivalent to multiplying each term with some value<1 which will DECREASE the SD by the coefficient c/e

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Re: Set J consists of the terms {a,b,c,d,e}, where e>d>c>b>a>1. Which of t [#permalink]
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