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Manager  G
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Set S contains five different positive integers a, b, c, d, and e, whe  [#permalink]

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Difficulty:   95% (hard)

Question Stats: 19% (01:40) correct 81% (01:54) wrong based on 158 sessions

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Set S contains five different positive integers a, b, c, d, and e, where a < b < c < d < e. Is c − b equal to d − c?
1) The mean of S is equal to the median of S
2) When any element of S is divided by 5, the remainder is 4

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Re: Set S contains five different positive integers a, b, c, d, and e, whe  [#permalink]

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Why is A not sufficient? S Contains only positive integers and from A we know that the integers are evenly spaced. Is there something I'm missing or overlooking? Thank you

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Manager  S
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Set S contains five different positive integers a, b, c, d, and e, whe  [#permalink]

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3
Saurabhminocha wrote:
Why is A not sufficient? S Contains only positive integers and from A we know that the integers are evenly spaced. Is there something I'm missing or overlooking? Thank you

Posted from my mobile device

Consider 2 sets of numbers:

Set 1:
a = 2
b = 3
c = 5
d = 7
e = 8
Sum = 25
Mean = 5
Median = 5
In this case, A works as mean = median and c-b = d-c = 2

Set 2:
a = 2
b = 3
c = 5
d = 6
e = 9
Sum = 25
Mean = 5
Median = 5
I changed the bolded numbers. Now mean = median, but d-c = 1 and c-b = 2.

Hence, A is not sufficient. I guess I should ask you this: what implies that numbers are evenly spaced? Just because numbers are positive and have a mean = median, it does not mean they are evenly spaced.
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Re: Set S contains five different positive integers a, b, c, d, and e, whe  [#permalink]

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can you post the OE please?
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Re: Set S contains five different positive integers a, b, c, d, and e, whe  [#permalink]

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1
for statement 1) The mean of S is equal to the median of S, we can conclude:
$$(a+b+c+d+e)/5 = c$$
$$a+b+c+d+e = 5c$$
$$a+b+d+e = 4c$$

the only constrain is that a<b<c<d<e,
so if we let c = 19,
then $$a+b+d+e = 19*4 = 76$$, which can be (1,2,20,53) or (1,3,20,52) or others.

so statement 1 is not sufficient.

for statement 2) When any element of S is divided by 5, the remainder is 4:
we have to choose numbers with unit digit of 4 or 9 so that the remainder remains 4 when divided by 5.

the set can be (4,9,14,19,24) or (4,9,14,99,199) or others.

so statement 2 is not sufficient.

if we tried combining:

let c = 19, and try choosing numbers with unit digit of 4 or 9 and their sum is $$4*19=76$$

set can be (4,9,19,24,39) where (d-c) not equal (c-b)
or can be (4,9,19,29,34) where (d-c) equal (c-b).

so still insufficient

so the answer is E.
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Re: Set S contains five different positive integers a, b, c, d, and e, whe  [#permalink]

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Raksat wrote:
Set S contains five different positive integers a, b, c, d, and e, where a < b < c < d < e. Is c − b equal to d − c?
1) The mean of S is equal to the median of S
2) When any element of S is divided by 5, the remainder is 4

chetan2u Requesting your expert inputs.
Why is 1) and 2) together not sufficient?

1) tells us that mean and median = c

2) tells us that all numbers are of the type 5k+4
I'm unable to proceed from this point.
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Re: Set S contains five different positive integers a, b, c, d, and e, whe  [#permalink]

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ScottTargetTestPrep Can you please provide an explaination?
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Set S contains five different positive integers a, b, c, d, and e, whe  [#permalink]

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Darshi04 wrote:
Raksat wrote:
Set S contains five different positive integers a, b, c, d, and e, where a < b < c < d < e. Is c − b equal to d − c?
1) The mean of S is equal to the median of S
2) When any element of S is divided by 5, the remainder is 4

chetan2u Requesting your expert inputs.
Why is 1) and 2) together not sufficient?

1) tells us that mean and median = c

2) tells us that all numbers are of the type 5k+4
I'm unable to proceed from this point.

Ok ..
it means are b and d equidistant from c, or whether c is the average of b, c and d.
(I) statement I tells us mean=median=c.
So if a and e are equidistant from c, Ans will be yes otherwise no..
But we can't say anything about d and e.
3,4,5,6,7.. 5 is mean and median ..Ans is yes..
1,2,5,6,11...5 is median and mean ...Ans is No
Insufficient

(II) statement II tells us that 4 is the remainder when divided by 5 ..
As correctly mentioned by you the numbers are 5k+4..
But we do not know anything about numbers..
9,14,19,24,29...yes
9,14,24,29,34...no
Insufficient

Combined..
We can still have many ways..
We can decrease the first two, a and b, and just increase only the largest e..
9,14,19,24,29..... mean=median=19...yes
4,9,19,24,29.... Mean=median=19......no
Still insufficient..
Or
The numbers are 5(k-x)+4; 5(k-y)+4, 5k+4; 5(k+a)+4; 5(k+b)+4..
Mean and median is 5k+4..
So 5a+5b-5x-5y=0...a+b=x+y
We are looking for y=b, which cannot be guaranteed here.
E
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Set S contains five different positive integers a, b, c, d, and e, whe  [#permalink]

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chetan2u wrote:
Darshi04 wrote:
Raksat wrote:
Set S contains five different positive integers a, b, c, d, and e, where a < b < c < d < e. Is c − b equal to d − c?
1) The mean of S is equal to the median of S
2) When any element of S is divided by 5, the remainder is 4

chetan2u Requesting your expert inputs.
Why is 1) and 2) together not sufficient?

1) tells us that mean and median = c

2) tells us that all numbers are of the type 5k+4
I'm unable to proceed from this point.

Ok ..
it means are b and d equidistant from c, or whether c is the average of b, c and d.
(I) statement I tells us mean=median=c.
So if a and e are equidistant from c, Ans will be yes otherwise no..
But we can't say anything about d and e.
3,4,5,6,7.. 5 is mean and median ..Ans is yes..
1,2,5,6,11...5 is median and mean ...Ans is No
Insufficient

(II) statement II tells us that 4 is the remainder when divided by 5 ..
As correctly mentioned by you the numbers are 5k+4..
But we do not know anything about numbers..
9,14,19,24,29...yes
9,14,24,29,34...no
Insufficient

Combined..
We can still have many ways..
We can decrease the first two, a and b, and just increase only the largest e..
9,14,19,24,29..... mean=median=19...yes
4,9,19,24,29.... Mean=median=19......no
Still insufficient..

E

chetan2u Many thanks for your inputs.

This is pretty much how I solved it. I was wondering if there is any algebraic method / using variables to solve and arrive at a conclusive answer, rather than using numbers (trial/error).

Or is it recommended to solve mean/median questions using real numbers?

Posted from my mobile device
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Originally posted by Darshi04 on 24 Dec 2018, 02:49.
Last edited by Darshi04 on 24 Dec 2018, 05:09, edited 1 time in total.
Math Expert V
Joined: 02 Aug 2009
Posts: 7672
Re: Set S contains five different positive integers a, b, c, d, and e, whe  [#permalink]

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Darshi04 wrote:
chetan2u wrote:
Raksat wrote:
Set S contains five different positive integers a, b, c, d, and e, where a < b < c < d < e. Is c − b equal to d − c?
1) The mean of S is equal to the median of S
2) When any element of S is divided by 5, the remainder is 4

Ok ..
it means are b and d equidistant from c, or whether c is the average of b, c and d.
(I) statement I tells us mean=median=c.
So if a and e are equidistant from c, Ans will be yes otherwise no..
But we can't say anything about d and e.
3,4,5,6,7.. 5 is mean and median ..Ans is yes..
1,2,5,6,11...5 is median and mean ...Ans is No
Insufficient

(II) statement II tells us that 4 is the remainder when divided by 5 ..
As correctly mentioned by you the numbers are 5k+4..
But we do not know anything about numbers..
9,14,19,24,29...yes
9,14,24,29,34...no
Insufficient

Combined..
We can still have many ways..
We can decrease the first two, a and b, and just increase only the largest e..
9,14,19,24,29..... mean=median=19...yes
4,9,19,24,29.... Mean=median=19......no
Still insufficient..

E

chetan2u Many thank for your inputs.

This is pretty much how I solved it. I was wondering if there is any algebraic method / using variables to solve and arrive at a conclusive answer, rather than using numbers (trial/error).

Or is it recommended to solve mean/median questions using real numbers?

Posted from my mobile device

A solution which is a bit more algebraic is...
The numbers are 5(k-x)+4; 5(k-y)+4, 5k+4; 5(k+a)+4; 5(k+b)+4..
Mean and median is 5k+4..
So 5(k-x)+4 + 5(k-y)+4 + 5k+4 + 5(k+a)+4 + 5(k+b)+4 = 5*(5k+4)..
5*(5k+4)-5x-5y+5a+5b=5*(5k+4)......
So 5a+5b-5x-5y=0......
a+b=x+y
We are looking for y=b, which cannot be confirmed as it will depend on a and x too..
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Re: Set S contains five different positive integers a, b, c, d, and e, whe  [#permalink]

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Raksat wrote:
Set S contains five different positive integers a, b, c, d, and e, where a < b < c < d < e. Is c − b equal to d − c?
1) The mean of S is equal to the median of S
2) When any element of S is divided by 5, the remainder is 4

We need to determine whether c – b = d – c in a set of five positive integers a, b, c, d, and e where a < b < c < d < e.

Statement One Only:
The mean of S is equal to the median of S.

Knowing the mean of a set is equal to the median of the set is not sufficient to determine the answer to the question.

For example, if a = 1, b = 2, c = 3, d = 4 and e = 5, then c – b = d – c since c – b = 1 and d – c = 1.

However, if a = 1, b = 2, c = 4, d = 5 and e = 8, then c – b  d – c since c – b = 2 and d – c = 1.

Statement Two Only:

When any element of S is divided by 5, the remainder is 4.

That is, each element is in the form of 5k + 4 where k is a nonnegative integer. However, knowing each element in a specific format is not sufficient to determine the answer to the question.

For example, if a = 4, b = 9, c = 14, d = 19 and e = 24, then c – b = d – c since c – b = 5 and d – c = 5.

However, if a = 4, b = 9, c = 19, d = 24 and e = 39, then c – b  d – c since c – b = 10 and d – c = 5.

Statements One and Two Together:

Knowing both statements are still not sufficient to answer the question. We can use the same two examples in statement two since both have their respective means equal to their medians and both have elements in the form of 5k + 4.

Answer: E
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