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M18 Q14 needs clarification

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M18 Q14 needs clarification [#permalink] New post 17 Sep 2008, 13:27
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What is the mean of four consecutive even integers a , b , c , d ? a is the smallest integer among these.

1. a + d = b + c
2. b + c = d - a

[Reveal] Spoiler: OA
B

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It is assumed that "a" is the first term in this sequence.

"a", however can be the first and the last term in the sequence. This is not specified in the question stem. Then answer is E.

If, "a" = first term, then we have b+c = d-a, a is first term then d is last and d = a+6, so d-a = 6. From there we find the sequence to be equal to a = 0, b = 2, c = 4, d = 6.

If, "a" = last term, then we have b+c = d-a, a is last term then d is first and d = a-6, so d-a = -6. From there we find the sequence to be equal to d = -6, b = -4, c = -2, a = 0.

a, b, c, d = 0, 2, 4, 6
d, b, c, a = -6, -4, -2, 0.

They both satisfy S1 and S2 but their means are 3 and -3 respectively. So unless we know what the first term of the sequence, "a" or "d" is, we can't define the mean.

REVISED VERSION OF THIS QUESTION WITH A SOLUTION IS HERE: m18-q14-needs-clarification-70373.html#p1195801
[Reveal] Spoiler: OA
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Re: M18 Q14 needs clarification [#permalink] New post 14 Mar 2013, 01:35
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BELOW IS REVISED VERSION OF THIS QUESTION WITH A SOLUTION:

What is the mean of four consecutive even integers a, b, c and d, where a<b<c<d?

(1) a + d = b + c. That statement is true for ANY four consecutive even integers. Not sufficient.

(2) b + c = d - a. Let a=2k, for some integer k. Now, since a, b, c and d are consecutive even integers then b=2k+2, c=2k+4 and d=2k+6. So, we have that (2k+2)+(2k+4)=(2k+6)-2k --> k=0 --> a=0, b=2, c=4 and d=6 --> mean=3. Sufficient.

Answer: B.
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Re: M18 Q14 needs clarification [#permalink] New post 06 Mar 2010, 09:02
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As the numbers are even consecutive assume them to be :
a
b = a + 2
c = a + 4
d = a + 6

Stmt 1: a + d = b + c
substitute a + a + 6 = a + 2 + a + 4
2a + 6 = 2a + 6
Not sufficient

Stmt 2: b + c = d - a
substitute a + 2 + a + 4 = a + 6 - a
2a + 6 = 6
a = 0
SUFF
so B.
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Re: M18 Q14 needs clarification [#permalink] New post 04 Mar 2010, 08:44
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A tricky question.

Stmt 1: no use as it does not give anything new what can be said from problem stmt.

Stmt 2: b+c = d-a
From problem stmt it is known that d-a=6 as a is first number in the sequence and d is fourth consecutive even integer.

Now plugging values,
1. 0,2,4,6 condition(b+c = d-a) satisfied
2. -2,0,2,4 condition(b+c = d-a) not satisfied
3. 2,4,6,8 condition(b+c = d-a) not satisfied

so it can be said a=0,b=2,c=4,d=6 hence their mean can be found, so answer is B.
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Re: M18 Q14 needs clarification [#permalink] New post 04 Mar 2010, 12:13
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gee, absolutely.
as they are consecutive the condition b+c = d-a is equal to a+2 + a+4 = a+6 - a -> a=0
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Re: M18 Q14 needs clarification [#permalink] New post 04 Mar 2010, 18:31
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What is the mean of four consecutive even integers a , b , c , d? a is the smallest of the four integers.

1. a + d = b + c
2. b + c = d - a

It's given: four consecutive even integers a , b , c , d.
The word consecutive means that the numbers are a, b=a+2, c=a+4, d=a+6. In this case the Stm1 is useless to determine the mean, but Stm2 gives us a=0 and mean=3.

But(!) the condition 'a is the smallest of the four integers' is a bell which rings " d is NOT necessarily the largest, that being written in increasing order only the four numbers will be consecutive".
Then the numbers may be a, b=a+2, c=a+6, d-a+4.
From the Stm2 2a+8=4 and a=-2 and the mean is 1. -2;0;4;2
Or, the numbers are a, b=a+4, c=a+6, d=a+2.
From the Stm2 2a+10=2, a=-4, mean is -1. -4;0;2;-2
Stm1 determines that d is the largest, d=a+6, Stm2 determines a=0, mean is 3.
Finally, answer is C. Both statements (1) and (2) TOGETHER are sufficient,but NEITHER statement ALONE is sufficient.

Last edited by nvgroshar on 06 Mar 2010, 12:03, edited 1 time in total.
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Re: M18 Q14 needs clarification [#permalink] New post 13 Mar 2013, 07:09
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What is the mean of four consecutive even integers a , b , c , d ? a is the smallest integer among these.

1. a + d = b + c
2. b + c = d - a

As given, a, b, c, d are consecutive even integers and a is the smallest one...
so, others integers will be a+2, a+4, a+6

Mean will be \frac{4a+12}{3} = a+3
so to find mean, we have to find a

I- a + d = b + c => 2a+ some integer = 2a+some integer
we can not conclude value of a from here, so this is not sufficient condition...

II- b + c = d - a => a+2 + a+4 = a+6 -a
here, we can calculate value of a

So this is sufficient.

Ans is B
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Re: M18 Q14 needs clarification [#permalink] New post 04 Mar 2010, 06:41
Exactly, my point too.

a is given as smallest integer in the set of four consecutive even integers. so, d = -6 and a = 0 cannot be the case at all. Consider flipping the set to a = -6, b = -4, c = -2 and d = 0. Then, stmt 2 becomes b + c = -6 but d - a = 6, so the set cannot satisfy the condition. I think B is right, when given that a is smallest integer, it HAS to be the first term of the consecutive even integers set.
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Re: M18 Q14 needs clarification [#permalink] New post 04 Mar 2010, 18:47
What is the mean of four consecutive even integers a,b ,c ,d ? a is the smallest integer among these.

1. a+d=b+c
2. b+c= d-a

1. 0,2,4,6 satisifies
2,4,6,8 satisifies so insufficient

2. 0,2,4,6 satisfies
-2, 0,2,4 not satisfies
-6,-2,-4,0 not satisfies

so ans B is sufficient
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Re: M18 Q14 needs clarification [#permalink] New post 13 Mar 2010, 06:57
Yes, no need to substitute numbers here. Just use a, a+2, a+4 and a+6 as the four consecutive even integers. We don't have to worry whether they're negative or positive. We just know that a is the least.
Statement 1 is insufficient because we get 2a+6=2a+6
Statement 2 is sufficient because it gives us the value of a=6 using which we can find the mean.
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Re: M18 Q14 needs clarification [#permalink] New post 13 Mar 2010, 07:48
sidhu4u wrote "Yes, no need to substitute numbers here. Just use a, a+2, a+4 and a+6 as the four consecutive even integers. We don't have to worry whether they're negative or positive. We just know that a is the least".
If you consider these four consecutive even integers as a, a+2,a+4,a+6 then there is NO NEED to mention that a is the least. But it's MENTIONED and there is no information that d is the largest. The four numbers a=-2,b=0,c= 4,d= 2 are four consecutive even integers and a is the least, but they are not written in increasing order.
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Re: M18 Q14 needs clarification [#permalink] New post 09 Mar 2011, 03:28
Can't second statement be used like

d= a+b+c, hence (a+b+c)+d = 2d

hence mean = 2d/4 = d/2
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Re: M18 Q14 needs clarification [#permalink] New post 09 Mar 2011, 06:39
From question, mean is :

(a + b + c + d)/4

= (a + a + 2 + a + 4 + a + 6)/4

= (4a + 12)/4 = a + 3

From (1)
a + a + 6 = a + 2 + a + 4

no new info, so not suff.

From (2)

2a + 6 = a + 6 - a

=> a = 0, so suff.

So answer is B
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Re: M18 Q14 needs clarification [#permalink] New post 10 Mar 2011, 01:21
v001c wrote:
Why do you assume a=0?



Statement 1 is satisfied by all consecutive intezer.
In stat2- a+b+c=d, and this can be satisfied only if you will consider that series starts from 0. So, the series will be 0,2,4,6. Same can be true for -6,-4,-2,0 but here -6 is the smallest. So, answer is B.
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Re: M18 Q14 needs clarification [#permalink] New post 10 Mar 2011, 02:09
sidhu4u wrote:
Yes, no need to substitute numbers here. Just use a, a+2, a+4 and a+6 as the four consecutive even integers. We don't have to worry whether they're negative or positive. We just know that a is the least.
Statement 1 is insufficient because we get 2a+6=2a+6
Statement 2 is sufficient because it gives us the value of a=6 using which we can find the mean.


per your calculation, I get a=0
given: stm2 =>b+c=d-a
=>a+2+a+4=a+6-a
=>2a+6=6
=>a=0

so the nos. are 0,2,4,6. ONLY this matches b+c=d-a. hence B is sufficient.
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Re: M18 Q14 needs clarification [#permalink] New post 13 Mar 2012, 13:05
topmbaseeker wrote:
As the numbers are even consecutive assume them to be :
a
b = a + 2
c = a + 4
d = a + 6

Stmt 1: a + d = b + c
substitute a + a + 6 = a + 2 + a + 4
2a + 6 = 2a + 6
Not sufficient

Stmt 2: b + c = d - a
substitute a + 2 + a + 4 = a + 6 - a
2a + 6 = 6
a = 0
SUFF
so B.

This is the best answer in the thread.

with a being the LEAST of these we know:
a,b,c,d = a, a+2, a+4, a+6

so, if we know a we can find the mean.

a) 2a+6 = 2a+6 is insuff because we can't get the answer.
b) 2a+6 = a-a+6 --> 2a+6 = 6; a=0.
(2+4+6)/4
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Re: M18 Q14 needs clarification [#permalink] New post 11 Apr 2012, 04:38
Another set that satisfies both the premise and statement 2 is:
a= -4 (least)
b=0
c=2
d= -2
Since both sets (a,b,c,d)=(-4,0,2,-2) and (a,b,c,d)=(0,2,4,6) satisfy statement 2, why can statement 2 lead to a certain solution?

Edit:
There is not a clear statement that the values are placed in increasing order. Can we assume so?
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Re: M18 Q14 needs clarification [#permalink] New post 13 Mar 2013, 10:03
clearly.. answer is B

take the four nos. as ( a, a+2, a+4, a+6) and solve the two given statements.
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Re: M18 Q14 needs clarification [#permalink] New post 14 Mar 2013, 06:40
Agreed that statement 1 cannot provide the answer
and
statement 2 provides 2 sets of answers.
So
we combine both and get
a+d = d-a
or a = 0
so the set can only be 0,2,4 and 6
so average = 3
Solution C (combining both statements 1 & 2)
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Re: M18 Q14 needs clarification [#permalink] New post 14 Mar 2013, 06:42
Expert's post
shyamsunder wrote:
Agreed that statement 1 cannot provide the answer
and
statement 2 provides 2 sets of answers.
So
we combine both and get
a+d = d-a
or a = 0
so the set can only be 0,2,4 and 6
so average = 3
Solution C (combining both statements 1 & 2)


Please check here: m18-q14-needs-clarification-70373.html#p1195801

Hope it helps.
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Re: M18 Q14 needs clarification   [#permalink] 14 Mar 2013, 06:42
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