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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.

Re: M18 Q14 needs clarification [#permalink]
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. _________________

Re: M18 Q14 needs clarification [#permalink]
04 Mar 2010, 18:31

1

This post received KUDOS

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.

Re: M18 Q14 needs clarification [#permalink]
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. _________________

Re: M18 Q14 needs clarification [#permalink]
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.

Re: M18 Q14 needs clarification [#permalink]
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.

Re: M18 Q14 needs clarification [#permalink]
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.

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?

Re: M18 Q14 needs clarification [#permalink]
13 Mar 2013, 07:09

1

This post received KUDOS

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 \(a\)

Re: M18 Q14 needs clarification [#permalink]
14 Mar 2013, 01:35

4

This post received KUDOS

Expert's post

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.

Re: M18 Q14 needs clarification [#permalink]
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)

Re: M18 Q14 needs clarification [#permalink]
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)