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What is the mean of four consecutive even integers a ,b ,c ,d ? i) a+d=b+c ii) d-a=b+c... although it is given as B ,i have a query... do we take the four consecutive even integers a ,b ,c ,d to be written in ascending order... if not then the ans should be C..... if the int are in ascending order , shudn't it be mentioned? _________________

What is the mean of four consecutive even integers a ,b ,c ,d ? i) a+d=b+c ii) d-a=b+c... although it is given as B ,i have a query... do we take the four consecutive even integers a ,b ,c ,d to be written in ascending order... if not then the ans should be C..... if the int are in ascending order , shudn't it be mentioned?

Dont really understand your question.....(in red)

The question is asking for the mean... of a,b,c,d all even integers... this means... we need to find the value of (a+b+c+d)/4 which would be an even integer for sure. ST1: gives a+d = b+c... This gives mean as (a+d)/2 or (b+c)/2.... hence NOT SUFF...

ST2: gives d-a = b+c --> d = a+b+c Therefore mean = (a+b+c+d)/4 = (d+d)/4 = d/2 .... Hence SUFF.... and answer is B

Why would the order be required???

Cheers! JT _________________

Cheers! JT........... If u like my post..... payback in Kudos!!

|Do not post questions with OA|Please underline your SC questions while posting|Try posting the explanation along with your answer choice| |For CR refer Powerscore CR Bible|For SC refer Manhattan SC Guide|

What is the mean of four consecutive even integers a ,b ,c ,d ? i) a+d=b+c ii) d-a=b+c... although it is given as B ,i have a query... do we take the four consecutive even integers a ,b ,c ,d to be written in ascending order... if not then the ans should be C..... if the int are in ascending order , shudn't it be mentioned?

Dont really understand your question.....(in red)

The question is asking for the mean... of a,b,c,d all even integers... this means... we need to find the value of (a+b+c+d)/4 which would be an even integer for sure. ST1: gives a+d = b+c... This gives mean as (a+d)/2 or (b+c)/2.... hence NOT SUFF...

ST2: gives d-a = b+c --> d = a+b+c Therefore mean = (a+b+c+d)/4 = (d+d)/4 = d/2 .... Hence SUFF.... and answer is B

Why would the order be required???

Cheers! JT

u are giving ans as B... first why i think order is reqd... if the seq is a,b,c,d... x,x+2,x+4,x+6....SII gives d-a=b+c.... means x+6-x=x+4+x+2....x=0 so nos are 0,2,4,6 mean =3...... but the eq d-a=b+c can be satisfied if ascending order is a,b,d,c eg. -2,0,2,4... here d-a=2-(-2)=4=4+0=4... mean in this case is 1... so two diff ans not suff... SIn SII combined are suff..... i would like the moderators to clarify on the issue pl... _________________

What is the mean of four consecutive even integers a ,b ,c ,d ? i) a+d=b+c ii) d-a=b+c... although it is given as B ,i have a query... do we take the four consecutive even integers a ,b ,c ,d to be written in ascending order... if not then the ans should be C..... if the int are in ascending order , shudn't it be mentioned?

Dont really understand your question.....(in red)

The question is asking for the mean... of a,b,c,d all even integers... this means... we need to find the value of (a+b+c+d)/4 which would be an even integer for sure. ST1: gives a+d = b+c... This gives mean as (a+d)/2 or (b+c)/2.... hence NOT SUFF...

ST2: gives d-a = b+c --> d = a+b+c Therefore mean = (a+b+c+d)/4 = (d+d)/4 = d/2 .... Hence SUFF.... and answer is B

Why would the order be required???

Cheers! JT

u are giving ans as B... first why i think order is reqd... if the seq is a,b,c,d... x,x+2,x+4,x+6....SII gives d-a=b+c.... means x+6-x=x+4+x+2....x=0 so nos are 0,2,4,6 mean =3...... but the eq d-a=b+c can be satisfied if ascending order is a,b,d,c eg. -2,0,2,4... here d-a=2-(-2)=4=4+0=4... mean in this case is 1... so two diff ans not suff... SIn SII combined are suff..... i would like the moderators to clarify on the issue pl...

If the question says that a,b,c,d are 4 consecutive even integers... I don't think they would consider them to be as 8,2,6,4..... ! Anyway... lets wait for Bunuel or someone else to comment on this! For me... Ans as B looks pretty convincing! _________________

Cheers! JT........... If u like my post..... payback in Kudos!!

|Do not post questions with OA|Please underline your SC questions while posting|Try posting the explanation along with your answer choice| |For CR refer Powerscore CR Bible|For SC refer Manhattan SC Guide|

hi .. i also think it should be given in ascending order... want to confirm it so that not found wanting in future... i m not sure where but i had done a q somewhere ,where the ans took the possiblity of they not being in an order.. _________________

Re: What is the mean of four consecutive even integers a ,b ,c [#permalink]

Show Tags

08 Dec 2011, 12:25

chetan2u wrote:

What is the mean of four consecutive even integers a ,b ,c ,d ? i) a+d=b+c ii) d-a=b+c... although it is given as B ,i have a query... do we take the four consecutive even integers a ,b ,c ,d to be written in ascending order... if not then the ans should be C..... if the int are in ascending order , shudn't it be mentioned?

Statement 2: d-a=b+c

therefore d-c=a+b ; as we know c and d are consecutive even integers; d-c=2 which gives us a+b=2 there is only one pair of consecutive a and b that works; 0 and 2. Therefore statement 2 is sufficient alone.

What is the mean of four consecutive even integers a ,b ,c ,d ? i) a+d=b+c ii) d-a=b+c... although it is given as B ,i have a query... do we take the four consecutive even integers a ,b ,c ,d to be written in ascending order... if not then the ans should be C..... if the int are in ascending order , shudn't it be mentioned?

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.