Great question, finance!

I'll let everyone get a fair shot at this one before I solve it, but let me offer a strategic tip here, one we call "Why Are You Here?". If a statement on its own is clearly (and almost insultingly) insufficient, it's there for a reason. Either you need it with statement 1 or they're trying to make you think you need it. Either way, you have to consider it specifically because the author of the question wouldn't use one of only three items he gets to write as a total throwaway. It's there for a reason.

So my question is: Why does statement 2 only allow for one even number?

IMO C

Let's say the 5 integers are 1, a, b, c and d. ( in ascending order).
Also, a, b, c and d all are positive distinct integers.

Statement 1:

(1 + a + b + c + d)/5 = 4
(1 + a + b + c + d) = 20

As median is also 4 we get b = 4.
So a can be either 2 or 3
Case 1: a = 2
(1 + 2 + 4 + c + d) = 20
c + d = 20 - ( 1 + 2 + 4) = 13
We know c has to be greater than 4.
if c = 5 then d = 8
if c = 6 then d = 7
if c = 7 then d = 6 ( Invalid as d > c)

Case 2: a = 3
(1 + 3 + 4 + c + d) = 20
c + d = 20 - ( 1 + 3 + 4) = 12
We know c has to be greater than 4.
if c = 5 then d = 7
if c = 6 then d = 6 ( Invalid as c and d are different integer)

So, St 1 is Insufficient.

Statement 2:
St 2 is insufficient as the only thing we know is that there is only 1 even integer.

Combining 1 and 2 :
Case 1 of St 1 explained above ( a = 2) is not possible as we know that c = 4 and no other integer in the set can be even.
So a = 3.
for a = 3, c must be 5 and d must be 7 (refer case 2 of statement 1 above) . Sufficient.

finance wrote:What is the greatest integer in a set of
five different positive integers if 1 is the
smallest integer in the set?
(1) The average (arithmetic mean) of
all integers is 4 and the median of
the set is also 4.
(2) The set contains only 1 even
number.

for sure i am wrong but my pick is A
we have set of integers
1.a2.a3.a4.a5, and need to find a5 -greatest
(1)1+a2+a3+a4+a5=20, with median a3=4,
to make a5 greatest, first least value for a2 is 2, and least for a4=5, the remainder for a5=20-(1+2+4+5)=8,so given set looks like
1,2,4,5,8
as for st 2 really don`t know how to apply, but want to replenish my mind

Brian@VeritasPrep wrote:Great question, finance!

I'll let everyone get a fair shot at this one before I solve it, but let me offer a strategic tip here, one we call "Why Are You Here?". If a statement on its own is clearly (and almost insultingly) insufficient, it's there for a reason. Either you need it with statement 1 or they're trying to make you think you need it. Either way, you have to consider it specifically because the author of the question wouldn't use one of only three items he gets to write as a total throwaway. It's there for a reason.

So my question is: Why does statement 2 only allow for one even number?

Hi Brian!

Thank you for your approach! I was mistaken to choose A because of the well-known statement"when numbers in a set are evenly distributed, then mean equal to median" but I see it now that the reverse of this statement does not have to be true...even though there is still somehow a symmetrical distribution of the numbers..

I am tryin my luck here, I wud look at it this way:

Statement 1:

If we look at the same mean and median we can assume it to be an evenly spaced series and go ahead and solve. This gives the number as 7.

It seems Statement 1 is sufficient

Statement 2:

But as soon as you read statement 2 u realize it cant be an evenly spaced series, so your earlier assumption was wrong.
Statement 2 on its own is INSUFFICIENT. But it helps you determine that statement 1 also was Insufficient.

Statement 1 & 2 together:

The mean and median is 4. The first term is 1. Since the median is 4 it has to be the 3rd term.

Now the terms are 1,a,4,b,c.

Now since their is just 1 even number, a has to be 3 (the only odd between 1 and 4).

So the terms are 1,3,4,b,c.

Now the mean is (8+b+c)/5=4.

i.e b+c = 12;

two possibilities 5,7 or 6,6. Since only one odd allowed and that was 4. it has to be 5,7.

So the largest integer is 7.

TOGETHER THEY ARE SUFFICIENT.

Kindly let me know if my approach was fine.. or where did i miss out.

Thanks

Great work on this one, everyone - and akhilsuhag you're right on with your approach once you got to statement 2.

For both finance and akhilsuhag - you nailed it in your explanations; while it's true that for an evenly spaced set the mean and median are the same, that doesn't mean that the inverse is also true.

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One way to think about this one is that, to average 4, you need an equal amount above 4 and below 4. And you're already at -3 by using 1 (you're 3 below the average). Your options for x are 2 and 3, and you can note those as -2 and -1 if you're just gauging distance from the mean.

So on the left hand side of the average, 4, you have either -4 or -5, meaning that you have 4 or 5 numbers to play with on the right. So you can either have +1 and +3 (because you can't have them equal), +2 and +3, or +1 and +4.

NOTE: This is pretty convoluted in writing but if you get the logic, what I really did much more quickly and intuitively in my head was say that my options are:

1, 2, 4 -- I need to distribute 5 places above 4, so:

1, 2, 4, 5, 8
or
1, 2, 4, 6, 7

1, 3, 4 -- I need to distribute 4 places above 4, so:

1, 3, 4, 5, 7

And statement 2 tells me that I can't use the first option, so I'm stuck with the last one and the high end of 7.

Strategically, you can use statement 2 to make statement 1 pretty clear. Say you thought statement 1 was clearly not sufficient - statement 2 forces you to try numbers just to get a feel for what is even possible. It puts a strange restriction on what is even possible, so it makes you work to make the options more concrete. The fact that they said you couldn't use even numbers means you just have to try a few numbers to see whether that's a severe restriction or not. Could you have used even numbers? (Yes.) Are there many options with only odd numbers? (Nope. Just one, so it's sufficient with both statements together)

finance wrote:What is the greatest integer in a set of
five different positive integers if 1 is the
smallest integer in the set?
(1) The average (arithmetic mean) of
all integers is 4 and the median of
the set is also 4.
(2) The set contains only 1 even
number.

the answer should be C.
for eg. if there are 5 integers in the set than the mean is 20 (5*4=20)
than the only option would be 1,3,4,5,7 (Thus the mean and median are both 4). The greatest integer is 7
But if there are only 3 integer than the mean is 12 (3*4= 12).
than the only option would be 1,4,7 (Thus the mean and median are both 4). These are the only two option possible . we can't create another set in which there is only 1 even no and the median and mean to be 4 because if we try to take digit 5 than the condition of 4 being a median will not be possible