Six numbers are randomly selected and placed within a set.

Question

Six numbers are randomly selected and placed within a set. If the set has 
a range of 16, a median of 6, a mean of 7 and a mode of 7, what is the 
greatest of the six numbers?
(1) The sum of the two smallest numbers is one-fifth of the sum of the two greatest numbers
(2) The middle two numbers are 5 and 7

Please give explanations of your answers.

saikadelik · Answer

let the set s =  

(c+d)/2 = 6
c+d = 12 -----Eq 1

f-a = 16 ------Eq 2

(a+b+c+d+e+f)/6 = 7
(a+b+c+d+e+f) = 42 -----Eq 3

Three equations, six unknowns.

(1) (a+b) = (e+f)/5
 5a+5b = e+f -----Eq 4
 insuff

(2) c=5 -----Eq 5
 d=7------Eq 6
 either e or f can be 7 since the mode is 7.

(1) & (2), sufficient. 6 equations, 6 unknowns.

Answer is C.

trivikram · Answer

E for me

-2,7,5,7,11,14

2,3,5,7,7,18

Caas · Answer

Try again guys. OA is neither C (wich BTW I got too) nor E.

Caas · Answer

-2, 7,5,7,11,14 The median is not 6 here. It is 7. Reorder the set from the smallest to the greatest

vikramjit_01 · Answer

Then it's D.
My first answer was E but after reading this....

Caas

OA is neither C (wich BTW I got too) nor E.

... I solved again and it took some (well, a lot more than that) time to get D.

from the stem:
Median is 6 & mode is 7 so series and the total is 42 (7*6 ie mean into no. of terms) could be
_ _ _77_
it can be nothing else ...figuring this took a lot of time..maybe it'll be better to guess in the GMAT.
If all terms were 7, the median wont be 6 and the if the last 3 terms were 7, the total won't add up to 42.
since the median is 6 the series will be
_ _ 577_ no other way to get a 6 median
now, lets chk the statements

I:
the sume of the middle 2 no's is 12 (7+5) so the other terms add up to 30
and the first 2 terms are 1/5,
or the last 2 terms are 4/5 of 30
and thats 24
One of these terms is 7 so the other is 17
.....................suff

II:
this was tougher to figure out and took longer.
so the series is
_ _577_
we know range = 16
so let the series be
x y 577 (16+x)
so
x + y + 16 + x = 23
2x + y = 7

so the first 2 number's of the series can only be
1,5 or
2,3

nothing else satisfies the condition.
but if the first 2 numbers were 1,5
then the series would be bi-modal (2 5's & 2 7's)
so the first 2 numbers are
2,3
we can find out the largest number.....................suff :shocked

Caas · Answer

Vikramjit, your explanation is very good. But you missed one small point. It is not said that all the numbers in the set are poisitve integers

GMAThopeful · Answer

Let the numbers be a,b,c,d,e,f arranged in ascending order
Average is 7, so a+b+c+d+e+f = 42
Since median is 6, average of c and d is 6, i.e. c+d = 12
hence a+b+e+f = 42 -12 = 30

St 1: (a+b) = (e + f) / 5
hence (e + f)/5 + (e+f) = 30
Solving we get (e+f) = 25

Now mode is 7. That means at least 2 numbers in the set are equal to 7. 
As the median is 6, a & b cannot be equal to 7. 
AS median is 6, c=5 & d = 7 or c=6 & d=6
If c=6 & d=6, then as the mode is 7, e = 7 and f = 7. But this violates (e+f) = 25. So c and d both cannot be equal to 6. As a result, we have c=5 & d = 7. As mode is 7, e = 7. f cannot be equal to 7 as (e+f) = 25. Hence f = 25-7 = 18

SUFFICIENT

St 2: c =5, d = 7. 
We cannot deduce anything from this.
Not sufficient.

HENCE A.

seofah · Answer

I agree with Gmathopeful.
A is correct.


On statement 1, Gmat Hopeful is to the point.

On statement 2, from the fact that d=7, we can deduce e=7 as well.
In the end we will remain with 1 equation where variables are b and f, hence INSUFFICIENT.

vshaunak@gmail.com · Answer

I agree with GMAThopeful's explanation.

Make the equation 
a+b+c+d+e+f = 42 ------------------------------1
a+b = (e+f)/5 ===> e+f = 5(a+ b) --------------------2
range = 16 ==> f - a = 16 ------------------------------3
median = 6 ==> (c+d)/2 = 6 ==> c+d = 12 ---------------4
Putting this values in eq 1

a+b = 5 ===> e+f = 25

lets try out numbers: range is 16. Possible values of a = 2 and b = 18.
lets put the numbers:
2,3,5,7,7,18
now try out all the equations whether they are satisfied. Yes they are satisfied. there may be another set .
2,4,5,7,7,17 all conditions except the e+f is not satisfied. 
Hence there will be the unique set to satisfy all the equations. 
Hence SUFF.

Statement 2 gives two sets:
2,3,5,7,7,18
2,4,5,7,7,17
hence NOT suff.

So answer is 'A'

trivikram · Answer

Yep..It should be A..... :sleep

Caas · Answer

Good job! OA is A indeed

Six numbers are randomly selected and placed within a set.

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