gabriel wrote:
1.) 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

Lets call the numbers arranged in ascending order as below

x a b c d y

Median = 6, so (b+c)/2 = 6

Mode = 7, so this number should repeat atleast twice. Since median is
6, b cannot be 7. That leaves us with c = d = 7.

If c = 7, b = 5 as the median is 6.

Mean = 7, so x + a + b + c + d + y = 42

In effect, the sequence now is

x a 5 7 7 y

1 - sufficient. x+a = 1/5 (d+y)

We know mean = 7, so
(x + a) + b + c + (d + y) = 42
1/5 (d+y) + 5 + 7 + (d+y) = 42
6/5 (d+y) = 30
d+y = 25
Since d = 7, y = 18

So, x = 18-16 = 2 (using range)

Using mean, 2 + a + 5 + 7 + 7 + 18 = 42, so a = 3

Sequence is 2 3 5 7 7 18

Sufficient.


2 - insufficient. We already know that 2 numbers are 5 and 7

x a 5 7 7 y sequence can take any positive or negative values
for x, a and y that satisfies all the constraints.


Hence, I think the answer is A.

jayhawk2001 wrote:

gabriel wrote:
1.) 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

Lets call the numbers arranged in ascending order as below

x a b c d y

Median = 6, so (b+c)/2 = 6

Mode = 7, so this number should repeat atleast twice. Since median is
6, b cannot be 7. That leaves us with c = d = 7.

If c = 7, b = 5 as the median is 6.

Mean = 7, so x + a + b + c + d + y = 42

In effect, the sequence now is

x a 5 7 7 y

1 - sufficient. x+a = 1/5 (d+y)

We know mean = 7, so
(x + a) + b + c + (d + y) = 42
1/5 (d+y) + 5 + 7 + (d+y) = 42
6/5 (d+y) = 30
d+y = 25
Since d = 7, y = 18

So, x = 18-16 = 2 (using range)

Using mean, 2 + a + 5 + 7 + 7 + 18 = 42, so a = 3

Sequence is 2 3 5 7 7 18

Sufficient.


2 - insufficient. We already know that 2 numbers are 5 and 7

x a 5 7 7 y sequence can take any positive or negative values
for x, a and y that satisfies all the constraints.


Hence, I think the answer is A.

hmmm... more takers ?

jayhawk2001 wrote:
Lets call the numbers arranged in ascending order as below

x a b c d y

Median = 6, so (b+c)/2 = 6

Mode = 7, so this number should repeat atleast twice. Since median is
6, b cannot be 7. That leaves us with c = d = 7.

If c = 7, b = 5 as the median is 6.

Mean = 7, so x + a + b + c + d + y = 42

In effect, the sequence now is

x a 5 7 7 y

1 - sufficient. x+a = 1/5 (d+y)

We know mean = 7, so
(x + a) + b + c + (d + y) = 42
1/5 (d+y) + 5 + 7 + (d+y) = 42
6/5 (d+y) = 30
d+y = 25
Since d = 7, y = 18

So, x = 18-16 = 2 (using range)

Using mean, 2 + a + 5 + 7 + 7 + 18 = 42, so a = 3

Sequence is 2 3 5 7 7 18

Sufficient.


2 - insufficient. We already know that 2 numbers are 5 and 7

x a 5 7 7 y sequence can take any positive or negative values
for x, a and y that satisfies all the constraints.


Hence, I think the answer is A.

excellent dude ... thats the answer..

.. for the other members look at jays solution and u will see how u r supposed to corelate the various information given in a DS .. also see how different concepts are being tested for the same q..

I have to go and check what is "mode".

Great explanation there Jayhawk!

and nice to see somebody with avatar

Nice Question and good explanation