ok guyz a very good question ... source - the net
giv it a try ... i assure u that u will learn something new
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
good DS problem
This topic has expert replies
- jayhawk2001
- Community Manager
- Posts: 789
- Joined: Sun Jan 28, 2007 3:51 pm
- Location: Silicon valley, California
- Thanked: 30 times
- Followed by:1 members
Lets call the numbers arranged in ascending order as belowgabriel 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
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.
- gabriel
- Legendary Member
- Posts: 986
- Joined: Wed Dec 20, 2006 11:07 am
- Location: India
- Thanked: 51 times
- Followed by:1 members
jayhawk2001 wrote:Lets call the numbers arranged in ascending order as belowgabriel 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
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 ?
- gabriel
- Legendary Member
- Posts: 986
- Joined: Wed Dec 20, 2006 11:07 am
- Location: India
- Thanked: 51 times
- Followed by:1 members
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..
- aim-wsc
- Legendary Member
- Posts: 2469
- Joined: Thu Apr 20, 2006 12:09 pm
- Location: BtG Underground
- Thanked: 85 times
- Followed by:14 members
I have to go and check what is "mode".
Great explanation there Jayhawk!
and nice to see somebody with avatar![Smile :)](./images/smilies/smile.png)
Great explanation there Jayhawk!
and nice to see somebody with avatar
![Smile :)](./images/smilies/smile.png)
Getting started @BTG?
Beginner's Guide to GMAT | Beating GMAT & beyond
Please do not PM me, (not active anymore) contact Eric.
Beginner's Guide to GMAT | Beating GMAT & beyond
Please do not PM me, (not active anymore) contact Eric.