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
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- jayhawk2001
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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
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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
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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
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I have to go and check what is "mode".
Great explanation there Jayhawk!
and nice to see somebody with avatar
Great explanation there Jayhawk!
and nice to see somebody with avatar
Getting started @BTG?
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Beginner's Guide to GMAT | Beating GMAT & beyond
Please do not PM me, (not active anymore) contact Eric.