good DS problem

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gabriel: Legendary Member; Posts: 986; Joined: Wed Dec 20, 2006 11:07 am; Location: India; Thanked: 51 times; Followed by:1 members

good DS problem

by gabriel » Sat Apr 07, 2007 10:04 am

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

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Source: Beat The GMAT — Data Sufficiency | Sat Apr 07, 2007 10:04 am

jayhawk2001: Community Manager; Posts: 789; Joined: Sun Jan 28, 2007 3:51 pm; Location: Silicon valley, California; Thanked: 30 times; Followed by:1 members

Re: good DS problem

by jayhawk2001 » Sat Apr 07, 2007 11:55 am

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.

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gabriel: Legendary Member; Posts: 986; Joined: Wed Dec 20, 2006 11:07 am; Location: India; Thanked: 51 times; Followed by:1 members

Re: good DS problem

by gabriel » Sun Apr 08, 2007 6:52 am

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 ?

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gabriel: Legendary Member; Posts: 986; Joined: Wed Dec 20, 2006 11:07 am; Location: India; Thanked: 51 times; Followed by:1 members

Re: good DS problem

by gabriel » Mon Apr 09, 2007 12:51 pm

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..

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