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integers----> nice question

This topic has 4 expert replies and 5 member replies

integers----> nice question

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a, b, and c are integers and a < b < c. S is the set of all integers from a to b, inclusive. Q is the set of all integers from b to c, inclusive. The median of set S is (3/4) b. The median of set Q is (7/8) c. If R is the set of all integers from a to c, inclusive, what fraction of c is the median of set R?

(A) 3/8 (B) ½ (C) 11/16 (D) 5/7 (E) ¾

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atulmangal wrote:
a, b, and c are integers and a < b < c. S is the set of all integers from a to b, inclusive. Q is the set of all integers from b to c, inclusive. The median of set S is (3/4) b. The median of set Q is (7/8) c. If R is the set of all integers from a to c, inclusive, what fraction of c is the median of set R?

(A) 3/8 (B) ½ (C) 11/16 (D) 5/7 (E) ¾
We can plug in a value for c.

Let c=16.
Median of Q = (7/8)*c = (7/8)*16 = 14.
Since 14 is halfway between b and 16, b=12.
Median of S = (3/4)*b = (3/4)*12 = 9.
Since 9 is halfway between a and 12, a=6.
Thus, one combination that satisfies all the conditions given is a=6, b=12, c=16.

Since R is the set of all the integers from a to c, inclusive, R = {6,7,8,9,10,11,12,13,14,15,16}.
Median of R = 11.
(Median of R)/c = 11/16.

The correct answer is C.

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Last edited by GMATGuruNY on Mon Dec 05, 2011 6:58 pm; edited 4 times in total

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hi Mitch,

As usual a great solution from your side. However I dont understand 1 thing: In set R , the median should be the middle value i.e 12..rite..??

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GMATGuruNY wrote:
atulmangal wrote:
a, b, and c are integers and a < b < c. S is the set of all integers from a to b, inclusive. Q is the set of all integers from b to c, inclusive. The median of set S is (3/4) b. The median of set Q is (7/8) c. If R is the set of all integers from a to c, inclusive, what fraction of c is the median of set R?

(A) 3/8 (B) ½ (C) 11/16 (D) 5/7 (E) ¾
We can plug in a value for c.

Let c=16.
Median of Q = (7/8)*c = (7/8)*16 = 14.
Since 14 is halfway between b and 16, b=12.
Median of S = (3/4)*b = (3/4)*12 = 9.
Since 9 is halfway between a and 12, a=6.
Thus, one combination that works for set R is a=6, b=12, c=16.

Given any set of consecutive integers:
Median = average = (biggest+smallest)/2.
Thus, median of R = (16+6)/2 = 11.

(Median of R)/c = 11/16.

The correct answer is C.
I did not get the point. Is the rule " Given any set of consecutive integers:
Median = average = (biggest+smallest)/2." valid for any consecutive numbers. What about the set of numbers:4,5,7,10???

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finance wrote:
Quote:
a, b, and c are integers and a < b < c. S is the set of all integers from a to b, inclusive. Q is the set of all integers from b to c, inclusive. The median of set S is (3/4) b. The median of set Q is (7/8) c. If R is the set of all integers from a to c, inclusive, what fraction of c is the median of set R?

(A) 3/8 (B) ½ (C) 11/16 (D) 5/7 (E) ¾
I did not get the point. Is the rule " Given any set of consecutive integers:
Median = average = (biggest+smallest)/2." valid for any consecutive numbers. What about the set of numbers:4,5,7,10???
Hi!

4, 5, 7 and 10 are not consecutive numbers, so the rule doesn't apply.

"Consecutive" means that the numbers are a fixed distance apart. Some examples of sets of consecutive numbers to which the rule applies:

{1, 2, 3, 4}
{3, 6, 9, 12}
{-10, -5, 0, 5, 10}
{11, 14, 17, 20}

In your set, there's a space of 1 between 4&5, a space of 2 between 5&7 and a space of 3 between 7&10; since the spacing varies (even though it follows a rule), the numbers aren't considered consecutive.

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Stuart Kovinsky wrote:
finance wrote:
Quote:
a, b, and c are integers and a < b < c. S is the set of all integers from a to b, inclusive. Q is the set of all integers from b to c, inclusive. The median of set S is (3/4) b. The median of set Q is (7/8) c. If R is the set of all integers from a to c, inclusive, what fraction of c is the median of set R?

(A) 3/8 (B) ½ (C) 11/16 (D) 5/7 (E) ¾
I did not get the point. Is the rule " Given any set of consecutive integers:
Median = average = (biggest+smallest)/2." valid for any consecutive numbers. What about the set of numbers:4,5,7,10???
Hi!

4, 5, 7 and 10 are not consecutive numbers, so the rule doesn't apply.

"Consecutive" means that the numbers are a fixed distance apart. Some examples of sets of consecutive numbers to which the rule applies:

{1, 2, 3, 4}
{3, 6, 9, 12}
{-10, -5, 0, 5, 10}
{11, 14, 17, 20}

In your set, there's a space of 1 between 4&5, a space of 2 between 5&7 and a space of 3 between 7&10; since the spacing varies (even though it follows a rule), the numbers aren't considered consecutive.
Thank you for your explanation! I know "consecutive integers" just as you described above, but I got a little bit confused from the provided solution because the given numbers a=6, b=12 and c=16 are said to be consecutive, but they have not equal distances among each other??

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finance wrote:
Stuart Kovinsky wrote:
finance wrote:
Quote:
a, b, and c are integers and a < b < c. S is the set of all integers from a to b, inclusive. Q is the set of all integers from b to c, inclusive. The median of set S is (3/4) b. The median of set Q is (7/8) c. If R is the set of all integers from a to c, inclusive, what fraction of c is the median of set R?

(A) 3/8 (B) ½ (C) 11/16 (D) 5/7 (E) ¾
I did not get the point. Is the rule " Given any set of consecutive integers:
Median = average = (biggest+smallest)/2." valid for any consecutive numbers. What about the set of numbers:4,5,7,10???
Hi!

4, 5, 7 and 10 are not consecutive numbers, so the rule doesn't apply.

"Consecutive" means that the numbers are a fixed distance apart. Some examples of sets of consecutive numbers to which the rule applies:

{1, 2, 3, 4}
{3, 6, 9, 12}
{-10, -5, 0, 5, 10}
{11, 14, 17, 20}

In your set, there's a space of 1 between 4&5, a space of 2 between 5&7 and a space of 3 between 7&10; since the spacing varies (even though it follows a rule), the numbers aren't considered consecutive.
Thank you for your explanation! I know "consecutive integers" just as you described above, but I got a little bit confused from the provided solution because the given numbers a=6, b=12 and c=16 are said to be consecutive, but they have not equal distances among each other??
The problem states that R is the set of all the integers from a to c, inclusive.
Thus, if a=6 and c=16, R = {6,7,8,9,10,11,12,13,14,15,16}.

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Private Tutor for the GMAT and GRE
GMATGuruNY@gmail.com

If you find one of my posts helpful, please take a moment to click on the "UPVOTE" icon.

Available for tutoring in NYC and long-distance.
For more information, please email me at GMATGuruNY@gmail.com.
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Last edited by GMATGuruNY on Tue May 10, 2011 12:03 pm; edited 1 time in total

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Thank you GMATGuruNY! I enjoyed your solution, but contrary to what you stated in your last reply, R is given as the set of all integers from a to c, inclusive, without mentioning it to be "consecutive". I thought that you found out that those terms are consecutive, because it is not mentioned in the question.

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finance wrote:
Thank you GMATGuruNY! I enjoyed your solution, but contrary to what you stated in your last reply, R is given as the set of all integers from a to c, inclusive, without mentioning it to be "consecutive". I thought that you found out that those terms are consecutive, because it is not mentioned in the question.
Hi,

if R is a set of all integers between two specific integers, inclusive, then by definition R will be a set of consecutive integers.

For example, if a=2 and c=10, then R = {2, 3, 4, 5, 6, 7, 8, 9, 10}.

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Stuart Kovinsky wrote:
finance wrote:
Thank you GMATGuruNY! I enjoyed your solution, but contrary to what you stated in your last reply, R is given as the set of all integers from a to c, inclusive, without mentioning it to be "consecutive". I thought that you found out that those terms are consecutive, because it is not mentioned in the question.
Hi,

if R is a set of all integers between two specific integers, inclusive, then by definition R will be a set of consecutive integers.

For example, if a=2 and c=10, then R = {2, 3, 4, 5, 6, 7, 8, 9, 10}.
Ohh I see! I have to read the questions more carefully. Thank you Stuart!

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