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KarishmaB
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a, b, and c are integers and a < b < c. S is the set of all integers 07 Mar 2016, 23:00
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goldgoldandgold
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) 1/2
(C) 11/16
(D) 5/7
(E) 3/4
Show more


Since the question and options give all data in terms of fractions, we can assume values. The point is for which variable we should assume a value?
a < b < c
S = { a, ..., b } Median = (3/4)b
Q = {b, ..., c } Median = (7/8)c

Assume a value of c (which will lead to a value of b) such that c will get divided by 8 and 4 successively so that both medians are integers.
Say c = 32

Median of Q = (7/8)*32 = 28
b = 28 - 4 = 24

Median of S = (3/4)*24 = 18
a = 18 - 6 = 12

R = { 12, 13, 14, ... 32}
Median of R = 22 which as a fraction of c is 22/32 = (11/16)

Answer (C)
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a, b, and c are integers and a < b < c. S is the set of all integers 26 Mar 2016, 05:08
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Hi Bunuel, VeritasPrepKarishma


I think that assumption in this question is to have distinct numbers, numbers can not duplicate, for example:

6 , 7 , 8 , 9 , 10 , 11 , 12 , 13 , 14 , 15 , 16 here a=6 b= 12 c= 16 and conditions are fulfiled for 3/4*b and 7/8*c so final answer would be 11/16.

BUT if we include more 9's in set S and more 14's in set Q we will still have all numbers inclusive from a to c in the joint set, but than the median would not be 11, can be different numebr depending how many 9's and 14' are duplicating.

Do you think that is possable or I'm wrong

Thanks in advance for your respnce
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a, b, and c are integers and a < b < c. S is the set of all integers 29 Mar 2016, 00:05
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kzivrev
Hi Bunuel, VeritasPrepKarishma


I think that assumption in this question is to have distinct numbers, numbers can not duplicate, for example:

6 , 7 , 8 , 9 , 10 , 11 , 12 , 13 , 14 , 15 , 16 here a=6 b= 12 c= 16 and conditions are fulfiled for 3/4*b and 7/8*c so final answer would be 11/16.

BUT if we include more 9's in set S and more 14's in set Q we will still have all numbers inclusive from a to c in the joint set, but than the median would not be 11, can be different numebr depending how many 9's and 14' are duplicating.

Do you think that is possable or I'm wrong

Thanks in advance for your respnce
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By definition, a set is a collection of distinct objects. So we can easily say that all integers in the sets must be distinct.
Sometimes, some questions do specify sets with identical elements but strictly speaking, sets have distinct elements.
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a, b, and c are integers and a < b < c. S is the set of all integers 30 Aug 2016, 17:11
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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. 1/2
C. 11/16
D. 5/7
E. 3/4

Show more


Another option is to PLUG in numbers.

We're told that the median of set Q is (7/8)c
Let's let c = 8
This means the median = (7/8)(8) = 7
For the median to equal 7, the other integer in set Q must be 6
So, set Q = {6, 7, 8}

This means that b = 6

We're told that the median of set S is (3/4)b
This means the median of set S = (3/4)(6) = 4.5
Starting with b = 6, let's keep adding integers LESS THAN 6 to set S until we get a median of 4.5
Set S = {5, 6} (nope, the median is not 4.5)
Set S = {4, 5, 6} (nope, the median is not 4.5)
Set S = {3, 4, 5, 6} (YES, the median is 4.5!)
So, set S = {3, 4, 5, 6}

This means set R = {3, 4, 5, 6, 7, 8}
Here, c = 8 and the median = 5.5
What fraction of c is the median of set R?
5.5/8 = 11/16

Answer:
Show Spoiler
C

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a, b, and c are integers and a < b < c. S is the set of all integers 26 Dec 2016, 01:25
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Bunuel
Hello, can you please clarify how it is clear from basic condition that sets are evenly spaced or consecutive?
Thanks.
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a, b, and c are integers and a < b < c. S is the set of all integers 26 Dec 2016, 02:53
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Bunuel
Hello, can you please clarify how it is clear from basic condition that sets are evenly spaced or consecutive?
Thanks.
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We are told that S is the set of all integers from a to b, inclusive. For example, S can be from 1 to 10, inclusive: 1, 2, 3, 4, 5, 6, 7, 8, 9 and 10. In any way S will be set of consecutive integers.
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a, b, and c are integers and a < b < c. S is the set of all integers 04 Jul 2017, 18:29
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VeritasPrepKarishma
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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) 1/2
(C) 11/16
(D) 5/7
(E) 3/4


Since the question and options give all data in terms of fractions, we can assume values. The point is for which variable we should assume a value?
a < b < c
S = { a, ..., b } Median = (3/4)b
Q = {b, ..., c } Median = (7/8)c

Assume a value of c (which will lead to a value of b) such that c will get divided by 8 and 4 successively so that both medians are integers.
Say c = 32

Median of Q = (7/8)*32 = 28
b = 28 - 4 = 24

Median of S = (3/4)*24 = 18
a = 18 - 6 = 12

R = { 12, 13, 14, ... 32}
Median of R = 22 which as a fraction of c is 22/32 = (11/16)

Answer (C)
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Where does 28 come from?
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a, b, and c are integers and a < b < c. S is the set of all integers 05 Jul 2017, 09:50
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VeritasPrepKarishma
goldgoldandgold
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) 1/2
(C) 11/16
(D) 5/7
(E) 3/4


Since the question and options give all data in terms of fractions, we can assume values. The point is for which variable we should assume a value?
a < b < c
S = { a, ..., b } Median = (3/4)b
Q = {b, ..., c } Median = (7/8)c

Assume a value of c (which will lead to a value of b) such that c will get divided by 8 and 4 successively so that both medians are integers.
Say c = 32

Median of Q = (7/8)*32 = 28
b = 28 - 4 = 24

Median of S = (3/4)*24 = 18
a = 18 - 6 = 12

R = { 12, 13, 14, ... 32}
Median of R = 22 which as a fraction of c is 22/32 = (11/16)

Answer (C)

Where does 28 come from?
Show more

We assumed c = 32
Median of Q = (7/8)*c = (7/8)*32 = 28

Since all data is given in fractions and we are asked to give the answer in fraction too, assuming any suitable value of c will help get the answer.
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a, b, and c are integers and a < b < c. S is the set of all integers 06 Jul 2017, 10:49
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shaselai
is it C?

I kinda drawn it out like:

3< 4.5(M) < 6 < 7(M) <8
I assumed C was 8 and since M(median) is 7, b has to be 6 and then figured out a being 3. Now add them all and found to be 33/6 = 5.5 average and then found which answer multiplies by C(8) gives me 5.5 Obviously not an efficient way of doing it but at least it got some kinda of result...

Kudos for the question!
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Hi VeritasPrepKarishma

Is the way above where the person does not rely on the medians given but simply assumes numbers that fit the set correct?

Also, as long as the values are integers, the median does not have to be a whole number, right? I can have set S consist of [3,4,5,6] which would meet the requirements but the median in this case would be 4.5, i.e. not an integer.

Thanks.
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a, b, and c are integers and a < b < c. S is the set of all integers 07 Jul 2017, 02:57
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azelastine
shaselai
is it C?

I kinda drawn it out like:

3< 4.5(M) < 6 < 7(M) <8
I assumed C was 8 and since M(median) is 7, b has to be 6 and then figured out a being 3. Now add them all and found to be 33/6 = 5.5 average and then found which answer multiplies by C(8) gives me 5.5 Obviously not an efficient way of doing it but at least it got some kinda of result...

Kudos for the question!

Hi VeritasPrepKarishma

Is the way above where the person does not rely on the medians given but simply assumes numbers that fit the set correct?

Also, as long as the values are integers, the median does not have to be a whole number, right? I can have set S consist of [3,4,5,6] which would meet the requirements but the median in this case would be 4.5, i.e. not an integer.

Thanks.
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Yes, it is correct. The person does rely on the medians to get the elements of the set.
You start with c = 8, get median as (7/8)*c so median is 7. This would mean that the set Q would be {6, 7, 8}. This means b = 6.
So median of (3/4)b = 4.5. Set S would need median of 4.5 which means its elements will be {3, 4, 5, 6}. Median of even number of consecutive integers will be a non-integer.
Set R = {3, 4, 5, 6, 7, 8}
Its median is 5.5.
Fraction = 5.5/8 = 11/16

Yes, median of a set of integers can certainly be a non integer.
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a, b, and c are integers and a < b < c. S is the set of all integers 15 Sep 2020, 11:43
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KocharRohit
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) 1/2
(C) 11/16
(D) 5/7
(E) 3/4
Show more

Q is the set of all integers from b to c, inclusive.
The median of set Q is (7/8)c.
Let \(c=16\).
Median of Q \(= \frac{7}{8}*c = \frac{7}{8}*6 = 14\).
Since the median of 14 must be halfway between \(b\) and c=16, \(b=12\).

S is the set of all integers from a to b, inclusive.
The median of set S is (3/4)b.
Median of S \(= \frac{3}{4}*b = \frac{3}{4}*12 = 9\).
Since the median of 9 must be halfway between \(a\) and b=12, \(a=6\).

The median of set R -- which is composed of all of the integers from a=6 to c=16, inclusive -- is equal to the the average of 6 and 16:
\(\frac{6+16}{2} = 11\)

What fraction of c is the median of set R?
\(\frac{median-of-R}{c} = \frac{11}{16}\)

Show Spoiler
C
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a, b, and c are integers and a < b < c. S is the set of all integers 15 Sep 2020, 13:21
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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) 1/2
(C) 11/16--> correct: median and mean for consecutive integers are same. so \(\frac{a+b}{2} =\frac{3c}{4} => b=2a \) & \(\frac{b+c}{2} =\frac{7c}{8} => 4b=3c => c = \frac{8a}{3}\), a, b & c are integers, so let's say a=3, so b = 2*3=6 & c = 8, so median of \(R = \frac{a+c}{2} =\frac{3+8}{2} =\frac{11}{2} = \frac{11}{16}*8 = \frac{11}{16}*c\)
(D) 5/7
(E) 3/4
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a, b, and c are integers and a < b < c. S is the set of all integers 18 Dec 2020, 00:18
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VeritasKarishma
Hello,

Can we take the series as S= (1,3,4), Q= (4,7,8) and R= (1,3,4,7,8),
In this case the median of R = 4 is 1/2 of the value of 8.
Also all given conditions are fulfilled by the set of numbers.
How can we consider the series is in A.P?
Please help.
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a, b, and c are integers and a < b < c. S is the set of all integers 20 Dec 2020, 23:57
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Sonalshrivastav
VeritasKarishma
Hello,

Can we take the series as S= (1,3,4), Q= (4,7,8) and R= (1,3,4,7,8),
In this case the median of R = 4 is 1/2 of the value of 8.
Also all given conditions are fulfilled by the set of numbers.
How can we consider the series is in A.P?
Please help.
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Sonalshrivastav

The questions specifies this.

"S is the set of all integers from a to b, inclusive."
So if a = 1 and b = 4, set S has to be {1, 2, 3, 4}

Similarly for Q and R too.
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a, b, and c are integers and a < b < c. S is the set of all integers 24 Jan 2022, 09:29
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KocharRohit
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) 1/2
(C) 11/16
(D) 5/7
(E) 3/4

Show Spoiler
\sqrt{I solved it lke this...
for set S (a+b)/ 2 = (3/4 ) B
for set Q (b+c)/2 = (7c)/8
after solving we got
3c = 4b = 8a
median for set R is b so
4b = 3c
b = (3c)/4..

got E ..plasde tell me am i wrong somewhere?
Rohit}
Show more

median of consecutive integers a to b = (a+b)/2
median of consecutive integers b to c = (b+c)/2
median of consecutive integers a to c = (a+c)/2

(a+b)/2 = 3/4 b => b = 2a

(b+c)/2 = 7/8c => 4b = 3c

=> 8a = 3c

median of consecutive integers a to c = (a+c)/2 = (3c/8+c)/2 = 11c/16

Answer = 11/16
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a, b, and c are integers and a < b < c. S is the set of all integers 07 Oct 2025, 10:50
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Hello, Can we use a shortcut like looking at the numbers, the numerator is increasing by 4 and denominator is multiplied by 2 so the next answer would be 11c/16 or is it just a coincedence?
Bunuel


Given that S is the set of all integers from a to b, inclusive, Q is the set of all integers from b to c, inclusive and R is the set of all integers from a to c, inclusive, so sets S, Q and R have to be consecutive integers sets. For any set of consecutive integers (generally for any evenly spaced set) median (also the mean) equals to the average of the first and the last terms.

So we have:
Median of \(S=\frac{a+b}{2}=b*\frac{3}{4}\) --> \(b=2a\);

Median of \(Q=\frac{b+c}{2}=c*\frac{7}{8}\) --> \(b=c*\frac{3}{4}\) --> \(2a=c*\frac{3}{4}\) --> \(a=c*\frac{3}{8}\);

Median of \(R=\frac{a+c}{2}=\frac{c*\frac{3}{8}+c}{2}=c*\frac{11}{16}\)

Answer: C (\(\frac{11}{16}\)).
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