Problem Solving

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

Lets take the integers in each list have a common difference between the elements since there is not a restriction

Median S = a+b/2 (integers are consecutive or in an AP Mean=Median=average of lowest and highest element)

Median Q = b+c/2

a+b/2 = 3/4b

a = b/2

b+c/2 = 7/8c

b = 3c/4

Median of a to c is a+c/2

= b/2 + c / 2
= 3c/8+c/2
= 11c/16

C

Hi Cramya,

That was a good solution, my work was similar to yours but I arrived at a different solution.

S = [a,b]
Q = [b,c]

Since set S has 2 numbers, the median in this set = the mean.

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

(3/2)b = a+b; a = (1/2)b

Same with set Q

(7/8)c = (b+c)/2

(7/4)c = b+c; b = (3/4)c; c = (4/3)b

Now set R has [a, b, c] which reproducing from above [1/2b, b, 4/3b]

here b is the median and since b = 3/4c, I got the answer as (E)

Am I doing something wrong here?

a+b/2 = 3/4b

a = b/2

Cramya, I think a= b/4, if we are using the above relationship for calculating a


Similarly,

b+c/2 = 7/8c

b = 3c/4

b = 3c/8

Please explain how you got the answer as a= b/2 and b=3c/4

Thanks

ssmiles08 wrote:Hi Cramya,

That was a good solution, my work was similar to yours but I arrived at a different solution.

S = [a,b]
Q = [b,c]

Since set S has 2 numbers, the median in this set = the mean.

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

(3/2)b = a+b; a = (1/2)b

Same with set Q

(7/8)c = (b+c)/2

(7/4)c = b+c; b = (3/4)c; c = (4/3)b

Now set R has [a, b, c] which reproducing from above [1/2b, b, 4/3b]

here b is the median and since b = 3/4c, I got the answer as (E)

Am I doing something wrong here?

I think you're assuming that the two sets have the same number of elements and then concluding that the median is b. According to your set

R ={1/2b,....,b, ...., 4/3b} if you calculate the mean ---> (1/2b + 4/3b )/2 = 11/12b that is not b, but if you substitute b=3/4c the median is 11/16c

As always, I liked the solutions that you guys have given..

I also have one solution.. different approach.. see if you feel comfy with this.. I am not choosing number of elements in any set here at all.. .. here we go:

Given

1] Median of a and b is 3/4b: If you just think about this.. when is this possible? only when a = b/2.. isnt it? after all its the median between a and b
So a = b/2 --------------- (1)

2] Median of b and c is 7/8c: Again, If you just think about this.. when is this possible? only when b = 6c/8.. isnt it? after all its the median between b and c -------------- (2)

All I have done above is to reduce the numerator by 1 in the fractions.. . That's all. But then its logical.. coz as we know median is the middle number in the sequence.. So if my largest number is b and middle number is 3/4b... this implies my smallest number has to be 2/4b.. similarly if my largest number is c and middle number is 7/8c.. this implies that my smallest number has to be 6/8c

so what do I have?

a = b/2 and b = 6/8c

if you just choose c = 8, I get the following numbers:

a = 3, b = 6 and c = 8

Full set is then - 3,4,5,6,7,8 Median = 11/2

and so fraction = Median/C = 11/16

Hey dtweah

From where you get all these questions i mean what is the source.
All your questions are mostly difficult.

abhinav85 wrote:Hey dtweah

From where you get all these questions i mean what is the source.
All your questions are mostly difficult.

I only practice the difficult stuff. If you are to de well on the GMAT you will not see too many easy stuff at least before the test questions hit a plateau. By Question number 15 to 20 I believe GMAT CAT has figured your mean and Standard deviation, and according to this Dreaded IRON Law, Humans don't stray too much from their Mean. So if you ever Guessed the first 20 questions correct on the GMAT, just know that you have won the game. No matter what you do beyond, you can't get below 700!!. I have proven this with GMAT Prep many times. Since the most difficult questions will come first, that is if you are doing well, I love them.

About source: Go to the Beat the GMAT section and read Eldar's post. Scroll thru to the end. And Feast on some good stuff.

Thanks for your opinion!!!

"Go to the Beat the GMAT section and read Eldar's post."

were is this post, i cant find it??

abhinav85 wrote:Thanks for your opinion!!!

"Go to the Beat the GMAT section and read Eldar's post."

were is this post, i cant find it??

I said go to "I just Beat the GMAT " section and look for Eldar's post. It is there. I will not post the link here.