Tricky one- Got stumped

I got C
the seven numbers i took are
m-3, m-2, m-1, m, 2m-2,2m-1,2m


10m/7 - 9/7... what is the OA

IMO C.

Max number = 2m

median = m

[m-3 , m-2 , m-1 , m , 2m-2 , 2m-1 , 2m ]

sum of all 7 numbers = 10m - 9
mean = (10m - 9 )/ 7

may i knw the source?

ankur.agrawal wrote:Set S contains seven distinct integers. The median of set S is the integer m, and all values in set S are equal to or less than 2m. What is the highest possible average (arithmetic mean) of all values in set S ?

m

10m/7

10m/7 – 9/7

5m/7 + 3/7

5m

What is the highest possible average (arithmetic mean) of all values in set S ?

[m-3 , m-2 , m-1 , m , 2m-2 , 2m-1 , 2m ]



If it would have been lowest possible then i guess the below mentioned whud be the numbers :-

[m-3 , m-2 , m-1 , m , m+1 , m+2 , 2m ]

sum = 8m - 3
mean = (8m-3)/7

Guys let me know if my approach is wrong.

what if the first element is (m-12345), m-1,m,m+1...blah balh

how did you come to lowest possible conclusion... havent understood AIM

HSPA wrote:what if the first element is (m-12345), m-1,m,m+1...blah balh

how did you come to lowest possible conclusion... havent understood AIM

For highest avg we need to maximize sum so the terms tht are gonna make diff are 5th and 6th terms.

1st : 2nd : 3rd : 4th : 5th : 6th : 7th


m-3 : m-2 : m-1: m :2m-2 : 2m-1 : 2m ---- To maximize sum

m-3 : m-2 : m-1: m :m+1 : m+2 : 2m ---- To minimize sum

Hope that helps.

ankur.agrawal wrote:Set S contains seven distinct integers. The median of set S is the integer m, and all values in set S are equal to or less than 2m. What is the highest possible average (arithmetic mean) of all values in set S ?

m

10m/7

10m/7 – 9/7

5m/7 + 3/7

5m

Make the situation concrete by plugging in a value for m.

Let m=5.

For the median to be 5, we need to choose 3 distinct integers that are smaller than 5 and 3 distinct integers that are larger than 5. Since we're trying to maximize the average, we want each integer to be as large as possible.

The largest allowed value is 2m = 2*5 = 10. So the 3 largest integers must be 8, 9, 10.
For the 3 distinct integers smaller than 5, the largest possible are 2, 3, 4.
So our 7 integers are 2, 3, 4, 5, 8, 9, 10.
Average = (2 + 3 + 4 + 5 + 8 + 9 + 10)/7 = 41/7. This is our target.

Now we plug m=5 into all the answers to which yields our target of 41/7.

Only answer choice C works:
10m/7 – 9/7 = (10*5)/7 - 9/7 = 50/7 - 9/7 = 41/7.

The correct answer is C.

may i knw the source?

Hi rohu,

The Source is MGMAT CAT 1 . Any specific reasons for asking about the source?

HSPA wrote:I got C
the seven numbers i took are
m-3, m-2, m-1, m, 2m-2,2m-1,2m


10m/7 - 9/7... what is the OA

OA is C. You nailed it.

yes ankur. i somehow felt it must be from the MGMAT CAT as i did get a similar sort of question.
can i reuqst u to mention the source, especially whn u r posting qustions from CAT's as many of us would not have taken them and reading it before hand would do us no good. hope u dnt mind.

ankur.agrawal wrote:

may i knw the source?

Hi rohu,

The Source is MGMAT CAT 1 . Any specific reasons for asking about the source?

can i reuqst u to mention the source, especially whn u r posting qustions from CAT's as many of us would not have taken them and reading it before hand would do us no good. hope u dnt mind.

I am extremely sorry for being so careless. I would take care of this from now. Thanks rohu.

no need for an apology buddy.glad tht u liked my suggestion

ankur.agrawal wrote:

can i reuqst u to mention the source, especially whn u r posting qustions from CAT's as many of us would not have taken them and reading it before hand would do us no good. hope u dnt mind.

I am extremely sorry for being so careless. I would take care of this from now. Thanks rohu.