Statistics

[contact.sumeshn]

Set R contains five numbers that have an average value of 55. If the median of the set is equal to the mean, and the largest number in the set is equal to 20 more than three times the smallest number, what is the largest possible range for the numbers in the set?

A. 78
B. 77 1/5
C. 66 1/7
D. 55 1/7
E. 52

given that mean = median, should we also consider the fact that set R is an evenly spaced set? Or is it true that whenever a set has the same mean and median, the set is an equally spaced set?

Thanks!

[kalyan_tcs08]

Set R contains five numbers that have an average value of 55. If the median of the set is equal to the mean, and the largest number in the set is equal to 20 more than three times the smallest number, what is the largest possible range for the numbers in the set?

A. 78
B. 77 1/5
C. 66 1/7
D. 55 1/7
E. 52

given that mean = median, should we also consider the fact that set R is an evenly spaced set? Or is it true that whenever a set has the same mean and median, the set is an equally spaced set?

Thanks!

You need not worry about mean and median while considering this problem...
say the numbers in set R as a, b, c, d, e.

a+b+c+d+e = 5*55= 275
since mean = median=55,
a+b+55+d+e = 275
a+b+d+(3a+20) = 220
4a+b+d = 200

Now do some thinking,, if the range needs to be maximum...the end numbers a, e must be as far as possible from each other.

so the numbers b ,d should be as minimum as possible. b is nearer to a , so it can take a min value of A. similarly, d is nearer to c so can take value of C.

so the equation becomes,

4a + (a) + 55 = 220

5a = 145 ; a=29, e= 107

range = 78

cheers!!!

[debmalya.dutta]

R need not be an event spaced set
for example
53,54,55,56,57 vs say 50,54,55,56,60

Set R contains five numbers that have an average value of 55. If the median of the set is equal to the mean, and the largest number in the set is equal to 20 more than three times the smallest number, what is the largest possible range for the numbers in the set?

A. 78
B. 77 1/5
C. 66 1/7
D. 55 1/7
E. 52

given that mean = median, should we also consider the fact that set R is an evenly spaced set? Or is it true that whenever a set has the same mean and median, the set is an equally spaced set?

Thanks!

[mschwrtz]

that's right debmalya.dutta.

evenly distributed values-->(mean=median)
BUT
(mean=median)-/->evenly distributed values.

if you take the difference between each number less than the mean and the mean itself and add all those differences together, the sum will be exactly the same as if you take the difference between each number greater than the mean and the mean itself and add all those differences together.

Huh? Well consider debmalya.dutta's example, 50,54,55,56,60. The mean is 55. (55-50)+(55-54)=(56-55)+(60-55).

This implies that median=mean for every set of evenly distributed values, and also for every set of values distributed symmetrically about the mean (in this case, 5 less, 1 less, mean, 1 more, 5 more.. We could generate other, even less obvious cases too, where there's no evident symmetry, e.g. 50, 52, 53, 55, 56, 59, 60.

[debmalya.dutta]

Quoting Tim Sanders's(MGMAT staff) explanation on this from a different post

If you want to maximize the highest element in a set of fixed sum, you need to make the other elements as small as possible. If we have a, _, 55, _, max in ascending order, you need to push each of the blanks as low as possible. This can be done by making the first blank a and the second one 55..

[mschwrtz]

Thanks debmalya.dutta, that's correct. I only spoke to the narrow point about evenly distributed values.