## Set R contains five numbers that have an average value of 55

Math questions from any Manhattan Prep GMAT Computer Adaptive Test.
jigar24
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### Set R contains five numbers that have an average value of 55

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?

1) 78
2) 77 1/5
3) 66 1/7
4) 55 1/7
5) 52

option (1) = 78

My question is: Mean is equal to median only when we have a set of consecutive no.s (difference between any two consecutive no.s is the same). So in this case the largest no. in the set is fixed. How can we manipulate the no.s when we know that the difference between 2 no.s is the same. I think the question is slightly flawed. If we were to manipulate other no.s to find largest no. in the set, then we should not be given the fact that mean = median.

Pls some instructor help me with this.

Jigar
nitin_prakash_khanna
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### Re: Set R contains five numbers that have an average value of 55

Hi "jigar24"

First thing to note is that if median = mean doesnt neccessarily mean that a set consists of consequetive numbers. It is true for all equaly spaces sets and consequetive numbers are a special case of Equally spaces set.

What i mean is that the set R can be

53,54,55,56,57 (which is was you probably thought)

or it can be

20,30,55,80,90

Both the above sets have mean = 55 and median = 55 also.
And there can be many more infinite possibilities for mean = median.

With that in mind lets get to the actual q.
With mean = median , we are also told that largest number of the set = 20 + 3* smallest number of set. And our objective is to maximize the range.

I will replicate explanation provided by MGMAT CAT.
let the set be

a,b,55,c,d
& its given that d= 3a+20

also given mean = 55
So a+b+55+c+3a+20 = 55*5 = 275------Eq.1

now range of set = 3a+20=a= 2a+20

Now to maximize range we need to maximize a.

And we can maximize the range if the set is

a,a,55,55,3a+20
5a+130=275 (from Eq.1)
5a= 145 ---> a=29

Range = 2a+20= 78.

And note the set becomes
29,29,55,55,107.

Which is not equally spaced. So the biggest takeaway from the question is that

1. For an equally spaced set (including consecutive number set), Mean = Median
2. But if Mean=Median doesnt mean that the set is Equally spaced, as we just saw above.
Ben Ku
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### Re: Set R contains five numbers that have an average value of 55

nitin_prakash_khanna wrote:So the biggest takeaway from the question is that

1. For an equally spaced set (including consecutive number set), Mean = Median
2. But if Mean=Median doesnt mean that the set is Equally spaced, as we just saw above.

This is exactly true and well-stated.
Ben Ku
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jigar24
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### Re: Set R contains five numbers that have an average value of 55

yes, sorry I forgot to post a reply.. Thanks a lot Nitin.. this surely clears the air for me... nice explanation..

Cheers!
Jigar
Ben Ku
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### Re: Set R contains five numbers that have an average value of 55

Ben Ku
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george.marinov
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### Re: Set R contains five numbers that have an average value of 55

Hi,

I have a question about this. Since the set 29,29,55,55,107 is NOT evenly spaced (increments are not the same) why do you use the formula: sum = average*n? Does this formula apply to all sets whose mean = median even though the sets could NOT be evenly spaced as in this case?
Students

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### Re: Set R contains five numbers that have an average value of 55

And we can maximize the range if the set is

a,a,55,55,3a+20

I did-not understand how you reached the above..can you please explain

Thank you
pranab.banerjee82
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### Re: Set R contains five numbers that have an average value of 55

george.marinov wrote:Hi,

I have a question about this. Since the set 29,29,55,55,107 is NOT evenly spaced (increments are not the same) why do you use the formula: sum = average*n? Does this formula apply to all sets whose mean = median even though the sets could NOT be evenly spaced as in this case?

I would like to answer this question .

First of all @nitin , lovely explanation .

@george.marinov ,

We are given that the average of the set of 5 numbers is 55 .

So we know for any set of numbers average = (sum of the numbers) / (total numbers in set )

And that is why irrespective of weather the set is evenly spaced or not we can find out the sum as we know the average and the number of elements in the set .
pranab.banerjee82
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### Re: Set R contains five numbers that have an average value of 55

nav.adi wrote:And we can maximize the range if the set is

a,a,55,55,3a+20

I did-not understand how you reached the above..can you please explain

Thank you

This is the concept for maximizing and minimizing . We are given that the mean in 55 .

Note : we have 5 numbers in the set so 55 must be the 3rd element in the set . To find the max range we would need the smallest element to be repeated max times .

We dont know the 2 nd and 4 th element in the set .

a,a,55,55,3a+20
5a+130=275 (from Eq.1)
5a= 145 ---> a=29

Lets suppose we consider 2 nd element to be say 33 (we need to consider this element to be smaller or equal to mean.)

And the 4 th element to be say 65(this value has to be equal to or greater than the mean) .

See what effect it has on Eq 1 .
a + 33 + 55 + 65 + 3a+20 = 275
4a + 173 = 275
a = 25.5

so the range = 2a + 20 = 71 . That is much smaller than 78 .

Now if we decrease the values of 2 nd and 4 th elements we would max value for a and thus the max value for the range .
Because the only variable on which the value of range depends is a (2a + 20) . So max the value of a max is the range .

Having said that the thing to note is the value of 2 nd element can not be lower than a or greater than mean , similarly the value of 4 th element can not be more than 3a+20 and less than mean . Because all elements in the set are arranged in ascending order .

Hope this clear things up a bit .
tim
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### Re: Set R contains five numbers that have an average value of 55

Thanks pranab. Yes 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..
Tim Sanders
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https://www.manhattanprep.com/gmat/forums/a-few-tips-t31405.html
sachin.w
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### Re: Set R contains five numbers that have an average value of 55

Hi ,

Here's how I did..

smallest no: s
largest no: 3s+20

since mean = median,

thought that numbers are in AP.

so average= (last no+first no)/2

therefore 55=(s+3s+20)/2
=> s=22.5

now l=20+3s
=> l=87.25

range =l-s=65..
Please let me know where I am going wrong.
jnelson0612
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### Re: Set R contains five numbers that have an average value of 55

sachin.w wrote:Hi ,

Here's how I did..

smallest no: s
largest no: 3s+20

since mean = median,

thought that numbers are in AP.

so average= (last no+first no)/2

therefore 55=(s+3s+20)/2
=> s=22.5

now l=20+3s
=> l=87.25

range =l-s=65..
Please let me know where I am going wrong.

Hi there,
Sure! You are right that 55 is the average but it is the average of all five numbers. We don't know that the largest and smallest themselves average to 55. Does this help?
Jamie Nelson
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sachin.w
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### Re: Set R contains five numbers that have an average value of 55

thanks jamie.. does this mean that
we cannot conclude that numbers are in A.P if the mean= median..

but if the numbers are in AP, mean will be equal to median..

Hope I got it right.
jnelson0612
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### Re: Set R contains five numbers that have an average value of 55

sachin.w wrote:thanks jamie.. does this mean that
we cannot conclude that numbers are in A.P if the mean= median..

but if the numbers are in AP, mean will be equal to median..

Hope I got it right.

Sorry, I must be blanking. What is "A.P."?
Jamie Nelson
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sachin.w
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### Re: Set R contains five numbers that have an average value of 55

Arithmetic Progression. .

Am I missing something ?