List R contains five numbers that have an average value of 55. If the

i did it like this:

given: (a+b+55+d+e)/5 = 55
and e = 2a+20
so, a+b+55+d+e = 275
a+b+d+e = 220
3a+b+d = 200
now, to minimize a we must maximize b and d
b can be a maximum of 55 (since c is 55)
=> 3a+55+d = 200
=> 3a+d = 145
d can be maximum when d=e=2a+20
=> 3a+2a+20 = 145
=> a = 25

This problem requires some logical thinking and careful planning.

Set up the numbers on your page as follows: x, A, 55, B, 2x+20

We want to minimize A, so we have to maximize the other choices. The max value for A is 55 because otherwise it would be larger than the median, and we know that the median is equal to the mean is equal to 55. And the max value for B is equal to 2x+20.

Now we have the following list, x, 55, 55, 2x+20, 2x+20. The sum total is 110+40+5x or 150+5x. The sum of the set is equal to 5*55 or 275. 150+5x = 275 then 5x = 125 or x = 25.

For an evenly spaced set (arithmetic progression), the median equals to the mean. Though the reverse is not necessarily true. Consider {0, 1, 1, 2} --> median = mean = 1 but the set is not evenly spaced.

Hi Bunuel,

I used the following approach.

Since mean is equal to median, the series is equally spaced.

So using the following formula :-

Sum= Average * number of items in the series= 55* 5= ((First number + Second Number)/2)*Number of items.
If x is the largest number and s the smallest number.
x=2s+ 20.

Therefore,
((2s+20+s)/2)*5=275.
Solving this I got smallest number=s=30.
What am I missing here. Please help.

Hi All,

While this question can be solved with Algebra, it can also be solved by TESTing THE ANSWERS.

We're given a lot of little pieces of information to work with, so we have to stay organized:
1) 5 numbers with an average of 55. This means that 5(55) = 275 is the SUM of these 5 numbers.
2) The question does NOT state that the numbers need to be distinct, so DUPLICATES ARE ALLOWED.
3) The median = mean = 55. The 3rd value of the 5 numbers MUST be 55.
4) The largest number = 2(smallest number) + 20.

The question asks us for the smallest POSSIBLE value in the list. Let's start by TESTing the smallest answer….

Answer E: 15

If the smallest number was 15, then the largest would be 50. This is NOT possible (the average and median are both 55).

Answer D: 20

If the smallest number was 20, then the largest would be 60. This is NOT possible (the sum of the other 3 values = 195; there's no way to get to a total of 195 with the 3 remaining values and the above restrictions).

Answer C: 25

If the smallest number was 25, then the largest would be 70. This IS POSSIBLE. The sum of the other 3 values would be 180 (with a 70, a 55 and another 55, we would get a total of 180).

Final Answer:

GMAT assassins aren't born, they're made,
Rich

OA: C
Average of List R\(=\)Median of List R \(=55\)

Given: List R :\([x,A,55,B,2x+20]\)

Average\(=\frac{x+A+55+B+(2x+20)}{5}\)

\(55=\frac{x+A+55+B+(2x+20)}{5}\)

\(x =\frac{55*5 -55-20-A-B}{3}= \frac{200-A-B}{3}\).........(1)

For \(x\) to be minimum, \(A\) and \(B\) have to maximum
There are following constraints on the value of \(A\) and \(B\)

\(x≤A≤55\) and \(55≤B≤2x+20\)

The maximum value of \(A\) can be \(55\) and \(B\) can be \(2x+20\). putting these in (1), we get

\(x= \frac{200-55-(2x+20)}{3}=\frac{125-2x}{3}\)

\(3x=125-2x\)

\(5x=125,x=\frac{125}{5}=25\)

Hi asgopala,

To answer your immediate question: NO - if the mean = median of a group of numbers, then that does NOT mean that the group is necessarily "evenly spaced." For example,

the group 1, 2, 3 has a mean = 2 and a median = 2 and is evenly spaced out.
the group 0, 0, 2, 3, 5 has a mean = 2 and a median = 2 and is NOT evenly spaced out.

IF you are dealing with a sequence of CONSECUTIVE INTEGERS (for example: 1, 2, 3, 4 and 5), then the average of the largest and smallest numbers will equal the average of the group.

GMAT assassins aren't born, they're made,
Rich

Quality Question.
Here is my solution to this one =>

Let the 5 numbers in set R in increasing order be
R1
R2
R3
R4
R5


Now Using

\(Mean = \frac{Sum}{#}\)


Sum(5) = 55*5 = 275

The Question stem Also says that Mean= Median
#=5=Odd
Hence Median = 3rd element => R3

Hence R3=55
R5=20+2R1(Given)

Now,Here we are asked the minimum possible value of R1

R1 would be minimum when all other elements are maximum
R1=R1
R2=Median =55
R3=55
R4=R5=2R1+20

Hence adding them up we get =>

R1+55+55+2R1+20+2R1+20=275
Hence 5R1=125=> R1=25



Hence C


Great Question.

Hi dyogendrarao,

To answer your immediate question: NO - if the mean = median of a group of numbers, that does NOT mean that the group is necessarily "evenly spaced." For example,

the group 1, 2, 3 has a mean = 2 and a median = 2 and is evenly spaced out.
the group 0, 0, 2, 3, 5 has a mean = 2 and a median = 2 and is NOT evenly spaced out.

In addition, an 'evenly spaced' group of numbers could have an odd OR an even number of terms. The group 1, 2, 3 is 'evenly spaced' and so is the group 1, 2, 3 4.

GMAT assassins aren't born, they're made,
Rich

The property you're referring to only goes in one direction.

That is, we can say: if a set of numbers are equally spaced, then the mean equals the median

However, the reverse is not true. That is, we cannot say that: If mean equals median then we are dealing with evenly spaced integers

For example take the set of numbers {1,1,5,8,10}
Here, the mean and median are both 5, but we can't conclude that the five numbers are equally spaced

Mean = Median does not mean an evenly spaced sequence.

For example.
2, 3, 3, 3, 4.....median =mean=3
2, 5, 6, 6, 11.....again median =mean=6
2, 5, 6, 8, 9....

Given:
1. List R contains five numbers that have an average value of 55.
2. The median of the numbers in the list is equal to the mean
3. The largest number is equal to 20 more than two times the smallest number

Asked: What is the smallest possible value in the list?

1. List R contains five numbers that have an average value of 55.
2. The median of the numbers in the list is equal to the mean
3. The largest number is equal to 20 more than two times the smallest number

Let the numbers be {s, 55, 55, 2s+20, 2s+20} For s to be smallest, all other number need to be largest possible since average (m=55) is constant
5*55 = s + 55 + 55 + 2s +20 + 2s + 20
3*55 = 5s + 40
5s = 165 - 40 = 125
s = 25

IMO C

Just a doubt, for a list which may has its mean equal to the median, it's wrong to assume that the list is evenly spaced, right ? Also when we talk about evenly spaced, the list should have odd number of elements, right ?

I have a query in a logic that I had applied to solve this.

Because the median = mean , isnt the sequence considered an equally spaced one - the difference between the every consecutive number in the sequence is the same.

Also for such sequence isnt this true: Mean of all 5 numbers in sequence = Arithmetic mean ( 1st , 5th) = Arithmetic mean (2nd, 4th)

e.g. 1,2,3,4,5

So using this above logic I calculated Arithmetic Mean ( a , 2a+20) = 55 but this way i solved for a (smallest # in the sequence ) as 30 ( which is not right ).

I wanted to know where I went wrong. Pls help

hey chetan2u VeritasKarishma BrentGMATPrepNow

why my solution is wrong? here is my reasoning

if mean equals median then we are dealing with evenly spaced integers ....

so let smallest number be S

then largest one is 20+2s

so ...

\(\frac{S+(20+2s)}{2} = 55\)

\(3s =90 \)

\(s =30 \)

What`s wrong with it ?
thanks!

One of the first things to look for in the question is whether we can repeat the numbers. Since there is no constraint, repeating the numbers is possible.


Given 5 numbers.

Sum of 5 Numbers = (5) * (55 Avg.) = 275

We are also given that the Median = Mean

Since it is an Odd count of 5 Numbers, the Median will be the 3rd number (middle) listed in Ascending order.

___ ; ____ ; 55 ; ____ ; _____


Call the smallest valued number X.

The greatest value is then 20 more than TWICE X ———-> 2X + 20


X < ____ < 55 < ____ < 2X + 20

And all 5 numbers must sum = 275


In order to minimize X, we want to maximize the 2 other numbers.

Since the 2nd smallest number can no exceed the Median, the most we can make it is 55 = Median

Lastly, since we are told the greatest number is (2X + 20) ——-> we can make the 4th number in ascending order equal to the greatest.


X < 55 < 55 < (2X + 20) < (2X + 20)

And all 5 numbers must Sum = 275

Can set up the equation and solve for X, which will be the MINIMUM Value we can make the Smallest Number.

X + 55 + 55 + 2X + 20 + 2X + 20 = 275

5X + 150 = 275

5X = 125

X = 25

(C)

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Hey Chetanu,

If a set consists of the same number multiple times, will that set's mean=median = number in the set ?