A set of numbers contains 7 integers and has an average

Question

A set of numbers contains 7 integers and has an average (arithmetic mean) value as well as a median value of 23. If the largest value is equal to 15 more than 4 times the smallest number, what is the largest possible range for the numbers in the set?

A. 33
B. 35
C. 38
D. 48
E. 75

A. 33
B. 35
C. 38
D. 48
E. 75

Bunuel · Accepted Answer

The average of 7 numbers is 23 --> the sum of these numbers is 7*23=161;
The median of 7 numbers is 23 --> 23 is the middle term: {*, *, *, 23, *, *, *};
The largest value is equal to 15 more than 4 times the smallest number --> say the smallest number is x then the largest number would be 4x+15, so our set is {x, *, *, 23, *, *, 4x+15};

Now, in order to maximize the range we need to make the second and the third numbers equal to x and the fifth and sixth numbers equal to 23, so the set should be {x, x, x, 23, 23, 23, 4x+15}.

Since the sum is 161 then x+x+x+23+23+23+4x+15=161 --> x=11.

The range is (4x+15)-x=3x+15=48.

Answer: D.

Joy111 · Answer

for a set of 7 numbers if the median is 23 then we can have 2 cases 1) all numbers are 23 { 23,23,23,23,23,23,23}, in this case too average is 23 2) {# <= 23, # <= 23, # <= 23, 23( center digit equal to 23) , # >= 23, >= 23, >= 23}, median of 7 numbers is 23 and average is 23, means that half of the numbers are below or equal to the median and half of the numbers are equal to or above the median . Now the given condition in the question is that the maximum = 4X + 15 , where x is the smallest one .So we know that all numbers are not equal . now in order to maximize the range we have to keep the other 6 as small as possible . { x,x,x,23,23,23,4X+15} now if the smallest is x then largest is 4x+15, this is given, so we cannot change these two. the center has to be 23 as this is the median. so the smallest value for the 2, and 3, digit has to be equal to the smallest one which is equal to x , we cannot take anything larger then x as that could make the range smaller , but want to find the largest range , so we have to take the smallest numbers. similarly: the smallest value for 4th and 5th digit cannot be smaller than 23, as 23 is the median , and cannot be larger the 23 as again we are looking for the largest range, so we have the keep the 4th and 5th digits as small as possible , and the smallest we can get for the 4th and 5th digit is 23. hence Bunuel's Statement Now, in order to maximize the range we need to make the second and the third numbers equal to x and the fifth and sixth numbers equal to 23, so the set should be {x, x, x, 23, 23, 23, 4x+15}. Hope it is clear to you now .Please ask if any doubt remains .

gmihir · Answer

Bunnel,

Isn't it true that when mean & median are same for the given set of integers, the nubmers have to be equally spaced ? i had read this somewhere so tried solving the problem in that direction. Can you please clarify? Thanks!

Bunuel · Answer

In any evenly spaced set the arithmetic mean (average) is equal to the median . For example for a set {1, 3, 5, 7, 9} mean=median=5.

But  the reverse of this property is not always true . For example a set {0, 1, 1, 1, 2} has the median as well as mean equal to 1, but this set is not evenly spaced.

Hope it's clear.

gmihir · Answer

Also, can you please explain this part: " Now, in order to maximize the range we need to make the second and the third numbers equal to x and the fifth and sixth numbers equal to 23, so the set should be {x, x, x, 23, 23, 23, 4x+15}.  " - Even if 2nd, 3rd, numbers had been anything between x & 23, vice versa, if 4th and 5th numbers had been any numbers between 23 and 4x+15, the range (4x + 15 - x) would have been same. So, the question is how did we guess 2nd, 3rd, 5th and 6th numbers of this set ? Thanks!

mazerath · Answer

Hi ,

I have same query.

Lets say we have a series ( x, x, x , 23 , 23 , 23 , 4x+15) the range in this case 4x+15 - x = 3x+15

The range is the same even for ( x , x+1 , x+ 3 , 23 , y , y+ 1 , 4x + 15) where 23 < y < 4x because from my understanding range is the difference between the largest and the smallest number in a set . 
The values in between do not have any significance on the range.

Is this a correct understanding , because from you answer range seems to mean something else.

Thanks in advance.

Regards
bharath

Joy111 · Answer

Hi

Remember that the total has to 161 , and we need one variable to sort this out , so by taking x and y , it would be difficult to find the range.

To find the answer the most efficient way is to take smallest one as x and largest one as 4x+15 and , and keep the others as small as possible .

Technically { 11, 11, 12, 23, 23, 24, 59 }this set is also possible , but to arrive at 11 and 59 , we have to proceed as shown by bunuel, one cannot solve the sum 
assuming the set to be { x, x, x+1, 23, y, y+1,4x+15}.

solving this way will be complicated and time consuming . Hope it helps .

kaka1989 · Answer

Isn't the average of first and last term supposed to equal the average hence (4x+15+x)/2 should equal 23 but then we get some other result for x??

Bunuel · Answer

The average of a set equals to the average of the first and the last terms if a set is evenly spaced, but if a set is not evenly spaced (like we have in original question) then this relationship is not necessary to be true.

bsahil · Answer

Hi Bunuel,

Can we take median ~= range/2.

Then median(23) ~= 46/2

So nearest value is 48 in the answers.

Thanks,
Sahil Bansal

udso · Answer

The best way to answer this query is to visualize this problem as having to balance 6 members of equal weights on a seesaw such that 3 members are on either side of the number 23, which acts as the fulcrum. The farther the members on one side are from the fulcrum the farther the the members on the other side have to be to counter the weight of these members. Hence, if you push all 3 weights on the left side to the lowest possible value (x), and keep 2 weights on the right at the fulcrum, then the 1 remaining weight on the right would need to be placed at the maximum possible distance from the fulcrum; thus maximizing the range. fulcrum.png

thefibonacci · Answer

let the numbers be a,b,c,23,e,f,g

(a+b+c+23+e+f+g)/7 = 23 
=> (a+b+c+23+e+f+g) = 161
=> (a+b+c+e+f+g) = 138 
given that, g = 4a+15 
range = 4a+15-a = 3a+15 = 3(a+5) 
now use the options. range should be a multiple of 3. hence, only option a,d,e stand. 
try with maximum (as we need to maximize) 
3(a+5) = 75
a = 20
=> g = 95
to maximize the range -> e=f=23 
numbers become -> 20,b,c,23,23,23,95 
the sum of these numbers > 161 , hence this isn't valid.

try for option d. 
the numbers come out to be -> 11,11,11,23,23,23,59 
which is valid. 
hence, d.

shypeleg · Answer

After being a little bit frustrated with the explanations provided I think I've finally reached an explanation that makes sense to me:

Reaching the formula:  X + y1 + y2 + 23 +y3 + y4 + (4X+15) = 161   was kind of clear for everyone. 
Now, how to pick y1-y4 in order to maximise the range?

Notice that the bigger x is , the bigger the range is:
X | 4x+15 | range
--------------------
1 | 19 | 18
2 | 23 | 21
3 | 27 | 24
This makes sense because 4x increases 4 times more than x when x grows : ] 
So - how to make X the largest given the constraints in the question?

X + y1 + y2 + 23 +y3 + y4 + (4X+15) = 161 
5x + y1 + y2 + y3 + y4 = 123 ( or x +( y1 + y2 + 23 +y3 + y4 + (4X+15))/5 = 161/5 )

If you want the X part of the equation to be the biggest, you want the y1+y2+y3+y4 part to be the smallest possible.
Under the constraints provided the smallest y1,y2 can be is x, and the smallest y3,y4 can be is 23.

So you can solve: 5x+x+x + 23 + 23 = 123 => x=11 , range = 48.

Does this seem right? 
I hope so because it took me a while to find some method that will make sense to me.

Shy Peleg.

testcracker · Answer

Bunuel hi your solution to the problem is very brilliant as usual I have, however, two very simple questions to your kind consideration 1. x _ _ 23 _ _ 4x + 15 here, in order to maximize the range, we comfortably can set 2nd and 3rd numbers equal to "x", but this change will not distort the mean. can you, please, comment something on this...? 2. here, in this question, the test maker has kindly provided us the arithmetic mean. If there was, however, no mention of the average, and if the numbers were needed not to be distinct, can the following sequence be established as long as the largest value is equal to 15 more than 4 times the smallest number...? 23, 23, 23, 23, 24, 24, 107 range = 84 And, if the numbers were needed to be distinct, then is the following pattern legitimate...? 20, 21, 22, 23, 24, 25, 95 range = 75 maybe they are very obvious, but I am badly in need of your help

thanks in advance, man

Bunuel · Answer

We are told that 23 is not only the mean but also the median.

The fact that 23 is the mean helps us to get the sum --> 7*23=161;
The fact that 23 is the median means that 23 is the middle value, so we can adjust numbers less than the median and more than the median.

testcracker · Answer

thank you very much for your quick reply, man

yes, 23 is median here, so we can easily adjust numbers, but here we are also given the mean

say,

3 numbers have a median of 4, so we can plot any number as below

3, 4, 8 
2, 4 , 5
3, 4, 6 and many more.

Look here we are free to chose any number, but when we are given the average, for example 5, we are not free to chose any number
in my exampe above, only 1st one fits in

in this line of reasoing, I asked to you how you set 2 numberes equal to "x". Obvously you are very right, but I wanted to know the science behind it

yes, now I can understand, here in this question you have provided the solution for, mean and median are the same number, so this mechanixm is permissible

in the exapmple, however, I have cited, mean and median are not the same, so only 1st one fits in

thanks to you again for your most precise time you took to write to me
take care

Bunuel · Answer

We are told that:
1. A set of numbers contains 7 integers. 
2. Mean = median = 23. 
3. The largest value is equal to 15 more than 4 times the smallest number.

Now, there could be many sets satisfying this requirements but we are asked to find the one which has the largest possible range. The solution provided adjusts the set in a way to find the set with the largest possible range.

Do we change the mean while doing that? NO, because we are using the actual mean of 23 (the sum of 161) when finding x:  Since the sum is 161  then x+x+x+23+23+23+4x+15=161 --> x=11.

EMPOWERgmatRichC · Answer

Hi All,

For these types of questions, you have some options when it comes to approaching the "math" behind the question. As long as you're not breaking any of the rules that the prompt describes, you can use whatever examples you want (to help you prove what the correct answer is).

We're asked for the largest possible range of this group of 7 numbers, which means that we have to make the largest value as large as possible and make the smallest value as small as possible (and by extension, make everything as small as possible too).

First, let's deal with the median, which equals 23 (when the numbers are ordered, the "middle number" is 23):

_ _ _ 23 _ _ _

This means that the 3 numbers to the "left" of the 23 can be 23 or less and the the 3 numbers to the "right" of the 23 can be 23 or more.

We also know that the average is 23, which means that the sum of the 7 numbers = sum/7 = 23…….sum = 161

We have one more piece of info to implement: the largest number is 15 more than 4 times the smallest number:

X = smallest number 
4X + 15 = largest number

X _ _ 23 _ _ (4X+15)

To make the largest number as large as possible, we have to make EVERYTHING ELSE as small as possible. Here's how to do it:

X X X 23 23 23 (4X+15)

Summing all of these values gives us:

7X + 84 = 161 
7X = 77 
X = 11

So, the smallest number = 11 and the largest number = 4(11)+15 = 59

The range = 59 - 11 = 48

Final Answer:  D

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

uzumakinaruto444 · Answer

hello,
can anyone please help me with why the set can't be {x,x,x,x,23,4x+15,4x+15,4x+15,4x+15} for maximising the range? why have we taken particularly 3 repetitions of 23 in the set?
thanks.

A set of numbers contains 7 integers and has an average

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