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Intern  Joined: 21 Dec 2015
Posts: 4
Re: A set of numbers contains 7 integers and has an average  [#permalink]

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Why does this maximise the range {x, x, x, 23, 23, 23, 4x+15}? I thought that {x, 23, 23, 23, 23, 23, 4x+15} would maximise the range. I am pretty confused. Thanks in advance for the help.

Bunuel wrote:
Joy111 wrote:
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

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.

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Re: A set of numbers contains 7 integers and has an average  [#permalink]

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zurich wrote:
Why does this maximise the range {x, x, x, 23, 23, 23, 4x+15}? I thought that {x, 23, 23, 23, 23, 23, 4x+15} would maximise the range. I am pretty confused. Thanks in advance for the help.

Bunuel wrote:
Joy111 wrote:
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

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.

Range is same for both the cases. till you keep the smallest and the largest elements are the same, the range will be the same irrespective of what are the other elements.

Range of 1,2,3,4,5,600 and 1,100,200,300, 400,600 is the same.

Hope this helps.
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Re: A set of numbers contains 7 integers and has an average  [#permalink]

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1
Joy111 wrote:
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

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.
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Posts: 1
Re: A set of numbers contains 7 integers and has an average  [#permalink]

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1
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.
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Re: A set of numbers contains 7 integers and has an average  [#permalink]

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Can somebody help me, why in this case, we are not considering the negative value. If we consider negative value the range can be expanded to as much as we want.

Appreciate any help. Thanks.
Math Expert V
Joined: 02 Sep 2009
Posts: 64951
Re: A set of numbers contains 7 integers and has an average  [#permalink]

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Vikas10 wrote:
Can somebody help me, why in this case, we are not considering the negative value. If we consider negative value the range can be expanded to as much as we want.

Appreciate any help. Thanks.

When solving this question we are not considering cases. The solutions presented are based on the info we have in the stem and equations we built from that info. These equations give an answer of 48.
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Re: A set of numbers contains 7 integers and has an average  [#permalink]

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Bunuel wrote:
Vikas10 wrote:
Can somebody help me, why in this case, we are not considering the negative value. If we consider negative value the range can be expanded to as much as we want.

Appreciate any help. Thanks.

When solving this question we are not considering cases. The solutions presented are based on the info we have in the stem and equations we built from that info. These equations give an answer of 48.

I found the reason that if we choose the negative value for the lowest integer the highest value will get limited because of the equation - a7 = 4 * a1 + 15
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GRE 1: Q169 V154 Re: A set of numbers contains 7 integers and has an average  [#permalink]

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Here is what i did in this Question -->

Let the smallest value in set be a.
The largest value in set => 4a+15

Now, Aa mean= median=23 => Sum of all value =23*7=> 161

Now to maximise the range,we should maximise the largest value for the constant value a.
The set will be => {a,a,a,23,23,23,4a+15}

Hence 7a+84=161
so a=11
Now largest value =4a+15=44+15=59

Range = 59-11 = 48

Hence D

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Re: A set of numbers contains 7 integers and has an average  [#permalink]

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Hi

In this why cant we apply L+S=23 as the average? (First term+Last term)/2

Bunuel wrote:
Joy111 wrote:
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

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.

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Status: love the club...
Joined: 24 Mar 2015
Posts: 257
Re: A set of numbers contains 7 integers and has an average  [#permalink]

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Bunuel wrote:
Joy111 wrote:
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

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.

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 Math Expert V
Joined: 02 Sep 2009
Posts: 64951
Re: A set of numbers contains 7 integers and has an average  [#permalink]

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gmatcracker2017 wrote:
Bunuel wrote:
Joy111 wrote:
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

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.

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 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.
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Re: A set of numbers contains 7 integers and has an average  [#permalink]

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gmatcracker2017 wrote:
Bunuel wrote:
Joy111 wrote:
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

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.

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 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
Math Expert V
Joined: 02 Sep 2009
Posts: 64951
Re: A set of numbers contains 7 integers and has an average  [#permalink]

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1
gmatcracker2017 wrote:

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

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.
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Re: A set of numbers contains 7 integers and has an average  [#permalink]

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1
1
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

GMAT assassins aren't born, they're made,
Rich
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Re: A set of numbers contains 7 integers and has an average  [#permalink]

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Bunuel wrote:
Joy111 wrote:
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

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.

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.
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Re: A set of numbers contains 7 integers and has an average  [#permalink]

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1
Hi applebear,

Your example has a couple of 'problems' with it:
1) It has 9 terms - and there should only be 7 terms
2) You assume that the LARGEST number shows up repeatedly.

Since the average of the 7 terms is 23, we know that the SUM of those terms is 161. That sum is a big 'limiter' in terms of what is possible. The other limiter is that if the smallest number is X, then the largest number is 4X+15 (meaning that the value of the smallest number limits how big the value of the largest number can get).

In these types of situations, to make one number as big as possible, we must make ALL of the other numbers as small as possible (based on the restrictions described in the prompt). We need those three "23s" to get to the solution because that is the smallest that we can make THOSE three values.

GMAT assassins aren't born, they're made,
Rich
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Re: A set of numbers contains 7 integers and has an average  [#permalink]

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Hello EMPOWERgmatRichC,
Thank you so much for clearing my doubt...however i have one more doubt , why can’t we make the fourth term of the set as “x”? Why 23 ?

Posted from my mobile device
EMPOWERgmat Instructor V
Status: GMAT Assassin/Co-Founder
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Re: A set of numbers contains 7 integers and has an average  [#permalink]

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1
Hi applebear,

The prompt tells us that the Average of the group is 23 AND the Median of the group is 23. The Median of a group of 7 numbers will the '4th' number (when the numbers are ordered from least to greatest). That is the one value that is 'locked in', so we can't change it.

GMAT assassins aren't born, they're made,
Rich
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Re: A set of numbers contains 7 integers and has an average  [#permalink]

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hello EMPOWERgmatRichC,
Alright, I got it now. thank you so much for your help. Intern  Joined: 10 May 2018
Posts: 33
Re: A set of numbers contains 7 integers and has an average  [#permalink]

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Joy111 wrote:
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

The terms, in increasing order, are-

x, a, b, 23, c, d, 4x+15

Sum = 23x7 = 161

Range = 4x+15 - x = 3x+15

For maximum range, we need maximum value of x and thus, we need minimum values of a,b,c,d.

The minimum values of a, b can be x each and those for c, d can be 23 each.

Hence, sum = x + x + x + 23 + 23 + 23 + 4x + 15 = 161
7x + 84 = 161
7x = 77
x = 11

Range = 3x + 15 = 48 Re: A set of numbers contains 7 integers and has an average   [#permalink] 22 Nov 2018, 13:01

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