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

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01 Jan 2016, 07:00

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.

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

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01 Jan 2016, 08:21

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.

Answer: D.

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.

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

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03 Jan 2016, 03:17

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.
_________________

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

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06 Sep 2016, 11:29

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?

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.

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

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09 Nov 2016, 08:14

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.

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.

Can you please give an example of your claim?

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.
_________________

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

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09 Nov 2016, 09:01

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.

Can you please give an example of your claim?

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.

Thanks Bunuel for the reply. 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

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

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03 Nov 2017, 08:16

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.