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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. The range is (4x+15)x=3x+15=48. Answer: D.



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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. Hope this helps.



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Re: A set of numbers contains 7 integers and has an average [#permalink]
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03 Jan 2016, 03:17
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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 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+15a = 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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Re: A set of numbers contains 7 integers and has an average [#permalink]
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06 Sep 2016, 11:29
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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 y1y4 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.
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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.
Appreciate any help. Thanks.



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Re: A set of numbers contains 7 integers and has an average [#permalink]
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09 Nov 2016, 08:40



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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



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Re: A set of numbers contains 7 integers and has an average [#permalink]
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09 Dec 2016, 22:41
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 = 5911 = 48
Hence D
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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. The range is (4x+15)x=3x+15=48. Answer: D.



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Re: A set of numbers contains 7 integers and has an average [#permalink]
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04 Dec 2017, 05:23
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. 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



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Re: A set of numbers contains 7 integers and has an average [#permalink]
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04 Dec 2017, 05:29
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. Answer: D. 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 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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04 Dec 2017, 06:12
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. Answer: D. 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 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



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Re: A set of numbers contains 7 integers and has an average [#permalink]
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04 Dec 2017, 06:22
gmatcracker2017 wrote: 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 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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08 Mar 2018, 12:30
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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: GMAT assassins aren't born, they're made, Rich
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