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Set R contains five numbers that have an average value of 55 27 May 2014, 16:24
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Bunuel
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Hi Bunuel,

Since the statement says that the median = mean, aren't we supposed to assume that it's an evenly spaced set? If so, wouldn't a2 and a4 be different from a1 and a3?

For evenly spaced set mean = median, but the reverse is not necessarily true. Consider {1, 1, 2, 2, 4} --> mean = median = 2, but the set is not evenly spaced.
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thanks for clarifying.
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Set R contains five numbers that have an average value of 55 24 Aug 2018, 03:50
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Orange08
Set R contains five numbers that have an average value of 55. If the median of the set is equal to the mean, and the largest number in the set is equal to 20 more than three times the smallest number, what is the largest possible range for the numbers in the set?

A. 78
B. 77 1/5
C. 66 1/7
D. 55 1/7
E. 52
Show more

To maximize the range, we want to make the smallest number as small as possible and the largest number as large as possible. So the five numbers would be:

x, x, 55, 55, y

The set above will minimize the smallest number (x) because the two smallest values are the same. It will maximize the largest number (y) because the next largest number (55) is the same as the median.

Since the largest number is 20 more than 3 times the smallest number, y = 20+3x. So our list of numbers can be rewritten as:

x, x, 55, 55, 20+3x

Since the average is 55, the sum of the five numbers = 5*55 = 275.
Thus, x + x + 55 + 55 + 20+3x = 275.
5x = 145
x= 29.

Thus, y = 20+3x = 20 + 3*29 = 107.

Thus, the largest possible range is y-x = 107-29 = 78.

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Set R contains five numbers that have an average value of 55 23 Feb 2019, 13:11
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Bunuel
Orange08
Set R contains five numbers that have an average value of 55. If the median of the set is equal to the mean, and the largest number in the set is equal to 20 more than three times the smallest number, what is the largest possible range for the numbers in the set?


78
77 1/5
66 1/7
55 1/7
52

{\(a_1\), \(a_2\), \(a_3\), \(a_4\), \(a_5\)}
As mean of 5 numbers is 55 then the sum of these numbers is \(5*55=275\);
The median of the set is equal to the mean --> \(mean=median=a_3=55\);
The largest number in the set is equal to 20 more than three times the smallest number --> \(a_5=3a_1+20\).

So our set is {\(a_1\), \(a_2\), \(55\), \(a_4\), \(3a_1+20\)} and \(a_1+a_2+55+a_4+3a_1+20=275\).

The range of a set is the difference between the largest and smallest elements of a set.

\(Range=a_5-a_1=3a_1+20-a_1=2a_1+20\) --> so to maximize the range we should maximize the value of \(a_1\) and to maximize \(a_1\) we should minimize all other terms so \(a_2\) and \(a_4\).

Min possible value of \(a_2\) is \(a_1\) and min possible value of \(a_4\) is \(median=a_3=55\) --> set becomes: {\(a_1\), \(a_1\), \(55\), \(55\), \(3a_1+20\)} --> \(a_1+a_1+55+55+3a_1+20=275\) --> \(a_1=29\) --> \(Range=2a_1+20=78\)

Answer: A.
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Dear,

I have one doubt, I tried to understand the question which says that the Mean of the set is equal to the median, and this happens in case of Arithmetic progression.
When I followed that process to apply Sum = N/2(a1 + 20+3a1) and calculated, the range came 66, which was in the answers.
Could you please help me with the catch as where did I go wrong ??

Thanks in advance.
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Set R contains five numbers that have an average value of 55 28 Aug 2019, 09:17
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Orange08
Set R contains five numbers that have an average value of 55. If the median of the set is equal to the mean, and the largest number in the set is equal to 20 more than three times the smallest number, what is the largest possible range for the numbers in the set?

A. 78
B. 77 1/5
C. 66 1/7
D. 55 1/7
E. 52
Show more

Given:
1. Set R contains five numbers that have an average value of 55.
2. The median of the set is equal to the mean, and the largest number in the set is equal to 20 more than three times the smallest number

Asked: What is the largest possible range for the numbers in the set?

mean = median = 55
Since 3x + 20 increase more than x with changes in x, x has to be maximised.
Let the numbers be = {x,x,55,55,3x+20}
2x+ 55 + 55+ 3x + 20 = 55*5 = 275
5x = 275 - 130 = 145
x = 29
The numbers are = {29,29,55,55,107}
Range = 107 -29 = 78

IMO A
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Set R contains five numbers that have an average value of 55 20 Nov 2019, 07:57
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Orange08
Set R contains five numbers that have an average value of 55. If the median of the set is equal to the mean, and the largest number in the set is equal to 20 more than three times the smallest number, what is the largest possible range for the numbers in the set?

A. 78
B. 77 1/5
C. 66 1/7
D. 55 1/7
E. 52
Show more

n=5, (avg,med)=55, sum=275
smallest=a, largest=tn, tn=3a+20
the median is the middle number 3, so what follows must be ≥ median;
(a,a,md,md,tn)/5=55…(a,a,55,55,tn)=275…(a,a,3a+20)=275-110…a=29
range: tn-a…3a+20-a…2a+20…2(29)+20…78

Ans (A)
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Set R contains five numbers that have an average value of 55 28 Feb 2021, 13:06
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Let us say R = {a,b,c,d,e}

We know that the median should be 55 so automatically we can set c = 55.

We also know that the largest (e) is equal to 20 more than 3 times the smallest (a). Therefore we can set e = 3a+20

Now how what should be the values of b and d so that we have a maximum range? To maximize the range, we must minimize the value for both b and d. The minimum value allowable for b must be equal to a. And the minimum allowable value for d must be the median 55.

Therefore we can say that b = a, and d = 55.

Let us then solve for the value of a.

a+b+c+d+e = 55(5) = 275.

b = a
c = 55
d = 55
e = 3a + 20

a + a + 55 + 55 + 3a + 20 = 275

5a = 145

a = 29.
e = 3(29) + 20 = 107

Range = e - a = 107 - 29 = 78

The answer is A
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Set R contains five numbers that have an average value of 55 17 Mar 2021, 09:36
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Bunuel
Orange08
Set R contains five numbers that have an average value of 55. If the median of the set is equal to the mean, and the largest number in the set is equal to 20 more than three times the smallest number, what is the largest possible range for the numbers in the set?


78
77 1/5
66 1/7
55 1/7
52

{\(a_1\), \(a_2\), \(a_3\), \(a_4\), \(a_5\)}
As mean of 5 numbers is 55 then the sum of these numbers is \(5*55=275\);
The median of the set is equal to the mean --> \(mean=median=a_3=55\);
The largest number in the set is equal to 20 more than three times the smallest number --> \(a_5=3a_1+20\).

So our set is {\(a_1\), \(a_2\), \(55\), \(a_4\), \(3a_1+20\)} and \(a_1+a_2+55+a_4+3a_1+20=275\).

The range of a set is the difference between the largest and smallest elements of a set.

\(Range=a_5-a_1=3a_1+20-a_1=2a_1+20\) --> so to maximize the range we should maximize the value of \(a_1\) and to maximize \(a_1\) we should minimize all other terms so \(a_2\) and \(a_4\).

Min possible value of \(a_2\) is \(a_1\) and min possible value of \(a_4\) is \(median=a_3=55\) --> set becomes: {\(a_1\), \(a_1\), \(55\), \(55\), \(3a_1+20\)} --> \(a_1+a_1+55+55+3a_1+20=275\) --> \(a_1=29\) --> \(Range=2a_1+20=78\)

Answer: A.
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Hello! I had one doubt. Since median = mean. we can conclude that this is an evenly spaced set. Then can we take 5 values as x, x+y, x+2y and so on. Please help.
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Set R contains five numbers that have an average value of 55 17 Mar 2021, 10:21
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Bunuel
Orange08
Set R contains five numbers that have an average value of 55. If the median of the set is equal to the mean, and the largest number in the set is equal to 20 more than three times the smallest number, what is the largest possible range for the numbers in the set?


78
77 1/5
66 1/7
55 1/7
52

{\(a_1\), \(a_2\), \(a_3\), \(a_4\), \(a_5\)}
As mean of 5 numbers is 55 then the sum of these numbers is \(5*55=275\);
The median of the set is equal to the mean --> \(mean=median=a_3=55\);
The largest number in the set is equal to 20 more than three times the smallest number --> \(a_5=3a_1+20\).

So our set is {\(a_1\), \(a_2\), \(55\), \(a_4\), \(3a_1+20\)} and \(a_1+a_2+55+a_4+3a_1+20=275\).

The range of a set is the difference between the largest and smallest elements of a set.

\(Range=a_5-a_1=3a_1+20-a_1=2a_1+20\) --> so to maximize the range we should maximize the value of \(a_1\) and to maximize \(a_1\) we should minimize all other terms so \(a_2\) and \(a_4\).

Min possible value of \(a_2\) is \(a_1\) and min possible value of \(a_4\) is \(median=a_3=55\) --> set becomes: {\(a_1\), \(a_1\), \(55\), \(55\), \(3a_1+20\)} --> \(a_1+a_1+55+55+3a_1+20=275\) --> \(a_1=29\) --> \(Range=2a_1+20=78\)

Answer: A.

Hello! I had one doubt. Since median = mean. we can conclude that this is an evenly spaced set. Then can we take 5 values as x, x+y, x+2y and so on. Please help.
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In 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.
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Set R contains five numbers that have an average value of 55 15 Oct 2024, 09:43
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STATS81

when we have found the maximum of the largest, we have effectively minimized all the other values, the smallest value too!

so when we want to find the maximum range, we just have to find the maximum of the largest. how do we do that?

just take into consideration that we have to minimize the values before and after the median.

the values before the median are minimized when they are equal to a value S (PAY ATTENTION TO THIS!! IT IS NOT ALWAYS THE CASE). NORMALLY WE WOULDN'T say that because obviously the range is maximized when the lowest value is as small as possible (example -10.000) but we MUST find a value that is the minimum AND must also meet the condition imposed by the problem.

the values above the median are minimized when equal to the median.

and the largest value will be named L


so we set the first 2 values to be the smallest values S, the largest value will be L while the third and fourth values will be equal to the MEDIAN.
thus we have
n=5
mean = 55
median = 55 (we call it M)
L = 3S+20

now we know that the mean will be 55

so we have
(S + S + M + M + L)/5= MEAN = 55

substitute and we obtain
2S+L = 165

the problem also tells us that
L = 3S+20


thus by solving the linear system:
2S+L = 165
L = 3S+20

we obtain that the maximum of the largest value is L = 107, while the minimum of the smaller value is S = 29

the range will be 107-29 = 78


so now notice this:

X1 X2 X3 X4 X5
29 29 55 55 107
range = 78
we can surely increase the range by decreasing the value of the lowest and still respecting the "AVERAGE" condition:

2S+L = 165

example:
X1 X2 X3 X4 X5
0 0 55 55 165
range = 165!


but in that case we no longer respect the other condition imposed by the problem:
L = 3S+20

in fact, if we plug S = 0 and L = 165 we get 165 = 20 that is obviously false.


IN GENERAL we have most of the times, 2 conditions to solve it:

--> average condition
--> condition imposed by the problem statement
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Set R contains five numbers that have an average value of 55 26 Nov 2025, 22:35
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