Set R contains five numbers that have an average value of 55 : Problem Solving (PS)

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03 Oct 2010, 13:21

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I took the set to be m, m, 55, 55, 3m+20. (second value has to be minimum possible - m, and fourth value has to be minimum possible - 55).

now average is 55 so 55 = (6m + 130)/5 which gives m = 29, and 3m+20=107

so range is largest - smallest = 107-29 = 78

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07 Mar 2011, 20:20

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Let smallest # = x, Largest = 3x + 20

So range = 2x + 20

x, x, 55, 55, 3x+20, For Max range lowest should be as low as possible and highest should be as high as possible

also, the 2nd value has to be minimized, so it is x, the fourth value also ahs to be kept at minimum, so it is 55

3x + 20 + 110 + 2x = 275

=> 5x = 275 - 130 = 145 => x = 29 , so range = 29*2 + 20 = 78

So answer is A.

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08 Mar 2011, 15:51

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gmat1220

Backsolving : Range = 2a + 20 where a = first number.

Hence the answer is even and highest options. It should be A.

That's an awesome application of number properties to solve this question is seconds.
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Bunuel

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08 Mar 2011, 16:06

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Yalephd

gmat1220

Backsolving : Range = 2a + 20 where a = first number.

Hence the answer is even and highest options. It should be A.

That's an awesome application of number properties to solve this question is seconds.
Kudos

That's not correct. Yes, the range equals to 2a+20 but without any further calculation we cannot say whether it must be even, for example if a is not an integer then 2a+20 can be odd or not an integer at all. Also the answer is not necessarily the highest option, it just happened to be so in this particular case.

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08 Mar 2011, 16:19

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Bunuel

Yalephd

gmat1220

Backsolving : Range = 2a + 20 where a = first number.

Hence the answer is even and highest options. It should be A.

That's an awesome application of number properties to solve this question is seconds.
Kudos

That's not correct. Yes, the range equals to 2a+20 but without any further calculation we can not say whether it must be even, for example if a is not an integer then 2a+20 can be odd or not an integer at all. Also the answer is not necessarily the highest option, it just happened to be so in this particular case.

Thanks. Assuming that A is an integer is where I erred.

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08 Mar 2011, 17:54

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I was back solving - to confirm the answer.

x^2 = 4
Implies x is not necessarily 2. It can be -2 :-D

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01 May 2011, 23:15

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max range will be when 55*3 = 165 will give 110 as range.But the value isn't present.
Hence go for two small numbers , 55*2 and largest number combination.
thus 2x+110 + 3x+20 = 275
will give, x= 29 and 3x+20 = 97.
Range = 78.

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01 May 2011, 23:16

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max range will be when 55*3 = 165 will give 110 as range.But the value isn't present.
Hence go for two small numbers , 55*2 and largest number combination.
thus 2x+110 + 3x+20 = 275
will give, x= 29 and 3x+20 = 97.
Range = 78.

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18 Nov 2011, 04:18

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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.

my approach was like yours, but it took me 6 min!!!

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01 Nov 2012, 08:01

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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.

Bunuel sir..

few questions that cums in my mnd like ..y did bunuel take A1 is equal to A2..and y didnt he take a2=55 instead of A4=55?

i got lots of questions like this and i cant give ans correctly..

Thank u in advance bunuel..

Bunuel

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02 Nov 2012, 05:17

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sanjoo

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.

Bunuel sir..

few questions that cums in my mnd like ..y did bunuel take A1 is equal to A2..and y didnt he take a2=55 instead of A4=55?

i got lots of questions like this and i cant give ans correctly..

Thank u in advance bunuel..

After some steps we have that our set in ascending order is {\(a_1\), \(a_2\), \(55\), \(a_4\), \(3a_1+20\)} and \(Range=2a_1+20\).

We need to maximize \(Range=2a_1+20\), thus we need to maximize \(a_1\) and to maximize \(a_1\) we should minimize all other terms so \(a_2\) and \(a_4\) (remember the sum of the terms is fixed, so we cannot just make \(a_1\) as large as we want).

Now, since the set is in ascending order min possible value of \(a_2\) is \(a_1\) (it cannot be less than the first term) and min possible value of \(a_4\) is \(median=a_3=55\) (it cannot be less than the third term).

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Hope it helps.

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02 Nov 2012, 06:13

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Thanks alot Bunuel..now i got that

..

i think in REAL GMAT these type of question cum frequenlty..!!.

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24 Jan 2013, 03:38

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Hi ,

Here's how I did..

smallest no: s
largest no: 3s+20

since mean = median,

thought that numbers are in AP.

so average= (last no+first no)/2

therefore 55=(s+3s+20)/2
=> s=22.5

now l=20+3s
=> l=87.25

range =l-s=65..
Please let me know where I am going wrong.

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24 Jan 2013, 13:29

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Sachin9

Hi ,

Here's how I did..

smallest no: s
largest no: 3s+20

since mean = median,

thought that numbers are in AP.

so average= (last no+first no)/2

therefore 55=(s+3s+20)/2
=> s=22.5

now l=20+3s
=> l=87.25

range =l-s=65..
Please let me know where I am going wrong.

Sachin, you assumed that the numbers are in AP, but problem doesn't state that.
This set S = {29, 29, 55, 55, 107} has the maximum range i.e. 78 and mean/median 55.

Note that these numbers are not in AP/sequence. Hence you cannot take average of last & first to find the mean.

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24 Jan 2013, 17:22

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PraPon

Sachin9

Hi ,

Here's how I did..

smallest no: s
largest no: 3s+20

since mean = median,

thought that numbers are in AP.

so average= (last no+first no)/2

therefore 55=(s+3s+20)/2
=> s=22.5

now l=20+3s
=> l=87.25

range =l-s=65..
Please let me know where I am going wrong.

Sachin, you assumed that the numbers are in AP, but problem doesn't state that.
This set S = {29, 29, 55, 55, 107} has the maximum range i.e. 78 and mean/median 55.

Note that these numbers are not in AP/sequence. Hence you cannot take average of last & first to find the mean.

Thanks mate..
I thought that the numbers would be in AP since their median and mean were same.

I now understand that if the nos are in AP , then their median and mean will be same but the vice versa is not necessarily true.

russ9

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26 May 2014, 13:17

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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.

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?

Bunuel

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27 May 2014, 01:16

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russ9

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.

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.

Set R contains five numbers that have an average value of 55

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