Averages

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pbahugun: Newbie | Next Rank: 10 Posts; Posts: 3; Joined: Sat Jan 01, 2011 7:22 pm

by pbahugun » Tue Jan 11, 2011 7:38 pm

I came across this eg in Kaplan Premier 2011 .

Example:- What is the sum of all 27 three-digit integers that can be created with digits 1,2 and 3.
Solution:- [As Mentioned in Kaplan]

Step1] Concept used sum=Average* Number of Values

Step2] Lowest possible number=111

Step 3] Highest possible number=333

Step4] Average=(111+333)/2.....................[I have a doubt in this step because in the book it is clearly mentioned that this concept is only applicable for consecutive integers but in our case I dont see the series as Consecutive integers.]

Step 5] sum=222* 27

Please I need clarification on Step 4.

Thanks

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Source: Beat The GMAT — Problem Solving | Tue Jan 11, 2011 7:38 pm

anshumishra: Legendary Member; Posts: 543; Joined: Tue Jun 15, 2010 7:01 pmFollowed by: 3 members; Thanked: 147 times

by anshumishra » Tue Jan 11, 2011 7:44 pm

pbahugun wrote:I came across this eg in Kaplan Premier 2011 .

Example:- What is the sum of all 27 three-digit integers that can be created with digits 1,2 and 3.
Solution:- [As Mentioned in Kaplan]

Step1] Concept used sum=Average* Number of Values

Step2] Lowest possible number=111

Step 3] Highest possible number=333

Step4] Average=(111+333)/2.....................[I have a doubt in this step because in the book it is clearly mentioned that this concept is only applicable for consecutive integers but in our case I dont see the series as Consecutive integers.]

Step 5] sum=222* 27

Please I need clarification on Step 4.

Thanks

Another approach :

sum of the units column is 9(1 + 2 + 3) =54
sum of the tens column is 9(1 + 2 + 3) =54
sum of the hundred column is 9(1 + 2 + 3) =54

So, sum = 54(100+10+1) = 54*111 = 5994

Thanks
Anshu

(Every mistake is a lesson learned )

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anshumishra: Legendary Member; Posts: 543; Joined: Tue Jun 15, 2010 7:01 pmFollowed by: 3 members; Thanked: 147 times

by anshumishra » Tue Jan 11, 2011 7:53 pm

pbahugun wrote:I came across this eg in Kaplan Premier 2011 .

Example:- What is the sum of all 27 three-digit integers that can be created with digits 1,2 and 3.
Solution:- [As Mentioned in Kaplan]

Step1] Concept used sum=Average* Number of Values

Step2] Lowest possible number=111

Step 3] Highest possible number=333

Step4] Average=(111+333)/2.....................[I have a doubt in this step because in the book it is clearly mentioned that this concept is only applicable for consecutive integers but in our case I dont see the series as Consecutive integers.]

Step 5] sum=222* 27

Please I need clarification on Step 4.

Thanks

Given Approach :

To understand this approach, lets try with a simpler example. Use these 3 digits (1,2,3) for two numbered digits only.
The idea is to see that 22 is the average (hence the sum would be equal to 22*no. of terms , if we want to find) :

11, 12, 13, 21, 22, 23, 31,32,33
Check out the left and right side of 22:
21 = 22-1, 23 = 22+1
13 = 22-9, 31 = 22+9
12 = 22-10, 32 = 22 + 10
11 = 22-11, 33 = 22 +11

Similarly, in the current problem 222 is the average, hence sum = 222*27

Thanks
Anshu

(Every mistake is a lesson learned )

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Anurag@Gurome: GMAT Instructor; Posts: 3835; Joined: Fri Apr 02, 2010 10:00 pmFollowed by: 523 members; Location: Milpitas, CA; Thanked: 1854 times; GMAT Score:770

by Anurag@Gurome » Tue Jan 11, 2011 8:54 pm

I have a doubt in this step because in the book it is clearly mentioned that this concept is only applicable for consecutive integers but in our case I dont see the series as Consecutive integers.

No.
This concept is also applicable for a set of numbers {a(1), a(2), a(3), ..., a(n - 1), a(n)} in increasing order with average A (say), such that a(1) and a(n), a(2) and a(n - 1), a(3) and a(n - 2) etc are equidistant from A. Which is equivalent to say that the average of each such pair is also equal to A. Which implies average of all such pairs, i.e. average of the set will be also equal to A. Which is again equivalent to the average of smallest and largest numbers in the set.

This is because the deviations from the average on either side of the average cancels each other. For example, take Anshu's example:

Set of integers : {11, 12, 13, 21, 22, 23, 31, 32, 33}
Average = 22
11 and 33 are equidistant from 22 --> Deviation from 22 on both side = 11
12 and 32 are equidistant from 22 --> Deviation from 22 on both side = 10
13 and 31 are equidistant from 22 --> Deviation from 22 on both side = 9
21 and 33 are equidistant from 22 --> Deviation from 22 on both side = 1
22 itself is equidistant from 22 --> Deviation from 22 on both side = 0

Average of each pair = 22

Now someone may be wondering what if the average is not on the set or the numbers of integers in the set is even! Well, just take out 22 from the above set. Everything remains same.

Also this is not only applicable for integers. It is applicable for any such set of real numbers. I think some of you have already noted that the case of consecutive integer is just a special situation for this generalized one.

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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Tue Jan 11, 2011 9:48 pm

Another way of stating the rule described above:

When a set of values is symmetrical about the median, sum = (biggest+smallest)/2 * (number of values).

For example:

The set of values {1, 2, 6, 10, 11} is symmetrical about the median (6).
2 and 10 are each 4 from the median; 1 and 11 are each 5 from the median.
Sum = (11+1)/2 * 5 = 30.

Since evenly spaced integers are symmetrical about the median, their sum is determined using the same formula.

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