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sum and averages

by sumasajja » Thu Jul 21, 2011 11:46 am
I could't understand an answer give to the following question of a book ............can any1 explain me how??????
What is the sum of all odd integers between 10 and 40?
(A) 250
(B) 325
(C) 375
(D) 400
(E) 450
The correct answer is (C). The average of the described numbers is
25-halfway between 10 and 40 (in other words, half the sum of 10 and 40). The
number of terms in the series is 15. (The first term is 11, and the last term is 39.)
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by GMATGuruNY » Thu Jul 21, 2011 12:43 pm
sumasajja wrote:I could't understand an answer give to the following question of a book ............can any1 explain me how??????
What is the sum of all odd integers between 10 and 40?
(A) 250
(B) 325
(C) 375
(D) 400
(E) 450
The correct answer is (C). The average of the described numbers is
25-halfway between 10 and 40 (in other words, half the sum of 10 and 40). The
number of terms in the series is 15. (The first term is 11, and the last term is 39.)
We need to determime the sum of all the odd integers between 11 and 39, inclusive.

Use the following:

Sum = number * average.

To count the number of evenly spaced integers in a set:

Number of integers = (Biggest - Smallest)/(distance between each successive pair) + 1
Since we're adding only the odd integers, the distance between each successive pair is 2.
Thus, the number of odd integers from 11 to 39 = (39-11)/2 + 1 = 15.

When numbers are evenly spaced, the average = the median = the average of the biggest and the smallest.
Given 11 to 39, the average = (39+11)/2 = 25.

Sum = number * average = 15*25 = 375.

The correct answer is C.
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by Anurag@Gurome » Thu Jul 21, 2011 8:32 pm
sumasajja wrote:I could't understand an answer give to the following question of a book ............can any1 explain me how??????
What is the sum of all odd integers between 10 and 40?
(A) 250
(B) 325
(C) 375
(D) 400
(E) 450
The correct answer is (C). The average of the described numbers is
25-halfway between 10 and 40 (in other words, half the sum of 10 and 40). The
number of terms in the series is 15. (The first term is 11, and the last term is 39.)

Solution:
Sum is 11 + 13 + 15 +.....+ 39 = (2 + 9) + (4 + 9) +......+(30 + 9).
= 2*(1 + 2 +...+ 15) + 9*15
= 2*15*16/2 + 135 = 240 + 135 = 375 {Using the formula that the sum of first n positive integers is n*(n+1)/2}
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