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Re: If the average of 10 consecutive odd integers is 224, what is the leas [#permalink]
Hoozan wrote:
Method 1

We know that there are 10 consecutive odd numbers. Leaving thr first number in the list ( the least one) we have 9 consecutive numbers. And we are given that the average is 224. Thus, 224-9 = 215 will be the least number in the list of odd numbers.


Method 2

10 consecutive odd numbers can be represented as a, a+2, a+3.... a+18 with "a" as the first number in the list.

The sum of these 10 numbers is 224×10 = 2240.

Thus we can say 10a + 90 = 2240. By solving we get a = 215.

Posted from my mobile device


Could you please eloborate more on your methods, thanks :please:
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Re: If the average of 10 consecutive odd integers is 224, what is the leas [#permalink]
satubus wrote:
Hoozan wrote:
Method 1

We know that there are 10 consecutive odd numbers. Leaving thr first number in the list ( the least one) we have 9 consecutive numbers. And we are given that the average is 224. Thus, 224-9 = 215 will be the least number in the list of odd numbers.


Method 2

10 consecutive odd numbers can be represented as a, a+2, a+3.... a+18 with "a" as the first number in the list.

The sum of these 10 numbers is 224×10 = 2240.

Thus we can say 10a + 90 = 2240. By solving we get a = 215.

Posted from my mobile device


Could you please eloborate more on your methods, thanks :please:

\(n + ( n + 2 ) + ( n + 4 ) + ( n + 6 ) + ( n + 8 ) + ( n + 10 ) + ( n + 12 ) + ( n + 14 ) + ( n + 16 ) + ( n + 18 ) = 2240\)

Or, \(10n + 90 = 2240\)

Or, \(n + 9 = 224\)

Or, \(n = 215\)

Hope this helps.....

Welcome to gmatclub and for your first post, happy learning!!
GMAT Club Bot
Re: If the average of 10 consecutive odd integers is 224, what is the leas [#permalink]
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