QR question-Do you know a quick approach?

[Doa110]

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If the average of 18 consecutive odd integers is 534, then the least of these integers is:
A-517
B-518
C-519
D-521
E-525

This is what I do for average of consecutive-odd or even # :
Odd....> (2n+1)+(2n+3)+(2n+5)+.../ Total # of consecutives=Mean

Does any one knows a quick way to solve this. 18 consecutive odd numers is long and time consuming! The answer is "A".

 

[klutzy1987]

If the average of 18 consecutive odd integers is 534, then the least of these integers is:
A-517
B-518
C-519
D-521
E-525

This is what I do for average of consecutive-odd or even # :
Odd....> (2n+1)+(2n+3)+(2n+5)+.../ Total # of consecutives=Mean

Does any one knows a quick way to solve this. 18 consecutive odd numers is long and time consuming! The answer is "A".

Well its 533-16=517. The reason for this is because the average is 534 that means that integer 9 is 533 and integer 10 is 535. Therefore 8 odd integers before 533 or 533-(8*2)=517.

 

[Doa110]

Thanks. I got it now.

 

[klutzy1987]

You welcome, although the way you were doing it, why did you have 2x as opposed x,x+2,x+2 etc..??