Rayovac
Advanced Member
United States
2,049 posts
Joined: Jan, 2008
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 Posted - 2010/03/11 : 23:22:29
Cheers Lilley! I don't know why but I messed up this one bad, my original equation was (x + 1) + (x + 3) = 54, which WAS correct (although it's faster if I just use yours, (x - 1) + (x + 1) = 54).
And yes I did spot the trick... 54/2 = 27, which is the answer. I trained myself not to even think about doing that kind of thing long ago as it almost always lead to wrong answer. 
quote:
Originally posted by Lilley:
quote:
Originally posted by Rayovac:
Tom's not online so I'll post this here.
I made some MAJOR progress over the last few days, although I've run into another problem. It's just a bonus problem but I really want to get it solved.
"The sum of the first and third of three consecutive integers is 54. What is the middle integer?
a. 27
b. 31
c. 35
d. 37"
It doesn't specify whether the consecutive integers are even or odd or anything like that, so I'm assuming 1 2 and 3. So I write it out like this:
(x + 1) + (x + 2) + (x + 3) = 54
Simplified: 3x + 6 = 54
Addition property of equality: 3x + 6 - 6 = 54 - 6, which gives 3x = 48
48/3 = 16
So what I get is 17 + 18 + 19 = 54, which is true, but 18 isn't an option. Where did I screw up?
you screwed up in reading the question. its only the first and third integers that make 54, not all three. 17+ 19 = 36. and yes consecutive means in a row. If it is odd consecutive then 3,5,7,9. if not specified then just the number directly next to it.
Just a hint. instead of (x+1) + (x+2), you can set your reference point (x) to a different number, even make it part of the sequence ie. (x) + (x+1) + (x+2) or (x-1) + (x) + (x+1). this should make things slightly easier.
Hint: there is a really really easy trick to this question if you can spot it. dont waste your time trying to spot it if you cant see it though.
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