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is m, p, and t are positive integers and m (1) t-p=p-m (2)

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Manager
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is m, p, and t are positive integers and m (1) t-p=p-m (2) [#permalink] New post 05 Oct 2003, 22:47
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A
B
C
D
E

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is m, p, and t are positive integers and m
(1) t-p=p-m
(2) t-m=16
answer: (E)

if ab=a, what is the value of (axb) (ayb)?
(1) ax=by=2
(2) 2xy=4
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 [#permalink] New post 31 Oct 2003, 10:04
How did you get C for the second one?please explain.
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 [#permalink] New post 26 Dec 2003, 07:36
Since ab = a b = 1
from condition 1 by = 2 so y = 2
from condition 2 2xy = 4 so x = 1
going back to condition 1 ax = by = 2 since x = 1 a = 2
we know a,b,x and y
So C is the answer
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 [#permalink] New post 26 Dec 2003, 08:18
I agree with C and anandak's explanation.
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 [#permalink] New post 26 Dec 2003, 08:39
I was not questioning his explanation. I didn't understand it at the first sight. Now, I see b = 1 from the given the problem. So, the answer should be C.
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 [#permalink] New post 30 Dec 2003, 15:21
Friends, I would agree with the answer but a little different opinion on the explanation.
Given,

ab = a => ab - a = 0 => a(b-1) = 0

=> a = 0 & b <> 1
OR
a = 0 & b = 1
OR
a <> 0 & b = 1


(Here <> stands for "not equal to")

What do we need to find? (axb) (ayb)

Since ab = a, this can be simplified to (ax) (ay) = (a^2)xy = ?

Statement 1:

ax=by=2 => Since ax = 2, a can NOT be 0. So the only possibility is the third one above (in bold). That is b = 1 => y = 2. This does not help us find the value of (a^2)xy.


Statement 2:

2xy = 4 => xy = 2 => This information along with the possible values of a and b above (in bold) will not help us find the value of (a^2)xy

TOGETHER

ax = 2
by = 2
xy = 2
=> (2/a) (2/b) = 2 => 4/ab = 2 => ab = 2 => a = 2 (because ab = a is given)

So we have value of a and xy hence,we can find out (a^2)xy
  [#permalink] 30 Dec 2003, 15:21
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