r + s + t even?

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r + s + t even?

by kayser » Mon Feb 09, 2009 12:08 pm
If r, s, and t are positive integers, is r + s + t even?
(1) r + s is even.
(2) s + t is even.

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Re : r + s + t even?

by ballaljoshi » Mon Feb 09, 2009 12:57 pm
Let's use the method of replacement (assumption) here.

r(2) + s(4) = 6
s(4) + t(8) = 12

So, r+s+t = 2+4+8 = 14

However, because r,s,t are postive integers, the value of any of these letters can be even or odd.

so if we replace the value of t as odd, then

s(4) + t(9) = 13

hence (s+t) + r = 13 + 2 = 15

Hence, the answer is E.

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by cramya » Mon Feb 09, 2009 5:21 pm
Each statement by itself is insufficient.

Stmt I

Take r,s to be both even
r+s+t is even if t is even or odd if t is odd

INSUFF


Stmt II

Take s,t to be both even

r+s+t is even if r is even or odd if r is odd

INSUFF

Together r,s,t can all be odd or can all be even satisfying stmt I and II which would make r+s+t odd or even.

Hence E
Last edited by cramya on Mon Feb 09, 2009 5:22 pm, edited 1 time in total.

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by mrsmarthi » Mon Feb 09, 2009 5:22 pm
THe question asked for is "Is r + s + t even? "

This can be answer if we know whether r,s and t are even / odd.

From stmt 1,

It is given that r + s is even ==> Both r and s should either be even or odd.

No informaiton related to t is given. So if t is even then the sum will be even. And if t is odd, then sum will be odd. Hence insufficient.

From stmt 2 -

It is given that s + t is even ==> Both s and t should either be even or odd.

No informaiton related to r is given. So if r is even then the sum will be even. And if r is odd, then sum will be odd. Hence insufficient.

Now combine both the clues.

r + s is even and s + t is even ==> We can derive that r,s,t are either all even or all odd. So their sum can be either be even or odd. Insifficient.

IMO E

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by Bidisha800 » Mon Feb 09, 2009 8:58 pm
(E)
Drill baby drill !

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