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a>b>c>0

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by xcusemeplz2009 » Fri Oct 23, 2009 7:50 pm
IMO A
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by Harbinder » Fri Oct 23, 2009 7:51 pm
it should be D
first statement is straight forward
it gives u a<3 and c is less than a then c has be < 3

Second statement try some numbers and fo only numbers where c <3 we can have the statement true e.g c=2, b=3 and a=6 if c is >or = 3 then b and c will hae to be greater than 3 and then 1/a + 1/b + 1/c can never be 1 because if 1/c = 3 then either both 1/a or 1/b will have to =3 or one of them need to be > 3 and other < 3 which is not true so c cannot be = or > 3.

I hope I was able to explain my logic...
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Re: a>b>c>0

by Stuart@KaplanGMAT » Fri Oct 23, 2009 8:00 pm
uptowngirl92 wrote:4)If a>b>c>0, is c <3?
(1) 1/a>1/3
(2) 1/a + 1/b + 1/c=1

EXPLANATIONS are welcome:)
(1) since a is positive, we can safely cross multiply to get:

3 > a

and, since we know that a > c from the original stem, we can answer "is c < 3?" with a definite YES... sufficient.

(2) we know that a, b and c are all positive. If a=b=c, then they'd all equal 3. However, we know that c is the smallest of the 3 numbers.

Since we're talking fractions, when we make a, b and c bigger we actually make the fractions smaller. Since a,b and c have to be different, at least one of the 3 numbers must be smaller than 3 to get a sum of 1.

Accordingly, since c is the smallest of the 3 numbers, c MUST be smaller than 3: sufficient.

Each of 1 and 2 is sufficient alone: choose D.
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by sudeeparies » Fri Oct 23, 2009 8:03 pm
(1) 1/a>1/3 => a<3 hence c<3 as a>>c
(2) 1/a + 1/b + 1/c=1 => only for c<3; for example take a=3, b=2, c=1 or a=6, b=3 and c=2 can this statement be true (as soon as you take c=3, b and c have to be greater than 3 and this means 1/a + 1/b + 1/c will be greated than 1 )

So, the answer should be option D
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x (y + z) ≥ 0?

by uptowngirl92 » Sat Oct 24, 2009 1:02 am
If xyz ≠ 0, is x (y + z) ≥ 0?


1) |y + z| = |y| + |z|
2) |x + y| = |x| + |y|

Have no clue as to how to proceed:(
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Re: x (y + z) ≥ 0?

by xcusemeplz2009 » Sun Oct 25, 2009 12:02 pm
uptowngirl92 wrote:If xyz ≠ 0, is x (y + z) ≥ 0?


1) |y + z| = |y| + |z|
2) |x + y| = |x| + |y|

Have no clue as to how to proceed:(
IMO C
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Re: x (y + z) ≥ 0?

by palvarez » Sun Oct 25, 2009 1:20 pm
uptowngirl92 wrote:If xyz ≠ 0, is x (y + z) ≥ 0?


1) |y + z| = |y| + |z|
2) |x + y| = |x| + |y|

Have no clue as to how to proceed:(
Note that this is not an equality. So, it is a bit easier to solve them.

(1) how many possible combos are there for y and z?
(positive, positive) (positive, negative), (negative, positive), (negative, negative).
Look out the possible combos that satisfy (1): (positive, positive) and (negative, negative)
In other words, yz >=0; given xyz <> 0, yz > 0

yz > 0 ---> (y+z) > 0 or (y+z) < 0
we don't know anything abt x. Insuff.

(2) Same logic.
xy > 0

Combining together:

xy > 0, yz > 0

Look out the possible combination of (x,y,z) that satisfy the above constraints.
(+ve, +ve, +ve) and (-ve, -ve, -ve)

(x, y+z) = (+ve, +ve) or (-ve, -ve)
x(y+z) = +ve

This shows both of em are required.
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Re: a>b>c>0

by aarok » Mon Oct 26, 2009 4:14 am
One nore vote for D
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