x and y

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x and y

by cmal_s » Wed Mar 24, 2010 10:17 am
QUESTION)

Is x+ y < 1

1) x< 8/9
2) y< 1/8

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by krbharat » Wed Mar 24, 2010 10:39 am
cmal_s wrote:QUESTION)

Is x+ y < 1

1) x< 8/9
2) y< 1/8



OA c?

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by kvamsy » Wed Mar 24, 2010 10:42 am
cmal_s wrote:QUESTION)

Is x+ y < 1

1) x< 8/9
2) y< 1/8
Is the answer is C ?

Stnt 1) is not sufficent alone as we dont know the y value ,simlar scenario for Stnt 2.

So with the combination we can determine the condition.

Please let me know whether this explanation is correct.

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by newton9 » Wed Mar 24, 2010 1:23 pm
take X = 7.9/9
Y = 0.9/8

now X + Y = 72.3/72 which is greater than 1.

For other values, it is less than 1.

So E

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by dxgamez » Wed Mar 24, 2010 8:21 pm
cmal_s, what is the answer?

i have my answer as C. but newton has a point.

if we were to let x to be 7.9/9 and y to be 0.9/9 then we might have an answer which is more than 1 as well as less than 1.

is E correct then? any other explanation would be really helpful.

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by tanviet » Wed Mar 24, 2010 8:48 pm
suppose A=8/9 +1/8>8/9+1/9=1 that means A>1

x+y<A
A>1

E is correct

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by sanju09 » Thu Mar 25, 2010 5:43 am
cmal_s wrote:QUESTION)

Is x+ y < 1

1) x< 8/9
2) y< 1/8
each statement alone is insufficient, fine

add the two inequalities and get

x + y < 8/9 + 1/8 or x + y < 73/72

now, a sum less than 73/72 can be less than, or equal to, or more than 1, so [spoiler]E[/spoiler]
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by el_torero » Thu Mar 25, 2010 4:56 pm
i agree with the above posts.

E is the right answer and here's why:

once you derive x+y < 73/72, it becomes clear you can come up with two numbers that are either < 1 (i.e. negative numbers for both x and y) or a pair of numbers whose sum is greater than 1 AND less than 73/72 (i.e. 729/720).

gg gl