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Expert replies
Source: — Data Sufficiency |

by MAAJ » Tue Apr 05, 2011 6:29 pm
Is it (x/3) + (3/x) > 2 ?
ccassel wrote:Hi,

How would you answer this question?

Is x/3+3/x>2

(1) x<3
(2) x>1

Cheers,
"There's a difference between interest and commitment. When you're interested in doing something, you do it only when circumstance permit. When you're committed to something, you accept no excuses, only results."
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by ccassel » Tue Apr 05, 2011 6:32 pm
It wasn't written with brackets but i would assume that is correct.
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by vineeshp » Tue Apr 05, 2011 6:58 pm
From 1) x < 3
If we choose x=2 --> x/3 + 3/x > 2
But if x = -2 (still satisfies x < 3) we can get that x/3 + 3/x < 0 hence !>2.

From 2)
x>1
At x=1 ( I know this is not valid, bt only for illustration)
x/3 + 3/x = 3 1/3

At x=2
x/3 + 3/x = 13/6 = 2 1/6

At x=3
x/3 + 3/x = 2

At x=4
x/3 + 3/x = 4/3 + 3/4 = 25/12 = 2 1/12

So expression attains minimum value of 2 at x = 3.

But it still does not prove that x/3 + 3/x > 2.

If you combine 1 and 2.
We just deduced that x/3 + 3/x decreases from 3 1/3 at x = 1 and stops at 2 when x = 3. So in between these values x/3 + 3/x > 2.

Hence C.
Vineesh,
Just telling you what I know and think. I am not the expert. :)
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by manpsingh87 » Tue Apr 05, 2011 8:00 pm
ccassel wrote:Hi,

How would you answer this question?

Is x/3+3/x>2

(1) x<3
(2) x>1

Cheers,
taking lcm we have (x^2+9)/3x >2;

1) if x<3; i.e. it includes all the value to the left of 3 on the number line..!!

if x=1 we have 10/3=3.33>2;
if x=-1 we have 10/-3, -3.33<2
as two different results are possible here for different values of x hence 1 is not sufficient to answer the question.

2)x>1; all the values to the right of 1 on the number line.. now consider x=3;
9+9/9=2 2 is not<2; and for all other values we have results greater than 2, hence 2 alone is also not sufficient to answer the question.

combining 1 and 2 we have
1<x<3;
for all the values between 1 and 3; x is greater than 2 hence both together are sufficient to answer the question...!!!!
C
O Excellence... my search for you is on... you can be far.. but not beyond my reach!
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by ccassel » Wed Apr 06, 2011 7:43 am
manpsingh87 - Why did you calculate the lcm of x/3+3/x>2? How do you get that result?

Thanks in advance for the tips.

Cheers,
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by manpsingh87 » Wed Apr 06, 2011 8:26 am
ccassel wrote:manpsingh87 - Why did you calculate the lcm of x/3+3/x>2? How do you get that result?

Thanks in advance for the tips.

Cheers,
my approach here is to combine two different fractions into one...!!for example if we have to solve the question like 1/3+1/2+1/4+1/6 then our approach should be to combine four different fractional terms into one i.e. to minimize the number of terms in the question..!!!
O Excellence... my search for you is on... you can be far.. but not beyond my reach!
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by ccassel » Wed Apr 06, 2011 8:40 am
I see. Thanks,
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