Inequalities: If r + s > 2t, is r > t ?

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Inequalities: If r + s > 2t, is r > t ?

by II » Thu Feb 14, 2008 3:05 am
If r + s > 2t, is r > t ?

(1) t > s

(2) r > s
Last edited by II on Tue Apr 29, 2008 5:26 am, edited 1 time in total.

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by its_me07 » Thu Feb 14, 2008 4:31 am
Is the OA C?

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by II » Thu Feb 14, 2008 2:43 pm
Sorry dont have the OC for this one.

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by arawamis » Thu Feb 14, 2008 5:54 pm
Lets see, it looks like A to me.

r+s>2t ==> r>2t-s

(1) t>s ==> 2t-s>t (if t=s then 2t-s=t)
r>2t-s>t ==> r>t sufficient

(2) r>s , so what? byitself INSUFFICIENT

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by gmatguy16 » Sat Feb 16, 2008 4:44 pm
imo d.

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by camitava » Sat Feb 16, 2008 8:46 pm
Guys, i m agree with arawamis! It should be A
Correct me If I am wrong


Regards,

Amitava

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by TkNeo » Sun Feb 17, 2008 9:39 am
I think the OA is D

Lets look at the (2)

We know R+S > 2T
-> S > 2T-R .. (1)

(2) S <R

From 1 and 2

2T - R < S <R

therefore
2T - R <R

therefore
T < R

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by ash g » Thu Mar 13, 2008 10:46 am
Is the answer A or D ?

My solution:

Given:
r + s > 2t OR
r + s > t + t


Question:
Is r > t


S1:
t > s i.e t = s + a where a is small positive number

r + s > t + t
r + s > s + a + t
r > t + a ---> this means r in fact is bigger then sum of t and another positve number so definitely r > t
SUFF

S2:
r > s i.e r = s + a where a is small positive number

r + s > t + t
r + r - a > t + t
r - a/2 > t ---> which means even after subtracting a small +ve number, r continues to remain bigger then t, so r > t
SUFF

Please let me know official answer as well as a shortcut/approach to such problems.

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by II » Wed Mar 19, 2008 8:10 am
(1) SUFFICIENT: We can combine the given inequality r + s > 2t with the first statement by adding the two inequalities:

r + s > 2t
t > s
r + s + t > 2t + s
r > t

(2) SUFFICIENT: We can combine the given inequality r + s > 2t with the second statement by adding the two inequalities:

r + s > 2t
r > s
2r + s > 2t + s
2r > 2t
r > t

The correct answer is D.

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by hengirl03 » Sat Sep 06, 2008 9:41 am
How do you know when it is a good idea to add inequality equations together?

II wrote:(1) SUFFICIENT: We can combine the given inequality r + s > 2t with the first statement by adding the two inequalities:

r + s > 2t
t > s
r + s + t > 2t + s
r > t

(2) SUFFICIENT: We can combine the given inequality r + s > 2t with the second statement by adding the two inequalities:

r + s > 2t
r > s
2r + s > 2t + s
2r > 2t
r > t

The correct answer is D.

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by hengirl03 » Thu Sep 11, 2008 9:19 pm
How do you know when it is a good idea to add inequality equations together?

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by shubhamkumar » Sun Apr 08, 2012 11:21 am
II wrote:If r + s > 2t, is r > t ?

(1) t > s

(2) r > s
Given that r+s>2t
1)t>s or -s> -t
add this to the given equation

r+s>2t
-s>-t
---------
r >t
Sufficient
2)r>s or r-s>0
add this to the given equation

r+s>2t
r-s>0
-------
2r>2t
r>t
Sufficient.
This is a Mgmat question and the OA is D