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Here, you can manipulate inequalities.

From the givens, we know that:

r+s > 2t

therefore:
(1)
r+s > 2t
t > s

Inequalities with the same sign can be added / subtracted

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

SUFF


(2)
r + s > 2t
r > s

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

SUFF

Answer: D
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Wayxi
If r + s > 2t, is r > t ?

(1) t > s

(2) r > s

r + s > 2t

(1) t > s
So, r+t > r+s > 2t
So, r+t > 2t
So, r>t
Sufficient

(2) r > s
So, r+r > s+r > 2t
So, 2r > 2t
So, r > t
Sufficient

Answer is (d)
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thx lagomez. just forgot that we could add inequalities and equations to help simplify an equation.
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Hi Bunnel, since we don't know the signs of t and s, how can we subtract s on both sides to simplify the inequality R+s+t>S+2T? Am I missing something here?

Bunuel
If r + s > 2t, is r > t ?

(1) t > s --> since the signs of two equations (t > s and r + s > 2t) are the same direction we can sum them: \(t+(r+s)>s+2t\) --> \(r>t\). Sufficient.

(2) r > s --> the same here: since the signs of two equations (r > s and r + s > 2t) are the same direction we can sum them: \(r+(r+s)>s+2t\) --> \(2r>2t\) --> \(r>t\). Sufficient.

Answer: D.

THEORY:
You can only add inequalities when their signs are in the same direction:

If \(a>b\) and \(c>d\) (signs in same direction: \(>\) and \(>\)) --> \(a+c>b+d\).
Example: \(3<4\) and \(2<5\) --> \(3+2<4+5\).

You can only apply subtraction when their signs are in the opposite directions:

If \(a>b\) and \(c<d\) (signs in opposite direction: \(>\) and \(<\)) --> \(a-c>b-d\) (take the sign of the inequality you subtract from).
Example: \(3<4\) and \(5>1\) --> \(3-5<4-1\).

Hope it helps.
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Hi Bunnel, since we don't know the signs of t and s, how can we subtract s on both sides to simplify the inequality R+s+t>S+2T? Am I missing something here?

Bunuel
If r + s > 2t, is r > t ?

(1) t > s --> since the signs of two equations (t > s and r + s > 2t) are the same direction we can sum them: \(t+(r+s)>s+2t\) --> \(r>t\). Sufficient.

(2) r > s --> the same here: since the signs of two equations (r > s and r + s > 2t) are the same direction we can sum them: \(r+(r+s)>s+2t\) --> \(2r>2t\) --> \(r>t\). Sufficient.

Answer: D.

THEORY:
You can only add inequalities when their signs are in the same direction:

If \(a>b\) and \(c>d\) (signs in same direction: \(>\) and \(>\)) --> \(a+c>b+d\).
Example: \(3<4\) and \(2<5\) --> \(3+2<4+5\).

You can only apply subtraction when their signs are in the opposite directions:

If \(a>b\) and \(c<d\) (signs in opposite direction: \(>\) and \(<\)) --> \(a-c>b-d\) (take the sign of the inequality you subtract from).
Example: \(3<4\) and \(5>1\) --> \(3-5<4-1\).

Hope it helps.

You are mixing subtraction/addition with multiplication/division. We are only concerned with sign when we multiply/divide an inequality.
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\(r+s>2t \rightarrow \frac{r+s}{2}>t\)

So the midpoint of \(r\) and \(s\) is greater than \(t\)

The orange line represents possible locations on the number line for \(t\)

Statement 1)

If \(t>s\) then only the second option is possible, and \(r\) must be greater than \(t\)

Statement 2)

If \(r>s\) then only the second option is possible, and again \(r\) must be greater than \(t\)


Answer D
Attachments

r,s,t.PNG
r,s,t.PNG [ 7.67 KiB | Viewed 27313 times ]

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One way to do this is to manipulate the original stem from r-s > 2t (given) into r-t > t-s (another way to see the stem info).

Qn asks if r-s is > 0 --> two ways to know if we can answer yes:

a) if (r-s) > 0 directly, or
b) if (t-s) > 0 --> therefore (r-t) MUST be > 0

Statement 1: (t-s) > 0 (sufficient)
Statement 2: (r-s) > 0 (sufficient - the question answers its own question)

Answer: D
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Sir Bunuel, do you have a quick list of similar questions at hand?
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Video solution from Quant Reasoning starts at 10:00
Subscribe for more: https://www.youtube.com/QuantReasoning? ... irmation=1
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CAN SOMEONE HELP ME BY DOING THIS BY PLUG IN VALUES?
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CAN SOMEONE HELP ME BY DOING THIS BY PLUG IN VALUES?


To be honest, I tried doing this by plugging numbers, but since there is very little info about the nature of r,s,t (integer, fraction, +ve, -ve etc) I took a lot of time.

Therefore I then chose to use the algebraic method.

Never hurts to have many times of weapon in your arsenal!

Let me know if you did manage to plug in values for this
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I didnt realize that we could get it by adding the equations. i was trying to work with the question stem trying to get it similar to 1 and 2. What particular clue in the question would tell us to add?

Quote:
Bunuel
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viswanath1
I didnt realize that we could get it by adding the equations. i was trying to work with the question stem trying to get it similar to 1 and 2. What particular clue in the question would tell us to add?

Quote:
Bunuel
The clue is this:

The direction of the inequalities is the same.

  • Both statements also give inequalities:
  • (1) t > s
    (2) r > s

Since these inequalities are all in the “greater than” direction, you can think of adding them together to combine information.

If the signs had been in opposite directions (one > and one <), adding wouldn't work, and you’d look to subtract instead.

That’s the key clue:

  • Same direction? Try adding.
  • Opposite direction? Try subtracting.
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