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# Is rst > 1?

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Joined: 02 Sep 2009
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14 Jun 2017, 06:49
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35% (medium)

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79% (01:42) correct 21% (01:49) wrong based on 69 sessions

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Is rst > 1?

(1) rs > t > 1

(2) (rs)^2 + t^2 > rst

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Re: Is rst > 1?  [#permalink]

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14 Jun 2017, 07:05
Bunuel wrote:
Is rst > 1?

(1) rs > t > 1

(2) (rs)^2 + t^2 > rst

(1) if t>1 and rs>t>1 then rst will be greater than 1..sufficient

(2) rst<(rs)^2 + t^2

since (rs)^2 + t^2>0 (Both (rs)^2 & t^2 are >0)
rst<0...sufficient

D
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Re: Is rst > 1?  [#permalink]

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14 Jun 2017, 08:09
1
Bunuel wrote:
Is rst > 1?

(1) rs > t > 1

(2) (rs)^2 + t^2 > rst

Statement 1 - from this option we may infer that both T and RS are greater than 1. Hence Sufficient
Statement 2 -
Scenario 1 - RS =0.2 & T = 0.1
RST<1

Scenario 1 - RS =2 & T = 0.1
RST>1

Hence Insufficient

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Re: Is rst > 1?  [#permalink]

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18 Jun 2017, 02:29
rst>1?
Statement 1 - rs>t>1 -> rs>1 and t>1 --- > rst>1 (multiplying both equations as we know that both inequalities are +ve's and have same sign) - sufficient.
Statement 2 - rs^2 + t^2 > rst --> rs^2 +t^2 -rst>0 ---> plugging in numbers r= 1/2, s= 2, t= 1/2 ---> (1/2*2)^2 +
(1/2)^2 - (1/2*2*1/2) = 1+1/4-1/2 = 3/4 which is >0 but RST = 1/2*2*1/2 = 1/2 <1 so the answer is no.
for r= 6 s= 5 t=5 --> 30^2 +25 - 150 = 925-150>0 and rst>1 i.e. ans is yes . therefore statement 2 is insufficient.

Thus the answer should be A.
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Re: Is rst > 1?  [#permalink]

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18 Jun 2017, 07:19
Top Contributor
Bunuel wrote:
Is rst > 1?

(1) rs > t > 1

(2) (rs)² + t² > rst

Target question: Is 1 < rst
ASIDE: I prefer to rearrange inequalities so that the right side is greater than the left side.

Statement 1: 1 < t < rs
Let's use some number sense.
This tells us that rs is greater than 1
IF t were to EQUAL 1, then rst would be greater than 1 (since rs is already greater than 1)
So, since t is GREATER THAN 1, we can be certain that rst is greater than 1
In other words, 1 < rst
Since we can answer the target question with certainty, statement 1 is SUFFICIENT

Statement 2: rst < (rs)² + t²
There are several values of r, s and t that satisfy statement 2. Here are two:
Case a: r = 2, s = 2 and t = 2. This satisfies the condition that rst < (rs)² + t². In this case rst = (2)(2)(2) = 8. So, 1 < rst (i.e., rst is greater than 1)
Case b: r = 1, s = 1 and t = 1. This satisfies the condition that rst < (rs)² + t². In this case rst = (1)(1)(1) = 1. So, 1 = rst (i.e., rst is NOT greater than 1)
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Cheers,
Brent
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Re: Is rst > 1?  [#permalink]

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20 Jun 2017, 00:59
1
mynamegoeson wrote:
Bunuel wrote:
Is rst > 1?

(1) rs > t > 1

(2) (rs)^2 + t^2 > rst

(1) if t>1 and rs>t>1 then rst will be greater than 1..sufficient

(2) rst<(rs)^2 + t^2

since (rs)^2 + t^2>0 (Both (rs)^2 & t^2 are >0)
rst<0...sufficient

D

hi, putting my 2 cents into your solution , as you are correct about statement 1 but in statement 2 it is not necessary that rst>1, take r=s=t= -1/2 then the statement will be false or insufficient .

hope its clear

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Re: Is rst > 1?  [#permalink]

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20 Jun 2017, 01:12
mynamegoeson wrote:
Bunuel wrote:
Is rst > 1?

(1) rs > t > 1

(2) (rs)^2 + t^2 > rst

(1) if t>1 and rs>t>1 then rst will be greater than 1..sufficient

(2) rst<(rs)^2 + t^2

since (rs)^2 + t^2>=0 (Both (rs)^2 & t^2 are >=0)
rst<=0...sufficient

D

Can someone explain the flaw in the reasoning in the second part i.e.

(2) rst<(rs)^2 + t^2

since (rs)^2 + t^2>=0 (Both (rs)^2 & t^2 are >=0)
rst<=0...sufficient
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Re: Is rst > 1?  [#permalink]

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06 Jul 2017, 18:30
Bunuel wrote:
Is rst > 1?

(1) rs > t > 1

(2) (rs)^2 + t^2 > rst

We need to determine whether rst > 1.

Statement One Alone:

rs > t > 1

Since the product of rs is greater than 1 and since t is greater than 1, the product of r, s, and t must be greater than 1. Statement one is sufficient to answer the question.

Statement Two Alone:

(rs)^2 + t^2 > rst

If r = s = t = 1, then rst is not greater than 1. However, if r = s = t = 2, then rst is greater than 1. Statement two alone is not sufficient to answer the question.

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Re: Is rst > 1?   [#permalink] 06 Jul 2017, 18:30
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