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# If t#0, is r greater than zero?

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If t#0, is r greater than zero?  [#permalink]

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20 Feb 2014, 01:19
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The Official Guide For GMAT® Quantitative Review, 2ND Edition

If $$t\neq{0}$$, is r greater than zero?

(1) rt = 12
(2) r + t = 7

Data Sufficiency
Question: 105
Category: Arithmetic; Algebra Arithmetic operations; Properties of numbers; Simultaneous equations
Page: 160
Difficulty: 650

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Re: If t#0, is r greater than zero?  [#permalink]

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20 Feb 2014, 01:20
1
SOLUTION

If $$t\neq{0}$$, is r greater than zero?

(1) rt = 12 --> if r=1 and t=12, then r>0 but if r=-1 and t=-12, then r<0. Not sufficient.

(2) r + t = 7 --> if r=1 and t=6, then r>0 but if r=-1 and t=8, then r<0. Not sufficient.

(1)+(2) From (2) t = 7- r, thus from (1) r(7-r) = 12 --> solving gives r=3 or r=4. In ether case r>0. Sufficient.

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Re: If t#0, is r greater than zero?  [#permalink]

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20 Feb 2014, 03:22
is r greater than zero?

(1) rt = 12

both r=-2 and t=-6 and r=2 and t=6 satisfy this equation

hence not sufficient

(2) r + t = 7

if r=0, t=7 also r=1 and t=6 satisfy this equation. hence not sufficient

combining 1 and 2
we have t=(7-r)
r(7-r)=12
since product is positive thus r and 7-r must have the same sign. if r= -ve then 7-r will be positive. If r= +ve then 7-r will be positive for all values less than seven. hence r will be greater than zero and its value will be less than 7 hence sufficient
therefore C.
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Re: If t#0, is r greater than zero?  [#permalink]

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20 Feb 2014, 04:08
1
If $$t\neq{0}$$, is r greater than zero?

(1) rt = 12
(2) r + t = 7

1 - insufficient. If t<0 -> r<0, if t>0 -> r>0.

2 - insufficient. Could be r=-1 and t=8, or r=1 and t=6.

2 and 2:
$$r+\frac{12}{t}-7=0$$
$$r^2 -7r+12=0$$
$$(r-1)(r-6)=0$$

So r=1 or 6. That are >0.
Ans C
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Re: If t#0, is r greater than zero?  [#permalink]

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21 Feb 2014, 00:32
Option C
From S1:r * t=12
r=3;t=4 could be one sol.
r=-3;t=-4 could be another sol.Not suff.

From S2:r+t=7.Infinite no. of both pos. and neg. sol possible for a linear equation.
Not suff.

Combining,we get r*t=12 and r+t=7.We can solve these two to get unique values of r and t.
t=12/r
Replacing t in other equation;
r^2-7r+12=0
(r-3)(r-4)=0
r=3/4>0
Hence suff.
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Re: If t#0, is r greater than zero?  [#permalink]

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21 Feb 2014, 04:11
t#0; Is r>0?

Statement (1):
rt = 12;
Here, either both r & t need to be positive or both need to be negative in order to get a positive number.
There are 6 possibilities each in both positive and negative:
r = 1 & t =12;
r = 2 & t = 6;
r = 3 & t = 4;
....

Not sufficient, as negative value for r is possible.

Statement (2):
r + t = 7;

Yes, if r = 4 & t = 3;
No, if r = -3 & t = 10;

Insufficient;

Combining 2 statements, substitute t = 7 - r in (1);
rt = 12;
r(7 - r) = 12;
7r - r^2 = 12;
r^2 - 7r + 12 = 0;
(r - 4)(r - 3) = 0;
r = 3, 4;

So, r is positive when 2 statements are combined.

Ans is (C).
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Re: If t#0, is r greater than zero?  [#permalink]

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02 Jun 2014, 18:05
1
No calculations approach:

If t#0, is r greater than zero?

(1) rt = 12
(2) r + t = 7

St1: Both r and t could be positive or both r and t could be negative. Not Suff
St2: Both r and t could be positive or either of r and t could be negative. Not Suff

1+2 --> common possibility is both r and t are positive. So r is positive i.e. greater than 0.
Ans:C
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Re: If t#0, is r greater than zero?  [#permalink]

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10 Nov 2017, 07:11
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