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Is the product st negative?

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Is the product st negative? [#permalink]

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Is the product st negative?

(1) s^2 – s < 0

(2) \(\frac{s-4}{t-3}=1\)


Bunuel could you shed some light on the fastest approach? I'd like to understand the mechanics.
Thanks in advance.
[Reveal] Spoiler: OA

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Is the product st negative?

(1) s^2 – s < 0 --> \(s(s-1)<0\) --> \(0<s<1\) --> s is positive. We know nothing about t. Not sufficient.

(2) \(\frac{s-4}{t-3}=1\) --> \(s=t+1\). If \(s=11\) and \(t=10\), then \(st>0\) but if \(s=0.5\) ans \(t=-0.5\), then \(st<0\). Not sufficient.

(1)+(2) Since from (2) \(s=t+1\), then from (1) we have that \(0<t+1<1\) --> \(-1<t<0\), so we have that t is negative. Therefore st = positive*negative = negative. Sufficient.

Answer: C.

Hope it's clear.
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Re: Is the product st negative? [#permalink]

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Bunuel wrote:
Is the product st negative?

(1) s^2 – s < 0 --> \(s(s-1)<0\) --> \(0<s<1\) --> s is positive. We know nothing about t. Not sufficient.

(2) \(\frac{s-4}{t-3}=1\) --> \(s=t+1\). If \(s=11\) and \(t=10\), then \(st>0\) but if \(s=0.5\) ans \(t=-0.5\), then \(st<0\). Not sufficient.

(1)+(2) Since from (2) \(s=t+1\), then from (1) we have that \(0<t+1<1\) --> \(-1<t<0\), so we have that t is negative. Therefore st = positive*negative = negative. Sufficient.

Answer: C.

Hope it's clear.



Well I did this other way around.

St 1 - s^2 -s <0 ( we need both s and t ) not sufficient

St 2- s-4/t-3 =1, from this we get => s-4=t-3 => s-1=t, not sufficient

From both we get, from St 1 s(s-1)<0 and St 2 says s-1=t, if we replace s-1 with t, we get

St<0

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Re: Is the product st negative? [#permalink]

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New post 18 Jan 2014, 02:48
Bunuel wrote:
Is the product st negative?

(1) s^2 – s < 0 --> \(s(s-1)<0\) --> \(0<s<1\) --> s is positive. We know nothing about t. Not sufficient.

(2) \(\frac{s-4}{t-3}=1\) --> \(s=t+1\). If s=10 and t=11, then st>0 but if s=-0.5 ans t=0.5, then st<0. Not sufficient.

(1)+(2) Since from (2) \(s=t+1\), then from (1) we have that \(0<t+1<1\) --> \(-1<t<0\), so we have that t is negative. Therefore st = positive*negative = negative. Sufficient.

Answer: C.

Hope it's clear.


Bunuel,

I am confused with the one...

In Stm2

S-t = 1

I thought both S and T will have same sign to get 1

E.g- S= 5, t= 4

S-T = 4


S= -4, T= -5

-4-(-5) = 1

In both the case S and T are having same sign and definitely ST is not negative....

Pls explain what am I missing??

Thanks
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Re: Is the product st negative? [#permalink]

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New post 18 Jan 2014, 03:11
Mountain14 wrote:
Bunuel wrote:
Is the product st negative?

(1) s^2 – s < 0 --> \(s(s-1)<0\) --> \(0<s<1\) --> s is positive. We know nothing about t. Not sufficient.

(2) \(\frac{s-4}{t-3}=1\) --> \(s=t+1\). If s=10 and t=11, then st>0 but if s=0.5 ans t=-0.5, then st<0. Not sufficient.

(1)+(2) Since from (2) \(s=t+1\), then from (1) we have that \(0<t+1<1\) --> \(-1<t<0\), so we have that t is negative. Therefore st = positive*negative = negative. Sufficient.

Answer: C.

Hope it's clear.


Bunuel,

I am confused with the one...

In Stm2

S-t = 1

I thought both S and T will have same sign to get 1

E.g- S= 5, t= 4

S-T = 4


S= -4, T= -5

-4-(-5) = 1

In both the case S and T are having same sign and definitely ST is not negative....

Pls explain what am I missing??

Thanks


In post you are quoting there is an example giving negative product: \(s=0.5\) ans \(t=-0.5\), then \(st<0\).
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Re: Is the product st negative? [#permalink]

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I have a question..

How did s(s-1)<0 become s>0 and s<1 ..

shouldn't it be s<0 and s<1 ?

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amz14 wrote:
I have a question..

How did s(s-1)<0 become s>0 and s<1 ..

shouldn't it be s<0 and s<1 ?


First of all: \(s(s - 1) < 0\) is true for \(0<s<1\): the "roots" are 0 and 1, "<" sign indicates that the solution is between the "roots": \(0<s<1\).

Please check the links below to study this technique:


Hope this helps.
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Re: Is the product st negative? [#permalink]

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New post 31 Jan 2014, 05:38
Thanks Bunuel ! I have read them now, this is very helpful.

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Re: Is the product s*t negative? [#permalink]

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New post 10 Aug 2014, 13:53
chatterjee06 wrote:
Is the product st negative?

(1)\(s^2\) – s < 0
(2) s-4 =t-3
.


(1) It means that \(0<s<1\). Nothing about \(t\). Insufficient.
(2) \(t=s-1\), don't know signs of \(s\) and \(t\). Insufficient.

(1)+(2) Since \(0<s<1\) and \(t=s-1\), then \(-1<t<0\). Hence, \(st<0\). Sufficient.
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Re: Is the product st negative? [#permalink]

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New post 16 Aug 2014, 08:57
Bunuel wrote:
Is the product st negative?

(1) s^2 – s < 0 --> \(s(s-1)<0\) --> \(0<s<1\) --> s is positive. We know nothing about t. Not sufficient.

(2) \(\frac{s-4}{t-3}=1\) --> \(s=t+1\). If \(s=11\) and \(t=10\), then \(st>0\) but if \(s=0.5\) ans \(t=-0.5\), then \(st<0\). Not sufficient.

(1)+(2) Since from (2) \(s=t+1\), then from (1) we have that \(0<t+1<1\) --> \(-1<t<0\), so we have that t is negative. Therefore st = positive*negative = negative. Sufficient.

Answer: C.

Hope it's clear.



For last 30 mins i am pondering on this B statement ...superb ,
& when i searched it over here, i was sure you must have already address it...& now i understand what i am missing !!!

I want to Kill DS....10 days to Exam only !!! :oops: :oops: :oops:
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Re: Is the product st negative? [#permalink]

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New post 04 May 2016, 06:09
Bunuel wrote:
Is the product st negative?

(1) s^2 – s < 0 --> \(s(s-1)<0\) --> \(0<s<1\) --> s is positive. We know nothing about t. Not sufficient.

(2) \(\frac{s-4}{t-3}=1\) --> \(s=t+1\). If \(s=11\) and \(t=10\), then \(st>0\) but if \(s=0.5\) ans \(t=-0.5\), then \(st<0\). Not sufficient.

(1)+(2) Since from (2) \(s=t+1\), then from (1) we have that \(0<t+1<1\) --> \(-1<t<0\), so we have that t is negative. Therefore st = positive*negative = negative. Sufficient.

Answer: C.

Hope it's clear.


Bunuel,

This might be a stupid question but for some reason I struggle with this.

in the following step:
(s−4)/(t−3)=1 so s-4=t-3 so s=t+1

How can you multiply by an unknown variable if you don't know if the variable is zero? Does this rule only apply if the variable is the only figure in the denominator? or does it only apply when dividing by an unknown variable?


Thanks in advance!

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Re: Is the product st negative? [#permalink]

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New post 04 May 2016, 06:52
grainflow wrote:
Bunuel wrote:
Is the product st negative?

(1) s^2 – s < 0 --> \(s(s-1)<0\) --> \(0<s<1\) --> s is positive. We know nothing about t. Not sufficient.

(2) \(\frac{s-4}{t-3}=1\) --> \(s=t+1\). If \(s=11\) and \(t=10\), then \(st>0\) but if \(s=0.5\) ans \(t=-0.5\), then \(st<0\). Not sufficient.

(1)+(2) Since from (2) \(s=t+1\), then from (1) we have that \(0<t+1<1\) --> \(-1<t<0\), so we have that t is negative. Therefore st = positive*negative = negative. Sufficient.

Answer: C.

Hope it's clear.


Bunuel,

This might be a stupid question but for some reason I struggle with this.

in the following step:
(s−4)/(t−3)=1 so s-4=t-3 so s=t+1

How can you multiply by an unknown variable if you don't know if the variable is zero? Does this rule only apply if the variable is the only figure in the denominator? or does it only apply when dividing by an unknown variable?


Thanks in advance!


We know that t-3 is not 0, because if it were, then (s−4)/(t−3) would be undefined and not equal to 1.
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Re: Is the product st negative? [#permalink]

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New post 11 Aug 2017, 05:01
amz14 wrote:
I have a question..

How did s(s-1)<0 become s>0 and s<1 ..

shouldn't it be s<0 and s<1 ?



hi

for your ease understanding you can see the inequality as under:

s(1-0)(s-1) < 0

here you can see the 2 roots easily ..

hope this helps ..
cheers... :lol:

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Re: Is the product st negative?   [#permalink] 11 Aug 2017, 05:01
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