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Is the product st negative? [#permalink]
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03 Dec 2013, 03:33
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Is the product st negative? (1) s^2 – s < 0 (2) \(\frac{s4}{t3}=1\) Bunuel could you shed some light on the fastest approach? I'd like to understand the mechanics. Thanks in advance.
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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 > \(s(s1)<0\) > \(0<s<1\) > s is positive. We know nothing about t. Not sufficient.
(2) \(\frac{s4}{t3}=1\) > \(s=t+1\). 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 s4/t3 =1, from this we get => s4=t3 => s1=t, not sufficient From both we get, from St 1 s(s1)<0 and St 2 says s1=t, if we replace s1 with t, we get St<0



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Re: Is the product st negative? [#permalink]
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18 Jan 2014, 02:48
Bunuel wrote: Is the product st negative?
(1) s^2 – s < 0 > \(s(s1)<0\) > \(0<s<1\) > s is positive. We know nothing about t. Not sufficient.
(2) \(\frac{s4}{t3}=1\) > \(s=t+1\). 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 St = 1 I thought both S and T will have same sign to get 1 E.g S= 5, t= 4 ST = 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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18 Jan 2014, 03:11
Mountain14 wrote: Bunuel wrote: Is the product st negative?
(1) s^2 – s < 0 > \(s(s1)<0\) > \(0<s<1\) > s is positive. We know nothing about t. Not sufficient.
(2) \(\frac{s4}{t3}=1\) > \(s=t+1\). 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 St = 1 I thought both S and T will have same sign to get 1 E.g S= 5, t= 4 ST = 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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31 Jan 2014, 01:15
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I have a question..
How did s(s1)<0 become s>0 and s<1 ..
shouldn't it be s<0 and s<1 ?



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Re: Is the product st negative? [#permalink]
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31 Jan 2014, 01:51



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Re: Is the product st negative? [#permalink]
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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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10 Aug 2014, 13:53
chatterjee06 wrote: Is the product st negative?
(1)\(s^2\) – s < 0 (2) s4 =t3 . (1) It means that \(0<s<1\). Nothing about \(t\). Insufficient. (2) \(t=s1\), don't know signs of \(s\) and \(t\). Insufficient. (1)+(2) Since \(0<s<1\) and \(t=s1\), then \(1<t<0\). Hence, \(st<0\). Sufficient.
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Re: Is the product st negative? [#permalink]
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16 Aug 2014, 08:57
Bunuel wrote: Is the product st negative?
(1) s^2 – s < 0 > \(s(s1)<0\) > \(0<s<1\) > s is positive. We know nothing about t. Not sufficient.
(2) \(\frac{s4}{t3}=1\) > \(s=t+1\). 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.
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Re: Is the product st negative? [#permalink]
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04 May 2016, 06:09
Bunuel wrote: Is the product st negative?
(1) s^2 – s < 0 > \(s(s1)<0\) > \(0<s<1\) > s is positive. We know nothing about t. Not sufficient.
(2) \(\frac{s4}{t3}=1\) > \(s=t+1\). 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 s4=t3 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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04 May 2016, 06:52
grainflow wrote: Bunuel wrote: Is the product st negative?
(1) s^2 – s < 0 > \(s(s1)<0\) > \(0<s<1\) > s is positive. We know nothing about t. Not sufficient.
(2) \(\frac{s4}{t3}=1\) > \(s=t+1\). 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 s4=t3 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 t3 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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11 Aug 2017, 05:01
amz14 wrote: I have a question..
How did s(s1)<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(10)(s1) < 0 here you can see the 2 roots easily .. hope this helps .. cheers...




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