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Question Stats:
63% (01:02) correct
36% (01:14) wrong based on 76 sessions
\(s^2 + t^2 < 12st\)
Quantity A 
Quantity B 
1S 
T 
A)The quantity in Column A is greater. B)The quantity in Column B is greater. C)The two quantities are equal. D)The relationship cannot be determined from the information given.
Last edited by Carcass on 20 Jul 2019, 12:37, edited 3 times in total.
Editing the post and adding the OA and Tags




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Re: Manhattan Prep GRE Algebra Question S^2 + T^2 < 12st [#permalink]
05 Jul 2016, 23:37
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GREhelp wrote: Hi, Any help on the following question would be greatly appreciated. It is a question from the Manhattan Prep GRE Algebra Book. s^2 + t^2 < 12st Quantity A: 1S Quantity B: t We have \(s^2 + t^2 < 12st\) addind 2st on both sides \(s^2 + t^2 +2st < 12st +2st\) \((s + t)^2< 1\) or \(s + t <1\) or \(t< 1s\) Hence option Option A is correct. Now both s and t can be negative quantities. like 10 and 100 respectively. So A is greater. Or one of them can be negative and other positive such that they satisfy the above in equality s=2 t= 2.5. Both of them are positive fractions s=99/100 t= 9/1000. Still Quantity A is greater.
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Re: S^2 + T^2 < 12st [#permalink]
06 Jul 2016, 00:50
Please use the rh rules for posting  a proper title for the question
 using the tags to identify it
 well formatted in its text
 adding the OA
Thank you
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Re: S^2 + T^2 < 12st [#permalink]
06 Jul 2016, 10:45
Hi Sandy, Thanks for your respond. I was a little confused.
In the solution below you said that:
(s + t)^2< 1 or s + t <1 or t< 1s
How would that be possible since there are two (s+t) and anything plugged into them would result in a positive number. (s + t)^2< 1 = (s+t) (s+t) < 1
If you to plug in negative quantities for s and t can, like 10 and 100 respectively. you will get a positive number. For example, S = 10 T = 100 (10+ 100) ( 10 + 100) = 110 * 110 = 12100
Please advise, for some reason I keep getting this question wrong.



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Re: S^2 + T^2 < 12st [#permalink]
06 Jul 2016, 14:40
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You cannot choose any value for s and t. s+t has to be less than 1. That is a constraint so if you put s= 10 then t= 10.99999999..... or 9.000000000000001 or something like that. Note that s and t are not independent. They are constrained. Hence they must follow s+t < 1. If you put any value that just means that there are no constraints. You cannot substitute "any" value.
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Re: S^2 + T^2 < 12st [#permalink]
29 Aug 2017, 04:56
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GREhelp wrote: \(s^2 + t^2 < 12st\)
Quantity A 
Quantity B 
1  s 
t 
First notice that we have a perfect square, (s + t)², "hiding" in the given inequality. The perfect square is in the form s² + 2st + t² Here's what I mean: Given: s² + t² < 1  2st Add 2st to both sides to get: s² + 2st + t² < 1 Factor the left side: (s + t)² < 1 IMPORTANT: If (something)² < 1, then it must be the case that 1 < something < 1Since we now know that (s + t)² < 1, we can say that 1 < s + t < 1Given: Quantity A: 1  s Quantity B: t Add s to both sides to get: Quantity A: 1 Quantity B: s + t Since 1 < s + t < 1, we can conclude that Quantity A is greater. Cheers, Brent
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Re: S^2 + T^2 < 12st [#permalink]
23 Aug 2018, 17:43
Hi Brent, can you explain why " If (something)² < 1, then it must be the case that 1 < something < 1"?



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Re: S^2 + T^2 < 12st [#permalink]
23 Aug 2018, 23:28
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msawicka wrote: Hi Brent, can you explain why " If (something)² < 1, then it must be the case that 1 < something < 1"? Let me try. It is because square, square root and absolute value go hand in hand. Let's say we have A^2=B, A can be any value, but B must be >=0 (i.e. nonnegative, or positive inclusion of zero). We can rewrite it as A= + or  square root of B, which means any number under square root is >=0. In turn, if we are given square root of B = A, then B must be nonnegative, so to make sure it satisfies this, we absolute value B => if we square both side, we get absolute value of B=A^2. In case of inequality, we have absolute value of B < A^2. And we have our familiar inequality: absolute value of B < A^2 <=> A^2 < B < A^2.



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Re: S^2 + T^2 < 12st [#permalink]
18 Jul 2019, 08:59
I'm guessing the reasoning behind the interval is that: square root of a^2 is the same as a and if a< x then x < a < x This is based on this page: https://www.mathsisfun.com/algebra/abso ... lving.htmlSo, I'm guessing that's why we can say: (s + t)² < 1 s + t < 1 so: 1 < s + t < 1 or maybe at this point: (s + t)² < 1 one could say we are taking the square root of both sides which leaves us with s+t < 1 and s+t <1 We have + and  because the square root of 1 can be +1 or 1, and we have to switch the direction of an inequality sign when we multiply by a negative. So we end up with: 1 < s + t < 1 I may be wrong.



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Re: S^2 + T^2 < 12st [#permalink]
18 Jul 2019, 09:10
arc601 wrote: I'm guessing the reasoning behind the interval is that: square root of a^2 is the same as a and if a< x then x < a < x This is based on this page: https://www.mathsisfun.com/algebra/abso ... lving.htmlSo, I'm guessing that's why we can say: (s + t)² < 1 s + t < 1 so: 1 < s + t < 1 or maybe at this point: (s + t)² < 1 one could say we are taking the square root of both sides which leaves us with s+t < 1 and s+t <1 We have + and  because the square root of 1 can be +1 or 1, and we have to switch the direction of an inequality sign when we multiply by a negative. So we end up with: 1 < s + t < 1 I may be wrong. Your reasoning (in both cases) is perfect. Cheers, Brent
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Re: S^2 + T^2 < 12st
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