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Question Stats:
31% (01:52) correct 69% (01:25) wrong based on 227 sessions
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If \(S\) and \(T\) are nonzero numbers and \(\frac{1}{S} + \frac{1}{T} = S + T\), which of the following must be true? A. \(ST = 1\) B. \(S + T = 1\) C. \(\frac{1}{S} = T\) D. \(\frac{S}{T} = 1\) E. none of the above
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16 Sep 2014, 01:03
Official Solution:If \(S\) and \(T\) are nonzero numbers and \(\frac{1}{S} + \frac{1}{T} = S + T\), which of the following must be true? A. \(ST = 1\) B. \(S + T = 1\) C. \(\frac{1}{S} = T\) D. \(\frac{S}{T} = 1\) E. none of the above \(\frac{1}{S} + \frac{1}{T} = S + T\); \(\frac{T+S}{ST}=S+T\)\(\rightarrow\); Crossmultiply: \(S+T=(S+T)*ST\); \((S+T)(ST1)=0\). Either \(S+T=0\) or \(ST=1\). Now, notice that if \(S+T=0\) is true then none of the options must be true. Answer: E
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Re: M1819
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13 Nov 2014, 21:16
Hello Bunuel,
I have one doubt in this question. when we say "non zero numbers", does it always mean non zero counting numbers (i.e > 0) or non zero real numbers (+ve and ve)? as in later scenatio ST can be 1, S = +/1 and T =+/1 and statement holds true. and if its former scenario then "none" remains the only choice when S != T.
Can you help me make sense of inferences, as I think I am over thinking and confused with many inferences.
rgds,



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Re: M1819
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14 Nov 2014, 02:34
Alaukik wrote: Hello Bunuel,
I have one doubt in this question. when we say "non zero numbers", does it always mean non zero counting numbers (i.e > 0) or non zero real numbers (+ve and ve)? as in later scenatio ST can be 1, S = +/1 and T =+/1 and statement holds true. and if its former scenario then "none" remains the only choice when S != T.
Can you help me make sense of inferences, as I think I am over thinking and confused with many inferences.
rgds, A nonzero number is a number which is NOT 0, so it can be any real number except 0, positive or negative.
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Re: M1819
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14 Nov 2014, 03:31
Bunuel wrote: Alaukik wrote: Hello Bunuel,
I have one doubt in this question. when we say "non zero numbers", does it always mean non zero counting numbers (i.e > 0) or non zero real numbers (+ve and ve)? as in later scenatio ST can be 1, S = +/1 and T =+/1 and statement holds true. and if its former scenario then "none" remains the only choice when S != T.
Can you help me make sense of inferences, as I think I am over thinking and confused with many inferences.
rgds, A nonzero number is a number which is NOT 0, so it can be any real number except 0, positive or negative. so the answer has to be ST = 1? right?



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Re: M1819
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14 Nov 2014, 05:36
Alaukik wrote: Bunuel wrote: Alaukik wrote: Hello Bunuel,
I have one doubt in this question. when we say "non zero numbers", does it always mean non zero counting numbers (i.e > 0) or non zero real numbers (+ve and ve)? as in later scenatio ST can be 1, S = +/1 and T =+/1 and statement holds true. and if its former scenario then "none" remains the only choice when S != T.
Can you help me make sense of inferences, as I think I am over thinking and confused with many inferences.
rgds, A nonzero number is a number which is NOT 0, so it can be any real number except 0, positive or negative. so the answer has to be ST = 1? right? No. Consider S = 2 and T = 2. In this case none of the options are true. The correct answer is E.
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14 Nov 2014, 06:36
aah ok. got it thanks.



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Re: M1819
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02 Jan 2015, 10:04
Bunuel wrote: Official Solution:
If \(S\) and \(T\) are nonzero numbers and \(\frac{1}{S} + \frac{1}{T} = S + T\), which of the following must be true?
A. \(ST = 1\) B. \(S + T = 1\) C. \(\frac{1}{S} = T\) D. \(\frac{S}{T} = 1\) E. none of the above
\(\frac{1}{S} + \frac{1}{T} = S + T\); \(\frac{T+S}{ST}=S+T\)\(\rightarrow\); Crossmultiply: \(S+T=(S+T)*ST\); \((S+T)(ST1)=0\). Either \(S+T=0\) or \(ST=1\). Now, notice that if \(S+T=0\) is true then none of the options must be true.
Answer: E How can we cross multiply ST, ST can be negative as well
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Re: M1819
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05 Jan 2015, 05:43
him1985 wrote: Bunuel wrote: Official Solution:
If \(S\) and \(T\) are nonzero numbers and \(\frac{1}{S} + \frac{1}{T} = S + T\), which of the following must be true?
A. \(ST = 1\) B. \(S + T = 1\) C. \(\frac{1}{S} = T\) D. \(\frac{S}{T} = 1\) E. none of the above
\(\frac{1}{S} + \frac{1}{T} = S + T\); \(\frac{T+S}{ST}=S+T\)\(\rightarrow\); Crossmultiply: \(S+T=(S+T)*ST\); \((S+T)(ST1)=0\). Either \(S+T=0\) or \(ST=1\). Now, notice that if \(S+T=0\) is true then none of the options must be true.
Answer: E How can we cross multiply ST, ST can be negative as well Notice that we have an equation, not an inequality. So, it does not matter whether ST is negative. Only for inequalities we should flip the sign when multiplying by negative value.
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Re: M1819
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18 Apr 2015, 17:31
When we solve we get 'Either S+T=0 or ST=1'. So, any of these is correct. We have 'ST=1' as option (A), why it is not correct? Please let me know!



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19 Apr 2015, 03:45
reddyMBA wrote: When we solve we get 'Either S+T=0 or ST=1'. So, any of these is correct. We have 'ST=1' as option (A), why it is not correct? Please let me know! Because it's not necessary that st=1. If s+t=0, then st could be some other number but 1.
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Re: M1819
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14 Oct 2015, 14:19
I have doubts with this question.. have same problem as others here: after manipulations we get the same expression st=1 like in the answer choice, and we see such questions very often by GMAT, and the right answer is just a manipulated version of the question.
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Re: M1819
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20 Feb 2016, 21:03
I found the plugin numbers method to be more effective for this problem.
If you plug in 1 for S and 1 for T you'll see that none of the options work.



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26 May 2016, 19:29
When I did this question I noticed that if you reduce by S+T and ended up with ST = 1, you'll see that 1/S = T is the same solution. That tipped me off.



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27 Aug 2016, 06:12
I think this is a highquality question and I agree with explanation.



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05 Sep 2016, 11:57
when you have an equation (x3)(x2) = 0; x is either 3 or 2.
(S+T)(ST1) = 0; => S+T = 0; or ST = 1;
ST = 1 should be the answer. However, S = 1/T is the same. So looks like there is more than one correct answer to this question!!



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Re: M1819
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02 Mar 2017, 04:48
Hi Bunuel , I fell in the same pitfall as many other users did. Simplifying S+T is basically letting us get rid of a possible solution, which is exactly what the question wants us to do to take the wrong path. My question is: how do we notice that a certain algebraic operation is going to neglect a potential solution? Can we assume as a ruleofthumb that multiplying is always more reliable than dividing unknown values? Many thanks, great question by the way.



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27 Jun 2017, 10:05
Great question! I'm not happy to get it wrong, but I'm happy to have learned the proper way to do it correctly in the future.
I got as far as trying to rationalize the denominator of the 1/S + 1/T. But then after that, I had no idea what to do. I didn't think that bringing them both to one side and setting the other equal to zero was the way to solve it.
Nice~



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Re: M1819
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27 Jun 2017, 15:19
Bunuel wrote: If \(S\) and \(T\) are nonzero numbers and \(\frac{1}{S} + \frac{1}{T} = S + T\), which of the following must be true?
A. \(ST = 1\) B. \(S + T = 1\) C. \(\frac{1}{S} = T\) D. \(\frac{S}{T} = 1\) E. none of the above Another approach for this great problem Nonzero integers means negative or positive integers Let see apply some cases with numbers to DISAPPROVE answer choices S = T = 1............Equation is 2 = 2 S = T =1............Equation is 2 = 2 S = 1 & T =1....Equation is 0 = 0 or S = 1 & T =1....Equation is 0 = 0 A. \(ST = 1\) ............Case 3 does not apply.................... Eliminate AB. \(S + T = 1\)........No cases applies............................. Eliminate BC. \(\frac{1}{S} = T\) .......Case 3 does not apply........... Eliminate C D. \(\frac{S}{T} = 1\) .......Case 3 does not apply..... Eliminate D E. none of the above........... CorrectAnswer: E



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Re: M1819
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12 Dec 2017, 07:46
Hey,
Can someone please clarify how did we get (S+T)(ST−1)=0?
here is the explanation again:
Crossmultiply: S+T=(S+T)∗ST
(S+T)(ST−1)=0
Thank you!







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