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If S and T are nonzero numbers and
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Updated on: 04 Nov 2012, 14:06
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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 M1819 Explanation provided:
1/S + 1/T = S+T;
T+S/ST = S+T→;
Crossmultiply: S+T=(S+T)∗ST;
(S+T)(ST−1)=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.
The correct answer is E
Question: I understand the way of the equation, however, what I would have done is interfere at the following step: S+T = (S+T)*ST > (S+T)/(S+T)=ST > ST = 1
Is there some rule which forbids me to take this step? Or is the only option to realize so, that you perform the above given equation as well and realise that "S+T = 0" negates all other options than E.... ??
Thanks in advance, best regards
P.S. Sry if the format is terrible, this is the first question I am copying out of somewhere.
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Originally posted by SirGMAT on 04 Nov 2012, 07:20.
Last edited by Bunuel on 04 Nov 2012, 14:06, edited 1 time in total.
Renamed the topic and edited the question.




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Re: If S and T are nonzero numbers and
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04 Nov 2012, 14:08
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 OE:\(\frac{1}{S} + \frac{1}{T} = S + T\) > \(\frac{T+S}{ST}=S+T\) > crossmultiply: \(S+T=(S+T)*ST\) > \((S+T)(ST1)=0\) > either \(S+T=0\) or \(ST=1\). So, if \(S+T=0\) is true then none of the options must be true. Answer: E. P.S. Please read carefully and follow: rulesforpostingpleasereadthisbeforeposting133935.html Please pay attention to the rule #3. Thank you.
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Re: GMATClub M1819  Arithmetic: Equations
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04 Nov 2012, 07:46
SirGMAT wrote: If S and T are nonzero numbers and 1S+1T=S+T, which of the following must be true? A. ST=1 B. S+T=1 C. 1/S=T D. S/T=1 E. none of the above Explanation provided: 1/S + 1/T = S+T; T+S/ST = S+T→; Crossmultiply: S+T=(S+T)∗ST; (S+T)(ST−1)=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. The correct answer is E Question: I understand the way of the equation, however, what I would have done is interfere at the following step: S+T = (S+T)*ST > (S+T)/(S+T)=ST > ST = 1 Is there some rule which forbids me to take this step? Or is the only option to realize so, that you perform the above given equation as well and realise that "S+T = 0" negates all other options than E.... ?? Thanks in advance, best regards P.S. Sry if the format is terrible, this is the first question I am copying out of somewhere. Yes, this is not correct way of cancelling. I'll show you one example. 5* 0 = 3*0 if we cancel 0 on both side, we get 5=3. is it correct? No. The crux (and the rule) is, we can cancel out a term only when we know its not 0. So the way it is done in explanation is absolutely correct and the right method. Hope it helps



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Re: If S and T are nonzero numbers and
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27 Dec 2015, 07:51
ArunpriyanJ wrote: 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 OE:\(\frac{1}{S} + \frac{1}{T} = S + T\) > \(\frac{T+S}{ST}=S+T\) > crossmultiply: \(S+T=(S+T)*ST\) > \((S+T)(ST1)=0\) > either \(S+T=0\) or \(ST=1\). So, if \(S+T=0\) is true then none of the options must be true. Answer: E. P.S. Please read carefully and follow: rulesforpostingpleasereadthisbeforeposting133935.html Please pay attention to the rule #3. Thank you. Bunuel, Not able to understand the highlighted part. Kindly explain in detail... Thx, Arun \(S+T=(S+T)*ST\) \((S+T)*ST(S+T)=0\) Factor s+t: \((S+T)(ST1)=0\) Hope it's clear.
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Re: If S and T are nonzero numbers and
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07 Nov 2012, 01:43
Marcab wrote: 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
OE:
\(\frac{1}{S} + \frac{1}{T} = S + T\) > \(\frac{T+S}{ST}=S+T\) > crossmultiply: \(S+T=(S+T)*ST\) > \((S+T)(ST1)=0\) > either \(S+T=0\) or \(ST=1\). So, if \(S+T=0\) is true then none of the options must be true.
Answer: E.
Bunuel, in your solution I need to ask one thing. Inequalities, there is a rule that if you dont know the sign of denominator, then dont cross multiply. In your solution, how can you be so sure of the sign of ST. Please let me know if i am missing something We are concerned with the sign when we crossmultiply an inequality because this operation might affect (flip) its sign (> to <, for example) but it's always safe to crossmultiply an equation.
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Re: If S and T are nonzero numbers and
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11 Feb 2013, 07:44
Sachin9 wrote: I get it Vippss. But in all other cases, the generalization will hold good right? Rules are good, generalizations are not Enjoy and practice kudos :p



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Re: If S and T are nonzero numbers and
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18 Mar 2013, 10:06
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 OE:\(\frac{1}{S} + \frac{1}{T} = S + T\) > \(\frac{T+S}{ST}=S+T\) > crossmultiply: \(S+T=(S+T)*ST\) > \((S+T)(ST1)=0\) > either \(S+T=0\) or \(ST=1\). So, if \(S+T=0\) is true then none of the options must be true. Answer: E. P.S. Please read carefully and follow: rulesforpostingpleasereadthisbeforeposting133935.html Please pay attention to the rule #3. Thank you. Hello Bunuel, How did u get S+T=(S+T)*ST > (S+T)(ST1)=0 ?? What did i miss in he below equation?? how come you got (S+T)(ST1)=0 ?? Ca you please explain 1/S+1/T = S+T (S+T) = (S+T) (ST) divide both side by (S+T) we get 1=1(ST) therefore ST=1 Thank you.



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Re: If S and T are nonzero numbers and
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27 Dec 2015, 07:46
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 OE:\(\frac{1}{S} + \frac{1}{T} = S + T\) > \(\frac{T+S}{ST}=S+T\) > crossmultiply: \(S+T=(S+T)*ST\) > \((S+T)(ST1)=0\) > either \(S+T=0\) or \(ST=1\). So, if \(S+T=0\) is true then none of the options must be true. Answer: E. P.S. Please read carefully and follow: rulesforpostingpleasereadthisbeforeposting133935.html Please pay attention to the rule #3. Thank you. Bunuel, Not able to understand the highlighted part. Kindly explain in detail... Thx, Arun



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Re: If S and T are nonzero numbers and
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07 Sep 2016, 06:06
also the OA can be crosschecked by plugging value of S=1 and T=1 in this case 1/S+1/T= S+T but \(ST\neq{1}\), \(1/S\neq{T}\) since S+T=0 here.



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Re: GMATClub M1819  Arithmetic: Equations
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04 Nov 2012, 08:43
SirGMAT wrote: If S and T are nonzero numbers and 1S+1T=S+T, which of the following must be true?
A. ST=1 B. S+T=1 C. 1/S=T D. S/T=1 E. none of the above
Explanation provided:
1/S + 1/T = S+T;
T+S/ST = S+T→;
Crossmultiply: S+T=(S+T)∗ST;
(S+T)(ST−1)=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.
The correct answer is E
Question: I understand the way of the equation, however, what I would have done is interfere at the following step: S+T = (S+T)*ST > (S+T)/(S+T)=ST > ST = 1
Is there some rule which forbids me to take this step? Or is the only option to realize so, that you perform the above given equation as well and realise that "S+T = 0" negates all other options than E.... ??
Thanks in advance, best regards
P.S. Sry if the format is terrible, this is the first question I am copying out of somewhere. There was this fun derivation that my math teacher showed us in school. Just to demonstrate how cancelling of 0 could yield wrong results. He claimed that he could prove that 1=2 and hence all numbers are equal. It goes as below: Let, \(a=b\) Multiplying both sides by a, We get \(a^2 = ab\) Subtracting \(b^2\) from both sides \(a^2  b^2 = ab  b^2\) \((a+b)(ab) = b(ab)\) Cancelling \((ab)\) on both sides, \(a+b = b\) Since \(a=b\) \(a+a = a\) \(2a = a\) \(2=1\)



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Re: GMATClub M1819  Arithmetic: Equations
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04 Nov 2012, 14:06
Vips0000 wrote: Yes, this is not correct way of cancelling. I'll show you one example. 5* 0 = 3*0 if we cancel 0 on both side, we get 5=3. is it correct? No. The crux (and the rule) is, we can cancel out a term only when we know its not 0. So the way it is done in explanation is absolutely correct and the right method. Hope it helps Thanks  lol I am an idiot  ...! I even thought of the zero number things, but mistakenly memorized the prompt as telling me that the respective equation could not be "0", though it only said that each number alone is nonzero.... Thanks guys!
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Re: If S and T are nonzero numbers and
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07 Nov 2012, 00:12
Hi bunuel,
In this step: s+t = (s+t)st can't we just cancel (s+t) and get > st =1?
thanks, K



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Re: If S and T are nonzero numbers and
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07 Nov 2012, 00:19
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
OE:
\(\frac{1}{S} + \frac{1}{T} = S + T\) > \(\frac{T+S}{ST}=S+T\) > crossmultiply: \(S+T=(S+T)*ST\) > \((S+T)(ST1)=0\) > either \(S+T=0\) or \(ST=1\). So, if \(S+T=0\) is true then none of the options must be true.
Answer: E.
Bunuel, in your solution I need to ask one thing. Inequalities, there is a rule that if you dont know the sign of denominator, then dont cross multiply. In your solution, how can you be so sure of the sign of ST. Please let me know if i am missing something
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Re: If S and T are nonzero numbers and
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07 Nov 2012, 01:30
kartik222 wrote: Hi bunuel,
In this step: s+t = (s+t)st can't we just cancel (s+t) and get > st =1?
thanks, K Never reduce equation by variable (or expression with variable), if you are not certain that variable (or expression with variable) doesn't equal to zero. We can not divide by zero.So, if you divide (reduce) s+t = (s+t)st by (s+t), you assume, with no ground for it, that (s+t) does not equal to zero thus exclude a possible solution (notice that both st=1 AND (s+t)=0 satisfy the equation). Hope it's clear.
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Re: If S and T are nonzero numbers and
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07 Nov 2012, 02:05
Thanks. That cleared my doubt.
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Re: If S and T are nonzero numbers and
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10 Feb 2013, 05:03
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 OE:\(\frac{1}{S} + \frac{1}{T} = S + T\) > \(\frac{T+S}{ST}=S+T\) > crossmultiply: \(S+T=(S+T)*ST\) > \((S+T)(ST1)=0\) > either \(S+T=0\) or \(ST=1\). So, if \(S+T=0\) is true then none of the options must be true. Answer: E. P.S. Please read carefully and follow: rulesforpostingpleasereadthisbeforeposting133935.html Please pay attention to the rule #3. Thank you. Bunuel, Since S+T=0 OR ST=1 and the question asks what must be true, the answer is E ? Another way to answer the question. . Is my reasoning right?
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Re: If S and T are nonzero numbers and
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10 Feb 2013, 05:09
Sachin9 wrote: 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 OE:\(\frac{1}{S} + \frac{1}{T} = S + T\) > \(\frac{T+S}{ST}=S+T\) > crossmultiply: \(S+T=(S+T)*ST\) > \((S+T)(ST1)=0\) > either \(S+T=0\) or \(ST=1\). So, if \(S+T=0\) is true then none of the options must be true. Answer: E. P.S. Please read carefully and follow: rulesforpostingpleasereadthisbeforeposting133935.html Please pay attention to the rule #3. Thank you. Bunuel, Since S+T=0 OR ST=1 and the question asks what must be true, the answer is E ? Another way to answer the question. . Is my reasoning right? Yes, if for example s=1 and t=1 (s+t=11=0), then none of the options is true (none of the options MUST be true).
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Re: If S and T are nonzero numbers and
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10 Feb 2013, 06:04
I understand what you are trying to say bunuel.. my question is that since the equation results in 2 soln and we have a OR .. . that is soln 1 OR soln 2 and the question asks for MUST be true.. So based on this reasoning, can we say the answer is E..? For something must be true , we cannot have soln 1 OR soln 2.. we need to have 1 soln / soln1 AND soln2.. Hope you are getting what I am trying to ask..
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Re: If S and T are nonzero numbers and
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10 Feb 2013, 13:55
Sachin9 wrote: I understand what you are trying to say bunuel.. my question is that since the equation results in 2 soln and we have a OR .. . that is soln 1 OR soln 2 and the question asks for MUST be true..
So based on this reasoning, can we say the answer is E..?
For something must be true , we cannot have soln 1 OR soln 2.. we need to have 1 soln / soln1 AND soln2..
Hope you are getting what I am trying to ask.. I would say that don't generalize this point. You know that because there are two solutions, any of the given options need not be a MUST. But if u really have an option that says (s+t)(st1)=0 then that MUST be true.



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Re: If S and T are nonzero numbers and
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10 Feb 2013, 18:17
Vips0000 wrote: Sachin9 wrote: I understand what you are trying to say bunuel.. my question is that since the equation results in 2 soln and we have a OR .. . that is soln 1 OR soln 2 and the question asks for MUST be true..
So based on this reasoning, can we say the answer is E..?
For something must be true , we cannot have soln 1 OR soln 2.. we need to have 1 soln / soln1 AND soln2..
Hope you are getting what I am trying to ask.. I would say that don't generalize this point. You know that because there are two solutions, any of the given options need not be a MUST. But if u really have an option that says (s+t)(st1)=0 then that MUST be true. But if u really have an option that says (s+t)(st1)=0 then that MUST be true. Ididn;t understand this.. if (s+t)(st1)=0 then either s+t=0 or st1=0.. we still have a OR here
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Re: If S and T are nonzero numbers and
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