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M18-19

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Bunuel
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M18-19 [#permalink] New post   16 Sep 2014, 01:03
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Question Stats:

34% (01:50) correct 66% (01:23) wrong based on 404 sessions
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If \(s\) and \(t\) are non-zero 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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E
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Bunuel
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Re M18-19 [#permalink] New post   16 Sep 2014, 01:03
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Official Solution:

If \(s\) and \(t\) are non-zero 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


\(\frac{1}{s} + \frac{1}{t} = s + t\);

\(\frac{t+s}{st}=s+t\).

Notice that we cannot reduce the above by \(s+t\), because \(s+t\) can equal 0, ad we cannot divide by 0 But we can cross-multiply:

\(s+t=(s+t)*st\);

\(s+t - (s+t)*st=0\).

Factor out \(s+t\):

\((s+t)(1 - st)=0\).

\(s+t=0\) or \(st=1\).

Now, notice that if \(s+t=0\) is true, for example, if \(s=10\) and \(t=-10\), then none of the options must be true. Therefore, E is the answer.


Answer: E
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Re: M18-19 [#permalink] New post   18 Apr 2015, 17:31
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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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Bunuel
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Re: M18-19 [#permalink] New post   19 Apr 2015, 03:45
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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: M18-19 [#permalink] New post   27 Aug 2016, 06:12
I think this is a high-quality question and I agree with explanation.
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Re: M18-19 [#permalink] New post   27 Jun 2017, 15:19
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Bunuel wrote:
If \(S\) and \(T\) are non-zero 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

Non-zero 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 A

B. \(S + T = 1\)........No cases applies.............................Eliminate B

C. \(\frac{1}{S} = T\) .......Case 3 does not apply...........Eliminate C

D. \(\frac{S}{T} = 1\) .......Case 3 does not apply.....Eliminate D

E. none of the above...........Correct

Answer: E
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Re: M18-19 [#permalink] New post   16 Nov 2019, 10:41
Bunuel I don't fully understand how \(T+S=0\) is possible:

\(\frac{ST}{(T+S)}=\frac{1}{(T+S)}\) is what I got manipulating the original equation.

Then I thought, Okay, I can simply multiply by \((T+S)\) as \(T+S=0\) is not possible (otherwise we would divide by 0).

Where's the mistake in that approach?
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Re: M18-19 [#permalink] New post   16 Nov 2019, 14:59
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SchruteDwight wrote:
Bunuel I don't fully understand how \(T+S=0\) is possible:

\(\frac{ST}{(T+S)}=\frac{1}{(T+S)}\) is what I got manipulating the original equation.

Then I thought, Okay, I can simply multiply by \((T+S)\) as \(T+S=0\) is not possible (otherwise we would divide by 0).

Where's the mistake in that approach?


S + T = 0 IS possible, for example, if S = 1 and T = -1. In this case 1/S + 1/T = S + T = 0. The mistake is in your manipulation: when you get \(\frac{T+S}{ST}=S+T\), you cannot write this as \(\frac{ST}{(T+S)}=\frac{1}{(T+S)}\) specifically because S + T COULD be 0. You should proceed the way shown in the solution.
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Re: M18-19 [#permalink] New post   20 Jun 2020, 11:52
I think this is a high-quality question and I agree with explanation.
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Re: M18-19 [#permalink] New post   03 Mar 2023, 10:17
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I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
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Re: M18-19 [#permalink] New post   20 Mar 2024, 20:48
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Re: M18-19 [#permalink]
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