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M18-19

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Re: M18-19  [#permalink]

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New post 12 Dec 2017, 08:02
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Re: M18-19  [#permalink]

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New post 13 Jun 2018, 06:57
omavsp wrote:
Hey,

Can someone please clarify how did we get (S+T)(ST−1)=0?

here is the explanation again:

Cross-multiply: S+T=(S+T)∗ST

(S+T)(ST−1)=0

Thank you!



Not sure if u still want the clarification, posting it here anyway.
after you cross multiply S+T=(S+T)∗ST you bring all the expressions on one side.
ST (S+T) - (S+T) = 0
Factoring out S+T you get :
(S+T) (ST-1) = 0

Please kudo if that helped.
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Re: M18-19  [#permalink]

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New post 17 Oct 2018, 03:34
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manhasnoname wrote:
when you have an equation (x-3)(x-2) = 0; x is either 3 or 2.

(S+T)(ST-1) = 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!!


Hello manhasnoname and Nipson,
The below is one of the explanations by Bunuel in one of his posts. Reading and understanding them carefully will clear your doubts for the question in discussion, which is a MUST be true question.

“MUST BE TRUE" questions:
These questions ask which of the following MUST be true, or which of the following is ALWAYS true no matter what set of numbers you choose. Generally for such kind of questions if you can prove that a statement is NOT true for one particular set of numbers, it will mean that this statement is not always true and hence not a correct answer.

As for "COULD BE TRUE" questions:
The questions asking which of the following COULD be true are different: if you can prove that a statement is true for one particular set of numbers, it will mean that this statement could be true and hence is a correct answer.

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New post 30 Jan 2019, 02:04
I think the problem can be solved by observation without much substitution:

If you closely observe, Option A and Option C are the same.
When you substitute numbers in Option B and Option D, none of them fit in right except for 1. Option B can't be right as neither S nor T can be 0.
If I assume the number to be 1, Options A and C are equal contenders. Thus, incorrect.

What remains is Option E which is the right answer. :)
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New post 21 Feb 2019, 14:06
Bunuel wrote:
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


\(\frac{1}{S} + \frac{1}{T} = S + T\);

\(\frac{T+S}{ST}=S+T\)\(\rightarrow\);

Cross-multiply: \(S+T=(S+T)*ST\);

\((S+T)(ST-1)=0\). Either \(S+T=0\) or \(ST=1\). Now, notice that if \(S+T=0\) is true then none of the options must be true.


Answer: E



I think this is a HIGH QUALITY question.

Bunuel, just a question - how do we decide when to cancel out numbers/variables and when not to?
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New post 21 Feb 2019, 21:47
siddharthsinha123 wrote:
Bunuel wrote:
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


\(\frac{1}{S} + \frac{1}{T} = S + T\);

\(\frac{T+S}{ST}=S+T\)\(\rightarrow\);

Cross-multiply: \(S+T=(S+T)*ST\);

\((S+T)(ST-1)=0\). Either \(S+T=0\) or \(ST=1\). Now, notice that if \(S+T=0\) is true then none of the options must be true.


Answer: E



I think this is a HIGH QUALITY question.

Bunuel, just a question - how do we decide when to cancel out numbers/variables and when not to?


Say you have ax = ay. Only when you know that a ≠ 0, you can divide by a and get x = y.
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Re: M18-19  [#permalink]

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New post 21 Feb 2019, 23:02
Bunuel wrote:
siddharthsinha123 wrote:
Bunuel wrote:
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


\(\frac{1}{S} + \frac{1}{T} = S + T\);

\(\frac{T+S}{ST}=S+T\)\(\rightarrow\);

Cross-multiply: \(S+T=(S+T)*ST\);

\((S+T)(ST-1)=0\). Either \(S+T=0\) or \(ST=1\). Now, notice that if \(S+T=0\) is true then none of the options must be true.


Answer: E



I think this is a HIGH QUALITY question.

Bunuel, just a question - how do we decide when to cancel out numbers/variables and when not to?


Say you have ax = ay. Only when you know that a ≠ 0, you can divide by a and get x = y.


Thanks Bunuel,

Is there a link or a specific topic under "ALL YOU NEED FOR QUANT ! ! !" where i can find this concept?

Thanks
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New post 21 Feb 2019, 23:22
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siddharthsinha123 wrote:
Thanks Bunuel,

Is there a link or a specific topic under "ALL YOU NEED FOR QUANT ! ! !" where i can find this concept?

Thanks
Sid


7. Algebra



For more check Ultimate GMAT Quantitative Megathread

Hope it helps.
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New post 17 Mar 2019, 22:17
Answer choices A and C are same, can this happen in actual GMAT exam?
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Re: M18-19  [#permalink]

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New post 21 Mar 2019, 04:36
Bunuel wrote:
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


\(\frac{1}{S} + \frac{1}{T} = S + T\);

\(\frac{T+S}{ST}=S+T\)\(\rightarrow\);

Cross-multiply: \(S+T=(S+T)*ST\);

\((S+T)(ST-1)=0\). Either \(S+T=0\) or \(ST=1\). Now, notice that if \(S+T=0\) is true then none of the options must be true.


Answer: E




HI Bunuel

I didn't get the last part.
How did you get (S+T)(ST−1)=0 ???
Can you please elaborate?


Would really appreciate your help
Thanks
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New post 21 Mar 2019, 04:38
JIAA wrote:
Bunuel wrote:
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


\(\frac{1}{S} + \frac{1}{T} = S + T\);

\(\frac{T+S}{ST}=S+T\)\(\rightarrow\);

Cross-multiply: \(S+T=(S+T)*ST\);

\((S+T)(ST-1)=0\). Either \(S+T=0\) or \(ST=1\). Now, notice that if \(S+T=0\) is true then none of the options must be true.


Answer: E




HI Bunuel

I didn't get the last part.
How did you get (S+T)(ST−1)=0 ???
Can you please elaborate?


Would really appreciate your help
Thanks


This is explained here: https://gmatclub.com/forum/m18-184162-20.html#p1977569
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M18-19  [#permalink]

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New post 02 Jul 2019, 16:37
Bunuel Just to confirm, when we don't know the sign of a variable we don't cross multiply. This rule is only applicable with inequalities, right? Otherwise i'm confused as to how you did it in your explanation!

The easier way to solve this is to test numbers that satisfy the stem:

If S and T are non-zero then they can either be positive/ negative or both.

Case 1: S and T are both positive
S= 1
T = 1

1/1 + 1/1 = 1+1
2=2

Satisfies

Case 2: S and T are both negative
S=-1
T=-1
1/-1 +1/-1 = -1+-1
-1 -1 = -2
Yes, satisfies

Case 3: S = positive, T = Negative
S = 1
T= -1
1/1 + (1/-1)= 1 -1
1 -1 = 1-1
Yes, Satisfies

Case 4: S and T are positive proper fractions
1/(1/2) + 1/(1/2) = 1/2 + 1/2
2+2 = 1? NO Does not satisfy -->same for negative fractions

Using the above we can POE:

A - could be true but doesn't HAVE to be true as per case 3
B - Not necessarily, can also equal 0
C - No, as we learned S can be positive, T can be negative
D - No, same reasoning as C
E is our answer.
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Re: M18-19  [#permalink]

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New post 02 Jul 2019, 16:47
Bunuel wrote:
omavsp wrote:
Hey,

Can someone please clarify how did we get (S+T)(ST−1)=0?

here is the explanation again:

Cross-multiply: S+T=(S+T)∗ST

(S+T)(ST−1)=0

Thank you!


\(S+T=(S+T)*ST\);

Re-arrange: \(0=(S+T)*ST-(S+T)\);

Factor out S + T: \(0=(S+T)(ST - 1)\).

Hope it's clear.


This is much more clearer to me. Would be really helpful to add this step into the Explanation! My algebraic eyes didn't catch the factor until i thought of it like this:
(some expression)*ST - (some expression)
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Re: M18-19   [#permalink] 02 Jul 2019, 16:47

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