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Is x^3/y = x + x/y?

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Is x^3/y = x + x/y? [#permalink]

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Is x^3/y = x + x/y?

(1) y = 8
(2) y = x^2 - 1
[Reveal] Spoiler: OA
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Re: Is x^3/y = x + x/y? [#permalink]

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New post 21 Jun 2012, 02:07
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Is x^3/y = x + x/y?

Is \(\frac{x^3}{y} = x + \frac{x}{y}\)? --> \(\frac{x^3}{y} = \frac{x(y+1)}{y}\)?

(1) y = 8. The question becomes: is \(\frac{x^3}{8} = \frac{9x}{8}\)? --> is \(x^3=9x\)? --> is \(x(x^2-9)=0\)? --> is \(x=0\), \(x=3\) or \(x=-3\)? We don't know that. Not sufficient.

(2) y = x^2 - 1. The question becomes: is \(\frac{x^3}{x^2-1} = \frac{x(x^2-1+1)}{x^2-1}\)? --> is \(\frac{x^3}{x^2-1} = \frac{x^3}{x^2-1}\)? The answer to this question is YES. Sufficient.

Answer: B.

Hope it's clear.
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Re: Is x^3/y = x + x/y? [#permalink]

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New post 21 Jun 2012, 02:57
Thanks a lot,that really helps.I cancelled the x in A,silly mistake.
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Re: Is x^3/y = x + x/y? [#permalink]

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New post 21 Jun 2012, 03:03
Expert's post
shivanigs wrote:
Thanks a lot,that really helps.I cancelled the x in A,silly mistake.


Even if you could cancel \(x\) in (1) (for example if we were told that \(x\neq{0}\)) question would become: is \(x^2=9\)? Or is \(x=3\) or \(x=-3\)? And since we don't know whether that's true, then the statement still would be insufficient.
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Re: Is x^3/y = x + x/y? [#permalink]

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New post 08 Aug 2012, 16:56
Bunuel wrote:

Even if you could cancel \(x\) in (1) (for example if we were told that \(x\neq{0}\)) question would become: is \(x^2=9\)? Or is \(x=3\) or \(x=-3\)? And since we don't know whether that's true, then the statement still would be insufficient.


Why isn't this allowed?

\(x^3 = xy + x\)

\(\frac{x^3-x}{x}=y\)

\(x^2-1=y\)

:oops:
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Re: Is x^3/y = x + x/y? [#permalink]

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New post 08 Aug 2012, 17:07
superpus07 wrote:

\(x^3 = xy + x\)

\(\frac{x^3-x}{x}=y\)

\(x^2-1=y\)

:oops:


\(x^3 = xy + x\)
\(x^3=x(y+1)\)
\(x^2= y+1\) = Statement 2 (Answer B)
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Re: Is x^3/y = x + x/y? [#permalink]

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New post 08 Aug 2012, 17:40
SOURH7WK wrote:
superpus07 wrote:

\(x^3 = xy + x\)

\(\frac{x^3-x}{x}=y\)

\(x^2-1=y\)

:oops:


\(x^3 = xy + x\)
\(x^3=x(y+1)\)
\(x^2= y+1\) = Statement 2 (Answer B)


Sorry if I was unclear. I meant for statement 1. If you take this approach you get \(x^2-9=0 --> (x-3)(x+3)\). But according to the post above the potential x's should include a zero also. So this means that this approach is not allowed?
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Re: Is x^3/y = x + x/y? [#permalink]

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New post 09 Aug 2012, 03:01
Expert's post
superpus07 wrote:
SOURH7WK wrote:
superpus07 wrote:

\(x^3 = xy + x\)

\(\frac{x^3-x}{x}=y\)

\(x^2-1=y\)

:oops:


\(x^3 = xy + x\)
\(x^3=x(y+1)\)
\(x^2= y+1\) = Statement 2 (Answer B)


Sorry if I was unclear. I meant for statement 1. If you take this approach you get \(x^2-9=0 --> (x-3)(x+3)\). But according to the post above the potential x's should include a zero also. So this means that this approach is not allowed?


We cannot divide \(x^3=9x\) by \(x\) because \(x\) can be zero and division by zero is not allowed.

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) \(x^3=9x\) by \(x\), you assume, with no ground for it, that \(x\) does not equal to zero thus exclude a possible solution (notice that \(x=0\) satisfies \(x^3=9x\)).

Hope it's clear.
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If y ≠ 0, is x^3/y = x + x/y ? [#permalink]

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New post 17 Nov 2015, 23:35
If y ≠ 0, is x^3/y = x + x/y ?

(1) y = 8
(2) y = x^2 -1


[Reveal] Spoiler:
My answer is A as x= 3,-3 after solving. Although official answer is different.

Last edited by Bunuel on 17 Nov 2015, 23:36, edited 1 time in total.
Renamed the topic and edited the question.
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Re: Is x^3/y = x + x/y? [#permalink]

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New post 17 Nov 2015, 23:39
Expert's post
aby0007 wrote:
If y ≠ 0, is x^3/y = x + x/y ?

(1) y = 8
(2) y = x^2 -1


My answer is A as x= 3,-3 after solving. Although official answer is different.


Merging topics.

As for your doubt in red:

When a DS question asks about the value of some variable, then the statement(s) is sufficient ONLY if you can get the single numerical value of this variable.

In a Yes/No Data Sufficiency questions, statement(s) is sufficient if the answer is “always yes” or “always no” while a statement(s) is insufficient if the answer is "sometimes yes" and "sometimes no".


Hope it's clear.
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Re: Is x^3/y = x + x/y? [#permalink]

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New post 18 Nov 2015, 03:03
Bunuel wrote:
aby0007 wrote:
If y ≠ 0, is x^3/y = x + x/y ?

(1) y = 8
(2) y = x^2 -1


My answer is A as x= 3,-3 after solving. Although official answer is different.


Merging topics.

As for your doubt in red:

When a DS question asks about the value of some variable, then the statement(s) is sufficient ONLY if you can get the single numerical value of this variable.

In a Yes/No Data Sufficiency questions, statement(s) is sufficient if the answer is “always yes” or “always no” while a statement(s) is insufficient if the answer is "sometimes yes" and "sometimes no".



Hope it's clear.



Hey Bunuel,

Just want to clarify one thing on this comment of yours
"[b]When a DS question asks about the value of some variable, then the statement(s) is sufficient ONLY if you can get the single numerical value of this variable."

As in this question option 1 gives 3 values , 0 ,-3 and 3 which we obtained by solving it .If put them back in the equation LHS =RHS in every value then after we have to always look for one solution only, by which I mean single numerical value if so can you please elaborate why?
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Re: Is x^3/y = x + x/y? [#permalink]

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New post 18 Nov 2015, 11:43
Expert's post
aby0007 wrote:
Bunuel wrote:
aby0007 wrote:
If y ≠ 0, is x^3/y = x + x/y ?

(1) y = 8
(2) y = x^2 -1


My answer is A as x= 3,-3 after solving. Although official answer is different.


Merging topics.

As for your doubt in red:

When a DS question asks about the value of some variable, then the statement(s) is sufficient ONLY if you can get the single numerical value of this variable.

In a Yes/No Data Sufficiency questions, statement(s) is sufficient if the answer is “always yes” or “always no” while a statement(s) is insufficient if the answer is "sometimes yes" and "sometimes no".



Hope it's clear.



Hey Bunuel,

Just want to clarify one thing on this comment of yours
"[b]When a DS question asks about the value of some variable, then the statement(s) is sufficient ONLY if you can get the single numerical value of this variable."

As in this question option 1 gives 3 values , 0 ,-3 and 3 which we obtained by solving it .If put them back in the equation LHS =RHS in every value then after we have to always look for one solution only, by which I mean single numerical value if so can you please elaborate why?


When a DS question asks about the value of some variable, the answer cannot be 3 or -3. The answer MUST be one and only then a statement is sufficient.
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PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: Is x^3/y = x + x/y? [#permalink]

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New post 21 Nov 2015, 10:01
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Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

Is x^3/y = x + x/y?

(1) y = 8
(2) y = x^2 - 1

If we modify the original condition, the question is asking Is x^3/y = x + x/y?, or x^3=yx+x?, x(x^2-y-1)=0?, and from condition 2, x^2-y-1=0 . This is sufficient and the answer becomes (B).

Once we modify the original condition and the question according to the variable approach method 1, we can solve approximately 30% of DS questions.
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Re: Is x^3/y = x + x/y?   [#permalink] 21 Nov 2015, 10:01
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