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(x^2-3x+2)(y^2-5y+6)=?

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by Max@Math Revolution » Sun Oct 28, 2018 11:50 pm

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[Math Revolution GMAT math practice question]

(x^2-3x+2)(y^2-5y+6)=?

1) x=1
2) y=1
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Source: — Data Sufficiency |
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by GMATGuruNY » Mon Oct 29, 2018 2:19 am
Max@Math Revolution wrote:[Math Revolution GMAT math practice question]

(x^2-3x+2)(y^2-5y+6)=?

1) x=1
2) y=1
Statement 1:
Substituting x=1 into (x² - 3x + 2)(y² - 5x + 6), we get:
(1 - 3*1 + 2)(y² - 5x + 6) = (0)(y² - 5x + 6) = 0.
SUFFICIENT.

Statement 2:
Substituting y=1 into (x² - 3x + 2)(y² - 5x + 6), we get:
(x² - 3x + 2)(1² - 5*1 + 6) = (x² - 3x + 2)(2).
Since the value of x is unknown, INSUFFICIENT.

The correct answer is A.
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by fskilnik@GMATH » Mon Oct 29, 2018 2:34 am
Max@Math Revolution wrote:[Math Revolution GMAT math practice question]

(x^2-3x+2)(y^2-5y+6)=?

1) x=1
2) y=1
$$? = {{{x^2} - 3x + 2} \over {{y^2} - 5y + 6}}\,\,\,\,\,\,\,\,\,\left[ {\,{y^2} - 5y + 6 \ne 0\,\,{\rm{implicitly}}\,} \right]$$
$$\left( 1 \right)\,\,x = 1\,\,\,\, \Rightarrow \,\,\,\,? = {{1 - 3 + 2} \over {{y^2} - 5y + 6}} = 0\,\,\,\,\, \Rightarrow \,\,\,\,\,{\rm{SUFF}}.$$
$$\left( 2 \right)\,\,y = 1\,\,\,\, \Rightarrow \,\,\,\,? = {{{x^2} - 3x + 2} \over {1 - 5 + 6}} = {{{x^2} - 3x + 2} \over 2}\,\,\,\,\,\left\{ \matrix{
\,\left( {{\mathop{\rm Re}\nolimits} } \right){\rm{Take}}\,\,x = 1\,\,\,\, \Rightarrow \,\,\,\,? = 0 \hfill \cr
\,{\rm{Take}}\,\,x = 0\,\,\,\, \Rightarrow \,\,\,\,? = 1 \hfill \cr} \right.\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,{\rm{INSUFF}}.$$

This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
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by Max@Math Revolution » Wed Oct 31, 2018 1:22 am
=>

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question.

Factoring the question gives (x^2-3x+2)(y^2-5y+6) = (x-1)(x-2)(y-2)(y-3).

Condition 1)
If x = 1, then (x-1)(x-2)(y-2)(y-3) = 0.
Condition 1) is sufficient.

Condition 2)
If y = 1, then (x-1)(x-2)(1-2)(1-3) = 2(x-1)(x-2).
Since we don't know the value of x, condition 2) is not sufficient.

Note: Tip 4) of VA (Variable Approach) method states that if both conditions are trivial, they are not usually sufficient. Thus, conditions 1) & 2) are not sufficient by Tip 4).

Therefore, A is the answer.
Answer: A
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