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[Math Revolution GMAT math practice question]

(number properties) Can \(n\) be expressed as the difference of \(2\) prime numbers?

\(1) (n-17)(n-21) = 0\)
\(2) (n-15)(n-17)=0\)

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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.

Since we have \(1\) variable (\(n\)) and \(0\) equations, D is most likely to be the answer. So, we should consider each condition on its own first.

Condition 1)
\(n\) is a prime factor of \(21 = 3*7\) and \(n\) is \(3\) or \(7\).
Since it does not give a unique answer, condition 1) is not sufficient.

Condition 2)
If \(n\) is a factor of \(49 = 7^2\), then \(n\) is \(1, 7\) or \(49\).
Since it does not give a unique answer, condition 2) is not sufficient.

Conditions 1) & 2)
The unique integer satisfying both conditions is \(n = 7.\)
Both conditions are sufficient, when taken together.

Therefore, C is the answer.
Answer: C

If the original condition includes “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations” etc., one more equation is required to answer the question. If each of conditions 1) and 2) provide an additional equation, there is a 59% chance that D is the answer, a 38% chance that A or B is the answer, and a 3% chance that the answer is C or E. Thus, answer D (conditions 1) and 2), when applied separately, are sufficient to answer the question) is most likely, but there may be cases where the answer is A,B,C or E.

[Math Revolution GMAT math practice question]

(absolute value) Is \(\frac{x}{y}<0\)?

\(1) |x+y|<|x|+|y|\)
\(2) x+y<0\)

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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.

Since we have \(1\) variable (\(n\)) and \(0\) equations, D is most likely to be the answer. So, we should consider each condition on its own first.

Condition 1)
\((n-17)(n-21) = 0\) is equivalent to the statement \(n = 17\) or \(n =21\)
If \(n = 17\), then \(17 = 19 – 2\) is a difference of two prime numbers and the answer is ‘yes’.
If \(n = 21\), then \(21 = 23 – 2\) is a difference of two prime numbers and the answer is ‘yes’.
Since it gives a unique answer, condition 1) is sufficient.

Condition 2)
\((n-15)(n-17) = 0\) is equivalent to the statement \(n = 15\) or \(n = 17\)
If \(n = 15\), then \(15 = 17 – 2\) is a difference of two prime numbers and the answer is ‘yes’.
If \(n = 17\), then \(17 = 19 – 2\) is a difference of two prime numbers and the answer is ‘yes’.
Since it gives a unique answer, condition 2) is sufficient.

Therefore, D is the answer.
Answer: D

In cases where 3 or more additional equations are required, such as for original conditions with “3 variables”, or “4 variables and 1 equation”, or “5 variables and 2 equations”, conditions 1) and 2) usually supply only one additional equation. Therefore, there is an 80% chance that E is the answer, a 15% chance that C is the answer, and a 5% chance that the answer is A, B or D. Since E (i.e. conditions 1) & 2) are NOT sufficient, when taken together) is most likely to be the answer, it is generally most efficient to begin by checking the sufficiency of conditions 1) and 2), when taken together. Obviously, there may be occasions on which the answer is A, B, C or D.

[Math Revolution GMAT math practice question]

Is \(\frac{x}{y}<0\)?

\(1) x^4y^5<0\)
\(2) x^5y^3<0\)

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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.

The question is equivalent to asking if \(xy < 0\). This can be seen by multiplying both sides of the inequality by \(y^2.\)

Condition 1) is equivalent to \(xy < 0\) as shown below:
\(|x+y|<|x|+|y|\)
\(=> |x+y|^2<(|x|+|y|)^2\)
\(=> (x+y)^2<|x|^2+2|x||y|+|y|^2\)
\(=> x^2+2xy+y^2<x^2+2|xy|+y^2\)
\(=> 2xy<2|xy|\)
\(=> xy<|xy|\)
\(=> xy<0\)
Thus, condition 1) is sufficient.

Condition 2)
If \(x = -2\) and \(y = 1\), then the answer is ‘yes’.
If \(x = -1\) and \(y = -1\), then the answer is ‘no’.
Since it does not give a unique answer, condition 2) is not sufficient.

Therefore, the correct answer is A.
Answer: A

=>

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 question is equivalent to asking if \(xy < 0\). This can be seen by multiplying both sides of the inequality by \(y^2\).

Since we can ignore even exponents in inequalities like \(x^4y^5<0\), condition 1) is equivalent to the statement \(y < 0\) and condition 2) is equivalent to the statement \(xy < 0\).

Therefore, the answer is B.
Answer: B

[Math Revolution GMAT math practice question]

(algebra) If \(x≠y\), what is the value of \(\frac{( x^2y – xy^2 )}{( x^3 – y^3 )}\)?

\(1) \frac{xy}{( x^2 + xy + y^2)} = \frac{1}{3}\)
\(2) x^2y^2=9\)

[Math Revolution GMAT math practice question]

(inequality) If \(x\) is integer and \(3|x|+x<4\), what is the value of \(x\)?

\(1) x<0\)
\(2) x>-2\)

\(3\left| x \right| + x < 4\,\,\,\,\,,\,\,\,\,x\,\,{\mathop{\rm int}}\)

\(? = x\)


\(\left( 1 \right)\,\,\,x < 0\,\,\,\,\, \Rightarrow \,\,\,\,3\left( { - x} \right) + x < 4\,\,\,\,\, \Rightarrow \,\,\,\, - 2x < 4\,\,\,\,\,\mathop \Rightarrow \limits^{:\,\,\left( { - 2} \right)} \,\,\,\,\,x > - 2\)

\(x > - 2\,\,\,\,\,\,\mathop \Rightarrow \limits^{x\,\,{\mathop{\rm int}} } \,\,\,\,\,\,x = - 1\,\,\,\,\,\left( {x < 0} \right)\,\,\,\,\, \Rightarrow \,\,\,\,{\rm{SUFF}}.\)


\(\left( 2 \right)\,\,\,x > - 2\,\,\,\,\left\{ \matrix{\\
\,\,{\rm{Take}}\,\,x = - 1\,\,\,\,\,\left[ {\,3\left| { - 1} \right| + \left( { - 1} \right) < 4\,} \right]\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,? = - 1\,\, \hfill \cr \\
\,\,{\rm{Take}}\,\,x = 0\,\,\,\,\,\left[ {\,3\left| 0 \right| + \left( 0 \right) < 4\,} \right]\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,? = 0\,\, \hfill \cr} \right.\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,{\rm{INSUFF}}.\)


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

Regards,
Fabio.

=>

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.

The expression from the question is equivalent to \(\frac{xy}{( x^2 + xy + y^2)}\) as shown below, which is the same as condition 1):
\(\frac{( x^2y – xy^2 )}{( x^3 – y^3 )}\)
\(= \frac{xy(x-y)}{(x-y)(x^2+xy+y^2)}\)
\(= \frac{xy}{( x^2 + xy + y^2 )}\)
Thus, condition 1) is sufficient.

Condition 2)
If \(x = 3, y = 1\), then \(\frac{( x^2y – xy^2 )}{( x^3 – y^3 )} = \frac{( 9 – 3 )}{( 27 – 1)} = \frac{6}{26} = \frac{3}{13}.\)
If \(x = -3, y = 1\), then \(\frac{( x^2y – xy^2 )}{( x^3 – y^3 )} = \frac{( 9 + 3 )}{( - 27 – 1)} = \frac{-12}{28} = \frac{-3}{7}.\)
Since it does not yield a unique solution, condition 2) is not sufficient.

Therefore, A is the answer.
Answer: A

[Math Revolution GMAT math practice question]

(inequality) If \(x\) and \(y\) are positive, is \(x>y\)?

\(1) 2x>3y\)
\(2) -5x<-7y\)

\(x,y\,\, > 0\,\)

\(x\,\mathop > \limits^? \,\,y\)


\(\left( 1 \right)\,\,2x > 3y\,\,\,\,\,\,\mathop \Rightarrow \limits^{:\,\,2} \,\,\,\,\,\,x\,\, > \,\,{3 \over 2}y\,\,\,\mathop > \limits^{\left( * \right)} \,\,\,y\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{\rm{YES}}} \right\rangle\)

\(\left( * \right)\,\,\,{3 \over 2} > 1\,\,\,\,\,\mathop \Rightarrow \limits^{y\,\, > \,\,0} \,\,\,\,\,{3 \over 2}y > y\)


\(\left( 2 \right)\,\, - 5x < - 7y\,\,\,\,\,\,\mathop \Rightarrow \limits^{ \cdot \,\,\left( { - {1 \over 5}} \right)} \,\,\,\,\,\,x\,\, > \,\,\,{7 \over 5}y\,\,\,\mathop > \limits^{{\rm{idem!}}} \,\,\,y\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{\rm{YES}}} \right\rangle\)


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

Regards,
Fabio.

=>

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.

Modifying the original condition:
There are two cases to consider.
Case 1) \(x ≥ 0\):
\(3|x|+x < 4\)
\(=> 3x + x < 4\)
\(=> 4x < 4\)
\(=> x < 1\)
\(=> 0 ≤ x < 1\)

Case 2) \(x < 0\):
\(-3x+x < 4\)
\(=> -2x < 4\)
\(=> x > -2\)
\(=> -2 < x <0\)

Therefore, \(x\) is an integer with \(-2 < x < 1\). Thus, the original condition tells us that \(x = -1\) or \(0\).

Since we have \(1\) variable (\(x\)) and \(0\) equations, D is most likely to be the answer. So, we should consider each condition on its own first.

Condition 1)
Since \(x < 0\), we must have \(x = -1\) as the original condition tells us that \(x = 0\) or \(x = -1.\)
Condition 1) is sufficient, because it yields a unique solution.


Condition 2)
Both \(x = 0\) and \(x = -1\) satisfy condition 2).
Since it does not yield a unique solution, condition 2) is not sufficient.

Therefore, A is the answer.
Answer: A

If the original condition includes “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations” etc., one more equation is required to answer the question. If each of conditions 1) and 2) provide an additional equation, there is a 59% chance that D is the answer, a 38% chance that A or B is the answer, and a 3% chance that the answer is C or E. Thus, answer D (conditions 1) and 2), when applied separately, are sufficient to answer the question) is most likely, but there may be cases where the answer is A,B,C or E.

[Math Revolution GMAT math practice question]

(geometry) \(x, y\) and \(z\) are the sides of the triangle shown and \(h\) is its height. Is the perimeter, \(x + y + z\) of the triangle greater than \(1\)?

\(1) h = \frac{1}{2}\)
\(2) x = y = \frac{1}{3}\)

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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.

Since \(x\) and \(y\) are positive, condition 1) tells us that \(3x > 2x > 3y\) or \(x > y\).
Thus, condition 1) is sufficient.

Condition 2)
\(-5x < -7y\)
\(=> 5x > 7y\)
This implies that \(7x > 5x > 7y\) or \(x > y\), since \(x\) and \(y\) are positive.
Condition 2) is sufficient.

Therefore, D is the answer.
Answer: D

FYI, Tip 1) of the VA method states that D is most likely to be the answer if conditions 1) and 2) provide the same information.

[Math Revolution GMAT math practice question]

(function) In the \(xy\)-plane, a circle has center \((0,0)\) and radius \(5\). Is the point \((r,s)\) inside or on the circle?

\(1) -3 < r < 3\)
\(2) -4 < s < 4\)