[Math Revolution GMAT math practice question]

(number properties) If \(n\) is a positive integer, is \(\sqrt{17n}\) an integer?

1) \(68n\) is the square of an integer.
2) \(\frac{n}{68}\) is the square of an integer.
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[Math Revolution GMAT math practice question]

(integer) If \(m\) and \(n\) are positive integers, what is the greatest common divisor of \(m\) and \(n\)?

\(1) m=n+1\)
\(2) m*n\) is divisible by \(2\)
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[Math Revolution GMAT math practice question]

(number property) If \(k\) is a positive integer and \(n=(k-1)k(k+1)\), is \(n\) a multiple of \(8\)?

1) \(k\) is an odd number
2) \(k = 1\)
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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 two consecutive integers are always relatively prime, the greatest common divisor of m and n is 1. Thus, condition 1) is sufficient.

Condition 2)
If \(m = 2\) and \(n = 3\), then the greatest common divisor of \(m\) and \(n\) is \(1\).
If \(m = 2\) and \(n = 4,\) then the greatest common divisor of \(m\) and \(n\) is \(2\).
Thus, condition 2) is not sufficient since it does not yield a unique solution.

Therefore, the correct answer is A.
Answer: A
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[Math Revolution GMAT math practice question]

(inequality) Is \(x + \frac{1}{x} > 2\)?

\(1) x > 0\)
\(2) x ≠ 1\)
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Excellent problem, Max. Congrats! (kudos!)

\(x + {1 \over x}\,\,\mathop > \limits^? \,\,2\)

\(\left( 1 \right)\,\,x > 0\,\,\,\left\{ \matrix{
\,{\rm{Take}}\,\,x = 1\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\, \hfill \cr
\,{\rm{Take}}\,\,x = 2\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\, \hfill \cr} \right.\)

\(\left( 2 \right)\,\,x \ne 1\,\,\left\{ \matrix{
\,\left( {{\mathop{\rm Re}\nolimits} } \right){\rm{Take}}\,\,\,x = 2\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\, \hfill \cr
\,{\rm{Take}}\,\,x = - 1\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\, \hfill \cr} \right.\)

\(\left( {1 + 2} \right)\,\,\,0\,\,\mathop < \limits^{x\, \ne \,1} \,\,{\left( {x - 1} \right)^2}\,\, = \,\,\,{x^2} - 2x + 1\,\,\,\,\mathop \Leftrightarrow \limits^{\,x\, > \,\,0} \,\,\,\,0 < {{{x^2} - 2x + 1} \over x} = x - 2 + {1 \over x}\,\,\,\,\, \Leftrightarrow \,\,\,\,x + {1 \over x} > 2\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\,\,\)


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

Regards,
Fabio.
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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.

\(x + \frac{1}{x} > 2\)
\(=> x^3 + x > 2x^2\) after multiplying both sides by \(x^2\)
\(=> x^3 - 2x^2 + x > 0\)
\(=> x^3 - 2x^2 + x > 0\)
\(=> x(x^2 - 2x + 1) > 0\)
\(=> x(x-1)^2 > 0\)
\(=> x > 0\) and \(x ≠ 1\)

Thus, we need both conditions together for sufficiency.

Therefore, C is the answer.
Answer: C
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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 in the original condition, D is most likely to be the answer. So, we should consider each condition on its own first. It is suggested we plug in numbers when solving remainder problems.


Condition 1)
If \(n = 3\), then \(\frac{36}{3} = 12\) is divisible by \(3\), and \(n\) is divisible by \(3\). The answer is ‘yes’.
If \(n = 1\), then \(\frac{36}{1} = 36\) is divisible by \(3\), but \(n\) is not divisible by \(3\). The answer is ‘no’.
Thus, condition 1) is not sufficient, since it does not yield a unique solution.

Condition 2)
If \(n = 3\), then \(\frac{27}{3} = 9\) and \(n\) is divisible by \(3\). The answer is ‘yes’.
If \(n = 1\), then \(\frac{27}{1} = 27\) and \(n\) is not divisible by \(3\). The answer is ‘no’.
Thus, condition 2) is not sufficient, since it does not yield a unique solution.

Conditions 1) & 2)
Even if we consider both conditions together, we still have two possible values of \(n: n = 1\) and \(3\).
Thus, both conditions together are not sufficient, since they do not yield a unique solution.

Therefore, E is the answer.
Answer: E

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.
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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.

If \(x >1\), then \(√x < x < x^2\) and \(x\) is their median.
If \(0 < x <1\), then \(√x > x > x^2\) and \(x\) is their median.
If \(x = 1\), then \(√x = x = x^2\) and \(x\) is their median.
Thus, the question asks for the value of \(x\).

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)
\(x^2=x\)
\(=> x^2-x=0\)
\(=> x(x-1)=0\)
\(=> x = 0\) or \(x = 1\)
Since \(x\) is positive, \(x = 1.\)
Condition 1) is sufficient.

Condition 2)
\(x^2+x+1=3x\)
\(=> x^2-2x+1=0\)
\(=> (x-1)^2=0\)
\(=> x = 1\).
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.

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.
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(ex 1) A certain store, books are sold. Books are hard cover or soft cover and hard cover books sold $10 each and soft cover books sold $6. Is the number of hard cover books sold greater than that of soft cover books sold?
1) The average price sold of total books is $9
2) The number of hard cover books sold is 100