Math Revolution DS Expert - Ask Me Anything about GMAT DS

[Math Revolution GMAT math practice question]

(number property) \([x]\) is the greatest integer less than or equal to \(x\). \(<x>\) is the least integer greater than or equal to \(x\). What is the value of \(x\)?

\(1) [x] = 2\)
\(2) <x> = 2\)

[Math Revolution GMAT math practice question]

(number properties) \(x\) and \(y\) are positive integers. Is \(y\) an even integer?

\(1) x^2+x=y+2\)
\(2) x = 2\)

=>

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.

\([x]\) is analyzed as follows.
If \(n ≤ x < n + 1\) for some integer \(n\), then \([x] = n.\)
\(<x>\) is analyzed as follows.
If \(n – 1 < x ≤ n\) for some integer \(n\), then \(<x> = n.\)

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)
\([x] = 2\)
\(=> 2 ≤ x < 3\)
Thus, condition 1) is not sufficient, since it does not yield a unique solution.

Condition 2)
\(<x> = 2\)
\(=> 1 < x ≤ 2\)
Thus, condition 2) is not sufficient, since it does not yield a unique solution.

Conditions 1) & 2)
Only \(x = 2\) satisfies both conditions.
Since the answer is unique, both conditions together are sufficient.

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]

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

\(1) x>y\)
\(2) x-y>1\)

=>

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 \(2\) variables (\(x\) and \(y\)) and \(0\) equations, C is most likely to be the answer. So, we should consider conditions 1) & 2) together first. After comparing the number of variables and the number of equations, we can save time by considering conditions 1) & 2) together first.

Conditions 1) & 2):
Since \(y = x^2 + x – 2\) and \(x = 2\), we have \(y = 4\).
Since this answer is unique, both conditions together are sufficient.

Since this question is an integer question (one of the key question areas), CMT (Common Mistake Type) 4(A) of the VA (Variable Approach) method tells us that we should also check answers A and B.

Condition 1)
Since \(y = x^2 + x – 2 = (x-1)(x+2)\) and \(x\) is an integer, one of \(x – 1\) and \(x + 2\) is an even integer.
Thus, \(y\) is always an even integer and condition 1) is sufficient.

Condition 2)
Since it provides no information about y, condition 2) is not sufficient.


Therefore, A is the answer.
Answer: A

Normally, in problems which require 2 equations, such as those in which the original conditions include 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, each of conditions 1) and 2) provide an additional equation. In these problems, the two key possibilities are that C is the answer (with probability 70%), and E is the answer (with probability 25%). Thus, there is only a 5% chance that A, B or D is the answer. This occurs in common mistake types 3 and 4. Since C (both conditions together are sufficient) is the most likely answer, we save time by first checking whether conditions 1) and 2) are sufficient, when taken together. Obviously, there may be cases in which the answer is A, B, D or E, but if conditions 1) and 2) are NOT sufficient when taken together, the answer must be E.

[Math Revolution GMAT math practice question]

(absolute value) Is \(|m-n|=|m|-|n|\) ?

\(1) m-n = 0\)
\(2) n = 0\)

=>

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.

\(\frac{x}{y}>1\)
\(=> xy>y^2\)
\(=> xy-y^2>0\)
\(=> y(x-y)>0\)
By condition 2), \(x-y > 1 > 0\), but we can’t determine whether \(y\) is positive from condition 1).

Therefore, E is the answer.
Answer: E

[Math Revolution GMAT math practice question]

(integer) \(n\) is a positive integer. Is \(\frac{n(n+1)(n+2)}{4}\) an even integer?

1) \(n\) is an even integer
2) \(1238 ≤ n ≤ 1240\)

Please check the OA, Highlighted text is wrong. Stem alreay mention n as > 1.

Regards

=>

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.

Squaring both sides of \(|m – n| = |m| - |n|\) yields
\(|m-n|^2=(|m|-|n|)^2\)
\(=> (m-n)^2=(|m|-|n|)^2\)
\(=> m^2+n^2-2mn=|m|^2+|n|^2-2|mn|\)
\(=> m^2+n^2-2mn=m^2+n^2-2|mn|\)
\(=> -2mn=-2|mn|\)
\(=> mn=|mn|\)
\(=> mn ≥ 0\)

Condition 1):
\(m – n = 0\) implies that \(m = n.\) So \(mn = n^2 ≥ 0\) for all values of \(n\).
Condition 1) is sufficient.

Condition 2):
If \(n = 0\), then \(0 = mn ≥ 0.\)
Condition 2) is sufficient.
Therefore, the correct answer is D.
Answer: D

=>

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.

Asking for \(\frac{n(n+1)(n+2)}{4}\) to be an even integer is equivalent to asking for \(n(n+1)(n+2)\) to be a multiple of \(8\). If \(n\) is an even integer, \(n\) and \(n+2\) are consecutive even integers and a product of two consecutive even integers is a multiple of \(8\). Thus, condition 1) is sufficient.

Condition 2)
If \(n = 1238, n(n+1)(n+2)=1238*1239*1240\) is a multiple of \(8\) since \(1240\) is a multiple of \(8\).
If \(n = 1239, n(n+1)(n+2)=1239*1240*1241\) is a multiple of \(8\) since \(1240\) is a multiple of \(8\).
If \(n = 1240, n(n+1)(n+2)=1240*1241*1242\) is a multiple of \(8\) since \(1240\) is a multiple of \(8\).
Thus, condition 2) is sufficient.

Therefore, the answer is D.
Answer: D

Note: This question is a CMT4(B) question. Condition 1) is easy to understand and condition 2) is hard. When one condition is easy to understand, and the other is hard, D is most likely to be the answer.

[Math Revolution GMAT math practice question]

(number property) \(f(x)\) is the greatest prime factor of \(x\). If \(n\) is a positive integer less than \(10\), what is the value of \(n\)?

\(1) f(n) = f(1000)\)
\(2) f(n) = 5\)

You are right.
The question should be changed to the followings.

If \(n\) is a positive integer, what is the value of \(n\)?

Thank you for your comments.

=>

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)
Since \(1000 = 2^3*5^3, f(1000) = 5.\)
So, \(f(n) = 5\) and \(n = 5.\)
Thus, condition 1) is sufficient, since it gives a unique solution.

Condition 2)
Condition 2) is the same as condition 1).
Thus, condition 2) is also sufficient.

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.

Therefore, D is the answer.
Answer: D

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]

(number properties) If \(m\) and \(n\) are positive integers, is \(m + n\) an odd number?

1) \(\frac{m}{n}\) is an even number
2) \(m\) or \(n\) is an even number

\(m,n\,\,\, \geqslant 1\,\,\,{\text{ints}}\,\,\,\,\left( * \right)\)

\(m + n\,\,\,\,\mathop = \limits^? \,\,{\text{odd}}\,\,\,\,\,\mathop \Leftrightarrow \limits^{\left( * \right)} \,\,\,\,\boxed{\,\,\,?\,\,\,:\,\,\,\left( {m\,\,{\text{odd}}\,,\,\,n\,\,{\text{even}}} \right)\,\,\,{\text{or}}\,\,\,{\text{vice - versa}\,\,}\,\,}\)


\(\left( 1 \right)\,\,\,\frac{m}{n} = {\text{even}}\,\,\,\,\left\{ \begin{gathered}\\
\,{\text{Take}}\,\,\left( {m,n} \right) = \left( {2,1} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\text{YES}}} \right\rangle \,\, \hfill \\\\
\,{\text{Take}}\,\,\left( {m,n} \right) = \left( {4,2} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\text{NO}}} \right\rangle \,\, \hfill \\ \\
\end{gathered} \right.\)

\(\left( 2 \right)\,\,\,m\,\,{\text{even}}\,\,\,{\text{or}}\,\,\,n\,\,{\text{even}}\,\,\,\,\left\{ \begin{gathered}\\
\,\left( {\operatorname{Re} } \right){\text{Take}}\,\,\left( {m,n} \right) = \left( {2,1} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\text{YES}}} \right\rangle \,\, \hfill \\\\
\,\left( {\operatorname{Re} } \right){\text{Take}}\,\,\left( {m,n} \right) = \left( {4,2} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\text{NO}}} \right\rangle \,\, \hfill \\ \\
\end{gathered} \right.\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\left( {\text{E}} \right)\)


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

Regards,
Fabio.

P.S.: "A or B" means "only A", "only B" or BOTH.

Beautiful problem, Max. Congrats (and kudos)!

\(n \geqslant 1\,\,\,\operatorname{int}\)

\(\sqrt {n + 1} \,\,\,\mathop = \limits^? \,\,{\text{even}}\,\,\,\,\,\,\,\mathop \Leftrightarrow \limits^{\left( * \right)} \,\,\,\,\,\,\boxed{\,\,n + 1\,\,\,\mathop = \limits^? \,\,\,{{\left( {{\text{even}}} \right)}^2}\,\,}\)


\(\left( 1 \right)\,\,\,n = \left( {2M - 1} \right)\left( {2M + 1} \right) = {\left( {2M} \right)^2} - {\left( 1 \right)^2}\,\,\,\,\,\left[ {M\,\,\operatorname{int} \,} \right]\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,\,n + 1 = {\left( {2M} \right)^2}\,\,\,,\,\,\,\,M\,\,\operatorname{int} \,\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,\,\left\langle {{\text{YES}}} \right\rangle \,\)

\(\left( 2 \right)\,\,\,\left\{ \matrix{\\
\,{\rm{Take}}\,\,n = 1\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\, \hfill \cr \\
\,{\rm{Take}}\,\,n = 3\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\, \hfill \cr} \right.\,\,\)


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

Regards,
Fabio.