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Math Revolution DS Expert - Ask Me Anything about GMAT DS

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New post 07 Jan 2019, 23:21
[Math Revolution GMAT math practice question]

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

\(1) x<0\)
\(2) x>-2\)
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New post 08 Jan 2019, 03:21
MathRevolution wrote:
[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.
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New post 09 Jan 2019, 00:17
MathRevolution wrote:
[Math Revolution GMAT math practice question]

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

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


=>

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
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New post 09 Jan 2019, 00:18
[Math Revolution GMAT math practice question]

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

\(1) 2x>3y\)
\(2) -5x<-7y\)
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New post 09 Jan 2019, 08:14
MathRevolution wrote:
[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.
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Re: Math Revolution DS Expert - Ask Me Anything about GMAT DS  [#permalink]

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New post 09 Jan 2019, 23:27
MathRevolution wrote:
[Math Revolution GMAT math practice question]

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

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


=>

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.
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New post 09 Jan 2019, 23:28
[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 \(1\)?

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

Attachment:
1.10.png
1.10.png [ 8.42 KiB | Viewed 291 times ]

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Re: Math Revolution DS Expert - Ask Me Anything about GMAT DS  [#permalink]

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New post 11 Jan 2019, 03:45
MathRevolution wrote:
[Math Revolution GMAT math practice question]

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

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


=>

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.
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New post 11 Jan 2019, 03:47
[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?

\(1) -3 < r < 3\)
\(2) -4 < s < 4\)
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Re: Math Revolution DS Expert - Ask Me Anything about GMAT DS  [#permalink]

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New post 11 Jan 2019, 06:28
MathRevolution wrote:
[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?

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

\(\left( {r,s} \right)\,\,\,\mathop \in \limits^? \,\,\,\left\{ {\,\left( {x,y} \right)\,\,:\,\,{{\left( {x - 0} \right)}^2} + {{\left( {y - 0} \right)}^2} \leqslant {5^2}\,} \right\}\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\boxed{\,\,\,{r^2} + {s^2}\,\,\,\mathop \leqslant \limits^? \,\,\,25\,}\,\,\)

\(\left( 1 \right)\,\,\,\,\left| r \right| < 3\,\,\,\left\{ \matrix{
\,{\rm{Take}}\,\,\left( {r,s} \right) = \left( {0,0} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\, \hfill \cr
\,{\rm{Take}}\,\,\left( {r,s} \right) = \left( {0,6} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\, \hfill \cr} \right.\)

\(\left( 2 \right)\,\,\,\left| s \right| < 4\,\,\,\,\left\{ \matrix{
\,\left( {{\mathop{\rm Re}\nolimits} } \right){\rm{Take}}\,\,\left( {r,s} \right) = \left( {0,0} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\, \hfill \cr
\,{\rm{Take}}\,\,\left( {r,s} \right) = \left( {6,0} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\, \hfill \cr} \right.\)

\(\left( {1 + 2} \right)\,\,\,\,\left\{ \matrix{
\,{r^2} = {\left| r \right|^2} < {3^2} \hfill \cr
\,{s^2} = {\left| s \right|^2} < {4^2} \hfill \cr} \right.\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,{r^2} + {s^2} < 25\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\left\langle {{\rm{YES}}} \right\rangle\)


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

Regards,
Fabio.
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Math Revolution DS Expert - Ask Me Anything about GMAT DS  [#permalink]

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New post 11 Jan 2019, 07:09
MathRevolution wrote:
[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 \(1\)?

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

Image

\(x + y + z\,\,\mathop > \limits^? \,\,\,1\)


\(\left( 1 \right)\,\,\,\left\{ \matrix{
\,x\mathop > \limits^{\left( * \right)} \,\,\,h = {1 \over 2} \hfill \cr
\,z\mathop > \limits^{\left( * \right)} \,\,\,h = {1 \over 2} \hfill \cr
y > 0 \hfill \cr} \right.\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,x + y + z > 1\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\left\langle {{\rm{YES}}} \right\rangle\)


Image


\(\left( 2 \right)\,\,\left\{ \matrix{
\,{\rm{figure}}\,\,{\rm{on}}\,\,{\rm{the}}\,\,{\rm{left}}\,\,\, \Rightarrow \,\,\,\,\,x + y + z\,\,\,\, < \,\,\,\,3\left( {{1 \over 3}} \right)\,\, = \,\,1\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,\left\langle {{\rm{NO}}} \right\rangle \hfill \cr
\,{\rm{figure}}\,\,{\rm{on}}\,\,{\rm{the}}\,\,{\rm{right}}\,\,\, \Rightarrow \,\,\,\,\,x + y + z\,\,\,\,\mathop < \limits^{{\rm{as}}\,{\rm{near}}\,\,{\rm{as}}\,\,{\rm{desired}}!} \,\,\,\,{2 \over 3} + {1 \over 3}\left( {\sqrt 2 } \right)\,\, = \,\,{{\sqrt 2 + 2} \over 3}\,\,\,\,\,\left[ { > \,\,\,1} \right]\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,\left\langle {{\rm{YES}}} \right\rangle \hfill \cr} \right.\)



The correct answer is therefore (A).


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

Regards,
Fabio.

P.S.: it is interesting to study the case in which D does not belong to the side BC!
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Re: Math Revolution DS Expert - Ask Me Anything about GMAT DS  [#permalink]

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New post 13 Jan 2019, 16:52
MathRevolution wrote:
[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 \(1\)?

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

Attachment:
1.10.png


=>

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.

In a triangle, the sum of the lengths of two sides is always greater than the length of the other side.
Thus, from triangle \(ABD\), we must have \(AB + BD > AD = h\), and from triangle \(ABD\), we must have \(AC + CD > AD = h\). If \(h = \frac{1}{2}\), then \(AB + BC + CA = AB + BD + DC + CA > \frac{1}{2} + \frac{1}{2} = 1.\)
Condition 1) is sufficient.

Condition 2):
If \(AC = \frac{1}{2}\), then \(AB + BC + CA = \frac{1}{3} + \frac{1}{3} + \frac{1}{2} = \frac{7}{6} > 1\) and the answer is ‘yes’.
If \(AC = \frac{1}{4}\), then \(AB + BC + CA = \frac{1}{3} + \frac{1}{3} + \frac{1}{4} = \frac{11}{12} < 1\) and the answer is ‘no’.
Since it does not yield a unique solution, condition 2) is not sufficient.

Therefore, the correct answer is A.
Answer: A
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Re: Math Revolution DS Expert - Ask Me Anything about GMAT DS  [#permalink]

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New post 13 Jan 2019, 17:00
Attachment:
1.14.png
1.14.png [ 22.82 KiB | Viewed 340 times ]
MathRevolution wrote:
[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?

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


=>

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 inequality satisfied by points inside or on the circle is \(r^2+s^2≤5^2=25\).

Since we have \(2\) variables (\(r\) and \(s\)) and \(0\) equations, C is most likely to be the answer. So, we should consider conditions 1) & 2) together first.

Conditions 1) & 2):
Since \(-3<r<3\) and \(-4<s<4\), we have \(0≤r^2<3^2=9\) and \(0≤s^2<4^2=16\). Thus, \(0≤r^2+s^2<9+16=25\) and both conditions together are sufficient.

Attachment:
1.14.png
1.14.png [ 22.82 KiB | Viewed 340 times ]


Therefore, the answer is C.
Answer: C
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Re: Math Revolution DS Expert - Ask Me Anything about GMAT DS  [#permalink]

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New post 14 Jan 2019, 00:04
[Math Revolution GMAT math practice question]

(absolute value, geometry) Is \(a > b - c\)?

\(1) |c-b| < a\)
\(2) a, b\), and \(c\) are the lengths of the \(3\) sides of a triangle.
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Re: Math Revolution DS Expert - Ask Me Anything about GMAT DS  [#permalink]

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New post 14 Jan 2019, 04:51
MathRevolution wrote:
[Math Revolution GMAT math practice question]

(absolute value, geometry) Is \(a > b - c\)?

\(1) |c-b| < a\)
\(2) a, b\), and \(c\) are the lengths of the \(3\) sides of a triangle.

\(a\,\,\mathop > \limits^? \,\,b - c\)


\(\left( 1 \right)\,\,\,a\,\, > \,\,\left| {c - b} \right|\,\,\, = \,\,\,\left| {b - c} \right|\,\,\,\, \ge \,\,\,b - c\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\)


\(\left( 2 \right)\,\, \Rightarrow \,\,\left( 1 \right)\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\)


The correct answer is therefore (D).


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

Regards,
Fabio.
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Re: Math Revolution DS Expert - Ask Me Anything about GMAT DS  [#permalink]

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New post 14 Jan 2019, 23:58
[Math Revolution GMAT math practice question]

(number properties) If \(x, y\) are integers, is \(x^2+x+y\) an odd integer?

1) \(x\) is an odd integer
2) \(y\) is an odd integer
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Re: Math Revolution DS Expert - Ask Me Anything about GMAT DS  [#permalink]

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New post 15 Jan 2019, 05:51
MathRevolution wrote:
[Math Revolution GMAT math practice question]

(number properties) If \(x, y\) are integers, is \(x^2+x+y\) an odd integer?

1) \(x\) is an odd integer
2) \(y\) is an odd integer

\(x,y\,\,{\text{ints}}\,\,\,\left( * \right)\)

\({x^{\text{2}}} + x + y\,\, = \,\,\underbrace {x\left( {x + 1} \right)}_{\left( * \right)\,\,{\text{even}}} + y\,\,\mathop = \limits^? \,\,\,{\text{odd}}\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\boxed{\,\,y\,\,\mathop = \limits^? \,\,\,{\text{odd}}\,\,}\)

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

\(\left( 2 \right)\,\,y\,\,{\text{odd}}\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\left\langle {{\text{YES}}} \right\rangle\)


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

Regards,
Fabio.
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New post 16 Jan 2019, 00:14
MathRevolution wrote:
[Math Revolution GMAT math practice question]

(absolute value, geometry) Is \(a > b - c\)?

\(1) |c-b| < a\)
\(2) a, b\), and \(c\) are the lengths of the \(3\) sides of a triangle.


=>

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.

Condition 1) is sufficient since it yields \(b – c ≤ | b - c | = | c – b | < a.\)

Condition 2)
Since the sum of the lengths of two sides of a triangle is greater than the length of the third side, we must have \(a + c > b\). Thus, 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.
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Math Revolution DS Expert - Ask Me Anything about GMAT DS  [#permalink]

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New post Updated on: 16 Jan 2019, 08:02
MathRevolution wrote:
[Math Revolution GMAT math practice question]

(number properties) If \(x\) and \(y\) are integers, is \(x^2-y^2\) an even integer?

1) \(x^3-y^3\) is an even integer
2) \(x+y\) is an even integer

VERY beautiful problem, Max. Congrats (and kudos)!

\(x,y\,\,{\rm{ints}}\,\,\,\,\left( * \right)\)

\(\left( {x + y} \right)\left( {x - y} \right) = {x^{\text{2}}} - {y^2}\,\,\mathop = \limits^? \,\,{\text{even}}\,\,\,\,\,\,\,\mathop \Leftrightarrow \limits^{\left( * \right)} \,\,\,\,\,\,\boxed{\,\,x + y\mathop = \limits^? \,\,{\text{even}}\,\,\,\,\,\,\mathop \Leftrightarrow \limits^{\left( {**} \right)} \,\,\,\,\,\,x - y\mathop = \limits^? \,\,{\text{even}}\,\,}\)

\(\left( {**} \right)\,\,\,\left\{ \matrix{
\,x + y = {\rm{even}}\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,x - y = x + y - 2y = {\rm{even}} \hfill \cr
\,x - y = {\rm{even}}\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,x + y = x - y + 2y = {\rm{even}} \hfill \cr} \right.\)


\(\left( 1 \right)\,\,\,{x^3} - {y^3} = {\rm{even}}\,\,\,\,\,\mathop \Leftrightarrow \limits^{\left( * \right)} \,\,\,\,\,\left\{ \matrix{
\,x\,\,{\rm{even}}\,\,,y\,\,{\rm{even}} \hfill \cr
\,{\rm{OR}} \hfill \cr
\,x\,\,{\rm{odd}}\,\,,y\,\,{\rm{odd}} \hfill \cr} \right.\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,x + y = {\rm{even}}\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,\,\left\langle {{\rm{YES}}} \right\rangle\)


\(\left( 2 \right)\,\,x + y = {\rm{even}}\,\,\,\, \Rightarrow \,\,\,\,\left\langle {{\rm{YES}}} \right\rangle\)



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

Regards,
Fabio.
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Originally posted by fskilnik on 16 Jan 2019, 04:59.
Last edited by fskilnik on 16 Jan 2019, 08:02, edited 1 time in total.
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Re: Math Revolution DS Expert - Ask Me Anything about GMAT DS  [#permalink]

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New post 16 Jan 2019, 05:20
MathRevolution wrote:
[Math Revolution GMAT math practice question]

(number properties) If \(x, y\) are integers, is \(x^2+x+y\) an odd integer?

1) \(x\) is an odd integer
2) \(y\) is an odd integer



\(x^2+x+y\)
can be written as
x(x+1) +y
so in this irrespective of value of x ; y has to be an odd integer to get \(x^2+x+y\) = odd

#1:
value of y not know in sufficient

#2:
yes sufficient

IMO B
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Re: Math Revolution DS Expert - Ask Me Anything about GMAT DS   [#permalink] 16 Jan 2019, 05:20

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