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Math Revolution DS Expert - Ask Me Anything about GMAT DS

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New post 17 Jan 2019, 02:06
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


=>

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 parity of \(x^2+x+y = x(x+1) + y\) is same as the parity of \(y\), since \(x^2+x = x(x+1)\) is the product of two consecutive integers and so it is always an even integer.
Thus, asking whether \(x^2+x+y = x(x+1) + y\) is odd is equivalent to asking whether \(y\) is odd.

Therefore, B is the answer.
Answer: B
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New post 17 Jan 2019, 02:07
[Math Revolution GMAT math practice question]

(number properties) A sequence An satisfies An+3=An+8. What is the remainder when A99 is divided by 8?

1) A1=1
2) A2=2
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Re: Math Revolution DS Expert - Ask Me Anything about GMAT DS  [#permalink]

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New post 18 Jan 2019, 05:07
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


=>

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 whether \(x^2-y^2\) is even is equivalent to asking whether \((x+y)(x-y)\) is an even integer.

Condition 2) is sufficient since \(x^2-y^2 = (x+y)(x-y)\) is an even integer if \(x + y\) is an even integer.

Condition 1)
In order for \(x^3-y^3\) to be an even integer, \(x\) and \(y\) must both have the same parity.
There are only two cases to consider: both \(x\) and \(y\) are even integers or both are odd integers.
Since \(x\) and \(y\) have the same parity, \(x – y\) is always an even integer.
Thus, \(x^2-y^2\) is an even integer. Condition 1) is sufficient.

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

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New post 18 Jan 2019, 05:38
MathRevolution wrote:
[Math Revolution GMAT math practice question]

(absolute value) Is \(ab<0\)?

\(1) |a+b| = - ( a + b )\)
\(2) |a+b| + 1 = |a| + |b|\)



for ab<0
either of a or b has to be -ve

#1:
|a+b| = - ( a + b )
LHS would always be +ve
and RHS also has to be +ve
for RHS to be +ve the values of ( a+b) has to be -ve which would possible when
both a & b are -ve or when either of a & b are -ve and the -ve value is greater than +ve value
in that case ab<0 wont be sufficeint

#2:
|a+b| + 1 = |a| + |b|[/m]

this can be written as:
-a-b+1 = a+b
1= 2 ( a+b)
a+b= 1/2
this would be possible when either of values of a & b are either + & -ve or if one of the values of a & b is 0 and other is 1/2 ; so insufficient

combining 1 & 2 :
- ( a + b )+1 = |a| + |b|
or say 1/2 = a+b
again this would be possible when either of values of a & b are either + & -ve or if one of the values of a & b is 0 and other is 1/2 ; so insufficient

IMO E
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New post 18 Jan 2019, 15:09
MathRevolution wrote:
[Math Revolution GMAT math practice question]

(absolute value) Is \(ab<0\)?

\(1) |a+b| = - ( a + b )\)
\(2) |a+b| + 1 = |a| + |b|\)

Excellent problem, Max. Congrats (and kudos)!

\(ab\mathop {\,\, < }\limits^? \,\,0\)


\(\left( 1 \right)\,\,\,\left| {a + b} \right| = - \left( {a + b} \right)\,\,\,\, \Leftrightarrow \,\,\,\,\,a + b \le 0\)

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


\(\left( 2 \right)\,\,\left| {a + b} \right| + 1 = \left| a \right| + \left| b \right|\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,\,ab < 0\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{YES}}} \right\rangle\)

\(\left( * \right)\,\,ab \ge 0\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\left| {a + b} \right| = \left| a \right| + \left| b \right|\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\left| {a + b} \right| + 1 \ne \left| a \right| + \left| b \right|\,\,\,\,,\,\,\,{\rm{impossible}}\)


We follow 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 20 Jan 2019, 18:38
MathRevolution wrote:
[Math Revolution GMAT math practice question]

(number properties) A sequence An satisfies An+3=An+8. What is the remainder when A99 is divided by 8?

1) A1=1
2) A2=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.

The initial condition, An+3=An+8, tells us that every third term has the same remainder when it is divided by 8.
By condition 1), A1 = 1, A4 = 9, A7 = 17, … , A97 = 257, A100 = 265 …. Each of these terms has remainder 1 when it is divided by 8.
By condition 2), A2 = 2, A5 = 10, A8 = 18, … , A98 = 258, A101 = 266 …. Each of these terms has remainder 2 when it is divided by 8.

A99 does not appear in either of these lists. Thus, both conditions together do not provide enough information to find the remainder when A99 is divided by 8. Therefore, E is the answer.

Note: in order to find the remainder when A99 is divided by 8, we need the value of A3. Thus, we need the values of 3 variables, and we are given only 2, so E should be the answer.

Therefore, the correct answer is E.
Answer: E
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New post 20 Jan 2019, 18:39
MathRevolution wrote:
[Math Revolution GMAT math practice question]

(absolute value) Is \(ab<0\)?

\(1) |a+b| = - ( a + b )\)
\(2) |a+b| + 1 = |a| + |b|\)


=>

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.

You should remember that the inequality \(|x+y| < |x| + |y|\) is equivalent to the inequality \(xy < 0.\)

Condition 2) tells us that \(|a+b| + 1 = |a| + |b|.\) Thus, \(|a + b| < |a| + |b|\) and \(ab < 0\).
Thus, condition 2) is sufficient.

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

Therefore, the answer is B.
Answer: B
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New post 21 Jan 2019, 01:14
[Math Revolution GMAT math practice question]

(number properties) \(n\) is a \(3\) digit integer of the form \(ab6\). Is \(n\) divisible by \(4\)?

1) \(a+b\) is an even integer
2) \(ab\) is an odd integer.
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New post 21 Jan 2019, 05:24
MathRevolution wrote:
[Math Revolution GMAT math practice question]

(number properties) \(n\) is a \(3\) digit integer of the form \(ab6\). Is \(n\) divisible by \(4\)?

1) \(a+b\) is an even integer
2) \(ab\) is an odd integer.

\(n = \left\langle {ab6} \right\rangle \,\,\,\, \Rightarrow \left\{ \matrix{
\,n > 0\,\,\,\left( {{\rm{implicitly}}} \right) \hfill \cr
\,a \in \left\{ {1,2,3, \ldots ,9} \right\} \hfill \cr
\,b \in \left\{ {0,1,2,3, \ldots ,9} \right\} \hfill \cr} \right.\)

\({{\left\langle {ab6} \right\rangle } \over 4}\,\,\mathop = \limits^? \,\,{\mathop{\rm int}}\)

\(\left( 1 \right)\,\,\,a + b = {\rm{even}}\,\,\,\left\{ \matrix{
\,{\rm{Take}}\,\,\left( {a,b} \right) = \left( {1,1} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\, \hfill \cr
\,{\rm{Take}}\,\,\left( {a,b} \right) = \left( {2,2} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\, \hfill \cr} \right.\)

\(\left( 2 \right)\,\,ab = {\rm{odd}}\,\,\, \Rightarrow \,\,\,b = {\rm{odd}}\,\,\, \Rightarrow \,\,\,\left\langle {b6} \right\rangle \in \left\{ {16,36,56,76,96} \right\}\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\,\)


The correct answer is therefore (B).


We follow 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 21 Jan 2019, 07:44
MathRevolution wrote:
MathRevolution wrote:
[Math Revolution GMAT math practice question]

(absolute value) Is \(ab<0\)?

\(1) |a+b| = - ( a + b )\)
\(2) |a+b| + 1 = |a| + |b|\)


=>

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.

You should remember that the inequality \(|x+y| < |x| + |y|\) is equivalent to the inequality \(xy < 0.\)

Condition 2) tells us that \(|a+b| + 1 = |a| + |b|.\) Thus, \(|a + b| < |a| + |b|\) and \(ab < 0\).
Thus, condition 2) is sufficient.

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

Therefore, the answer is B.
Answer: B


MathRevolution

QUOTING YOUR LINE :
You should remember that the inequality \(|x+y| < |x| + |y|\) is equivalent to the inequality \(xy < 0.\)


BUT HERE #2
\(2) |a+b| + 1 = |a| + |b|\)

ITS NOT > BUT = .. SO HOW IS THAT RELATION VALID?

IMO

#2:
|a+b| + 1 = |a| + |b|[/m]

this can be written as:
-a-b+1 = a+b
1= 2 ( a+b)
a+b= 1/2
this would be possible when either of values of a & b are either + & -ve or if one of the values of a & b is 0 and other is 1/2 ; so insufficient

combining 1 & 2 :
- ( a + b )+1 = |a| + |b|
or say 1/2 = a+b
again this would be possible when either of values of a & b are either + & -ve or if one of the values of a & b is 0 and other is 1/2 ; so insufficient

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

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New post 22 Jan 2019, 00:29
[Math Revolution GMAT math practice question]

(absolute values) If \(y=|x-1|+|x+1|,\) then \(y=\)?

\(1) x>-1\)
\(2) x<1\)
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Re: Math Revolution DS Expert - Ask Me Anything about GMAT DS  [#permalink]

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New post 22 Jan 2019, 03:42
MathRevolution wrote:
[Math Revolution GMAT math practice question]

(number properties) \(n\) is a \(3\) digit integer of the form \(ab6\). Is \(n\) divisible by \(4\)?

1) \(a+b\) is an even integer
2) \(ab\) is an odd integer.



given n= ab6 is it divisible by 4

any no which is two times divisible by 2 would be divisible by 4

#1
a+b is an even integer
here a,b can both be odd or even
so in sufficient
#2
ab is an odd integer
means both ab are odd integers

so ab6 is not divisible by 4
sufficient IM O B
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Re: Math Revolution DS Expert - Ask Me Anything about GMAT DS  [#permalink]

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New post 22 Jan 2019, 03:47
MathRevolution wrote:
[Math Revolution GMAT math practice question]

(absolute values) If \(y=|x-1|+|x+1|,\) then \(y=\)?

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



IMO e
would be sufficeint
#1:
x>-1 ; x can be any no from -1 to infinity
#2
x<1 : x can be any value <1 to infinity

from 1 & 2
range of 1>x>-1 ; x can be 0 ,0.5 , -0.5 ..

combining both we get answer different values of y
IMO E
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New post 22 Jan 2019, 07:14
MathRevolution wrote:
[Math Revolution GMAT math practice question]

(absolute values) If \(y=|x-1|+|x+1|,\) then \(y=\)?

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

\(y = \left| {x - 1} \right| + \left| {x + 1} \right|\)

\(? = y\)


\(\left( 1 \right)\,\,x > - 1\,\,\,\left\{ \matrix{
\,{\rm{Take}}\,\,x = 0\,\,\,\, \Rightarrow \,\,\,? = 2\, \hfill \cr
\,{\rm{Take}}\,\,x = 2\,\,\,\, \Rightarrow \,\,\,? = 4\, \hfill \cr} \right.\)


\(\left( 2 \right)\,\,x < 1\,\,\,\left\{ \matrix{
\,\left( {{\mathop{\rm Re}\nolimits} } \right){\rm{Take}}\,\,x = 0\,\,\,\, \Rightarrow \,\,\,? = 2\, \hfill \cr
\,{\rm{Take}}\,\,x = - 2\,\,\,\, \Rightarrow \,\,\,? = 4\, \hfill \cr} \right.\)


\(\left( {1 + 2} \right) - 1 < x < 1\,\,\,\, \Rightarrow \,\,\,\,\left\{ \matrix{
\,\left| {x - 1} \right| = 1 - x \hfill \cr
\,\left| {x + 1} \right| = x + 1 \hfill \cr} \right.\,\,\,\,\,\, \Rightarrow \,\,\,\,\,? = 2\)


We follow 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 22 Jan 2019, 07:19
Archit3110 wrote:
from 1 & 2
range of 1>x>-1 ; x can be 0 ,0.5 , -0.5 ..

combining both we get answer different values of y

Hi, Archit3110 !

Please read my solution above, and take into account that -1<x<1 implies:

(a) -1<x hence x+1 > 0 , hence |x+1| = x+1
(b) x<1 hence x-1 < 0 , hence |x-1| = -(x-1) = 1-x

Then y = (x+1) + (1-x) = 2 , therefore x will have infinite possible values (between -1 and 1), but y will be always 2 (when -1<x<1).

I hope I could help you!

Regards and success in your studies!
Fabio.
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Re: Math Revolution DS Expert - Ask Me Anything about GMAT DS  [#permalink]

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New post 23 Jan 2019, 01:06
MathRevolution wrote:
[Math Revolution GMAT math practice question]

(number properties) \(n\) is a \(3\) digit integer of the form \(ab6\). Is \(n\) divisible by \(4\)?

1) \(a+b\) is an even integer
2) \(ab\) is an odd integer.


=>

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.

We can determine whether \(a\) number is divisible by \(4\) from its final two digits.
Numbers with the final digits \(16, 36, 56, 76\) and \(96\) are divisible by \(4\) and those with final digits \(06, 26, 46, 66\) and \(88\) are not divisible by \(4\). Thus, asking whether \(n\) is divisible by \(4\) is equivalent to asking whether \(b\) is odd.

Since it implies that both \(a\) and \(b\) are odd integers, condition 2) is sufficient.

Condition 1)
There are two cases to consider.
If \(a\) is an even integer and \(b\) is an odd integer, the answer is ‘yes’.
If \(a\) is an odd integer and \(b\) is an even integer, the answer is ‘no’.
Since it does not yield a unique solution, condition 1) is not sufficient.

Therefore, B is the answer.
Answer: B
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New post 23 Jan 2019, 01:08
[Math Revolution GMAT math practice question]

(algebra) \(x=\)?

\(1) x^3-x=0\)
\(2) x=-x\)
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Re: Math Revolution DS Expert - Ask Me Anything about GMAT DS  [#permalink]

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New post 24 Jan 2019, 06:24
MathRevolution wrote:
[Math Revolution GMAT math practice question]

(absolute values) If \(y=|x-1|+|x+1|,\) then \(y=\)?

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


=>

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.

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

There are three ranges of values of x to consider.
If \(x > 1\), then \(y = | x – 1 | + | x + 1 | = x – 1 + x + 1 = 2x\) and we don’t have a unique value of \(y\).
If \(-1 ≤ x ≤ 1\), then \(y = | x – 1 | + | x + 1 | = - ( x – 1 ) + x + 1 = 2\) and we have a unique value of \(y\).
If \(x < 1\), then \(y = | x – 1 | + | x + 1 | = -( x – 1 ) – ( x + 1 ) = -2x\) and we don’t have a unique value of \(y\).

Asking for the value of \(y\) is equivalent asking if \(-1 ≤ x ≤ 1\).
Both conditions yield the inequality \(-1 < x < 1\), when applied together. Therefore, both conditions are sufficient, when taken together.

In inequality questions, the law “Question is King” tells us that if the solution set of the question includes the solution set of the condition, then the condition is sufficient.
Therefore, C is the answer.
Answer: C
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Re: Math Revolution DS Expert - Ask Me Anything about GMAT DS  [#permalink]

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New post 24 Jan 2019, 06:27
[Math Revolution GMAT math practice question]

If \(x\) and \(y\) are non-zero numbers and \(x≠±y\), then \(\frac{( x^2 + y^2 )}{( x^2 - y^2 )}=?\)

\(1) |\frac{x}{y}|=\frac{1}{3}\)
\(2) y=-3x\)
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Re: Math Revolution DS Expert - Ask Me Anything about GMAT DS  [#permalink]

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New post 25 Jan 2019, 02:28
MathRevolution wrote:
[Math Revolution GMAT math practice question]

(algebra) \(x=\)?

\(1) x^3-x=0\)
\(2) x=-x\)


=>

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

Condition 1)

\(x^3-x = 0\)
\(=> x(x^2-1) = 0\)
\(=> x(x+1)(x-1) = 0\)
\(=> x = 0, x = -1\) or \(x = 1.\)
Since it does not yield a unique solution, condition 1) is not sufficient.

Condition 2)
\(x = -x\)
\(=> 2x = 0\)
\(=> x = 0.\)
Since it gives a unique solution, condition 2) is sufficient.

Therefore, B is the answer.
Answer: B
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Re: Math Revolution DS Expert - Ask Me Anything about GMAT DS   [#permalink] 25 Jan 2019, 02:28

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