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Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
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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.

_________________
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8243
GMAT 1: 760 Q51 V42 GPA: 3.82

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[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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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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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.

MathRevolution

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
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8243
GMAT 1: 760 Q51 V42 GPA: 3.82

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[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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GMAT Club Legend  V
Joined: 18 Aug 2017
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Location: India
Concentration: Sustainability, Marketing
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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
GMAT Club Legend  V
Joined: 18 Aug 2017
Posts: 5460
Location: India
Concentration: Sustainability, Marketing
GPA: 4
WE: Marketing (Energy and Utilities)

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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
GMATH Teacher P
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Joined: 12 Oct 2010
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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.
_________________
Fabio Skilnik :: GMATH method creator (Math for the GMAT)
Our high-level "quant" preparation starts here: https://gmath.net
GMATH Teacher P
Status: GMATH founder
Joined: 12 Oct 2010
Posts: 935

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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).

Regards and success in your studies!
Fabio.
_________________
Fabio Skilnik :: GMATH method creator (Math for the GMAT)
Our high-level "quant" preparation starts here: https://gmath.net
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8243
GMAT 1: 760 Q51 V42 GPA: 3.82

### Show Tags

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.

_________________
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8243
GMAT 1: 760 Q51 V42 GPA: 3.82

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[Math Revolution GMAT math practice question]

(algebra) $$x=$$?

$$1) x^3-x=0$$
$$2) x=-x$$
_________________
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8243
GMAT 1: 760 Q51 V42 GPA: 3.82

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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.
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Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8243
GMAT 1: 760 Q51 V42 GPA: 3.82

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[Math Revolution GMAT math practice question]

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

$$1) |\frac{x}{y}|=\frac{1}{3}$$
$$2) y=-3x$$
_________________
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8243
GMAT 1: 760 Q51 V42 GPA: 3.82

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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.

_________________
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
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GMAT 1: 760 Q51 V42 GPA: 3.82

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[Math Revolution GMAT math practice question]

(exponents) $$m+n=?$$

$$1) (4^m)(2^n)=16$$
$$2) (2^{2m})(4^n)=64$$
_________________
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
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MathRevolution wrote:
[Math Revolution GMAT math practice question]

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

$$1) |\frac{x}{y}|=\frac{1}{3}$$
$$2) y=-3x$$

=>

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 asks for the value of $$\frac{( x^2 + y^2 )}{( x^2 - y^2 )}= ( (\frac{x}{y})^2 + 1 ) / (\frac{x}{y})^2 – 1 ).$$

When a question asks for a ratio, if one condition includes a ratio and the other condition just gives a number, the condition including the ratio is most likely to be sufficient.

Condition 1)

Since $$|\frac{x}{y}|=\frac{1}{3}, \frac{x}{y} = ±(\frac{1}{3})$$, and $$\frac{( x^2 + y^2 )}{( x^2 - y^2 )}= ( (\frac{x}{y})^2 + 1 ) / ( (\frac{x}{y})^2 – 1 ) = ( (\frac{1}{3})^2 + 1 ) / ( (\frac{1}{3})^2 – 1) = (\frac{1}{9} + 1)/(\frac{1}{9}-1) = (\frac{10}{9})/(-\frac{8}{9}) = -\frac{10}{8} = -\frac{5}{4}.$$
Condition 1) is sufficient since it gives a unique solution.

Condition 2)
Since $$y = -3x, \frac{x}{y} = -\frac{1}{3}$$, and $$\frac{( x^2 + y^2 )}{( x^2 - y^2 )}= ( (\frac{x}{y})^2 + 1 ) / ( (\frac{x}{y})^2 – 1 ) = ( (-\frac{1}{3})^2 + 1 ) / ( (-\frac{1}{3})^2 – 1) = (\frac{1}{9} + 1)/(\frac{1}{9}-1) = (\frac{10}{9})/(-\frac{8}{9}) = -\frac{10}{8} = -\frac{5}{4}.$$
Condition 2) is sufficient since it gives a unique solution.

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 Instructor V
Joined: 16 Aug 2015
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GMAT 1: 760 Q51 V42 GPA: 3.82

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MathRevolution wrote:
[Math Revolution GMAT math practice question]

(exponents) $$m+n=?$$

$$1) (4^m)(2^n)=16$$
$$2) (2^{2m})(4^n)=64$$

=>

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 2) is equivalent to $$m + n = 3$$ as shown below:
$$(2^{2m})(4^n)=64$$
$$=> (2^{2m})(2^{2n})=2^6$$
$$=> 2^{2m+2n}=2^6$$
$$=> 2m+2n = 6$$
$$=> m + n = 3$$
Condition 2) is sufficient.

Condition 1)
$$(4^m)(2^n)=16$$
$$=> (2^{2m})(2^n)=2^4$$
$$=> 2^{2m+n}=2^4$$
$$=> 2m+n = 4$$
If $$m = 1$$ and $$n =2$$, then $$m + n = 3$$.
If $$m = 0$$ and $$n = 4,$$ then $$m + n = 4.$$
Since it does not yield a unique solution, condition 1) is not sufficient.

_________________
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8243
GMAT 1: 760 Q51 V42 GPA: 3.82

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[GMAT math practice question]

(number properties) If $$n$$ is positive integer, is $$4^n+n^2+1$$ divisible by $$2$$?

1) $$n$$ is a multiple of $$4$$
2) $$n$$ is a multiple of $$6$$
_________________
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
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GMAT 1: 760 Q51 V42 GPA: 3.82

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[GMAT math practice question]

(function) If operation $$#$$ represents one of addition, subtraction, multiplication, and division, what is the value of $$0#1$$?

$$1) 2#1 = 2$$
$$2) 4#2 = 2$$
_________________
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Location: India
Concentration: Sustainability, Marketing
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MathRevolution wrote:
[GMAT math practice question]

(function) If operation $$#$$ represents one of addition, subtraction, multiplication, and division, what is the value of $$0#1$$?

$$1) 2#1 = 2$$
$$2) 4#2 = 2$$

#1
2#1=2
# can be multiplication or division so sufficient
as doing both operations we would get answer as 0 for $$0#1$$

#2
4#2 =2
# can be subtraction of division; in sufficient

IMO A

Originally posted by Archit3110 on 29 Jan 2019, 03:07.
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# Math Revolution DS Expert - Ask Me Anything about GMAT DS  