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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.
_________________
Fabio Skilnik :: GMATH method creator (Math for the GMAT)
Our high-level "quant" preparation starts here: https://gmath.net
GMATH Teacher P
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Joined: 12 Oct 2010
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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 $$1$$?

$$1) h = \frac{1}{2}$$
$$2) x = y = \frac{1}{3}$$ $$x + y + z\,\,\mathop > \limits^? \,\,\,1$$

$$\left( 1 \right)\,\,\,\left\{ \matrix{ \,x\mathop > \limits^{\left( * \right)} \,\,\,h = {1 \over 2} \hfill \cr \,z\mathop > \limits^{\left( * \right)} \,\,\,h = {1 \over 2} \hfill \cr y > 0 \hfill \cr} \right.\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,x + y + z > 1\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\left\langle {{\rm{YES}}} \right\rangle$$ $$\left( 2 \right)\,\,\left\{ \matrix{ \,{\rm{figure}}\,\,{\rm{on}}\,\,{\rm{the}}\,\,{\rm{left}}\,\,\, \Rightarrow \,\,\,\,\,x + y + z\,\,\,\, < \,\,\,\,3\left( {{1 \over 3}} \right)\,\, = \,\,1\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,\left\langle {{\rm{NO}}} \right\rangle \hfill \cr \,{\rm{figure}}\,\,{\rm{on}}\,\,{\rm{the}}\,\,{\rm{right}}\,\,\, \Rightarrow \,\,\,\,\,x + y + z\,\,\,\,\mathop < \limits^{{\rm{as}}\,{\rm{near}}\,\,{\rm{as}}\,\,{\rm{desired}}!} \,\,\,\,{2 \over 3} + {1 \over 3}\left( {\sqrt 2 } \right)\,\, = \,\,{{\sqrt 2 + 2} \over 3}\,\,\,\,\,\left[ { > \,\,\,1} \right]\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,\left\langle {{\rm{YES}}} \right\rangle \hfill \cr} \right.$$

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!
_________________
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: 8235
GMAT 1: 760 Q51 V42 GPA: 3.82

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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.
_________________
Math Revolution GMAT Instructor V
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Attachment: 1.14.png [ 22.82 KiB | Viewed 312 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 [ 22.82 KiB | Viewed 312 times ]

_________________
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]

(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.
_________________
GMATH Teacher P
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Joined: 12 Oct 2010
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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.
_________________
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: 8235
GMAT 1: 760 Q51 V42 GPA: 3.82

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[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
_________________
GMATH Teacher P
Status: GMATH founder
Joined: 12 Oct 2010
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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.
_________________
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: 8235
GMAT 1: 760 Q51 V42 GPA: 3.82

### Show Tags

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.

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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1
[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
_________________
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Joined: 12 Oct 2010
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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.
_________________
Fabio Skilnik :: GMATH method creator (Math for the GMAT)
Our high-level "quant" preparation starts here: https://gmath.net

Originally posted by fskilnik on 16 Jan 2019, 05:59.
Last edited by fskilnik on 16 Jan 2019, 09:02, edited 1 time in total.
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Location: India
Concentration: Sustainability, Marketing
GPA: 4
WE: Marketing (Energy and Utilities)

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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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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

$$x^2-y^2$$ an even integer
would be valid for values when both x & y are either odd or even

#1:
$$x^3-y^3$$ is an even integer
for both x & y being either even or odd ; it would be valid and sufficient

#2:
$$x+y$$ is an even integer

again when both x & y are either even or odd then relation would be valid and sufficinet

for all values we can test with integers : even 2,4 and odd 1,3

IMO D
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8235
GMAT 1: 760 Q51 V42 GPA: 3.82

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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.

_________________
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8235
GMAT 1: 760 Q51 V42 GPA: 3.82

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[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
_________________
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8235
GMAT 1: 760 Q51 V42 GPA: 3.82

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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.

_________________
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8235
GMAT 1: 760 Q51 V42 GPA: 3.82

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1
[Math Revolution GMAT math practice question]

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

$$1) |a+b| = - ( a + b )$$
$$2) |a+b| + 1 = |a| + |b|$$
_________________
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Location: India
Concentration: Sustainability, Marketing
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MathRevolution wrote:
[Math Revolution GMAT math practice question]

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

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

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
GMATH Teacher P
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Joined: 12 Oct 2010
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MathRevolution wrote:
[Math Revolution GMAT math practice question]

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

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

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.
_________________
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: 8235
GMAT 1: 760 Q51 V42 GPA: 3.82

### Show Tags

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.
_________________ Re: Math Revolution DS Expert - Ask Me Anything about GMAT DS   [#permalink] 20 Jan 2019, 18:38

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# Math Revolution DS Expert - Ask Me Anything about GMAT DS   