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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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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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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

[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.

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

[Math Revolution GMAT math practice question]

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

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

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

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

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

\(? = y\)


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


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


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


We follow the notations and rationale taught in the GMATH method.

Regards,
Fabio.

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.

=>

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

[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.

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

[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.

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

[Math Revolution GMAT math practice question]

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

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