Max@Math Revolution wrote:[Math Revolution GMAT math practice question]

Is 1+x+x^2+x^3+x^4+x^5+x^6 < 1/(1-x)?

1) x>0
2) x<1

\[1 + x + {x^2} + {x^3} + {x^4} + {x^5} + {x^6}\,\,\mathop < \limits^? \,\,\,\frac{1}{{1 - x}}\]
Another beautiful problem, Max. Congrats!
\[\left( 1 \right)\,\,x > 0\,\,\,\left\{ \begin{gathered}
\,{\text{Take}}\,\,x = \frac{1}{2}\,\,\,\, \Rightarrow \,\,\,\left\langle {{\text{YES}}} \right\rangle \,\,\,\,\,\,\,\,\,\,\left[ {\,\,1 + \frac{1}{2} + \frac{1}{4} + \ldots + \frac{1}{{32}} + \frac{1}{{64}}\,\,\, < \,\,\,1\, + 1\,\, = 2 = \frac{1}{{1 - \frac{1}{2}}}\,\,} \right]\,\, \hfill \\
\,{\text{Take}}\,\,x = 2\,\,\,\, \Rightarrow \,\,\,\left\langle {{\text{NO}}} \right\rangle \,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {\,\,1 + 2 + \ldots + {2^6}\,\,\,\mathop < \limits^? \,\, - 1\,\,} \right]\,\,\,\, \hfill \\
\end{gathered} \right.\]



\[\left( 2 \right)\,\,x < 1\,\,\,\left\{ \begin{gathered}
\,\left( {\operatorname{Re} } \right){\text{Take}}\,\,x = \frac{1}{2}\,\,\,\, \Rightarrow \,\,\,\left\langle {{\text{YES}}} \right\rangle \hfill \\
\,{\text{Take}}\,\,x = - 1\,\,\,\, \Rightarrow \,\,\,\left\langle {{\text{NO}}} \right\rangle \,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {\,\,1 + \underbrace {\left( { - 1} \right) + {{\left( { - 1} \right)}^2}}_{ = \,\,0} + \underbrace {{{\left( { - 1} \right)}^3} + {{\left( { - 1} \right)}^4}}_{ = \,\,0} + \underbrace {{{\left( { - 1} \right)}^5} + {{\left( { - 1} \right)}^6}}_{ = \,\,0}\,\,\,\mathop < \limits^? \,\,\frac{1}{2}\,\,} \right]\,\,\,\, \hfill \\
\end{gathered} \right.\]
\[\left( {1 + 2} \right)\,\,\,\,0 < x < 1\]
\[1 + x + {x^2} + {x^3} + {x^4} + {x^5} + {x^6}\,\,\mathop < \limits^? \,\,\,\frac{1}{{1 - x}}\,\,\,\,\,\,\,\mathop \Leftrightarrow \limits^{1 - x\, > \,\,0} \,\,\,\,\,\,\left( {1 - x} \right)\left( {1 + x + {x^2} + {x^3} + {x^4} + {x^5} + {x^6}} \right)\,\,\,\mathop < \limits^? \,\,\,1\,\]
\[\left( {1 - x} \right)\left( {1 + x + {x^2} + {x^3} + {x^4} + {x^5} + {x^6}} \right)\,\,\, = \,\,\,1 - {x^7}\,\,\mathop < \limits^? \,\,\,1\,\,\,\,\,\mathop \Rightarrow \limits^{0\, < \,x\, < \,1} \,\,\,\,\,\,\left\langle {{\text{YES}}} \right\rangle \]


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

Regards,
Fabio.

Max@Math Revolution wrote:[Math Revolution GMAT math practice question]

Is 1+x+x^2+x^3+x^4+x^5+x^6 < 1/(1-x)?

1) x>0
2) x<1

Alternate way: ("Algebraic only" solution)
\[1 - {x^7} = \left( {1 - x} \right)\left( {1 + x + {x^2} + {x^3} + {x^4} + {x^5} + {x^6}} \right)\]
\[1 + x + {x^2} + {x^3} + {x^4} + {x^5} + {x^6}\,\,\mathop < \limits^? \,\,\,\frac{1}{{1 - x}}\,\,\,\,\,\mathop \Leftrightarrow \limits^{ \cdot \,\,\left( {1 - x} \right)} \,\,\,\,\,\,\left\{ \begin{gathered}
\,1 - {x^7}\,\,\mathop < \limits^? \,\,1\,\,\,,\,\,{\text{when}}\,\,\left( {1 - x} \right) > 0\,\,,\,\,{\text{i}}{\text{.e}}{\text{.}}\,,\,\,x < 1 \hfill \\
\,1 - {x^7}\,\,\mathop > \limits^? \,\,1\,\,\,,\,\,{\text{when}}\,\,\left( {1 - x} \right) < 0\,\,,\,\,{\text{i}}{\text{.e}}{\text{.}}\,,\,\,x > 1 \hfill \\
\end{gathered} \right.\,\,\,\,\,\,\,\left( * \right)\]
\[\left( 1 \right)\,\,x > 0\,\,\,\left\{ \begin{gathered}
\,{\text{Take}}\,\,\,x = \frac{1}{2}\,\,\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,\,1 - {x^7}\,\,\mathop < \limits^? \,\,1\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\left\langle {{\text{YES}}} \right\rangle \,\, \hfill \\
\,{\text{Take}}\,\,\,x = 2\,\,\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,\,1 - {x^7}\,\,\mathop > \limits^? \,\,1\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\left\langle {{\text{NO}}} \right\rangle \,\,\,\, \hfill \\
\end{gathered} \right.\]
\[\left( 2 \right)\,\,x < 1\,\,\,\left\{ \begin{gathered}
\,\left( {\operatorname{Re} } \right){\text{Take}}\,\,x = \frac{1}{2}\,\,\,\, \Rightarrow \,\,\,\left\langle {{\text{YES}}} \right\rangle \hfill \\
\,{\text{Take}}\,\,x = - 1\,\,\,\, \Rightarrow \,\,\,\,\,1 - {x^7}\,\,\mathop < \limits^? \,\,1\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\left\langle {{\text{NO}}} \right\rangle \,\,\,\, \hfill \\
\end{gathered} \right.\]
\[\left( {1 + 2} \right)\,\,\,\,0 < x < 1\,\,\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,\,\,\,1 - {x^7}\,\,\mathop < \limits^? \,\,1\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\left\langle {{\text{YES}}} \right\rangle \,\]

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

Regards,
Fabio.

Checking if the left hand side is < right hand side
LHS<RHS
$$LHS=1+x+x^2+x^3+x^4+x^5+x^6$$
$$RHS=\frac{1}{1-x}$$
$$1+x+x^2+x^3+x^4+x^5+x^6\ <\ \frac{1}{1-x}$$
$$STATEMENT\ 1\ -->\ x>0$$
$$if\ x=2,\
$$1+2+2^2+2^3+2^4+2^5+2^{6\ }<\ \frac{1}{1-2}$$
$$1+2+4+8+16+32+64<\frac{1}{-1}$$
$$127<-1$$
$$LHS>RHS\ \&\ 127\ >-1$$
$$If\ \ x=0.5$$
$$1+0.5\ +0.5^2+0.5^3+0.5^4+0.5^4+0.5^5+0.5^6<\frac{1}{1-0.5}$$
$$1+0.5+0.25+0.125+0.0625+0.03125+0.015625<\frac{1}{0.5}$$
$$1.984375<2$$
$$LHS<RHS\ \&1.984375<2$$
$$statement\ 1\ is\ NOT\ SUFFICIENT$$
$$STATEMENT\ 2\ =x<1
$$ $$if\ \ x\ =-1$$
$$1+\left(-1\right)+\left(-1\right)^2+\left(-1\right)^3+\left(-1\right)^4+\left(-1\right)^5+\left(-1\right)^6<\frac{1}{1-\left(-1\right)}$$
$$1-1+1-1+1-1+1<\frac{1}{1-\left(-1\right)}$$
$$1-1+1-1+1-1+1\ <\frac{1}{2}$$ $$1-1+1-1+1-1+1\ <\frac{1}{2}$$
$$1<\frac{1}{2}$$
$$LHS\ >RHS\ \ and\ 1>\frac{1}{2}$$

if x=0.5
$$1+0.5+0.5^2+0.5^3+0.5^4+0.5^5+0.5^6\ <\ \frac{1}{1-0.5}$$
$$1+0.5+0.25+0.125+0.0625+0.03125+0.015625\ <\ \frac{1}{0.5}$$
1.984375 < 2

LHS<RHS and 1.984375 < 2
Statement 2 is NOT SUFFICIENT.
Combine statement 1 and 2 together ---->>>
Statement 1 = x>0
statement 2= x<0
From this, we can deduce that
0<x<1

This means that 'x' is greater than 0 and less than 1. Definitely, 'x' will always be a positive integer in the range 0.1 and 0.9.

The question stem is to find cut if
$$1+x+x^2+x^3+x^4+x^5+x^6\ <\ \frac{1}{1-x}$$
if x=0.1
$$1+0.1+0.1^2+0.1^3+0.1^4+0.1^5+0.1^6\ <\ \frac{1}{1-0.1}$$
1.111111 < 1.1111111111

LHS is less than right hand side

if x=0.9
$$1+0.9+0.9^2+0.9^3+0.9^4+0.9^5+0.9^6\ <\ \frac{1}{1-0.9}$$
4.217031 < 10

Hence, the two statement together are SUFFICIENT

OPTION C IS THEREFORE CORRECT.....

Max@Math Revolution wrote:[Math Revolution GMAT math practice question]

Is 1+x+x^2+x^3+x^4+x^5+x^6 < 1/(1-x)?

1) x>0
2) x<1

CRITICAL POINTS in an equality occur when the two sides are EQUAL or the inequality is undefined.
Here, the two statements provide the critical points for the inequality in the prompt:
When x=0, the two sides are equal.
When x=1, the inequality is undefined.
To determine the range(s) where the left side is LESS than the right side, test one value to the left and right of each critical point.

Case 1: x > 1
If x=2, the left side is positive, while the right side is negative.
Since the right side is greater than the left side, x>0 is not a valid range.
Case 2: 0 < x < 1
If x=0.99, the left side is relatively small, while the right side = 1/(1-0.99) = 1/0.01 = 100/1 = 100.
Since the the left side is less than the right side, 0 < x < 1 is a valid range.
Case 3: x < 0
If x=-1, the left side = 1, while the right side = 1/2.
Since the left side is greater than the right side, x<0 is not a valid range.

The inequality is valid only when 0 < x < 1.
Question stem, rephrased:
Is 0 < x < 1?

Clearly, neither statement alone is sufficient.
When the statements are combined, 0 < x < 1.
SUFFICIENT.

The correct answer is C.

=>

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 1+x+x^2+x^3+x^4+x^5+x^6 < 1/(1-x) is equivalent to 0 < x < 1 as shown below:

For x ≠ 1,
1+x+x^2+x^3+x^4+x^5+x^6 < 1/(1-x)
=> (1+x+x^2+x^3+x^4+x^5+x^6)(1-x)^2 < (1-x)
=> (1 - x^7)(1 - x) < 1 - x
=> 1 - x^7 - x +x^8 < 1 - x
=> - x^7 + x^8 < 0
=> x^7( x - 1 ) < 0
=> x( x - 1 ) < 0
=> 0 < x < 1

Since both conditions must be applied together to obtain this inequality, both conditions 1) & 2) are sufficient, when applied together.

Therefore, C is the answer.
Answer: C