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GMAT Strategies: Articles from GMATQuantum Blog

Friday
Apr072017

GMAT Problem Solving 84: Absolute value identity

Try this GMAT problem solving on absolute values. This problem can certainly be done by plugging in specific values for \(x\). However, I would like you to also try solving it algebraically which involves using absolute value identity and how to remove the absolute value sign of an algebraic expression. These two steps are often seen on the GMAT, especially in data sufficiency where you need to be comfortable with algebraic manipulation.

Given \(x<0\), \(|x- \sqrt{(x-1)^2}|\) equals
  1. \(\quad 1\)

  2. \(\quad 1-2x\)

  3. \(\quad -2x-1\)

  4. \(\quad 1+2x\)

  5. \(\quad 2x-1\)

Tuesday
Mar212017

Exponents: All you need to know for GMAT test

In this post I first summarize all the exponent rules that you need to know for the GMAT, followed by how these exponent rules are tested on the GMAT.

Exponent Rules: Summary

$$ \begin{array}{|c|c|c|} \hline \textbf{Rule} & \textbf{Arithmetic example} & \textbf{Algebraic example} \\ \hline (a^m) (a^n) = a^{m+n} & (5^{3})(5^5) = 5^{3+5}=5^8 & (x^6)(x^{-4})=x^{6+(-4)} = x^2 \\ \hline (a^m)^n = a^{mn} = (a^n)^m &(2^2)^3=2^{6}=64 &(3z^2)^3=(3^3)(z^2)^3=27z^{6} \\ \hline \dfrac{a^m}{a^n} = a^{m-n} & \dfrac{7^{8}}{7^5} = 7^{8-5}=7^3 & \dfrac{x^5}{x^{-4}}=x^{5-(-4)} = x^9\\ \hline a^{-n} = \dfrac{1}{a^n} \quad (a \neq 0)& 2^{-3} = \dfrac{1}{2^3} = \dfrac{1}{8} & x^{-3} = \dfrac{1}{x^3} \\ \hline a^0=1 & (-5)^0=1 & x^0=1 \quad (x \neq 0) \\ \hline (a \times b)^n = (a^n)(b^n) & (2\times 5)^6=(2^6)(5^6) =10^6 & (2x)^3 = (2)^3(x)^3= 8x^3\\ \hline \left(\dfrac{a}{b}\right)^n = \dfrac{a^n}{b^n} & \left(\dfrac{3}{2}\right)^3 = \dfrac{3^3}{2^3}=\dfrac{27}{8} & \left(\dfrac{2x}{3y^{2}}\right)^3 = \dfrac{2^3 x^3}{3^{3} (y^{2})^3}= \dfrac{8x^3}{27 y^{6}}\\ \hline \end{array} $$

Exponents: Arithmetic and Algebraic Manipulation

You will be expected to rewrite and manipulate arithmetic and algebraic expressions containing exponent terms. Here I list examples of common exponent manipulations tested on the GMAT: $$8^x = (2^3)^x = 2^{3x}$$ $$27^x = (3^3)^x = 3^{3x}$$ $$3^{6x} = (3^2)^{3x} = 9^{3x}$$ $$3^{6x} = (3^3)^{2x} = 27^{2x}$$ $$9(3^x) = (3^2)(3^x) = 3^{x+2} \quad \text{Note:} \quad 9(3^x) \neq 27^{x}$$ $$8(2^x) = (2^3)(2^x) = 2^{x+3} \quad \text{Note:} \quad 8(2^x) \neq 16^{x}$$ $$ \dfrac{9^x}{27^y} = \dfrac{(3^2)^x}{(3^3)^y} = \dfrac{3^{2x}}{3^{3y}} = 3^{2x-3y} $$ $$\dfrac{3^{x+1}}{3} =\dfrac{3^{x+1}}{3^1}= 3^{(x+1)-1} = 3^x \quad \text{Note:} \quad \dfrac{3^{x+1}}{3} \neq x+1$$ $$2^{8} \times 5^{9} = 2^{8} \times 5^{8} \times 5^{1} = (2 \times 5)^8 \times 5 = 5(10^8) $$ $$3(6)^8 = 3(2 \times 3)^8 = 3 \times 2^8 \times 3^8 = 2^8 \times 3^8 \times 3^1 = 2^8 \times 3^9 $$ $$(2^{32})(3^{31}) + (2^{31})(3^{32}) + (2^{31})(3^{31}) = 2^{31}3^{31} (2^{1} + 3^{1} + 1) = (2 \times 3)^{31}(6) = (6)^{31}(6)^{1} = 6^{32}$$

Exponents: Sum and Difference of Powers

There is no general exponent rule when adding powers of numbers that have the same base, however, there are cases where simplification is possible using other rules of arithmetic. In general, if you see a question on the GMAT that asks you to add terms with the same bases, the best approach is to factor the largest common term, and in most cases the resulting terms will collapse to something simple. $$ 2^{22} + 2^{22} = 2^{22}(1 + 1) = 2^{22}(2) = (2^{22})(2^{1}) = 2^{22+1} = 2^{23} $$ $$ 2^4 + 2^4 + 2^4 + 2^4 = 2^4(1 + 1 + 1 + 1) = 2^4(4) = 2^4(2^2) = 2^{4+2} = 2^6 $$ $$ 3^{33} + 3^{33} + 3^{33} = 3^{33}(1 + 1+1) = 3^{33}(3) = (3^{33})(3^{1}) = 3^{33+1} = 3^{34} $$ $$ 7^9 - 7^8 = 7^8(7 - 1) = 6(7^8)$$ $$ \require{cancel} \displaystyle \frac{10^{11}+10^{12}+10^{13}}{10^6+10^7+10^8} = \frac{10^{11}(1+10+10^{2})}{10^{6}(1+10+10^2)} = \frac{10^{11}\cancel{(1+10+10^{2})}}{10^{6}\cancel{(1+10+10^2)} } =\frac{10^{11}}{10^6}=10^{11-6}=10^{5} $$ $$ 2^{x} + 2^{x+1} = 2^x + (2^{x})(2^1) = 2^{x}(1+2) =3(2^x)$$ $$ 5^{x} - 5^{x-2} = 5^{x-2}(5^2 - 1) = 24(5^{x-2}) $$ $$\dfrac{1}{2^{9}} - \dfrac{1}{2^{10}} = \dfrac{1}{2^{9}}\left(1 - \dfrac{1}{2}\right) = \dfrac{1}{2^{9}}\left(\dfrac{1}{2}\right) = \dfrac{1}{2^{10}} $$ $$\dfrac{1}{2^{8}} + \dfrac{1}{2^{9}} + \dfrac{1}{2^{9}} = \dfrac{1}{2^{9}}\left(2 + 1 + 1\right) = \dfrac{4}{2^{9}} = \dfrac{2^2}{2^{9}} = \dfrac{1}{2^7}$$ $$\dfrac{1}{2^{x}} + \dfrac{1}{2^{x}} = \dfrac{2}{2^{x}} = \dfrac{1}{2^{x-1}}$$

Exponents: Common Mistakes

  • \((2^4)(2^4) \neq 2^{16}\), instead \((2^4)(2^4)=2^{4+4} = 2^{8}\).
  • \((2^{2})(2^{x}) \neq 2^{2x} \), instead \((2^{2}) (2^{x}) = 2^{2+x} \).
  • \((3^{x})^2 \neq (3^{x^2})\), instead \((3^{x})^2 = 3^{2x}\).
  • \(10^8-10^7 \neq 10^1\), instead \(10^8 - 10^7 = 10(10^7) - 10^7 = 10^7(10-1) = 9(10^7)\).
  • \([3^{x-2}]^3 \neq 3^{(3x-2)}\), instead \([3^{x-2}]^3 = [3^{3(x-2)}]= 3^{3x-6}\).
  • \(\displaystyle \frac{2^{22} + 4^{11}}{2^{22}} \neq \frac{\cancel{2^{22}} + 4^{11}}{\cancel{2^{22}}} = 4^{11}\), instead \(\displaystyle \frac{2^{22} + 4^{11}}{2^{22}} = \frac{{2^{22}} + (2^2)^{11}}{2^{22}} = \frac{{2^{22}} + 2^{2\times 11}}{2^{22}}= \frac{2^{22} + 2^{22}}{2^{22}}= \frac{2^{22}(1 + 1)}{2^{22}} = \frac{\cancel{2^{22}}(2)}{\cancel{2^{22}}} = 2\)
  • \(20(20^{599}) \neq 400^{599}\), instead \(20(20^{599})=20^{600} = (20^{2})^{300} = 400^{300}\).
Thursday
Mar092017

GMAT Problem Solving 83: Hexagon and familiar triangles

Try this GMAT problem solving on hexagons and how to create familiar right triangles to find the length of sides. The video explanation is attached after the problem statement.