2015 Grade 9 MTAP Solution(1-5)

In order to help students preparing for 2016 MMC, we will be providing the solutions to the 2015 MMC for all year level in high school. In this post, solutions to problem 1 to 5 for grade 9 MMC will be illustrated.

Disclaimer: MTAP and Metrobank is not affiliated in this website. This is a standalone site aiming to help self-motivated students to review for the upcoming math challenge. Solutions presented here are solely from the author of this post.

Problem 1:

Simplify:       \sqrt{\displaystyle\frac{3}{2}}-\sqrt{\displaystyle\frac{2}{3}}

Solution:

We can rewrite the expression as follows,

=\displaystyle\frac{\sqrt{3}}{\sqrt{2}}-\displaystyle\frac{\sqrt{2}}{\sqrt{3}}

By cross multiplication,

=\displaystyle\frac{\sqrt{3\cdot 3}-\sqrt{2\cdot 2}}{\sqrt{3\cdot 2}}

=\displaystyle\frac{\sqrt{9}-\sqrt{4}}{\sqrt{6}}

=\displaystyle\frac{3-2}{\sqrt{6}}

=\displaystyle\frac{1}{\sqrt{6}}

Since the denominator is an irrational number, we rationalize it by multiplying the expression by 1 to preserve its identity.

=\displaystyle\frac{1}{\sqrt{6}}\cdot \displaystyle\frac{\sqrt{6}}{\sqrt{6}}

=\displaystyle\frac{\sqrt{6}}{\sqrt{36}}

=\boxed{\displaystyle\frac{\sqrt{6}}{6}}

Problem 2:

Evaluate:    \displaystyle\frac{2^0+2^{-1}}{2^{-2}+2^{-3}}

Solution:

By long method, we can convert negative exponents to fractions and simplify it using fractions.

Recall that  x^{-m}=\displaystyle\frac{1}{x^m}

We do the same in the problem,

\displaystyle\frac{2^0+2^{-1}}{2^{-2}+2^{-3}}

=\displaystyle\frac{1+\displaystyle\frac{1}{2}}{\displaystyle\frac{1}{2^2}+\displaystyle\frac{1}{2^3}}

Simplify numerator and denominator as follows,

=\displaystyle\frac{\displaystyle\frac{2+1}{2}}{\displaystyle\frac{2^2+2^3}{2^2\cdot 2^3}}

=\displaystyle\frac{\displaystyle\frac{3}{2}}{\displaystyle\frac{4+8}{4\cdot 8}}

=\displaystyle\frac{\displaystyle\frac{3}{2}}{\displaystyle\frac{12}{32}}

=\displaystyle\frac{\displaystyle\frac{3}{2}}{\displaystyle\frac{3}{8}}

=\displaystyle\frac{3}{2}\cdot\displaystyle\frac{8}{3}

=\displaystyle\frac{8}{2}

=\boxed{4}

Problem 3:

Simplify:  \sqrt{\displaystyle\frac{1}{9}+\displaystyle\frac{1}{16}}

Solution:

We can just do it by using fractions.

\sqrt{\displaystyle\frac{1}{9}+\displaystyle\frac{1}{16}}

=\sqrt{\displaystyle\frac{16+9}{9\cdot 16}}

=\sqrt{\displaystyle\frac{25}{9\cdot 16}}

Extracting the perfect squares,

=\displaystyle\frac{5}{3\cdot 4}

=\boxed{\displaystyle\frac{5}{12}}

Problem 4:

Simplify:     \displaystyle\frac{x^{-1}-y^{-1}}{x^\frac{1}{3}-y^\frac{1}{3}}

Solution:

rewrite the expression as follows,

=\displaystyle\frac{1}{\sqrt[3]{x}-\sqrt[3]{y}}\cdot (\displaystyle\frac{1}{x}-\displaystyle\frac{1}{y})

=\displaystyle\frac{y-x}{xy(\sqrt[3]{x}-\sqrt[3]{y})}

Rationalizing the denominator,

=\displaystyle\frac{y-x}{xy(\sqrt[3]{x}-\sqrt[3]{y})}\cdot\displaystyle\frac{\sqrt[3]{x^2}+\sqrt[3]{xy}+\sqrt[3]{y^2}}{\sqrt[3]{x^2}+\sqrt[3]{xy}+\sqrt[3]{y^2}}

=\displaystyle\frac{(y-x)(\sqrt[3]{x^2}+\sqrt[3]{xy}+\sqrt[3]{y^2})}{xy([\sqrt{x}]^3-[\sqrt{y}]^3)}

=\displaystyle\frac{(y-x)(\sqrt[3]{x^2}+\sqrt[3]{xy}+\sqrt[3]{y^2})}{xy(x-y)}

=\displaystyle\frac{-1(x-y)(\sqrt[3]{x^2}+\sqrt[3]{xy}+\sqrt[3]{y^2})}{xy(x-y)}

x-y cancels out leaving,

=\displaystyle\frac{-1(\sqrt[3]{x^2}+\sqrt[3]{xy}+\sqrt[3]{y^2})}{xy}

=\boxed{\displaystyle\frac{-\sqrt[3]{x^2}-\sqrt[3]{xy}-\sqrt[3]{y^2}}{xy}}

Problem 5:

If a, b, and c are real numbers such that         \dfrac{b}{a}=5     and     \dfrac{b}{c}=2 Find the value of      \displaystyle\frac{a+b}{b+c}.

Solution:

Given the ratio of b and a and b and c, we find ways that b must be of the same value in both proportions. By multiplying a number that is equal to 1, such condition can be attained.

  \dfrac{b}{a}=\dfrac{5}{1}

  \dfrac{b}{a}=\dfrac{5}{1}\cdot\dfrac{2}{2}

  \dfrac{b}{a}=\dfrac{10}{2}

For the other ratio:

\dfrac{b}{c}=2

\dfrac{b}{c}=\dfrac{2}{1}

\dfrac{b}{c}=\dfrac{2}{1}\cdot\dfrac{5}{5}

\dfrac{b}{c}=\dfrac{10}{5}

Since we already have a uniform values for a, b, and c. we can say that a:b:c=2:10:5

Solving for the value of   \displaystyle\frac{a+b}{b+c}.

\displaystyle\frac{a+b}{b+c}

=\displaystyle\frac{2+10}{10+5}

=\displaystyle\frac{12}{15}

=\boxed{\displaystyle\frac{4}{5}}

Dan

Dan

Blogger and a Math enthusiast. Has no interest in Mathematics until MMC came. Aside from doing math, he also loves to travel and watch movies.
Dan

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