## 2015 Grade 10 MTAP Sectoral Level

This is the 2015 Metrobank-MTAP-NCR Math Challenge Sectoral Level for  Grade 10. Please inform me if you see any errors.

More past tests can be found here. If you have old tests, kindly send it to mtapreviewers@gmail.com, so we can share it here with answers/solutions.

Easy

1.) What is the next term in the arithmetic sequence whose first two terms are −18 and then 19?

2.) Express $-3 + \log_2 x + 3 \log_2 y$ as a single logarithm with base $2$.

3.) What should be the value of $c$ if the point where $f(x) = -x^2 + 2x + c$ attains its maximum, is also a zero of $f$?

4.) What is the reciprocal of $\sqrt[3]{9}- 2$ in simplest terms?

5.) If $\log_x25 = 4$, what is $\log_25x$Continue reading “2015 Grade 10 MTAP Sectoral Level”

## 2015 Grade 9 MTAP Sectoral Level

This is the 2015 Metrobank-MTAP-NCR Math Challenge Sectoral Level for  Grade 9. Please inform me if you see any errors.

More past tests can be found here. If you have old tests, kindly send it to mtapreviewers@gmail.com, so we can share it here with answers/solutions.

Easy

1.) If $\displaystyle \frac {x}{y} = 2$, find $\frac {x^2 - y^2}{x^2 + y^2}$.

2.) In a cube with sides of length 1 cm, one vertex is denoted by A. What is the sum of the distances from A to each of the other vertices of the cube?

3.) Find two positive numbers in the ratio 7:12 so that the bigger number exceeds the smaller by 10.

## Grade 9 MTAP 2015 Elimination Questions with Solutions – Part 2

This is the second part (questions 11-20) of the solutions of the Grade 9 MTAP 2015 Elimination Questions. The first part can be read here.

Although reasonable care has been given to make the solution accurate as possible, the solver is also human. Please comment below if you see any errors.

11.) If $x \neq 1$, solve for $x$ in $2 \sqrt{x} + \frac{3}{\sqrt{x}} = 5$.

Solution
Multiplying both sides by $\sqrt{x}$, we obtain

$2x + 3 = 5\sqrt{x}$

Squaring both sides of the equation, we get

$4x^2 + 12x + 9 = 25x$

$4x^2 - 13x + 9 = 0$.

Factoring, we have

$(4x - 9)(x - 1) = 0$

So, $x = \frac{9}{4}$ or $x = 1$.

But from the given above, $x \neq 1$, so the only solution is $x = \frac{9}{4}$

## Grade 9 MTAP 2015 Elimination Questions with Solutions – Part 1

This is the first part (questions 1-10) of the solutions of the Grade 9 MTAP 2015 Elimination Questions. Although much care has been given to make the solution as accurate as possible, the solver is also human. Please comment below if you see any errors.

Also available: Questions 11-20

1.) Simplify $\sqrt{\frac{3}{2}} - \sqrt{\frac{2}{3}}$

Solution

$\sqrt {\dfrac{3}{2}} - \sqrt {\dfrac{2}{3}} = \dfrac {\sqrt {3}}{\sqrt {2}} - \dfrac {\sqrt {2}}{\sqrt {3}} = \dfrac {(\sqrt {3})^2-(\sqrt {2})^2}{\sqrt {2}\sqrt {3}} = \dfrac {1}{\sqrt {6}}$.

Rationalizing the denominator,

$\dfrac {1}{\sqrt {6}} \times \dfrac {\sqrt {6}}{\sqrt {6}} = \dfrac {\sqrt {6}}{6}$.

Answer: $\sqrt{6}/6$

2.) Evaluate $\dfrac{2^0 + 2{^-1}}{2^{-2} + 2^{-3}}$

Solution

$\dfrac {2^0 + 2^{-1}}{2^{-2} + 2^{-3}} = \dfrac {1 + \frac {1}{2}}{\frac {1}{4} + \frac {1}{8}} = \dfrac {\frac {3}{2}}{\frac {3}{8}} = \dfrac {3}{2} \times \dfrac {8}{3} = 4$

3.) Simplify $\sqrt{ \dfrac{1}{9} + \dfrac{1}{16}}$.
$\sqrt{\dfrac{1}{9} + \dfrac {1}{16}} = \sqrt {\dfrac{16 + 9}{(9)(16)}} = \sqrt{\dfrac{25}{(9)(16)}} =\dfrac{5}{(3)(4)} = \dfrac{5}{12}$.