Asian Science And Math Olympiad (ASMO) 2018 For Grade 10

Asian Science and Maths Olympiad (ASMO) is a competition platform designed to challenge and evaluate student’s knowledge in Mathematics and Science at their grade level. The questions in the Olympiad will stretch their knowledge and understanding of the concepts. Our syllabus fits nicely into the syllabus that concentrates on non-routine problem-solution to prepare the students for the competition. With the expansion of STEM education worldwide, ASMO certainly answers the need of it. Students will be well prepared with the skills to meet the science and technology challenges.

In Malaysia, ASMO is officially endorsed by Ministry of Education and all participants will obtain curriculum marks. In 2018 alone, Asian Science and Mathematics Olympiad has received 70,000 entries from across the ASEAN countries. We are targeting for the number to increase at 80,000 for 2019.

We are also proud to present that ASMO International is a new effort by ASMO Malaysia which started in 2017 in Pattaya, Thailand. When it was initially launched, the competition was setup via collaboration with ASMOPSS and ASMO Thai was the host for the competition. In 2018, Malaysia has become the host for the competition and it was participated by 10 Asian countries.

The idea of opening up a new competition platform which is ASMO International is to expand the level of competition and to provide more opportunities for primary and secondary school students to experience international engagement. (sc : http://asmo2u.com/about-us)

Berikut ini problems and solution ASMO 2018 grade 10

1. \(\log_{\frac{2}{3}}(\frac{9}{4}) + \log_4 (\sqrt 2) + \log_2(\sqrt[4]8)+2^{\frac{1}{\log_4 2}}\)

\(\log_{\frac{2}{3}}(\frac{9}{4}) = \log_{\frac{2}{3}}(\frac{3}{2})^2= \log_{\frac{2}{3}}(\frac{2}{3})^{−2}= −2\)

\(\log_4 \sqrt 2 = \log_{2^2} (2)^{\frac{1}{2}} =\frac{\frac{1}{2}}{2}=\frac{1}{4}\)

\(\log_2 \sqrt[4] 8 = \log_2 2^{\frac{3}{4}} =\frac{3}{4}\)

\(2^{\frac{1}{\log_4 2}} = 2^\frac{1}{\frac{1}{2}}=2^2=4\)

Jadi \(\log_{\frac{2}{3}}(\frac{9}{4}) + \log_4 (\sqrt 2) + \log_2(\sqrt[4]8)+2^{\frac{1}{\log_4 2}}= −2 +\frac{1}{4}+\frac{3}{4}+ 4 = 3\)

2. How many possible value(s) of X such that integer 23X8 is divisible by 8?

Syarat habis dibagi 8 adalah 3 angka terakhir habis dibagi 8. Tiga digit terakhir yaitu 3X8 habis dibagi 8, nilai X yang memenuhi adalah 2 dan 6. Jadi banyak bilangan X yang memenuhi adalah 2 bilangan.

3. If \(f(x)=x+2\sqrt x + 1\) calculate the value of \(\sqrt{f^{2016}(4)} \)

not yet available

4. How many solution for \(x\) in equation \(\log x = x^2 – 1\)?

\(\log 𝑥 = 𝑥^2 − 1\)
\(⇒ \log 𝑥 = 𝑥^2 − \log 10\)
\(⇒ \log 𝑥 + \log 10 = 𝑥^2\)
\(⇒ \log 10𝑥 = \log 10^{𝑥^2}\)
\(⇒ 10𝑥 = 10^{𝑥^2}\)

Jadi yang memenuhi hanya \(1\) nilai \(x\) yaitu \(𝑥 = 1\)

5. Nonlinear simultaneous equation \(\begin{cases}
x^2 + xy – 8x = 3\\
y^2 + xy -8y = 6\\
\end{cases}\) has 2 solution sets. What will be the value of \(\frac{y}{x}\) for those solution sets?

Jumlahkan kedua persamaan
\(𝑥^2 + 𝑦^2 + 2𝑥𝑦 − 8𝑥 − 8𝑦 = 9\)
\((𝑥 + 𝑦)^2 − 8(𝑥 + 𝑦) − 9 = 0\)
\(((𝑥 + 𝑦) − 9)((𝑥 + 𝑦) + 1) = 0\)
\((𝑥 + 𝑦) = 9 𝑎𝑡𝑎𝑢\; (𝑥 + 𝑦) = −1\)
Untuk \(𝑥 + 𝑦 = 9 ⇒ 𝑦 = 9 − 𝑥\) subtitusi kepersamaan pertama
\(𝑥^2 + 𝑥𝑦 − 8𝑥 = 3\)
\(𝑥^2 + 𝑥(9 − 𝑥) − 8𝑥 − 3 = 0\)
\(𝑥^2 + 9𝑥−𝑥^2 − 8𝑥 − 3 = 0\)
\(𝑥 = 3\)
Solusi \((x,y)\) yang menuhi adalah \((3, 6)\)
Untuk \(𝑥 + 𝑦 = −1 ⇒ 𝑥 = −1 − 𝑦\) subtitusi kepersamaan kedua
\(𝑦^2 + 𝑥𝑦 − 8𝑦 = 6\)
\(𝑦^2 + 𝑦(−1 − 𝑦) − 8𝑦 − 6 = 0\)
\(𝑦^2 − 𝑦−𝑦^2 − 8𝑦 − 6 = 0\)
\(−9𝑦 = 6\)
\(𝑦 = −\frac{6}{9}= −\frac{2}{3}\)
Solusi \((x,y)\) yang menuhi adalah \((−\frac{1}{3}, −\frac{2}{3})\)
Jadi nilai dari \(\frac{𝑦}{𝑥}\) adalah \(\frac{6}{3}=2\), atau \(-\frac{2}{3}÷(−\frac{1}{3})= 2\)

6. If \(\cos(x )+\cos(y)+cos(z)=0\) and \(\sin(x)+\sin(y)+\sin(z)=0\)
What is the value of \(\cos(x-y)+\cos(y-z)+\cos(z-x)\)?

Misalkan
\(\cos x = 𝑎, \cos y = 𝑏, \cos z = 𝑐\) dan \(\sin x = 𝑝 , \sin y = 𝑞, \sin z = 𝑟\)
berdasarkan identitas trigonometri \(𝑎^2 + 𝑝^2 = 𝑏^2 + 𝑞^2 = 𝑐^2 + 𝑟^2 = 1\)
\((𝑎 + 𝑏 + 𝑐)^2 = 𝑎^2 + 𝑏^2 + 𝑐^2 + 2(𝑎𝑏 + 𝑏𝑐 + 𝑎𝑐) = 0 …(1)\)
\((𝑝 + 𝑞 + 𝑟)^2 = 𝑝^2 + 𝑞^2 + 𝑟^2 + 2(𝑝𝑞 + 𝑞𝑟 + 𝑝𝑟) = 0 …(2)\)
Jumlahkan persamaan \((1)\) dan persamaan \((2)\)
\(𝑎^2 + 𝑝^2 + 𝑏^2 + 𝑞^2 + 𝑐^2 + 𝑟^2 + 2(𝑎𝑏 + 𝑝𝑞 + 𝑏𝑐 + 𝑞𝑟 + 𝑎𝑐 + 𝑝𝑟) = 0\)
\(3 + 2(𝑎𝑏 + 𝑝𝑞 + 𝑏𝑐 + 𝑞𝑟 + 𝑎𝑐 + 𝑝𝑟) = 0\)
\(𝑎𝑏 + 𝑝𝑞 + 𝑏𝑐 + 𝑞𝑟 + 𝑎𝑐 + 𝑝𝑟 =\frac{−3}{2}\)
Jadi nilai dari
\(\cos(x-y) + \cos(y-z) + \cos (z-x) = 𝑎𝑏 + 𝑝𝑞 + 𝑏𝑐 + 𝑞𝑟 + 𝑎𝑐 + 𝑝𝑟 =\frac{−3}{2}\)

7. Solve for \(x\) in the equation \(\log_3(1-2.3^x)=2x+1\)

\(\log_3(1 − 2 · 3^𝑥 ) = 2𝑥 + 1\)
\(\log_3(1 − 2 · 3^𝑥 ) = \log_3 3^{2𝑥+1}\)
\(1 − 2 · 3^𝑥 = 3^{2𝑥+1}\)
\(1 − 2 · 3^𝑥 = 3^{2𝑥}. 3\)

Misalkan \(𝑎 = 3^𝑥\) maka

\(3𝑎^2 + 2𝑎 − 1 = 0\)
\((3𝑎 − 1)(𝑎 + 1) = 0\)
\(𝑎 =\frac{1}{3}\; atau\; 𝑎 = −1(TM)\)

Jadi \(3^𝑥 =\frac{1}{3}= 3^{−1} ⇒ 𝑥 = −1\)

8. Triangle ABC with coordinate A (-1,2) , B(3,5) and C(5, -4). If a straight line pass through point (3,5) divided the area of triangle ABC equally. What is the intersection of that line with line AC?

Agar \([𝐴𝐵𝐷] = [𝐵𝐷𝐶]\) maka titik \(C\) merupakan titik tengah dari garis \(AC\)
Jadi titik \(D\) adalah \(\left(\frac{−1+5}{2},\frac{2+(−4)}{2}\right) = (2, −1)\)

9. A point X is randomly placed along on a straight line AB, divided that line into two line segments (AX and XB). If line AB is 10 cm in length, what is the probability to have point X such that the difference of length for AX and XB is less than 2 cm?

not yet available

10. As shown in the following diagram is a cube ABCDEFGH with length of edge being 4cm. What is the angle between plane ΔAHC and plane ΔAGE?

not yet available

Problem And Solution SEAMO 2017 Paper E
Problems And Solutions Future Intelligence Student Olympiad (9-10 Grades)

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