Asian Science And Math Olympiad (ASMO) 2019 For Grade 11

ASMO

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 2019 grade 11


1. Let \(a\) and \(b\) be the roots of \(x^2 + 2000x +1= 0\) and \(c\) and \(d\) be the roots of \(x^2 –  2008x +1= 0\). Determine the value of \((a + c)(b + c)(a – d)(b – d)\).


Berdasarkan dalil vieta pada \(𝑥^2 + 2000𝑥 + 1 = 0\)
\(𝑎 + 𝑏 = −2000\) dan \(𝑎𝑏 = 1\)
Berdasarkan dalil vieta pada \(𝑥^2 − 2008𝑥 + 1 = 0\)
\(𝑐 + 𝑑 = 2008\) dan \(𝑐𝑑 = 1\)
berlaku juga \(𝑐^2 − 2008𝑐 + 1 = 0\) dan \(𝑑^2 − 2008𝑑 + 1 = 0\)
Selanjutnya
\(\begin{align}
(𝑎 + 𝑐)(𝑏 + 𝑐)(𝑎– 𝑑)(𝑏– 𝑑)&= (𝑎𝑏 + 𝑎𝑐 + 𝑏𝑐 + 𝑐^2)(𝑎𝑏 − 𝑎𝑑 − 𝑏𝑑 + 𝑑^2)\\
&= (1 + 𝑐(𝑎 + 𝑏) + 𝑐^2)(1 − 𝑑(𝑎 + 𝑏) + 𝑑^2)\\
&= (1 − 2000𝑐 + 𝑐^2)(1 + 2000𝑑 + 𝑑^2)\\
&= (1 − 2008𝑐 + 𝑐^2 + 8𝑐)(1 − 2008𝑑 + 𝑑^2 + 4008𝑑)\\
&= (8𝑐)(4008𝑑)\\
&= 32064𝑐𝑑\\
&= 32064\\
\end{align}\)


2. Determine the value of the expression \(3\sqrt{5\sqrt{3 \sqrt{5 \sqrt{…}}}}\).


\(3\sqrt{5\sqrt{3 \sqrt{5 \sqrt{…}}}}=a\)
\(⇒3\sqrt{5\sqrt{a}}=a\)
\(⇒9(5\sqrt{a}=a^2\)
\(⇒45\sqrt{a}=a^2\)
\(⇒2025a=a^3\)
\(⇒2025=a^2\)
\(⇒a=\sqrt{2025}=45\)


3. What are all the two-digit positive integers in which the difference between the integer and the product of its two digits is 12?


Misalkan bilangan dua digitnya adalah \( \overline{ab}\), 

\(\overline{𝑎b} − 𝑎𝑏 = 12\) atau \(𝑎𝑏 − \overline{ab} = 12\)

Kemungkinan 1
\(\overline{𝑎b} − 𝑎𝑏 = 12\)
\(⇒10𝑎 + 𝑏 − 𝑎𝑏 = 12\)
\(⇒(10 − 𝑏)(𝑎 − 1) + 10 = 12\)
\(⇒(10 − 𝑏)(𝑎 − 1) = 2 = 1 × 2 = 2 × 1\)
Untuk \((10 − 𝑏)(𝑎 − 1) = 1 × 2\), diperoleh \(𝑏 = 9, 𝑎 = 3\), bilangan dua digit yang memenuhi adalah 39
Untuk \((10 − 𝑏)(𝑎 − 1) = 2 × 1\), diperoleh \(𝑏 = 8, 𝑎 = 2\), bilangan dua digit yang memenuhi adalah \(28\)
Kemungkinan 2
\(𝑎𝑏 − \overline{ab} = 12\)
\(⇒𝑎𝑏 − (10𝑎 + 𝑏) = 12\)
\(⇒𝑎𝑏 − 10𝑎 − 𝑏 = 12\)
\(⇒(𝑎 − 1)(𝑏 − 10) − 10 = 12\)
\(⇒(𝑎 − 1)(𝑏 − 10) = 22\)
Karena \(a\) dan \(b\) bilangan satu digit maka kemungkinan dua tidak ada yang memenuhi
Jadi banyak bilangan yang memenuhi hanya \(2\) yaitu \(28\) dan \(39\)


4. Determine there are how many positive integer \(x\) less than \(2007\) we can find such that \([\frac{x}{2}]+[\frac{x}{3}]+[\frac{x}{6}]=x\)
where \([n]\) is the greatest integer less than or equal to \(n\). (i.e., \([3.5]=3\);\([6]=6; [-3.5]=-4 \)etc.)


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5. A rectangle is made by placing together three smaller rectangles P, Q and R, without gaps or overlaps. Rectangle P measures 3 cm × 8 cm and Q measures 2 cm × 5 cm. Determine the number of possibilities are there for the measurements of R.


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6. Levine selects an integer, multiplies it by 8 then subtracts 4. She then multiplies her answer by 6 and finally subtracts 90. Her answer is a two-digit number. Determine the smallest integer she could select.


Misalkan bilangan yang dipilih adalah \(x\) hasilnya adalah \(y\) (\(y\) merupakan bilangan dua digit)
Berdasarkan perintah soal diperoleh persamaan:

\((8𝑥 − 4)6 − 90 = 𝑦\)
\(⇒ 48𝑥 − 24 − 90 = 𝑦\)
\(⇒ 48𝑥 − 114 = 𝑦\)
\(⇒ 48𝑥 = 𝑦 + 114\)
\(⇒ 𝑥 =\frac{𝑦 + 114}{48}\)

Karena \(y\) bilangan dua digit dan \(𝑦 + 114\) habis dibagi \(48\), maka bilangan terkecil \(y\) sehingga \(x\) terkecil adalah \(y=30\), diperoleh \(𝑥 = 3\).


7. Three positive integers are such that they differ from each other by at most 6. It is also known that the product of these three integers is 2808. Determine the smallest integer among them.


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8. We consider a line segment AB with the length c and all right-angled triangles with hypotenuse AB. For all such right-angled triangles, determine the maximum diameter of a circle with the centre on AB which is tangent to the other two sides of the triangle.


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9. Solve the equation \(|x-3|^{(\frac{x^2-8x+15}{x-2})}=1\)


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10. We have a red cube with side length 2 cm. What is the minimum number of identical cubes that must be adjoined to the red cube in order to obtain a cube with volume \(\left(\frac{12}{5}\right)^3\) cm³?


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Problem And Solution SEAMO 2017 Paper F
Asian Science And Math Olympiad (ASMO) 2018 For Grade 11


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