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 9

1. Given $$10^x ·100^{2x} =1000^5$$ . What is the value of $$x$$?

$$10^𝑥 ∙ 100^{2𝑥} = 1000^5$$
$$⇒ 10^𝑥 ∙ 10^{4𝑥} = (10^3)^5$$
$$⇒ 10^{𝑥+4𝑥} = 10^{15}$$
Diperoleh $$5𝑥 = 15 ⇒ 𝑥 = 3$$

2. $$200000^2 – 199999^2=…$$

$$200000^2– 199999^2 = (200000 − 199999)(200000 + 199999) = 399999$$

3.  A 5 digit number consists of 5 consecutive positive integers. For example 26534. How many such 5 digit numbers are there?

12345 permutasinya ada 5! = 120 bilangan
23456 permutasinya ada 5! = 120 bilangan
34567 permutasinya ada 5! = 120 bilangan
45678 permutasinya ada 5! = 120 bilangan
56789 permutasinya ada 5! = 120 bilangan

4. A rectangular box has integer side lengths in the ratio 1:3:4. Given the volume is less than 500 units³. What is the maximum volume of the box?

misalkan $$𝑝, 𝑙$$ dan $$𝑡$$ nya adalah $$𝑥 , 3𝑥, 4𝑥$$

$$𝑉 = 𝑝𝑙𝑡 < 500$$
$$⇒ 𝑥(3𝑥)(4𝑥) < 500$$
$$⇒ 12𝑥^3 < 500$$
$$⇒ 𝑥^3 <\frac{500}{12}= 41,67$$
Karena $$x$$ bilangan bulat maka nilai $$x$$ yang memenuhi adalah $$3$$. Jadi volume maksimum adalah $$12𝑥^3 = 12(27) = 324$$

5. Two different numbers are randomly selected from a set of numbers {0,1,2,3,4,5} and multiplied together. Find the probability that the product is greater than 10.(Note : (a,b) dan (b,a) are different)

Banyak sampel karena $$(a,b)$$ dan $$(b,a)$$ dihitung berbeda adalah $$6 × 5 = 30$$ Kemungkinan dua bilangan yang hasil kalinya melebihi 10 adalah $$\{(3,4),(4,3),(3,5),(5,3), (4,5), (5,4)\}$$ ada $$6$$ pasangan. Jadi peluangnya adalah $$\frac{6}{30}=\frac{1}{5}$$

6. What is the ratio of the area of shaded region to the area of not shaded region of the 8 x 6 rectangle below?

$$𝐿𝑢𝑎𝑠\; 𝐴 = \frac{1}{𝑆𝑅}[𝑆𝑂𝑅] =\frac{1}{8}(12) =\frac{3}{2}$$
$$𝐿𝑢𝑎s\; 𝐵 =\frac{1}{𝑆𝑃}[𝑆𝑂𝑃] =\frac{1}{6}(12) = 2$$
$$𝐿𝑢𝑎𝑠\; 𝑎𝑟𝑠𝑖𝑟𝑎𝑛\; = 2(𝐿𝑢𝑎𝑠\; 𝐴 + 𝐿𝑢𝑎𝑠\; 𝐵) = 2(\frac{3}{2}+ 2) = 3 + 4 = 7$$
$$𝐿𝑢𝑎𝑠\; 𝑃𝑄𝑅𝑆 = 8 × 6 = 48$$

Jadi perbandingan luas yang diarsir dan yang tidak diarsir adalah $$7: (48 − 7) = 7: 41$$

7. Given $$p+q+r=26$$ and $$\frac{1}{p}+\frac{1}{q}+\frac{1}{r}=31$$ Find

$$\frac{p}{q}+\frac{q}{r}+\frac{r}{p}+\frac{q}{p}+\frac{r}{q}+\frac{p}{r}=…?$$

$$(𝑝 + 𝑞 + 𝑟) \left(\frac{1}{𝑝}+\frac{1}{𝑞}+\frac{1}{𝑟}\right)=\frac{𝑝+𝑞+𝑟}{𝑝}+\frac{𝑝+𝑞+𝑟}{𝑟}+\frac{𝑝+𝑞+𝑟}{𝑞}$$
$$⇒26(31) = 1 +\frac{𝑞}{𝑝}+\frac{𝑟}{𝑝}+ 1 +\frac{𝑝}{𝑟}+\frac{𝑞}{𝑟}+\frac{𝑝}{𝑞}+ 1 +\frac{𝑟}{𝑞}$$
$$⇒806=3+\frac{𝑝}{𝑞}+\frac{𝑞}{𝑟}+\frac{𝑟}{𝑝}+\frac{𝑞}{𝑝}+\frac{𝑟}{𝑞}+\frac{𝑝}{𝑟}$$
$$⇒\frac{𝑝}{𝑞}+\frac{𝑞}{𝑟}+\frac{𝑟}{𝑝}+\frac{𝑞}{𝑝}+\frac{𝑟}{𝑞}+\frac{𝑝}{𝑟}=806-3=803$$

8. The mean, median, and mode of the 7 data values 60, 100, x, 40, 50, 200, 90 are all equal to x. What is the value of x?

karena rata-ratanya $$x$$ maka

$$𝑥 =\frac{60 + 100 + 𝑥 + 40 + 50 + 200 + 90}{7}$$
$$⇒7𝑥 = 540 + 𝑥$$
$$⇒6𝑥 = 540$$
$$⇒𝑥 = 90°$$
Urutkan datanya dari terkecil keterbesar
$$40, 50, 60, 90, 90, 100, 200$$
Diperoleh $$median = modus = 90$$
Jadi nilai $$x$$ adalah $$90$$

9. If $$a, b$$ and $$c$$ are three consecutive even numbers in increasing order, find the value of $$a^2 – 2b^2 + c^2$$.

misalkan bilangan genap berurutannya adalah $$𝑎 = 𝑏 − 2, 𝑏, 𝑐 = 𝑏 + 2$$

$$𝑎^2 − 2𝑏^2 + 𝑐^2 = (𝑏 − 2)^2 − 2𝑏^2 + (𝑏 + 2)^2$$
$$= 𝑏^2 − 4𝑎 + 4 − 2𝑏^2 + 𝑏^2 + 4𝑏 + 4$$
$$= 8$$

10. The lowest common multiple of a and b is 48 and the lowest common multiple of b and c is 108. What is the least possible value of the lowest common multiples of a and c?

not yet available

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