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 9

1. When two dice are rolled, find the probability of getting a sum that is divisible by 4.

Mata dadu berjumlah $$4 : \{(2, 2), (1, 3), (3, 1)\}$$ ada $$3$$
Mata dadu berjumlah $$8 : \{(4, 4), (2, 6), (3, 5), (5, 3), (6, 2)\}$$ ada $$5$$
Mata dadu berjumlah $$12 : \{(6, 6)\}$$ ada $$1$$
$$\frac{3+5+1}{36}=\frac{9}{36}=\frac{1}{4}$$

2. The tens digit of a two-digit number is two more than the units digit. When this two-digit number is divided by the sum of its digits, the answer is 6 remainder 3. Determine the sum of the digits of the two-digit number.

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3. The diagram shows a rectangle with length 9 cm and width 7 cm. One of the diagonals of the rectangle has been divided into seven equal parts. Determine the area of the shaded region. not yet available

4. If $$x + \sqrt{xy} + y = 9$$ and $$x^2 + xy + y^2=27$$, then determine the value of $$x – \sqrt{xy} + y$$.

Misalkan

$$𝑥 − \sqrt{𝑥𝑦} + 𝑦 = 𝐴$$

Kurangkan persamaan $$𝑥 + \sqrt{𝑥𝑦} + 𝑦 = 9$$ dan $$𝑥 − \sqrt{𝑥𝑦} + 𝑦 = 𝐴$$, diperoleh

$$2\sqrt{𝑥𝑦} = 9 − 𝐴 ⇒ 𝑥𝑦 = \frac{(9 − 𝐴)^2}{2}$$

Jumlahkan persamaan $$𝑥 + \sqrt{𝑥𝑦} + 𝑦 = 9$$ dan $$𝑥 − \sqrt{𝑥𝑦} + 𝑦 = 𝐴$$, diperoleh

$$2(𝑥 + 𝑦) = 9 + 𝐴$$

$$4(𝑥^2 + 𝑦^2 + 2𝑥𝑦) = (9 + 𝐴)^2$$
$$⇒4(27 + 𝑥𝑦) = (9 + 𝐴)^2$$
$$⇒4 (27 + (\frac{(9-A)^2}{4}) = (9 + 𝐴)^2$$
$$⇒108 + (9 − 𝐴)^2 = (9 + 𝐴)^2$$
$$⇒108 = (9 + 𝐴)^2 − (9 − 𝐴)^2$$
$$⇒18(2𝐴) = 108$$
$$⇒𝐴 = 3$$

5. There are 81 players taking part in a knock-out quiz tournament. Each match in the tournament involves 3 players and only the winner of the match remains in the tournament，the other two players are knocked out. Determine the number of matches that are required until there is an overall winner?

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6. Solve $$12x^4-56x^3+89x^2-56x+12= 0$$.

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7. The sizes in degrees of the interior angles of a hexagon are consecutive even numbers. Determine the size of the largest of these angles.

Misalkan keenam sudutnya adalah $$∠𝑎_1, ∠𝑎_2, ∠𝑎_3, ∠𝑎_4, ∠𝑎_5$$ dan $$∠𝑎_6$$ dimana $$∠𝑎_1 < ∠𝑎_2 < ∠𝑎_3 < ∠𝑎_4 < ∠𝑎_5 < ∠𝑎_6$$, karena keenam sudut membentuk pola bilangan genap berurutan maka diperoleh persamaan

$$∠𝑎_1, +∠𝑎_2 + ∠𝑎_3 + ∠𝑎_4 + ∠𝑎_5 + ∠𝑎_6 = 720°$$
$$⇒∠𝑎_6 − 10, +∠𝑎_6 − 8 + ∠𝑎_6 − 6 + ∠𝑎_6 − 4 + ∠𝑎_6 − 2 + ∠𝑎_6 = 720°$$
$$⇒6(∠𝑎_6) − 30° = 720°$$
$$⇒∠𝑎_6 =\frac{750}{6}= 125°$$

8. An integer is chosen from the set $$\{1, 2, 3, …, 499, 500\}$$. The probability that this integer is divisible by 7 or 11 is $$\frac{n}{m}$$ in its lowest terms. Determine the value of $$n+m$$.

Banyak himpunan bagian yang habis dibagi $$7 : ⌊\frac{500}{7}⌋ = 71$$
Banyak himpunan bagian yang habis dibagi $$11 : ⌊\frac{500}{11}⌋ = 45$$
Banyak himpunan bagian yang habis dibagi $$77 : ⌊\frac{500}{77}⌋ = 6$$
Jadi peluang terambilnya bilangan yang habis dibagi $$7$$ atau $$11$$ adalah $$\frac{71+45−6}{500}=\frac{110}{500}=\frac{11}{50}=\frac{𝑛}{𝑚}$$.
Nilai $$𝑛 + 𝑚 = 11 + 50 = 61$$

9. Determine all the possible three-digit numbers which are equal to 34 times the sum of their digits.

Misalkan bilangan 3 angka adalah $$\overline{abc}$$

$$\overline{abc}= 34(𝑎 + 𝑏 + 𝑐)$$
$$⇒100𝑎 + 10𝑏 + 𝑐 = 34𝑎 + 34𝑏 + 34𝑐$$
$$⇒66𝑎 = 24𝑏 + 33𝑐$$
$$22𝑎 = 8𝑏 + 11𝑐$$
$$22𝑎 − 11𝑐 = 8𝑏$$
$$11(2𝑎 − 𝑐) = 8𝑏$$

Bagian kiri adalah kelipatan $$11$$, maka hanya $$1$$ nilai $$b$$ yang mungkin yaitu $$0$$, dikarenakan nilai $$𝑏$$ selain $$0$$ hasil dari $$8𝑏$$ bukan kelipatan $$11$$.
karena nilai $$b=0$$ maka $$(2𝑎 − 𝑐) = 0$$, pasangan $$(𝑎, 𝑏)$$ yang memenuhi $$2𝑎 − 𝑐 = 0$$ adalah $$\{(1, 2), (2, 4), (3, 6), (4, 8)\}$$. Jadi banyak bilangan $$3$$ digit yang memenuhi ada $$4$$ bilangan yaitu $$102, 204, 306,$$ dan $$408$$.

10. How many integers between 1 and 1000, both 1 and 1000 inclusive, do not share a common factor with 1000?

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