Asian Science and Maths Olympiad (ASMO) is a competition platform designed to challenge and evaluate studentβ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 7

1. Jessica walks from location A to location B in seven days, and back in six days. Each day, she walks one kilometer more than on the preceding day. Determine the distance between the location A and location B.

Misalkan hari pertama Jessica dapat berjalan sejauh $$x$$ km
Dari $$A$$ ke $$B$$ ditempuh selama $$7$$ hari
Jarak $$A$$ ke $$B$$ adalah $$π₯ + (π₯ + 1) + (π₯ + 2) + (π₯ + 3) + (π₯ + 4) + (π₯ + 5) + (π₯ + 6) = 7π₯ + 21$$
Dari $$B$$ ke $$A$$ ditempuh selama $$6$$ hari
Jarak $$B$$ ke $$A$$ adalah $$(π₯ + 7) + (π₯ + 8) + (π₯ + 9) + (π₯ + 10) + (π₯ + 11) + (π₯ + 12) = 6π₯ + 57$$
Karena jarak $$A$$ ke $$B$$ dan jarak $$B$$ ke $$A$$ sama maka

$$7π₯ + 21 = 6π₯ + 57$$
$$β7π₯ β 6π₯ = 57 β 21$$
$$βπ₯ = 36$$

Jadi jarak $$A$$ ke $$B$$ adalah $$7π₯ + 21 = 7(36) + 21 = 252 + 21 = 273$$ km

2. Winson thought of a whole number and then multiplied it by either 5 or 6. Erica added 5 or 6 to Winson’s answer. Finally Rachael subtracted either 5 or 6 from Erica’s answer. The final result was 73. Determine the number Winson thought.

Misalkan bilangan yang dipikirkan Wilson adalah $$W$$

Kemungkinan yang bisa menghasilkan $$73$$ adalah $$6π + 1 = 73 β 6π = 72 β π = 12$$

3. Determine the largest integer $$x$$ such that $$x^{6021}< 2007^{2007}$$ .

$$π₯^{6021} < 2007^{2007}$$
Kedua ruas dipangkatkan $$\frac{1}{2007}$$, diperoleh
$$π₯^3 < 2007$$
Nilai $$x$$ terbesar yang memenuhi adalah $$12$$

4. Let $$x$$ and $$y$$ be two positive prime integers such that $$\frac{1}{x}-\frac{1}{y}=\frac{192}{2005^2 – 2004^2}$$. Determine the value of $$y$$ where $$y > x$$.

$$\frac{1}{π₯}β\frac{1}{π¦}=\frac{192}{2005^2 β 2004^2}$$
$$\frac{π¦ β π₯}{π₯π¦}=\frac{192}{(2005 β 2004)(2005 + 2004)}$$
$$\frac{π¦ β π₯}{π₯π¦}=\frac{192}{4009}=\frac{192}{19 Γ 221}$$
Jadi diperoleh nilai $$y$$ nya adalah $$192$$

5. In a certain triangle, the size of each of the angles is a whole number of degrees. Also, one angle is 60Β° larger than the average of the other two angles. Determine the largest possible size of an angle in this triangle.

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6. Find the sum

$$\frac{2019}{1Γ2}+\frac{2019}{2Γ3}+…+\frac{2019}{2018Γ2019}$$

$$\frac{2019}{1Γ2}+\frac{2019}{2Γ3}+…+\frac{2019}{2018Γ2019}$$
=$$2019\left(\frac{1}{1Γ2}+\frac{1}{2Γ3}+…+\frac{1}{2018Γ2019}\right)$$
=$$2019\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+…+\frac{1}{2018}-\frac{1}{2019}\right)$$
=$$2019\left(\frac{1}{1}-\frac{1}{2019}\right)$$
=$$2019\left(\frac{2019-1}{2019}\right)$$
=$$2019\left(\frac{2018}{2019}\right)$$
=$$2019$$

7. Ada thinks of a number. She adds 2 to it to get a second number. She then adds 3 to the second number to get a third number, adds 4 to the third to get a fourth, and finally adds 5 to the fourth to get a fifth number. Alan also thinks of a number but he subtracts 3 to get a second. He then subtracts 4 from the second to get a third, and so on until he too has five numbers. Finally, they discover that the sum of Adaβs five numbers is the same as the sum of Alanβs five numbers. Determine the difference between the two numbers of which they first thought.

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8. The figure shows part of a tiling, which extends indefinitely in every direction across the whole plane. Each tile is a regular hexagon. Some of the tiles are white, the others are black. Determine the fraction that the plane is white.

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9. In the following sum, O represent the digit 0. A, B, X and Y each represents distinct digit. Determine there are how many possible digits can A be.

Diketahui O = 0
karena XX adalah bilangan yang kembar maka nilai A+B hasilnya bilangan 2 digit yang kembar, Kemungkinannya hanya satu yaitu A + B = 11
Pasangan bilangan A dan B yang memenuhi adalah (2,9), (3,8), (4,7), (5,6), (6,5), (7,4), (8,3), (9,2). Banyak kemungkinan nilai A ada 8

10. The tens digit of a two-digit number is three more than the units digit. When this two- digit number is divided by the sum of its digits, the answer is 7 remainder 3. Determine the sum of the digits of the two-digit number.

Kemungkinan bilangan dua digit yang digit puluhannya lebih dari dari digit satuan adalah $$\{30, 41, 52, 63. 74, 85, 96\}$$, cek bilangan yang memenuhi jika dibagi dengan jumlah digitnya hasilnya adalah 7 dan bersisa 3.
$$\frac{30}{3+0}= 10$$ tidak memenuhi
$$\frac{41}{4+1}= 8$$ bersisa $$1$$ (Tidak memenuhi)
$$\frac{52}{5+2}= 7$$ bersisa $$3$$ (memenuhi)
Bilangan yang memenuhi adalah $$52$$, jumlah digitnya adalah $$7$$

11. Richard has four cubes all the same size: one blue, one red, one white and one yellow. He wants to glue the four cubes together to make the solid shape as shown in the figure below.
Determine the number of differently-coloured shapes can Richard make. [Two shapes are considered to be the same if one can be picked up and turned around so that it looks identical to the other.]

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12. Determine the sum of all corner angles of a 5-pointed star: $$a+b+c+d+e.$$

Jumlah sudut pada segitiga adalah $$180Β°$$, jadi

$$π + π + π + π + π = 180Β°$$

13. A 3-digit integer is called a βV-numberβ if the digits go βhigh-low-highβ β that is, if the tens digit is smaller than both the hundreds digit and the units (or βonesβ) digit. Determine the number of 3-digit βV-numbersβ that can be formed.

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14. Oscar wants to put the numbers 2, 3, 4, 5, 6 and 10 into the circles so that the products of the three numbers along each edge are the same, and as large as possible. Determine the product of the three numbers.

15. A two-digit number βmnβ is multiplied by its reverse βnmβ. The ones (units) and tens digits of the four-digit answer are both 0. Determine the value of the smallest such two-digit number βmnβ.

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16. Simplify $$\frac{2005^2(2004^2-2003)}{(2004^2-1)(2004^3+1)}Γ\frac{2003^2(2004^2+2005)}{2004^3-1}$$

Misalkan $$π = 2004$$

$$\frac{2005^2(2004^2 β 2003)}{(2004^2 β 1)(2004^3 + 1)}Γ\frac{2003^2(2004^2 + 2005)}{(2004^3 β 1)}$$

$$=\frac{(2004 + 1)^2(2004^2 β 2004 + 1)}{(2004^2 β 1)(2004^3 + 1)}Γ\frac{(2004 β 1)^2(2004^2 + 2004 + 1)}{(20043 β 1)}$$

$$=\frac{(π + 1)2(π^2 β π + 1)}{(π^2 β 1)(π^3 + 1)}Γ\frac{(π β 1)^2(π^2 + π + 1)}{(π^3 β 1)}$$

$$=\frac{(π + 1)^2(π^2 β π + 1)}{(π^2 β 1)(π + 1)(π^2 β π + 1)}Γ\frac{(π β 1)^2(π^2 + π + 1)}{(π β 1)(π^2 + π + 1)}$$

$$= 1$$

17. Determine the value of $$a+b+c$$ if $$a, b$$ and $$c$$ stand for different digits.

Dengan melakukan percobaan diperoleh

Diperoleh $$π, π$$ dan $$π$$ nya adalah $$3, 6$$ dan $$7$$, Jadi nilai dari $$π + π + π = 3 + 6 + 7 = 16$$

18. Simplify the expression $$\sqrt[n]{10\sqrt[n]{10\sqrt[n]{10\sqrt[n]{10\sqrt[n]{…}}}}}$$ given that $$n=2, 3, 4, 5, β¦ .$$

$$\sqrt[n]{10\sqrt[n]{10\sqrt[n]{10\sqrt[n]{10\sqrt[n]{…}}}}}=a$$

$$β10\sqrt[n]{10\sqrt[n]{10\sqrt[n]{10\sqrt[n]{…}}}}=a^n$$

$$β10a=a^n$$

$$β10=a^{n-1}$$

$$βa=\sqrt[n-1]{10}$$

19. Suppose that $$x^2 + a = 2006, x^2 +b = 2007$$ and $$x^2 + c = 2008$$ and $$abc = 3$$ . Determine the value of $$\frac{a}{bc}+\frac{b}{ac}+\frac{c}{ab}-\frac{1}{a}-\frac{1}{b}-\frac{1}{c}.$$

$$π₯^2 + π = 2006 β π = 2006 β π₯^2 = π β 1$$
$$π₯^2 + π = 2007 β π = 2007 β π₯^2$$
$$π₯^2 + π = 2008 β π = 2008 β π₯^2 = π + 1$$

selanjutnya

\begin{align} \frac{π}{ππ}+\frac{π}{ππ}+\frac{π}{ππ}β\frac{1}{π}β\frac{1}{π}β\frac{1}{π}&= \frac{π^2 + π^2 + π^2 β (ππ + ππ + ππ)}{πππ}\\ &=\frac{1}{3}[(π^2 β 2π + 1 + π^2 + π^2 + 2π + 1 β (π^2 + π + π^2 β 1 + π^2 β π)]\\ &=\frac{1}{3}[2 + 1] = 1\\ \end{align}

20. The difference between the highest common factor and the lowest common multiple of m and 18 is 120. Determine the value of m.

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21. The internal bisector of angle A of triangle ABC meets BC at D. The external bisector of angle A meets BC produced at E. If AB=6 units, AC=4 units and BC=6 units, determine the length DE.

$$nb=ma$$

Buat segitiga baru yaitu $$AFE$$ yang kongruen dengan segitiga $$ACE$$.

Gunakan rumus garis bagi pada segitiga $$BEF$$

$$4(6 + π§) = 6π§$$
$$24 + 4π§ = 6π§$$
$$π§ = 12$$

Gunakan rumus garis bagi pada segitiga $$ABC$$

$$6π¦ = 4π₯$$
$$π₯ : π¦ = 3: 2$$

Panjang $$π¦ =\frac{2}{5}(6) =\frac{12}{5}= 2\frac{2}{5}$$

Jadi panjang $$DE = 12 + 2\frac{2}{5}= 14\frac{2}{5}$$

22. In triangle ABC, D and E are points on AB and BC respectively. Given that AD:DB=2:3 and DE is parallel to AC, determine the ratio of the area of triangle BDE to the area of triangle ABC.

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23. In how many ways can the letters of the word MURMUR be arranged without letting two letters which are alike come together?

Banyak susunan seluruhnya adalah $$\frac{6!}{2!.2!.2!} = 90$$ susunan
Banyak susunan jika $$UU$$ berdekatan adalah $$\frac{5!}{2!.2!}= 30$$
Banyak susunan jika $$RR$$ berdekatan adalah $$\frac{5!}{2!.2!}= 30$$
Banyak susunan jika $$MM$$ berdekatan adalah $$\frac{5!}{2!.2!} = 30$$
Banyak susunan jika $$RR$$ dan $$UU$$ berdekatan adalah $$\frac{4!}{2!}= 12$$
Banyak susunan jika $$RR$$ dan $$MM$$ berdekatan adalah $$\frac{4!}{2!}= 12$$
Banyak susunan jika $$MM$$ dan $$UU$$ berdekatan adalah $$\frac{4!}{2!}= 12$$
Banyak susunan jika $$RR, MM$$ dan $$UU$$ berdekatan adalah $$3! = 6$$
Jadi banyak susunan seluruhnya agar tidak ada huruf yang kember berdekatan adalah $$90 β 30 β 30 β 30 + 12 + 12 + 12 β 6 = 30$$ susunan

24. Determine how many numbers less than 2013 are both satisfying condition (1) and condition (2).
(1) the sum of five consecutive positive integers ; and
(2) the sum of two consecutive positive integers.

Dari syarat $$(2)$$ : penjumlahan dua bilangan bulat berurutan selalu bernilai ganjil
Dari syarat $$(1)$$ : misalkan bilangan tengahnya adalah n, maka penjumlahan lima bilangan berurutan dapat ditulis :

$$(π β 2) + (π β 1) + π + (π + 1) + (π + 2) = 5π$$

Dipastikan dari syarat $$(2)$$ bilangan yang memenuhi adalah bilangan kelipatan $$5$$, kecuali $$5$$ karena penjumlahan lima bilangan bulat positif minimal adalah $$15$$.
Dari gabungan kedua syarat bilangan yang memenuhi adalah bilangan ganjil kelipatan $$5$$ selain $$5$$.
Banyak bilangan yang memenuhi adalah $$β\frac{2010}{5}β β β\frac{2010}{10}β β 1 = 402 β 201 β 1 = 200$$

25. We are given bases $$|AB |= 23$$ and $$|CD|= 5$$ of a trapezoid $$ABCD$$ with diagonals $$| AC |= 25$$ and $$|BD|=17$$. Determine the lengths of its sides BC and AD.

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