Asian Science And Math Olympiad (ASMO) 2018 For Grade 8

ASMO SMP

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 8


1. If \(a\%\) of \(b\) kg of a substance is 100kg, what is the mass of \(b\%\) of \(a\) kg of the same substance? (\(a\) and \(b\) are positive real numbers)


not yet available


2. Find the largest two-digit number that is:
i) Sum of THREE consecutive positive integers, AND
ii) Sum of TWO consecutive positive integers.


Misalkan bilangan tersebut adalah \(N\)
\(𝑁 = (𝑛 βˆ’ 1) + 𝑛 + (𝑛 βˆ’ 1) = 3𝑛 , 𝑛\) bilangan bulat
\(𝑁 = π‘š + (π‘š + 1) = 2π‘š + 1\)
KarenaΒ \(N\) dapat ditulis dalam bentuk \(3𝑛\) dan \(2π‘š + 1\) maka dipastikan \(𝑁\) adalah bilangan kelipatan 3 dan juga bilangan ganjil. Jadi bilangan \(𝑁\) terbesar yang memenuhi adalah \(99\).
\(𝑁 = 32 + 33 + 34 = 49 + 50 = 99\)


3. The sum of two positive numbers is three times of their difference. Find the simplest ratio of the larger number to the smaller number.


Misalkan bilangan yang terbesar adalah \(π‘Ž\) dan terkecil adalah \(𝑏\)

\(π‘Ž + 𝑏 = 3(π‘Ž βˆ’ 𝑏)\)
\(β‡’π‘Ž + 𝑏 = 3π‘Ž βˆ’ 3𝑏\)
\(⇒𝑏 + 3𝑏 = 3π‘Ž βˆ’ π‘Ž\)
\(β‡’4𝑏 = 2π‘Ž\)

Jadi perbandingan \(π‘Ž ∢ 𝑏\) adalah \(2 : 1\)


4. Given the following linear equations:

\(π‘Ž + 3𝑏 + 9𝑐 + 2𝑑 = 37\)
\(2π‘Ž + 5𝑏 βˆ’ 2𝑐 + 9𝑑 = 48\)
\(6π‘Ž + 𝑏 + 7𝑐 + 5𝑑 = 28\)
\(6π‘Ž + 6𝑏 + 𝑐 βˆ’ 𝑑 = 22\)

Find the value of the following

\(π‘Ž + 𝑏 + 𝑐 + 𝑑 = β‹―\)


Jumlahkan keempat persamaan di atas
\(π‘Ž + 3𝑏 + 9𝑐 + 2𝑑 = 37\)
\(2π‘Ž + 5𝑏 βˆ’ 2𝑐 + 9𝑑 = 48\)
\(6π‘Ž + 𝑏 + 7𝑐 + 5𝑑 = 28\)
\(6π‘Ž + 6𝑏 + 𝑐 βˆ’ 𝑑 = 22\)
____________________ +
\(15π‘Ž + 15𝑏 + 15𝑐 + 15𝑑 = 135\)
\(π‘Ž + 𝑏 + 𝑐 + 𝑑 = 9\)


5. Given three prime numbers \(a, b\) and \(c\) such that

\(a+b=c\)

AND

\(a≀b≀c\)

find the value of \(a\).


not yet available


6. How many trailing zeros are there in the following number?

\(2^6Γ—5^{15}Γ—6^3\)


\(2^6 Γ— 5^{15} Γ— 6^3\)
\(= 2^6 Γ— 5^{15} Γ— (2 Γ— 3)^3\)
\(= 2^6 Γ— 5^{15} Γ— 2^3 Γ— 3^3\)
\(= 2^9 Γ— 5^{15} Γ— 3^3\)
\(= 2^9 Γ— 5^9 Γ— 5^6 Γ— 3^3 = 10^9 Γ— 5^6 Γ— 3^3\)
Jadi banyaknya angka \(0\) ada \(9\)


7.


Figure above shows a pair of parallel lines. Given that angles \(a+b=75Β°\), find the angle \(x\) in degrees.


\(π‘Ž + 𝑏 + π‘₯ = 180\)
\(75 + π‘₯ = 180\)
\(π‘₯ = 180 βˆ’ 75 = 105Β°\)


8. A packet of marbles is given to 3 kids to be shared. They accidentally lost two of the marbles. The first kid takes \(\frac{4}{7}\) of what is remaining in the packet and passed the packet two the second kid. The second kid then takes \(\frac{1}{6}\) of the marbles in the packet. The remaining 35 marbles belong to the third kid. How many marbles are there in the packet initially?


Kerjakan dari bawah
Misalkan banyak kelereng yang tidak diambil anak kedua adalah A, karena sisanya 35 dan anak kedua mengambil \(\frac{1}{6}\) maka dapat ditulis \(\frac{5}{6}A=35β‡’A=\frac{6}{5}(35)=42\)

Selanjutnya karna anak pertama mengambil \(\frac{4}{7}\) dari mula-mula makaΒ  \(\frac{3}{7}\) dari mula-mula adalah 42. Misalkan mula-mula adalah N, maka \(N=\frac{7}{3}(42)=7(14)=98\)

Dikarenakan ada kehilangan dua kelereng maka banyak kelereng sebenarnya adalah \(98 + 2 = 100\)


9. The Lowest Common Multiple \((LCM)\) of \(a\) and \(b\) is \(48\) and the \(LCM\) of \(b\) and \(c\) is \(72\). Find the \(LCM\) of \(a\) and \(c\).


\(48 = 16 Γ— 3 = 2^4 Γ— 3\)
\(72 = 8 Γ— 9 = 2^3 Γ— 3^2\)
\(𝐿𝐢𝑀 (π‘Ž, 𝑏) = 2^4 Γ— 3, 𝐿𝐢𝑀(𝑏, 𝑐) = 2^3 Γ— 3^2\)
Nilai \(a, b\) dan \(c\) yang mungkin adalah
\(π‘Ž = 2^4 Γ— 3\)
\(𝑏 = 3\)
\(𝑐 = 2^3. 3^2\)
\(𝐿𝐢𝑀(π‘Ž, 𝑐) = 2^4 Γ— 3^2 = 144\)


10. A three-digit number is formed by three distinct integers. Find the number given the following conditions:
i) 702 – 1 correct digit in wrong position.
ii) 230 – 1 correct digit in wrong position.
iii) 413 – 1 correct digit in wrong position.
iv) 591 – 1 correct digit in correct position.
v) 120 – no correct digits.


Dari (v) sudah dipastikan ketiga digitnya tidak memuat digit 1, 2 dan 0
Dari (i) bilangan 7 memenuhi tapi posisinya salah, kemungkinan bilangan tiga digit adalah _7_ atau _ _ 7
Dari (ii) bilangan 3 memenuhi tapi posisi salah, kemungkinan bilangan tiga digit yang memenuhi adalah 37_ , _73 atau 3_7.
Dari (iii) bilangan 3 memenuhi tapi posisi salah maka kemungkinan 37_ atau 3_7
Dari (iv) karena tinggal 37_ atau 3_7 yang memenuhi dan 1 tidak termasuk, maka bilangan 9 yang memenuhi dan posisinya tepat.

Jadi bilangan tiga digit yang memenuhi adalah 397


11. Given that 100kg of wet potatoes contain 99% of water weight and 1 % of dry weight. After being dried under the sun, the potatoes now contain 98% of water weight. Find the total weight of the potatoes after being dried.


not yet available


12. Find the value of the following series:

\(\frac{2}{2017}+\frac{4}{2017}+\frac{6}{2017}+…+2=…\)


\(\frac{2}{2017}+\frac{4}{2017}+\frac{6}{2017}+…+2\)

\(=\frac{2}{2017}+\frac{4}{2017}+\frac{6}{2017}+…+\frac{4034}{2017}\)

\(=\frac{2+4+6+…+4034}{2017}\)

\(=\frac{2(1+2+3+…+2017}{2017}\)

\(=\frac{2(\frac{(1+2017)2017}{2})}{2017}=2018\)


13. The ratio of measures of two acute angles is 7:3. Given that the complement of one of them is three times as large as the compliment of the other. Find the difference between the two angles in degrees.


Misalkan sudut terbesar adalah \(7π‘Ž\) dan sudut terkecil adalah \(3π‘Ž\)
Komplemen dari \(7π‘Ž\) adalah \(90 – 7π‘Ž\)
Komplemen dari \(3π‘Ž\) adalah \(90 βˆ’ 3π‘Ž\)
Karena \(90 βˆ’ 3π‘Ž > 90 βˆ’ 7π‘Ž\), maka berdasarkan keterangan soal

\(90 βˆ’ 3π‘Ž = 3(90 βˆ’ 7π‘Ž)\)
\(90 βˆ’ 3π‘Ž = 270 βˆ’ 21π‘Ž\)
\(18π‘Ž = 180\)
\(π‘Ž =\frac{180}{18}= 10\)
Jadi selisih kedua sudut adalah \(7π‘Ž βˆ’ 3π‘Ž = 4π‘Ž = 4(10) = 40Β°\)


14.

Figure above shows two right angled triangles ABC and ADB. Given DC=1cm and BC=2cm find the length of AD in cm.


Gunakan rumus

\(𝐡𝐷^2 = 𝐴𝐷. 𝐷𝐢 = 𝐴𝐷. 1\)

Dengan menggunakan rumus Pythagoras

\(𝐡𝐷^2 = 𝐡𝐢^2 βˆ’ 𝐷𝐢^2 = 22 βˆ’ 12 = 4 βˆ’ 1 = 3\)

Jadi panjang \(𝐴𝐷 = 𝐡𝐷^2 = 3 π‘π‘š\)


15.

Figure above shows a square ABCD with 4 semicircles with each side of the square as diameters. If the perimeter of the square is 28cm, find the total area of the shaded regions in cm2. (Use \(Ο€=\frac{22}{7}\) )


Jari-jari lingkaran \(= sisi\; persegi\; : 2 =\frac{7}{2} π‘π‘š\)

\(\begin{align}
Luas\; daerah\; yang\; diarsir\; adalah &= 8(π‘™π‘’π‘Žπ‘ \; π‘—π‘’π‘Ÿπ‘–π‘›π‘” βˆ’ 𝐿\; π‘ π‘’π‘”π‘–π‘‘π‘–π‘”π‘Ž)\\
&= 8(\frac{1}{4}πœ‹π‘Ÿ^2 βˆ’\frac{1}{2}(π‘Ÿ)(π‘Ÿ))\\
&= 2(\frac{22}{7})(\frac{7}{2})(\frac{7}{2}) βˆ’ 4(\frac{7}{2})(\frac{7}{2})\\
&= 77 – 49\\
&= 28\ cm^2\\
\end{align}\)


16.

Figure above shows three rectangles. The width of the smallest rectangle is 1cm and each side of the bigger rectangles is 1cm away from the corresponding sides of the smaller rectangle. If the area of the shaded regions forms an arithmetic progression, find the length of the smallest rectangle in cm.


Luas arsiran uang kecil = Luas PP sedang – Luas PP kecil
\(= (𝑝 + 2)3 βˆ’ 𝑝. 1 = 3𝑝 + 6 βˆ’ 𝑝 = 2𝑝 + 6\)

Luas arsiran uang besar = Luas PP besar – Luas PP sedang
\(= (𝑝 + 4)5 βˆ’ (𝑝 + 2)3 = 5𝑝 + 20 βˆ’ 3𝑝 βˆ’ 6 = 2𝑝 + 14\)

Karna luas arsiran membentuk barisan aritmetika maka beda luasnya adalah sama

Luas daerah arsiran sedang – Luas persegi panjang kecil = Luas daerah arsiran besar – luas daerah arsiran sedang
\(2𝑝 + 6 βˆ’ 𝑝 = 2𝑝 + 14 βˆ’ (2𝑝 + 6)\)
\(𝑝 + 6 = 8\)
\(𝑝 = 2\)

Jadi panjang persegi panjang terkecil adalah 2 cm


17. Given \(2π‘₯ = 3𝑦 = 6𝑧, x, y, z\) non zero, find the value of

\(\frac{2π‘₯𝑦}{𝑧(π‘₯+𝑦)}\)


\(2^π‘₯ = 6^𝑧\) kedua ruas dipangkat dengan \(\frac{1}{π‘₯}\)
\((2^π‘₯)^{\frac{1}{π‘₯}} = (6^𝑧)^{\frac{1}{π‘₯}} β‡’ 2 = 6^{\frac{𝑧}{π‘₯}} … (1)\)

\(3^y = 6^𝑧\) kedua ruas dipangkat dengan \(\frac{1}{y}\)
\((3^y)^{\frac{1}{y}} = (6^𝑧)^{\frac{1}{y}} β‡’ 3 = 6^{\frac{𝑧}{y}} … (2)\)

Kalikan persamaan (1) dan (2)

\(2 Γ— 3 = 6^{\frac{𝑧}{π‘₯}} Γ— 6^{\frac{𝑧}{𝑦}}\)
\(β‡’6 = 6^{\frac{z}{x}+\frac{z}{y}}\)
\(β‡’1 =\frac{z}{x}+\frac{z}{y}\)
\(\frac{z}{x}+\frac{z}{y}=\frac{𝑦𝑧}{π‘₯𝑦}+\frac{π‘₯𝑧}{π‘₯𝑦}=\frac{𝑧(π‘₯ + 𝑦)}{π‘₯𝑦}\)
\(β‡’\frac{π‘₯𝑦}{𝑧(π‘₯ + 𝑦)}= 1 ⟹\frac{2π‘₯𝑦}{𝑧(π‘₯ + 𝑦)}= 2\)


18. Find the value of the following expression:

\(\frac{1}{2-\frac{1}{2-\frac{1}{2-\frac{1}{2-…}}}}\)


misalkan \(\frac{1}{2-\frac{1}{2-\frac{1}{2-\frac{1}{2-…}}}}=x\), maka

\(\frac{1}{2-x}=x\)
\(β‡’x=1\)


19. A small packet of sweets contains 4 sweets while a big packet of sweets contains 7 sweets. If the sweets can only be bought with packets, what is the maximum number of sweets that can’t be bought by combining big and small packets?


4 + 4 = 8
4 + 7 = 11
7 + 7 = 14
4 + 4 + 7 = 15
4 + 4 + 4 + 4 = 16
7 + 7 + 4 = 18
4 + 4 + 4 + 7 = 19
4 + 4 + 4 + 4 + 4 = 20
7 + 7 + 7 = 21
18 + 4 = 22
19 + 4 = 23
20 + 4 = 24
21 + 4 = …
….
Karena untuk banyak paket 18, 19, 20, dan 21 dapat dibentuk dari kombinasi keduanya maka untuk banyaknya lebih dari 17 selalu bisa diperoleh dari kombinasi paket kecil dan besar.Jadi maksimum banyak paket yang tidak dapat dikombinasikan adalah 17


20. How many positive integers below 50 have EXACTLY 4 factors?


Bilangan yang mempunyai tepat 4 faktor adalah bilangan yang berbentuk \(π‘π‘ž\) atau \(𝑝^3\) dimana \(𝑝\) dan \(π‘ž\) adalah bilangan prima.

  • Β Untuk \(π‘π‘ž < 50\)
    \(2\) dipasangkan dengan \(\{3, 5, 7, 11, 13, 17, 19, 23\}\) ada \(8\)
    \(3\) dipasangkan dengan \(\{5, 7, 11, 13\}\) ada \(4\)
    \(5\) dipasangkan dengan \(\{7\}\) ada \(1\)
  • Untuk \(𝑝^3 < 50\)
    nilai \(p\) yang mungkin \(\{2, 3\}\) ada \(2\)

Jadi banyaknya bilangan yang memenuhi adalah \(8 + 4 + 1 + 2 = 15\) bilangan


21. How many numbers can be formed by product of three elements from the set \(\{2,3,5,7,11,13\}\)? (Repetition of factors is allowed)


  • Ketiga bilangan sama ada 6 cara
  • Dua bilangan sama
    Angka dua yang sama maka banyaknya ada 5
    Angka tiga yang sama maka banyaknya ada 5
    Angka lima yang sama maka banyaknya ada 5
    Angka tujuh yang sama maka banyaknya ada 5
    Angka sebelas yang sama maka banyaknya ada 5
    Angka tigabelas yang sama maka banyaknya ada 5
  • Tiga bilangan berbeda banyak cara \({6\choose 3}\)\(=\frac{6!}{3!.3!}= 20\)

Jadi banyak bilangan yang memenuhi adalah \(6 + 30 + 20 =56\) bilangan


22. If there are 4 consecutive positive integers such that the smallest is divisible by 2, the second is divisible by 5, the third is divisible by 8, and the largest is divisible by 11. Find the lowest possible sum of the 4 integers.


Misalkan bilangan tersebut adalahΒ  \(π‘˜ , π‘˜ + 1, π‘˜ + 2, π‘˜ + 3\)
\(π‘˜ ≑ 0\; π‘šπ‘œπ‘‘\; 2 β‡’ π‘˜ ≑ 0\; π‘šπ‘œπ‘‘\; 2 … (1)\)
\(π‘˜ + 1 ≑ 0\; π‘šπ‘œπ‘‘\; 5 β‡’ π‘˜ ≑ βˆ’1\; π‘šπ‘œπ‘‘\; 5 β‡’ π‘˜ ≑ 4\; π‘šπ‘œπ‘‘\; 5 … (2)\)
\(π‘˜ + 2 ≑ 0\; π‘šπ‘œπ‘‘\; 8 β‡’ π‘˜ ≑ βˆ’2\; π‘šπ‘œπ‘‘\; 8 β‡’ π‘˜ ≑ 6\; π‘šπ‘œπ‘‘\; 8 … (3)\)
\(π‘˜ + 3 ≑ 0\; π‘šπ‘œπ‘‘\; 11 β‡’ π‘˜ ≑ βˆ’3\; π‘šπ‘œπ‘‘\; 11 β‡’ π‘˜ ≑ 8\; π‘šπ‘œπ‘‘\; 11 … (4)\)
Dengan menggunakan CRT, samakan persamaaan \((2)\) dan \((3)\)

\(π‘˜ = 8π‘Ž + 6 ≑ 4\; π‘šπ‘œπ‘‘\; 5\)
\(8π‘Ž ≑ βˆ’2\; π‘šπ‘œπ‘‘\; 5\)
\(4π‘Ž ≑ βˆ’1\; π‘šπ‘œπ‘‘\; 5\)
\(βˆ’π‘Ž ≑ βˆ’1\; π‘šπ‘œπ‘‘\; 5\)
\(π‘Ž ≑ 1\; π‘šπ‘œπ‘‘\; 5\)
\(π‘Ž = 5𝑏 + 1\)

Samakan persamaanΒ \((3)\) dan \((4)\)

\(8π‘Ž + 6 ≑ 8\; π‘šπ‘œπ‘‘\; 11\)
\(8π‘Ž ≑ 2\; π‘šπ‘œπ‘‘\; 11\)
\(4π‘Ž ≑ 1\; π‘šπ‘œπ‘‘\; 11\)
\(12π‘Ž ≑ 3\; π‘šπ‘œπ‘‘\; 11\)
\(π‘Ž ≑ 3\; π‘šπ‘œπ‘‘\; 11\)
\(5𝑏 + 1 ≑ 3\; π‘šπ‘œπ‘‘\; 11\)
\(5𝑏 ≑ 2\; π‘šπ‘œπ‘‘\; 11\)
\(10𝑏 ≑ 4\;π‘šπ‘œπ‘‘\; 11\)
\(βˆ’π‘ ≑ 4\; π‘šπ‘œπ‘‘\; 11\)
\(𝑏 ≑ βˆ’4\; π‘šπ‘œπ‘‘\; 11 = 7\; π‘šπ‘œπ‘‘\; 11\)
\(𝑏 = 11𝑐 + 7\)

Karena yang dicari nilai minimum \(k\) maka kita pilih \(c=0\), diperoleh \(𝑏 = 7 , π‘Ž = 5𝑏 + 1 = 5(7) + 1 = 36, π‘˜ = 8π‘Ž + 6 = 8(36) + 6 = 294\)
Jadi jumlah minimum yang mungkin dari keempat bilangan berurutan adalah
\(π‘˜ + (π‘˜ + 1) + (π‘˜ + 2) + (π‘˜ + 3) = 4π‘˜ + 6 = 4(294) + 6 = 1182\)


23. The positive integers 1 to 9 are filled into a 3 by 3 grid with the following restriction: consecutive integers must share a same edge (in simpler words consecutive integers must be side by side). If the numbers at the four corners sums up to 22, what is the number in the middle?


Dari bilangan 1 sampai dengan 9, bilangan genap 2, 4, 6 dan 8, sedangkan ganjil 1, 3, 5, 7, 9.Di karenakan kotak yang memiliki sisi persekutuan adalah bilangan berurutan maka posisi bilangan pada kotak diisi selang seling ganjil genap. Terdapat 5 ganjil dan 5 genap maka kemungkinan susunan bilangan adalah

Berdasarkan keterangan tabel maka posisi sudut ditempati oleh bilangan ganjil . Jumlah semua angka di sudut adalah 22. Misalkan bilangan yang di tengah adalah \(N\) maka

\(1 + 3 + 5 + 7 + 9 βˆ’ 𝑁 = 22\)
\(25 βˆ’ 𝑁 = 22\)
\(𝑁 = 3\)


24.

Figure above shows a square with sides of 8cm. inside the square are two semi circles and a small circle with their circumferences touching. Find the radius of the small circle.


Dengan menggunakan rumus Pythagoras
\(𝐴𝑂^2 = 𝐴𝐡^2 + 𝑂𝐡^2\)
\(β‡’(4 + π‘Ÿ)^2 = 42 + (4 βˆ’ π‘Ÿ)^2\)
\(β‡’(4 + π‘Ÿ)^2 βˆ’ (4 βˆ’ π‘Ÿ)^2 = 42\)
\(β‡’2π‘Ÿ(8) = 16\)
\(π‘Ÿ =\frac{16}{16}= 1\; π‘π‘š\)


25. What is the maximum radius of hemisphere that can be fit inside a cone with base radius and height of r=1cm and h=2cm. Give your answer in cm.


\(Δ𝐴𝐡𝑂 β‰ˆ Δ𝑂𝐡𝐢\)
Diperoleh perbandingan
\(\frac{𝑂𝐢}{𝑂𝐴}=\frac{𝑂𝐡}{𝐴𝐡}\)
\(\frac{π‘Ÿβ€²}{β„Ž}=\frac{π‘Ÿ}{\sqrt{β„Ž^2 + π‘Ÿ^2}}\)
\(\frac{π‘Ÿβ€²}{2}=\frac{1}{\sqrt{4 + 1}}\)
\(π‘Ÿβ€² =\frac{2}{\sqrt 5}Γ—\frac{\sqrt 5}{\sqrt 5}=\frac{2}{5}\sqrt 5\)



 

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