Asian Science And Math Olympiad (ASMO) 2018 For Grade 7

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 7


1. What is the value of \(2018×(-2)^3 – 2018^0\)


\(𝑎^0 = 1, 𝑎 ≠ 0\)

\(2018 × (−2)^3– 2018^0 = 2018 × (−8) − 1 = −16144 − 1 = −16145\)


2. The number of days already passed in February 2018 is 6 times of the number of days left. How many days have passed in the month?


Banyak hari pada bulan februari adalah 28 hari
Misalkan banyak hari yang tersisa adalah A karena Jumlah hari yang sudah dilalui pada bulan Februari 2018 adalah 6 kali lipat dari jumlah hari yang tersisa maka dapat ditulis dalam persamaan

\(6𝐴 + 𝐴 = 28\)
\(⇒7𝐴 = 28\)
\(⇒𝐴 = 4\)

Jadi banyak hari yang telah berlalu pada bulan tersebut adalah \(6𝐴 = 6(4) = 24\) hari


3. David is a football coach. He needs 30 footballs in training. He now has 5 balls and will buy the rest in half-dozen per package, P. How many packages David has to buy in order to have 30 footballs. Give your answer in inequality form.


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4. In a hexagon, first angle is 36°. The second angle is 4 times as large as the first. The third angle is 96° more than the second. The fourth angle is 50°. The ratio between the fifth and sixth angle is 4:1. What is the measure of the fifth angle, in degree?


Misalkan sudut segienam adalah \(∠𝐴_1, ∠𝐴_2, ∠𝐴_3, ∠𝐴_4, ∠𝐴_5, ∠𝐴_6\)
\(∠𝐴_1 = 36°\)
\(∠𝐴_2 = 4(∠𝐴_1) = 4(36) = 144°\)
\(∠𝐴_3 = 96 + ∠𝐴_2 = 96 + 144 = 240°\)
\(∠𝐴_4 = 50°\)
\(∠𝐴_5: ∠𝐴_6 = 4 : 1\)

Misalkan \(∠𝐴_5 = 4𝑥\) dan \(∠𝐴_6 = 𝑥\)

\(∠𝐴_1 + ∠𝐴_2 + ∠𝐴_3 + ∠𝐴_4 + ∠𝐴_5 + ∠𝐴_6 = 720°\)
\(36 + 144 + 240 + 50 + 4𝑥 + 𝑥 = 720\)
\(470 + 5𝑥 = 720\)
\(5𝑥 = 250\)
\(𝑥 = 50°\)

Jadi besar sudut kelima adalah \(∠𝐴5 = 4𝑥 = 4(50°) = 200°\)


5. Maria is playing a game. For each of her turn, she rolls a dice with numbers 1-6. She must also pick a card from a poker deck which excludes the 4 joker cards. For her to win the game, she needs to roll a ‘6’ on the dice and pick a ‘Diamond Q’ from the poker deck. What is the probability that Maria will win the game?


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6. Bryan eats supper every 5th day. If he ate supper on Monday (assume the date is 1 March 2018), what is the date that he will be eating his next supper again on a Monday?


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7. Mr. John wants to cut down a tall tree in front of his house. He separates the tree into 10 equal portions and cut them one by one. He does not plan to take a rest and starts cutting the tree at 7:55a.m. in the morning. If a portion of tree took him 33 minutes, at what time can he fully cut off the tree?


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8. There are 6 trains per day from Madrid to Rome and 5 buses per day from Rome to Prague. How many ways can a traveler arrive at Prague from Madrid?


\(6 × 5 = 30\) 𝑐𝑎𝑟𝑎


9. In a row of 50 students, Ali is 30th from the right end and Ralph is 32nd from the left. How many students are between them?


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10. How many ways are there to arrange the word ‘GERONA’ such that the vowels can only interchange with vowels, consonants can only interchange with consonants.


Bilangan konsonan hanya dapat ditukar dengan bilangan konsonan demikian juga dengan vocal. Cara menyusun konsonan adalah \(3!=6\) cara dan cara menyusun vocal adalah \(3!=6\) cara. Banyak cara seluruhnya adalah \(6 × 6 = 36\) cara


11. A grocery buys good at 30% discount from a wholesaler. If he wished to mark-up the selling price by 20% and still make a profit of RM 50, what is the price before discount offered by the wholesaler?


Modal selalu dianggap \(100%\), misalkan harga beli dari grosir adalah \(N\)
\(\frac{𝑁}{50}=\frac{100}{20}⇒ 𝑁 = 5 × 50 = 250\)

Selanjtnya harga awal grosir dianggap \(100\%\), misalakan harga grosir sebelum diskon adalah \(M\)
\(\frac{𝑀}{250}=\frac{100}{70}⇒M=\frac{25000}{70}=RM357.14\)


12. How many degrees are in the angle formed by the hands of a clock at 4:12?


sudut yang dibentuk jarum pendek dan panjang pada pukul 04 : 12 adalah
\(∠A -∠B+∠C=4 × 30 −\frac{12}{60}× 360 +\frac{12}{60}× 30 = 120 − 72 + 6 = 54° \)


13. Ms. Geetha brought 40 sweets to school as a gift to her students. She has 8 students and each of them has a plastic bag to collect the sweets. Each of the students gets different number of sweets. No one gets exactly 5 sweets, and no one gets nothing. What is the greatest number of sweets that a student can get?


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14. The diagram from a top view of a solid shows the combination of 6 equilateral triangles and a square. If the area of square EFGH is 36cm2, what is the area of the region shaded in blue?


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15. In a Sunday football league table, Blue FC has 8 points below league average. Red FC has 5 points above league average. Green FC has 95 points. The average points of these 3 teams were equal to the league average. What was the league average?


Misalkan rata-ratanya \(x\) maka

\(\frac{(𝑥 − 8) + (𝑥 + 5) + 95}{3}= 𝑥\)
\(⇒ 2𝑥 + 92 = 3𝑥\)
\(⇒ 3𝑥 − 2𝑥 = 92\)
\(⇒ 𝑥 = 92\)


16. In a competition, a participant scores 2 points for each correct answer and gets a deduction of 4 points for each incorrect answer. Jeffrey completed 80 questions in total and scored -2 points. How many questions did he answer wrongly?


Misalkan banyak benar adalah \(B\) dan banyak salah adalah \(S\)
\(𝐵 + 𝑆 = 80 … (1)\)
\(2𝐵 − 4𝑆 = −2 … (2)\)
Eliminasi persamaan \(2×(1)\) dan \((2)\)
\(2𝐵 + 2𝑆 = 160\)
\(2𝐵 − 4𝑆 = −2\)
______________ –
\(6𝑆 = 162\)
\(⇒ 𝑆 =\frac{162}{6}= 27\)
Jadi jawaban yang dijawab salah sebanyak \(27\) pertanyaan


17. The numbers from 1 through 9 we placed in the grid, exactly one per box without repeats. The number shown at the end of each row is the products of the number in the row. The number shown at the bottom of each column is the products of the numbers in that column. Complete the grid below.


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18. The diagram represents a cone of height 9 cm made in 2 parts. The top part (small cone) is mathematically similar to the whole cone. The volume of the large cone is 12π cm³. Find the volume of small cone. (volume of cone \(=\frac{1}{3}πr² h\)) Leave your answer in fraction and π .


\(𝑉_{𝐵𝑒𝑠𝑎𝑟} =\frac{1}{3}𝜋𝑅^2(9) = 12𝜋\)
\(3𝑅^2 = 12\)
\(𝑅^2 = 4\)
\(𝑅 = 2\)

Karena \(Δ𝐵𝐶𝐷 ≈ Δ𝐵𝑂𝐴\) maka berlaku perbandingan
\(\frac{𝐶𝐷}{𝑂𝐴}=\frac{𝐵𝐶}{𝐵𝑂}⇒\frac{𝐶𝐷}{2}=\frac{6}{9}\)
\(⇒ 𝐶𝐷 =\frac{12}{9}=\frac{4}{3}\)
Diperoleh jari-jari kerucut kecil adalah \(𝐶𝐷 = 𝑟 =\frac{4}{3}\)

Jadi volume kerucut kecil adalah \(\frac{1}{3}𝜋𝑟^26 =\frac{1}{3}𝜋(\frac{4}{3})^2 6 =\frac{1}{3}𝜋(\frac{16}{9})6 =\frac{32}{9}𝜋 𝑐𝑚^3\)


19. From 2,4,6,8 and 10, we choose 2 different numbers x and y.
What is the largest possible value of \(\frac{3(10-x)}{2(12-y)}\)?


untuk mendapatkan nilai maksimum pilih \(x\) minimum dan \(y\) maksimum, nilai \(x\) dan \(y\) yang mungkin adalah \(2\) dan \(10\)

Jadi \(\frac{3(10 − 𝑥)}{2(12 − 𝑦)}=\frac{3(10 − 2)}{2(12 − 10)}= 6\)


20. Billy brings his family for a 3 days trip to Perth. According to a weather forecast, the chance of raining on the first day is \(\frac{1}{2}\) , on the second day is \(\frac{2}{3}\) and third day is \(\frac{1}{3}\) . What is the probability that it will rain on at least 2 of 3 days.


Hari pertama , peluang hujan \(𝑃(𝐻_1)=\frac{1}{2}\) dan peluang tidak hujan \(𝑃(𝑇𝐻_1)=\frac{1}{2}\)
Hari kedua , peluang hujan \(𝑃(𝐻_2)=\frac{2}{3}\) dan peluang tidak hujan \(𝑃(𝑇𝐻_2)=\frac{1}{3}\)
Hari ketiga , peluang hujan \(𝑃(𝐻_3) =\frac{1}{3}\) dan peluang tidak hujan \(𝑃(𝑇𝐻_3)=\frac{2}{3}\)
Peluang setidaknya 2 hari hujan ada 2 kemungkinan yaitu 2 hari hujan 1 hari tidak hujan dan 3 hari hujan.

Peluang 2 hari hujan 1 hari tidak adalah :
\(𝑃(𝐻_1). 𝑃(𝐻_2). 𝑃(𝑇𝐻_3) + 𝑃(𝐻_1). 𝑃(𝑇𝐻_2). 𝑃(𝐻_3) + 𝑃(𝑇𝐻_1). 𝑃(𝐻_2). 𝑃(𝐻_3)\)
\(=\frac{1}{2}(\frac{2}{3})(\frac{2}{3})+\frac{1}{2}(\frac{1}{3})(\frac{1}{3})+\frac{1}{2}(\frac{2}{3})(\frac{1}{3})\)
\(=\frac{4 + 1 + 2}{18}=\frac{7}{18}\)

Peluang 3 hari hujan
\(𝑃(𝐻_1). 𝑃(𝐻_2). 𝑃(𝐻_3) =\frac{1}{2}(\frac{2}{3})(\frac{1}{3})=\frac{2}{18}\)

Jadi peluanya setidaknya dua hari hujan adalah
\(\frac{7}{18}+\frac{2}{18}=\frac{9}{18}=\frac{1}{2}\)


21. A restaurant with 16 tables can sit 88 people. Some tables can sit 4 people and some can sit 8 people. What is the ratio of the number of 4 person tables to the number of 8 person tables?


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22. \(X, Y\) and \(Z\) are 3 prime numbers. They are all larger than 5. What is the minimum value of \( x^2+ y^2+ z^2\).


\(𝑥^2 + 𝑦^2 + 𝑧^2 = 7^2 + 7^2 + 7^2 = 49 + 49 + 49 = 147\)


23. Find the range of the values of \(P\) if the equation \(x^2 + 2px -5p-6=0\) has \(2\) real roots.


Syarat suatu persamaan kuadrat memiliki \(2\) akar real adalah

\(𝐷 > 0\)
\(𝑏^2 − 4𝑎𝑐 > 0\)
\((2𝑝)^2 − 4(1)(−5𝑝 − 6) > 0\)
\(4𝑝^2 + 20𝑝 + 24 > 0\)
\(𝑝^2 + 5𝑝 + 6 > 0\)
\((𝑝 + 3)(𝑝 + 2) > 0\)

Range nilai \(p\) yang memenuhi \(𝑝 < −3\) atau \(𝑝 > −2\)


24. The area of rectangle ABFG is 50cm² and each side-length is a counting number of cm. F is the midpoint of EG. The area of square CDEF is between 17cm² and 48cm². Find the perimeter of the entire diagram, in cm.


Karena sisi-sinya bilangan cacah maka kemungkinan panjang sisi bangun di bawah ini berdasarkan keterangan soal di atas adalah

\(\begin{align}
\text{Keliling} &= 𝐴𝐺 + 𝐺𝐹 + 𝐸𝐷 + 𝐷𝐶 + 𝐶𝐵 + 𝐵𝐴\\
&= 10 + 10 + 5 + 5 + 5 + 5\\
&= 40\; 𝑐𝑚\\
\end{align}\)


25. A regular hexagon in inscribed in a circle ABCDEP so that each side of the hexagon touches the circle. AD 10 cm is the diameter of the big circle. AP, is the diameter of the small circle. Find the circumference of the small circle. (circumference \(=2πr\) .Leave your answer in π.


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