Problems and Solutions TIMO HEAT ROUND 2020-2021 Secondary 2

Thailand International Mathematical Olympiad (TIMO) is an annual Mathematical Olympiad competition organised by the Thailand Mathematics Society. As the majority of Mathematical Olympiad competitions are only focusing on 0.1% of the elite students, TIMO provides an opportunity for ALL students with strong interests in Mathematics to participate in Mathematical Olympiad competition.

What makes TIMO special is that after the competition, TIMO will release the general statistics of studentsβ performance which shows the award achievement and the percentage of Gold, Silver, Bronze winners on each grade. Winners can know the global ranking compared with all oversea students. In addition, TIMO consists of 5 main topics: Logical Thinking, Arithmetic / Algebra, Number Theory, Geometry & Combinatorics. (sc : https://www.thaiimo.com)

Berikut ini soal dan solusi TIMO Heat Round 2020-2021 Secondary 7

1.There are some chickens and rabbits in a cage. The number of rabbits is 3 times less 16 as the number of chickens. The total legs of rabbitβs is 186 more than that of chickenβs. How many rabbit(s) is / are there?

Misalkan banyak kelinci adalah $$πΎ$$ dan banyak ayam adalah $$π΄$$
$$πΎ = 3π΄ β 16 βΉ 4πΎ = 12π΄ β 64$$ β¦(1)
$$4πΎ = 2π΄ + 186$$ β¦(2)
Samakan persamaan (1) dan (2)

$$4πΎ = 4πΎ$$
$$β12π΄ β 64 = 2π΄ + 186$$
$$β10π΄ = 250$$
$$βπ΄ = 25$$

Subtitusi nilaiΒ $$A$$ ke persamaan (1), diperoleh
$$πΎ = 3π΄ β 16 = 3(25) β 16 = 75 β 16 = 59$$
Jadi banyaknya kelinci ada $$59$$ kelinci

2. There are 25 problems in a mathematics competition. The scores of each problem are allocated in the following ways: 2 marks will be given for a correct answer, 1 mark will be deducted from a blank answer and 2 marks will be deducted from a wrong answer. Find the minimum number of candidate(s) to ensure that 3 candidates will have the same scores in the competition.

Pertama kita cari dulu kemungkinan banyaknya nilai yang mungkin diperoleh siswa. 50, 47, 46, 44, 43, 42, 41, 40, 39, 38, 37, β¦β¦,2, 1, 0, β¦., – 50 . Banyak nilai yang mungkin diperoleh siswa ada 98 nilai. Agar terdapat 3 siswa yang nilainya sama maka dipastikan terdapat 2 siswa yang memiliki nilai sama dari semua nilai yang tersedia sehingga apabila ditambah satu orang apapun nilainya pasti ada 3 siswa yang nilainya sama. Jadi banyak minimum peserta untuk memastikan ada 3 peserta yang memiliki nilai yang sama dalam kompetisi tersebut adalah 98 Γ 2 + 1 = 197 siswa

3. According to the pattern shown below, what is the number in the blank?

2 γ 4 γ 18 γ 116 γ 802 γ 5604 γ __ γβ¦.

$$π_1 = 2$$
$$π_2 = 7(π_1) β 10 = 7(2) β 10 = 14 β 10 = 4$$
$$π_3 = 7(π_2) β 10 = 7(4) β 10 = 28 β 10 = 18$$
$$π_4 = 7(π_3) β 10 = 7(18) β 10 = 126 β 10 = 116$$
$$π_5 = 7(π_4) β 10 = 7(116) β 10 = 812 β 10 = 802$$
$$π_6 = 7(π_5) β 10 = 7(802) β 10 = 5614 β 10 = 5604$$
$$π_7 = 7(π_6) β 10 = 7(5604) β 10 = 39228 β 10 = 39218$$
Jadi bilangan selanjutnya adalah $$39218$$

4. There are six teams named A, B, C, D, E and F, participating a tournament. In 5 days, each team will play one game in each day. They play with another team once in the tournament. So there are 3 matches every day. Given that:
1) Team B is defeated by Team F on the first day.
2) Team A wins Team C on the second day.
3) Team E is defeated by Team B on the third day.
4) Team C wins Team F on the fourth day. Which team does Team D play with on the fifth dayοΌ

Pertandingan hari pertama : BF, AD, CE
Pertandingan hari kedua : AC, BD, FE
Pertandingan hari ketiga : BE, AF, CD
Pertandingan hari keempat : CF, AB, ED
Pertandingan dihari kelima tim D akan melawan tim F, karena tim yang lain sudah
dilawan hari sebelumnya

5. Terdapat 5 tersangka yaitu A, B, C, D, E dalam sebuah kasus. Mereka memberikan kesaksian secara berurutan di bawah ini. Diketahui hanya ada satu di antara mereka yang tidak berbohong dan hanya ada satu penjahat. Siapakah penjahat tersebut?
B : A berbohong dan saya bukanlah penjahatnya.
C : D dan E keduanya bukanlah penjahatnya.
D : C berbohong dan A bukanlah penjahatnya.

6. How many integral solutions is/are there for $$x$$ if $$β4 > \frac{7β2π₯}{4}> β7$$

$$β4 >\frac{7 β 2π₯}{4}> β7$$
$$β16 > 7 β 2π₯ > β28$$
$$β23 > β2π₯ > β35$$
$$\frac{23}{2}< π₯ <\frac{35}{2}$$

bilangan bulat $$x$$ yang memenuhi adalah $$\{12, 13, 14, 15, 16, 17\}$$, banyaknya ada $$6$$ bilangan

7. Factorize $$π₯^2 β π¦^2 β 4π₯ + 6π¦ β 5$$

$$π₯^2 β π¦^2 β 4π₯ + 6π¦ β 5$$
$$(π₯^2 β 4π₯) β (π¦^2 β 6π¦) β 5$$
$$= (π₯ β 2)^2 β 4 β ((π¦ β 3)^2 β 9) β 5$$
$$= (π₯ β 2)^2 β (π¦ β 3)^2 β 4 + 9 β 5$$
$$= (π₯ β 2)^2 β (π¦ β 3)^2$$
$$= (π₯ β 2 + π¦ β 3)(π₯ β 2 β π¦ + 3)$$
$$= (π₯ + π¦ β 5)(π₯ β π¦ + 1)$$

8. Given $$x$$ is a non-zero real number and $$π₯^2 β 6π₯ + 1 = 0$$, find the value $$x^2 + \frac{1}{x^2}$$

$$π₯^2 β 6π₯ + 1 = 0$$
Kedua ruas dibagi $$x$$
$$π₯ β 6 +\frac{1}{π₯} = 0 βΉ π₯ +\frac{1}{π₯}= 6$$
$$π₯^2 + 2π₯ (\frac{1}{π₯}) +\frac{1}{π₯^2} = 36$$
$$π₯^2 +\frac{1}{π₯^2} = 36 β 2 = 34$$