GUANGDONG-HONG KONG-MACAO GREATER BAY AREA MATHEMATICAL OLYMPIAD 2019 (MOCK EXAM) SECONDARY 8

Berikut ini soal dan solusi MOCK PAPER BBC 2019 Secondary 8, semoga bermanfaat

1. Find the remainder of \(2014^{504}÷7\) .

\(\begin{align}
2014^{504}\; mod\; 7 &= 5^{504}\; mod\; 7\\
&= (5^3)^{168}\; mod\; 7\\
&= (125)^{168}\; mod\; 7\\
&= (−1)^{168}\; mod\; 7\\
&= 1\; mod\; 7\\
&= 1
\end{align}\)

2. Refer to the figure below, \(AB = AC. AD = DE = EF = FB = BC\). Find the value of \(∠DEF\).

misalkan \(∠BAC=a\) karena \(AB = AC. AD = DE = EF = FB = BC\) maka diperoleh sudut yang lain seperti gambar berikut

\(∠𝐴 + ∠𝐵 + ∠𝐶 = 180° ⟹ 𝑎° + 4𝑎° + 4𝑎° = 180° ⟹ 𝑎 = 20°\)
\(∠𝐷𝐸𝐹 = 180° − 4𝑎° = 180° − 4(20°) = 180° − 80° = 100°\)

3. According to the following equation, find the value of \(a+b+c+d\)

\(a+b+c=17\)
\(a+b+d=15\)
\(a+c+d=10\)
\(b+c+d=15\)

Jumlahkan keempat persamaan
\(3𝑎 + 3𝑏 + 3𝑐 + 3𝑑 = 57\)
\(𝑎 + 𝑏 + 𝑐 + 𝑑 = 19\)

4. Solve the equation \(𝑥^3 – x^2 – 4x – 6 = 0\) .

\(𝑥^3 – x^2 – 4x – 6 = 0\)
\(⇒𝑥^2(𝑥 − 3) + 2𝑥^(𝑥 − 3) + 2(𝑥 − 3) = 0\)
\(⇒(𝑥 − 3)(𝑥2 + 2𝑥 + 2) = 0\)
Diperoleh \(𝑥 − 3 = 0\) atau \(𝑥^2 + 2𝑥 + 2 = 0\)
Untuk \(𝑥 − 3 = 0 ⇒ 𝑥 = 3\)
Untuk \(𝑥^2 + 2𝑥 + 2 = 0\) (tidak mempunyai solusi karena \(𝐷 < 0\))

5. Factorize \(𝑥^2 + 3𝑥𝑦 + 2𝑦^2 − 6𝑥 − 7𝑦 + 5\)

\((𝑥 + 2𝑦 − 5)(𝑥 + 𝑦 − 1)\)

6. Inside a class election, we have to choose 1 for chairperson, 3 for secretary and 1 for finance. If 30 students can take the above posts, how many post arrangements are there?

\({30\choose 1}{29\choose 3}{26\choose 1}= 2.850.120\) cara

7. If \(x\) and \(y\) are positive integers, find the number of solutions of

\(\frac{1}{𝑥}+ \frac{1}{𝑦} =\frac{1}{8}\)

\(\frac{1}{𝑥}+ \frac{1}{𝑦} =\frac{1}{8}\)
\(\frac{𝑥 + 𝑦}{𝑥𝑦}=\frac{1}{8}\)
\(𝑥𝑦 = 8𝑥 + 8𝑦\)
\(𝑥𝑦 − 8𝑥 − 8𝑦 = 0\)
\((𝑥 − 8)(𝑦 − 8) = 64\)
Jadi banyaknya pasangan \((𝑥, 𝑦)\) sama dengan banyaknya factor positif dari \(64\) yaitu sebanyak \(7\) pasang

8. For any positive integers \(n\), it is known that \(n -5\) and \(𝑛^2 + 10\) are prime numbers. Find the value of \(n\).

Karena \(𝑛^2 + 10 > 2\) dan bilangan prima lebih dari dua selalu ganjil maka nilai \(𝑛\) pasti ganjil.
Karena \(n\) ganjil maka \(𝑛 − 5\) genap, \(𝑛 − 5\) bilangan prima, bilangan prima genap yang memenuhi hanya angka \(2\), jadi nilai \(𝑛 = 7\)

9. Simplify \(\sqrt{3 + 2\sqrt 2} − \sqrt{3 − 2\sqrt 2}\)

\(\sqrt{3 + 2\sqrt 2} − \sqrt{3 − 2\sqrt 2}=\sqrt 2 + 1 − (\sqrt 2 − 1) = 2\)

10. From 1 to 100 to choose numbers such that any 2 numbers are not in an integral-multiple relationship, how many numbers can be chosen?

Bilangan yang diambil adalah bilangan prima \(\{2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97\}\) jadi banyaknya bilangan yang dipilih ada 25 bilangan.

11. The perimeter of a rectangle with integral lengths is 2014. Find the minimum value of the area of this rectangle.

\(𝐾 = 2(𝑝 + 𝑙) = 2014\)
\(𝑝 + 𝑙 = 1007\)
Agar luas minimal maka nilai \(𝑝 = 1006\) dan \(𝑙 = 1\), jadi luasnya adalah \(1006 × 1 = 1006\)

12. Find the value of \(\frac{13577}{13581×13578−13579^2}\)

Misalkan \(𝑎 = 13579\)

\(\frac{13577}{13581×13578−13579^2}\)\(=\frac{𝑎−2}{(𝑎+2)(𝑎−1)−𝑎^2}\)\(=\frac{a-2}{a^2+a-2-a^2}\)\(=\frac{𝑎−2}{𝑎−2}=1\)

13. Refer to the figure below, \(ABCD\) is a square. \(E\) and \(F\) lie on \(BC\) and \(CD\) respectively. It is
known that \(BE + FD =EF\) . Find the value of \(∠EAF\) .

Karena \(𝐵𝐸 + 𝐹𝐷 = 𝐸𝐹\) maka \(𝐸𝐺 =𝐵𝐷\) dan \(𝐺𝐹 = 𝐹𝐷\), diperoleh \(Δ𝐴𝐵𝐸 ≅Δ𝐴𝐸𝐺\) dan \(Δ𝐴𝐺𝐹 ≅ Δ𝐴𝐹𝐷\)
Karena \(Δ𝐴𝐵𝐸 ≅ Δ𝐴𝐸𝐺\) dan \(Δ𝐴𝐺𝐹 ≅Δ𝐴𝐹𝐷\) maka \(∠𝐵𝐴𝐸 = ∠𝐸𝐴𝐺\) dan \(∠𝐺𝐴𝐹 = ∠𝐹𝐴𝐷\)
\(∠𝐵𝐴𝐸 + ∠𝐸𝐴𝐺 + ∠𝐺𝐴𝐹 + ∠𝐹𝐴𝐷 = 90°\)
\(2(∠𝐸𝐴𝐺 + ∠𝐺𝐴𝐹) = 90°\)
\((∠𝐸𝐴𝐺 + ∠𝐺𝐴𝐹) = 45°\)
\(∠𝐸𝐴𝐹 = 45°\)

14. It is know that \(𝑥 +\frac{1}{𝑥}=3\). Find the value of \(𝑥^3 − 𝑥^2 − 5𝑥 + 5\).

\(𝑥 +\frac{1}{𝑥}= 3 ⇒ 𝑥^2 − 3𝑥 + 1 = 0\)
Selanjutnya

\(𝑥^3−𝑥^2−5𝑥+5\)
\(=𝑥(𝑥^2−3𝑥+1)+2𝑥^2−6𝑥+5\)
\(=2𝑥^2−6𝑥+5\)
\(=2(𝑥^2−3𝑥+1)+3\)
\(= 3\)

15. It is known that \(𝑎\) dan \(𝑏\) are positive integers and \(𝑎^2 − 𝑏^2 = 20140504\). Find the maximum value of \(𝑏\).

\(𝑎^2 − 𝑏^2 = 20140504\)
\((𝑎 + 𝑏)(𝑎 − 𝑏) = 5035126 × 4\)
Diperoleh \((𝑎 + 𝑏) = 5035126\) dan \(𝑎 − 𝑏 = 4\)
Kurangkan kedua persamaan diperoleh
\(2𝑏 = 5035122 ⟹ 𝑏 = 2.517.561\)

16. A sequence of positive integers \(𝑎_1, 𝑎_2, 𝑎_3, …\) satisfies the relation: \(𝑎_𝑛 = 𝑎_{𝑛−1} + 2𝑎_{𝑛−2}\) and \(𝑛 ≥ 3\). If \(𝑎_2 > 𝑎_1\) and \(𝑎_4 = 11\), what is the possible maximum value of \(𝑎_1\)?

Untuk \(𝑛 = 3\)
\(𝑎_3 = 𝑎_2 + 2𝑎_1\)
Untuk \(𝑛 = 4\)
\(𝑎_4 = 𝑎_3 + 2𝑎_2\)
\(11 = 𝑎_2 + 2𝑎_1 + 2𝑎_2 = 𝑎_2 + 4𝑎_1\)
Pasangan \((𝑎_1, 𝑎_2)\) yang memenuhi adalah \((1, 7), (2, 3)\)
Jadi nilai maksimum \(𝑎_1\) adalah \(3\)

17. Find the unit digit of \(11^2 + 13^3 + 15^3 + ⋯ + 199^3\)

Belum tersedia

18. Inside a right-anged triangle, \( sin\;θ=\frac{7}{25}\). Find the value of \(Tan\;θ\)

Belum tersedia

19. If \(𝑥 = 1,2,3, . . . ,200\) and substitute all \(x\) into \(𝑦 = | 𝑥 − 100|\) . Find the sum of all values of \(y\).

\(|1 − 100| + |2 − 100| + |3 − 100| + ⋯ + |100 − 100| + |101 − 100 + ⋯ + |200 − 100|\)
\(= (99 + 98 + ⋯ + 1) + 0 + (1 + 2 + 3 + ⋯ + 100)\)
\(= 4950 + 5050\)
\(= 10000\)

20. Solve the equation of \((𝑥2 − 3𝑥 + 2)^{𝑥−1} = 0\)

\(𝑥^2 − 3𝑥 + 2 = 0\)
\((𝑥 − 2)(𝑥 − 1) = 0\)
\(𝑥 = 2\) atau \(𝑥 = 1\)(tidak memenuhi karena \(\frac{0}{0} ≠ 0\))
Jadi solusinya adalah \(𝑥 = 2\)

21. Find the least positive integral solution of \(11x ≡7 (mod\;19)\)

\(11x ≡7 (mod\;19)\)
\(22x ≡14 (mod\;19)\)
\(3x ≡14 (mod\;19)\)
\(18x ≡84 (mod\;19)\)
\(-x ≡8 (mod\;19)\)
\(x ≡-8 (mod\;19)\)
\(x ≡11 (mod\;19)\)
\(𝑥 = 11\)

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