# Problem And Solution SEAMO 2017 Paper D

The Southeast Asia Mathematical Olympiad (SEAMO) is an international Math Olympiad competition that originated in Singapore and was founded by Mr. Terry Chew in 2016 in 8 Southeast Asian Countries. Since then, it is growing its popularity around the world. In 2019 it was recognized by 18 countries. In 2020 total number of participating countries increased to 22, including students from Indonesia, Brazil, China, Newzealand, and Taiwan students enrolled in SEAMO 2022-23.

Problem and Solution SEAMO 2017 paper D. Soal ini bersumber dari seamo-official.org

1. Find the value of x in $$\frac{x}{x+3}-\frac{x}{4-x}$$.
(A) – 12
(B) + 12
(C) – 24
(D) + 24
(E) + 30

$$\frac{𝑥}{3 + 𝑥}−\frac{𝑥}{4 − 𝑥}= 2$$
$$⇒\frac{𝑥}{3 + 𝑥}= 2 +\frac{𝑥}{4 − 𝑥}$$
$$⇒\frac{𝑥}{3 + 𝑥}=\frac{2(4 − 𝑥)}{4 − 𝑥}+\frac{𝑥}{4 − 𝑥}$$
$$⇒\frac{𝑥}{3 + 𝑥}=\frac{8 − 2𝑥}{4 − 𝑥}+\frac{𝑥}{4 − 𝑥}=\frac{8 − 𝑥}{4 − 𝑥}$$
$$⇒ 𝑥(4 − 𝑥) = (3 + 𝑥)(8 − 𝑥)$$
$$⇒ 4𝑥 − 𝑥^2 = 24 − 3𝑥 + 8𝑥 − 𝑥^2$$
$$⇒ 4𝑥 − 8𝑥 + 3𝑥 = 24$$
$$⇒−𝑥 = 24 ⇒ 𝑥 = −24$$

2. Find all positive values of $$n$$, such that $$2^𝑛 − 1$$ is divisible by $$7$$

(A) $$n$$ must be a multiple of $$2$$
(B) $$n$$ must be a multiple of $$3$$
(C) $$n$$ must be a multiple of $$4$$
(D) $$n$$ must be a multiple of $$5$$
(E) None of the above

$$2^𝑛 − 1 ≡ 0\; 𝑚𝑜𝑑\; 7$$
$$⇒(2^3)^{\frac{𝑛}{3}} − 1 ≡ 0\; 𝑚𝑜𝑑\; 7$$
$$⇒(8)^{\frac{𝑛}{3}} − 1 ≡ 0\; 𝑚𝑜𝑑\; 7$$
$$⇒(1)^{\frac{𝑛}{3}} − 1 ≡ 0\; 𝑚𝑜𝑑\; 7$$
$$⇒1 − 1 ≡ 0\; 𝑚𝑜𝑑\; 7$$
Karena $$\frac{𝑛}{3}$$ bilangan bulat positif maka nilai $$n$$ adalah semua bilangan bulat positif kelipatan 3

3. A certain grade of Colombian and Indonesian coffee are mixed in the ratio m : n. The Colombian coffee cost $40 and the Indonesian coffee cost$60 if the cost of Indonesian coffee is increased by 15% and the cost of Colombian coffee decreased by 15% the cost of the mixture remained unchanged. Find m : n.

(A) 1:2
(B) 2:3
(C) 3:2
(D) 2:1
(E) 5:2

$$40𝑚 + 60𝑛 = 1,15(60)𝑛 + 0,85(40)𝑚$$
$$40𝑚 + 60𝑛 = 69𝑛 + 34𝑚$$
$$6𝑚 = 9𝑛$$

Jadi perbandingan $$𝑚 ∶ 𝑛 = 3 ∶ 2$$

4. Evaluate

$$\frac{400^2×(254^2+246^2)×(254^4+246^4)}{(254^8−246^8)}$$

(A) 30
(B) 40
(C) 50
(D) 60
(E) 70

misalkan $$𝑎 = 254, 𝑏 = 246$$

$$\frac{400^2 × (254^2 + 246^2) × (254^4 + 246^4)}{(254^8 − 246^8)}$$

$$=\frac{400^2 × (𝑎^2 + 𝑏^2) × (𝑎^4 + 𝑏^4)}{(𝑎^8 − 𝑏^8)}$$

$$=\frac{400^2 × (𝑎^2 + 𝑏^2) × (𝑎^4 + 𝑏^4)}{(𝑎^4 − 𝑏^4)(𝑎^4 + 𝑏^4)}$$

$$=\frac{400^2 × (𝑎^2 + 𝑏^2)}{(𝑎^2 − 𝑏^2)(𝑎^2 + 𝑏^2)}$$

$$=\frac{400^2}{(𝑎^2 − 𝑏^2)}$$

$$=\frac{400^2}{(254^2 − 246^2)}$$

$$=\frac{400^2}{(254 − 246)(254 + 246)}$$

$$=\frac{400^2}{8(500)}$$

$$= 40$$

5. Evaluate

$$\frac{20172016^2}{20172015^2+20172017^2−2}$$

(A) $$\frac{1}{3}$$
(B) $$\frac{1}{2}$$
(C) $$1$$
(D) $$\frac{3}{2}$$
(E) None of the above

Misalkan $$20172016 = 𝑥$$

$$\frac{20172016^2}{20172015^2+20172017^2−2}$$

$$=\frac{𝑥^2}{(𝑥 − 1)^2 + (𝑥 + 1)^2 − 2}$$

$$=\frac{𝑥^2}{𝑥^2 − 2𝑥 + 1 + 𝑥^2 + 2𝑥 + 1 − 2}$$

$$=\frac{𝑥^2}{2𝑥^2} =\frac{1}{2}$$

6. It is known that $$𝑚 =\frac{𝑎}{𝑏+𝑐}=\frac{𝑏}{𝑎+𝑐}=\frac{𝑐}{𝑎+𝑏}$$
Given $$𝑎 + 𝑏 + 𝑐 ≠ 0$$, find the value of $$m$$

(A) $$\frac{1}{4}$$
(B) $$\frac{1}{3}$$
(C) $$\frac{1}{2}$$
(D) $$1$$
(E) None of the above

$$𝑚 =\frac{𝑎}{𝑏+𝑐}⟹ 𝑎 = 𝑚(𝑏 + 𝑐)$$
$$𝑚 =\frac{𝑏}{𝑎+𝑐}⟹ 𝑏 = 𝑚(𝑎 + 𝑐)$$
$$𝑚 =\frac{𝑐}{𝑎+𝑏}⟹ 𝑐 = 𝑚(𝑎 + 𝑏)$$
Jumlahkan ketiga persamaan diperoleh
$$𝑎 + 𝑏 + 𝑐 = 𝑚(2𝑎 + 2𝑏 + 2𝑐)$$
$$𝑎 + 𝑏 + 𝑐 = 2𝑚(𝑎 + 𝑏 + 𝑐)$$
karena $$𝑎 + 𝑏 + 𝑐 ≠ 0$$, maka
$$1 = 2𝑚$$
$$𝑚 =\frac{1}{2}$$

7. A rectangle is inscribed in a square as shown. It is known that the total area of the 4 isosceles right angled Δ is 98 cm². Find $$|xy|$$, the length of the diagonal of rectangle.

(A) 10
(B) 11
(C) 12
(D) 13
(E) 14

$$𝑚^2 + 𝑛^2 = 98$$
Dengan menggunakan sifat segitiga siku-siku sama kaki panjang persegi panjang biru adalah $$𝑚\sqrt 2$$ dan lebarnya adalah $$𝑛\sqrt 2$$.
Panjang $$|XY|$$ dicari dengan menggunakan rumus Pythagoras:

$$|𝑋𝑌| = \sqrt{(𝑚\sqrt 2)^2 + (𝑛\sqrt 2)^2}$$
$$= \sqrt {2𝑚^2 + 2𝑛^2}$$
$$= \sqrt{2(𝑚^2 + 𝑛^2)}$$
$$= \sqrt{2(98)}$$
$$= \sqrt{2⋅2⋅49}$$
$$= 2(7) = 14$$ cm

8. Evaluate

$$(\frac{2}{3}+\frac{3}{4}+…+\frac{49}{50})(\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+…+\frac{49}{50})(\frac{2}{3}+\frac{3}{4}+…+\frac{48}{49})$$

(A) $$\frac{49}{100}$$
(B) $$\frac{99}{100}$$
(C) $$\frac{49}{50}$$
(D) $$\frac{48}{49}$$
(E) $$1$$

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