SEAMO PAPER D 2019 [PROBLEM And SOLUTION]

SEAMO SMP

13. Evaluate

\(\sqrt{14+6\sqrt 5} – \sqrt{14-6\sqrt 5}\)

(A) 5
(B) 2√5
(C) √5
(D) 1
(E) None of the above


\(\sqrt{14+6\sqrt 5} – \sqrt{14-6\sqrt 5}\) 

\(=\sqrt{14+2\sqrt{9⋅5}} – \sqrt{14-2\sqrt{9⋅5}}\)

\(=\sqrt 9 + \sqrt 5 – (\sqrt 9 – \sqrt 5)\)

\(=2\sqrt 5\)


14. Evaluate

\(\frac{1}{2}+\frac{2}{4}+\frac{3}{8}+\frac{4}{16}+…+\frac{8}{2^8}\)

(A) \(\frac{249}{128}\)
(B) \(\frac{251}{128}\)
(C) \(\frac{449}{256}\)
(D) \(\frac{551}{256}\)
(E) None of the above


Misalkan:
\(A=\frac{1}{2}+\frac{2}{4}+\frac{3}{8}+\frac{4}{16}+…+\frac{8}{2^8}\)
\(2𝐴 =\frac{2}{2}+\frac{2}{2}+\frac{3}{4}+\frac{4}{8}+ ⋯ +\frac{8}{2^7}\)
______________________________________________________________________________ _
\(𝐴 = 1 +\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+ ⋯ +\frac{1}{2^7} −\frac{8}{2^8}\)

Selanjutnya misalkan
\(𝐵 = 1 +\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+ ⋯ +\frac{1}{2^7}\)
\(2𝐵 = 2 + 1 +\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+ ⋯ +\frac{1}{2^6}\)
__________________________________________________________________________  _
\(𝐵 = 2 −\frac{1}{2^7}\)

Kita peroleh
\(𝐴 = 1 +\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+ ⋯ +\frac{1}{2^7}−\frac{8}{2^7}\)
\(= 𝐵 −\frac{8}{2^7} = 2 −\frac{1}{2^7} −\frac{8}{2^8}\)
\(= 2 −\frac{1}{2^7} −\frac{4}{2^7}\)
\(= 2 −\frac{5}{2^7} =\frac{2^8 − 5}{2^7}\)
\(=\frac{256 − 5}{128}\)
\(=\frac{251}{128}\)


15. Find a positive integer \(x\), such that

\(2^6 + 2^{10} + 2^x\)

is a perfect square.
(A) 7
(B) 8
(C) 9
(D) 10
(E) None of the above


\(2^6+2^{10}+2^𝑥=2^6(1+2^4+2^{𝑥−6})\)
Karena \(2^6\) kuadrat sempurna maka \((1+2^4+2^{𝑥−6})\) juga merupakan kuadrat sempurna
\((1+2^4+2^{𝑥−6})=17+2^{𝑥−6}\)
Nilai \(2^{𝑥−6}\) yang memenuhi adalah
\(2^{𝑥−6}=8 ⟹2^{𝑥−6}=2^3⟹𝑥=9\)


16. Given that a, b and c are positive real numbers and \(a+ b + c = 1\), evaluate

\(\sqrt{4a + 1} + \sqrt{4b + 1} + \sqrt{4c + 1}\)

(A) 4
(B) 5
(C) < 4
(D) > 5
(E) None of the above



17. A bag contains 20 balls numbered 1 to 20. If 2 balls are randomly chosen from the bag, find the probability that the sum of the two numbers is prime.

(A) \(\frac{23}{40}\)
(B) \(\frac{29}{40}\)
(C) \(\frac{31}{380}\)
(D) \(\frac{17}{190}\)
(E) None of the above


Ruang sampel \(𝑛(𝑆)={20\choose 2}=\frac{20!}{18!.2!}=\frac{20.19.18!}{18!.2}=190\) cara
Banyak kemungkin jumlah angka pada \(2\) bola adalah bilangan prima.
• Berjumlah \(3, (1, 2)\) ada \(1\)
• Berjumlah \(5, (1, 4), (2, 3)\) ada \(2\)
• Berjumlah \(7, (1, 6), (2, 5), (3, 4)\) ada \(3\)
• Berjumlah \(11, (1,10), (2, 9), (3, 8), (4, 7), (5, 6)\) ada \(5\)
• Berjumlah \(13, (1, 12), (2, 10), …, (6, 7)\) ada \(6\)
• Berjumlah \(17, (1, 16), … (8, 9)\) ada \(8\)
• Berjumlah \(19, (1, 18), …, (9, 10)\) ada \(9\)
• Berjumlah \(23, (4, 19), …, (11, 12) ada [latex]8\)
• Berjumlah \(29, (9, 20), (10, 19), …., (14,15)\) ada \(6\)
• Berjumlah \(31, (11, 20), …., (15, 16)\) ada \(5\)
• Berjumlah \(37, (17, 20), (18, 19)\) ada \(2\)
Banyaknya ada \(55\)
Jadi peluangnya adalah \(\frac{55}{190}=\frac{11}{38}\)


18. Mr. Woodley paid $650,000 for two commercial properties. A year later, the market value of the 1st property rose by 20%, while that of the 2nd dropped by 25%. The two properties then had the same market value. Of what market value was the 1st property at first?
(A) $220,000
(B) $230,000
(C) $240,000
(D) $250,000
(E) None of the above


Misalkan harga Property pertama adalah \(A\) dan harga Property kedua adalah \(B\)

\(𝐴 + 𝐵 = 650000\)

Setelah setahun

\(120\%𝐴 = 75\%𝐵\)
\(\frac{𝐴}{𝐵}=\frac{75}{120}=\frac{5}{8}\)

Jadi harga properti pertama \(𝐴 =\frac{5}{13}(650000) = $250000\)


Pages ( 3 of 4 ): « Previous12 3 4Next »

Leave a Reply

Your email address will not be published. Required fields are marked *