SEAMO PAPER C 2019 [PROBLEM And SOLUTION]

Problem and Solution SEAMO 2019 paper C. Soal ini bersumber dari seamo-official.org

1.A new operation is defined as

$m*n=\frac{m+n}{2}$

Evaluate $3 * (6 * 8)$

(A) 2
(B) 3
(C) 4
(D) 5
(E) None of the above

$3 * (6 * 8)=3 * \frac{6+8}{2}$
$⇒3 * 7=\frac{3+7}{2}=5$

2. 10 identical circles are arranged as shown.

AB is a straight line that cuts through the centres of 2 circles.
What is the ratio of the area above line AB to the area below line AB ?
(A) 2 : 7
(B) 2 : 3
(C) 7 : 13
(D) 9 : 11
(E) None of the above

daerah lingkaran kuning bisa ditukar sehingga masing-masing bagian membentuk lingkaran utuh. Bagian atas garis AB terdiri dari 4 bagian lingkaran sedangkan bagian bawah garis AB terdiri dari 6 bagian lingkaran.

Jadi perbandingannnya 4 : 6 = 2 : 3

3. John writes the following numbers on the whiteboard:

8, 9, 10, 11, 12, 13, 14

Each time, he selects 2 numbers to erase. Then, he replaces them with
another number that is 1 less than their sum. What will be the last number remaining on the whiteboard?
(A) 23
(B) 29
(C) 69
(D) 71
(E) None of the above

ada beberapa alternativ jawaban, namun yang kita cari apakah adalah salah satu dari pilihan jawaban di atas?
kita coba memilih secara random dua bilangan. Setelah berapa kali percobaan, yang mungkin menjadi bilangan yang tersisa di whiteboard adalah 71. Langkahnya sebagai berikut:
– menghapus bilangan 8 dan 9 lalu mengantikannya dengan 8+9-1=16, tersisa bilangan 16, 10, 11, 12, 13, 14.
– menghapus bilangan 10 dan 11 lalu mengantikannya dengan bilangan 10+11-1=20, tersisa bilangan 16, 20, 12, 13, 14
– menghapus bilangan 12 dan 13 lalu mengantikannya dengan bilangan 12+13-1=24, tersisa bilangan 16, 20, 24, 14
– menghapus bilangan 16 dan 14 lalu mengantikannya dengan bilangan 16+14-1=29, tersisa bilangan 29, 20, 24.
– menghapus bilangan 20 dan 24 lalu mengantikannya dengan bilangan 20+24-1=43, tersisa bilangan 29, 43.
– menghapus bilangan 29 dan 43 lalu mengantikannya dengan bilangan 29+43-1=71, tersisa bilangan 71.

4. At 2:45 PM, the angle formed by the hour and minute hands is $x°$ , where $0° < 3 < 180°$. What is the value of $x°$ ?
(A) 163°
(B) 169°
(C) 172.5°
(D) 175°
(E) None of the above

perhatikan gambar, setiap loncatan berwarna hitam bernilai $30º$ dan yang berwarna merah adalah $\frac{9}{12}(30º)=\frac{3}{4}(30º)$
sehingga diperoleh,
$ x = 5\times 30°+\frac{3}{4}\times 30°$
$x = 150°+22,5°=172,5º$

5. There were 4 Thursdays and 5 Fridays in the month of October some years ago. On which day of the week was the $20^{th}$ of October that year?
(A) Monday
(B) Tuesday
(C) Wednesday
(D) Thursday
(E) Sunday

Solution : C

6. $p$ and $q$ are prime numbers. Given that $2p+5q=2019$, find $p-q$
(A) 992
(B) 984
(C) 889
(D) 968
(E) None of the above

nilai $p$ dan $q$ yang memenuhi adalah $p=997$ dan $q=5$.
jadi nilai dari $p-q=997-5=992$

8. 1-, 2-, 4-, 8- and 16-gram weights are used on a balancing device. Given that the weight(s) can only be placed on one side of the device, how many different masses is the device capable of measuring?

(A) 28
(B) 29
(C) 30
(D) 31
(E) None of the above

Banyak kemungkinan berat berbeda yang bisa diukur adalah sama dengan mencari banyak himpunan bagian dari himpunan beranggotakan {1, 2, 4, 8, 16} selain himpunan kosong. Banyak cara adalah $ 2^5 – 1=31$

9. A rectangle paper is folded along the diagonal as shown in the diagram
below. Find the area of the shaded region.

(A) 11
(B) 12
(C) 13
(D) 14
(E) None of the above

$a+b=8⇒b=8-a$ … (1) dan
$a^2-b^2=16$…(2)
Subtitusi (1) ke (2)
$a^2-(8-a)^2=16$
$a^2-(64-16a+a^2)=16$
$-64+16a=16$
$a=\frac{80}{16}=5$

Jadi [BED]$=\frac{1}{2}(a)(4)=\frac{1}{2}(5)(4)=10 cm^2$

10. Group A consists of 16 numbers whose sum is 98.
Group B consists of some numbers whose average is 11.
The average of all the numbers in both groups is 8.
How many numbers are there in Group B?
(A) 9
(B) 10
(C) 11
(D) 12
(E) None of the above

misalkan banyak anggota grup B adalah $n$ maka

$\overline x=\frac{98+11n}{16+n}=8$
$⇒98+11n=8(16+n)$
$⇒98+11n=128+8n$
$⇒3n=30$
$⇒n=10$

11. In the diagram below, $∠ADB= 90°$ and $C$ is the midpoint of arc $AB$.Given that the area of the shaded region $x$ is $12 cm^2$ , find the area of the shaded region $y$ in $cm^2$.

(A) 12
(B) 14
(C) 12π
(D) 14π
(E) None of the above

Misalkan jari-jari lingkaran besar $R$ , Jari-jari lingkaran kecil adalah $\frac{R}{2}$
$y+z=\frac{1}{8}πR^2$ … (1)
$x+z=\frac{1}{2}π(\frac{R}{2})^2=\frac{1}{8}πR^2$…(2)
Persamaan (1) dan (2) bernilai sama, maka
$y+z=x+z$
$y=x=12 cm^2$

jadi luas $y$ adalah $12 cm^2$

12. 60% of the students at North Shore School were boys. 44% of the students took part in the sports meet, of which 52% were boys. What percentage of the North Shore girls did not participate?
(A) 60%
(B) 64%
(C) 68%
(D) 72%
(E) None of the above

banyak siswa laki-laki : $60\%$, banyak siswa perempuan : $40\%$.

Siswa yang ikut turnamen adalah $44\%$
Siswa laki-laki yang ikut turnamen:$\frac{52}{100}(44\%)=22,88\%$
Siswa perempuan yang ikut adalah $44\%-22,88\%=21,12\%$

jadi siswa perempuan yang tidak ikut turnamen adalah $40\%-21,12\%=18,88\%$

13. Cindy and Diane each have some savings. If Cindy spends $20 and Diane spends $10 a day, Cindy will have $3500 remaining when Diane finishes up her savings. If Cindy spends $10 and Diane spends $20 a day, Cindy will have $3950 remaining when Diane finishes up her savings. How much money does Cindy have at first?
(A) $3900
(B) $4100
(C) $4300
(D) $4500
(E) None of the above

misalkan uang Cindy mula-mula adalah C dan D, Diana menghabiskan tabungannya selama n hari. Berdasarkan keterangan pertama pada soal di peroleh :
$C – 20n=3500 ⇒ C = 20n+3500$
berdasrkan keterangan kedua pada soal di peroleh persamaan:
$C-10(\frac{n}{2})=3950 ⇒ C=5n+3950$
Samakan kedua persamaan:
$ 20n+3500=5n+3950$
$15n=450$
$n=30$

Jadi banyak uang Cindy adalah $5(30)+3950= $4100$

14. A bag contains 4 red and 4 black balls. When 3 balls are chosen at random, without replacement, the probability of getting 2 red balls and 1 black ball
is $\frac{n}{m}$.
Find $m+n$.
(A) 5
(B) 6
(C) 7
(D) 8
(E) None of the above

Kemungkinan pilihan yang terjadi adalah MMR, MRM, RMM

Peluangnya adalah
$ (\frac{2}{4})(\frac{1}{3})(\frac{1}{4})+(\frac{2}{4})(\frac{1}{4})(\frac{1}{3})+(\frac{1}{4})(\frac{2}{4})(\frac{1}{3})$
$=3\times \frac{1}{24}=\frac{1}{8}=\frac{n}{m}$

jadi nilai $n+m=9$

15. At least how many numbers from 1 to 30 must be chosen to ensure there always exists a number that is twice another?
(A) 17
(B) 18
(C) 19
(D) 20
(E) None of the above

Ambil Bilangan ganjil ada 15.
Ambil {4, 12, 16, 20, 28} sehingga pada saat pengambilan angka terakhir dipastikan ada yang nilainya 2 kali dari bilangan yang lain. Banyaknya pengambilan : 15 + 5 + 1 = 21 cara

16. Find the integer part of

$\frac{1}{\frac{1}{71}+\frac{1}{72}+\frac{1}{73}+\frac{1}{74}+\frac{1}{75}}$

(A) 14
(B) 13
(C) 12
(D) 15
(E) None of the above

$\frac{1}{\frac{1}{71}+\frac{1}{71}+\frac{1}{71}+\frac{1}{71}+\frac{1}{71}} <\frac{1}{\frac{1}{71}+\frac{1}{72}+\frac{1}{73}+\frac{1}{74}+\frac{1}{75}}<\frac{1}{\frac{1}{75}+\frac{1}{75}+\frac{1}{75}+\frac{1}{75}+\frac{1}{75}}$

$\frac{1}{\frac{5}{71}} <\frac{1}{\frac{1}{71}+\frac{1}{72}+\frac{1}{73}+\frac{1}{74}+\frac{1}{75}}<\frac{1}{\frac{5}{75}}$

$\frac{71}{5} <\frac{1}{\frac{1}{71}+\frac{1}{72}+\frac{1}{73}+\frac{1}{74}+\frac{1}{75}}<\frac{75}{5}$

$14.2 <\frac{1}{\frac{1}{71}+\frac{1}{72}+\frac{1}{73}+\frac{1}{74}+\frac{1}{75}}<15$

Jadi bagian bulatnya adalah 14

17. An equilateral triangle of side length 1 is rotated pivoting at B. Then it is turned at point C. Find the distance travelled by the point A.

(A) $\frac{7}{5}π$
(B) $\frac{6}{5}π$
(C) $\frac{5}{3}π$
(D) $\frac{4}{3}π$
(E) None of the above

Panjang putaran titik A adalah $\frac{240}{360}\times 2πr=\frac{2}{3}\times 2π=\frac{4}{3}π$

18. Find the value of $a$, when $(a-b)$ is minimum.

(A) 573
(B) 575
(C) 577
(D) 579
(E) None of the above

Misalkan suku ke-n pada barisan pertama adalah $𝑎$ dan suku ke-n pada barisan kedua adalah $𝑏$, diperoleh persamaan
$𝑎 = 1 + (𝑛 − 1)4 = 1 + 4𝑛 − 4 = 4𝑛 − 3$

$𝑏 = 1000 + (𝑛 − 1)(−3) = 1000 − 3𝑛 + 3 = 1003 − 3𝑛$

$(𝑎 − 𝑏) = 4𝑛 − 3 − (1003 − 3𝑛) = 7𝑛 − 1006$
$(𝑎 − 𝑏)$ minimum dicapai ketika $𝑛 = 144$
jadi nilai $𝑎 = 4(144) − 3 = 573$

19. Pipe A takes twice as long to fill a pool as compared with Pipe B.

It takes them 12 hours to fill a pool when turned on together.
If Pipe A alone is turned on for m hours and then turned off, Pipe B will take another 9 hours to fill the pool.
Find m.

(A) 14
(B) 16
(C) 18
(D) 20
(E) None of the above

20. Given that

$a=5^{39}, b=3^{52}, c=2^{91}$

which of the following statements is true?
(A) $a>b>c$
(B) $b>c>a$
(C) $b>a>c$
(D) $c>a>b$
(E) None of the above

21. Find the remainder if the following expression is divided by 11.

$\underbrace{2\times 2\times 2\times 2\times …\times 2\times 2\times 2}_{\mbox{2019}}$

22. $\overline{abc}$ is a 3-digit number with no repeated digits. Given that

$\overline{ab}+\overline{ba}+\overline{ac}+\overline{ca}+\overline{bc}+\overline{cb}=\overline{abc}$

the sum of digits of $\overline{abc}$ are multiples of ….

23. At 9:00 AM, Cars A and B left Towns X and Y, respectively, and travelled towards each other with their speeds in the ratio 5 : 4. After the two cars passed each other, Car A’s speed reduced by 20% while Car B’s speed increased by 20%.

Given that Car B was still 10 km away from Town X when Car A reached Town Y, find the distance between the towns.

24. The sum of a whole number $n$ and
125 is a square number. The sum of $n$ and 176 is also a square number. Find the value of $n$.

25. Evaluate

$\frac{1}{7}+\frac{3}{8}+\frac{7}{36}+\frac{29}{56}+\frac{37}{63}+\frac{41}{72}+\frac{53}{77}+\frac{29}{84}+\frac{3}{88}$

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