Elementary Mathematics International Contest (IMC 2008) Solutions

Solution 2 :
3x + 2y = 32, y = (32 - 3x) / 2
4x + 3y = 44
4x + 3(32 - 3x) /2 = 44
8x + 96 - 9x = 88
x = 8
y = (32 - 24) / 2 = 4
2x + y = 16 + 4 = 20

Solution 3 :
sit + stand = 100, stand = 100 - sit.
0.75 stand + 0.25 sit = 70
0.75(100 - sit) + 0.25 sit = 70
75 - 0.75 sit + 0.25 sit = 70
5 = 0.50 sit
sit = 10

Solution 4 :
(110000 - 100000) / 3600 x t = (17 + 3)
t = 20 x 3600 / 10000 =  7.2 seconds

Elementary Mathematics International Contest (IMC 2008)

1. Starting from the central circle, move between two tangent circles. What is the number of ways of covering four circles with the numbers 2, 0, 0 and 8 inside, in that order?

2. Each duck weighs the same, and each duckling weighs the same. If the total weight of 3 ducks and 2 ducklings is 32 kilograms, the total weight of 4 ducks and 3 ducklings is 44 kilograms, what is the total weight, in kilograms, of 2 ducks and 1 duckling?

3. If 25% of the people who were sitting stand up, and 25% of the people who werestanding sit down, then 70% of the people are standing. How many percent of the people were standing initially?

4. A sedan of length 3 metres is chasing a truck of length 17 metres. The sedan is
travelling at a constant speed of 110 kilometres per hour, while the truck is travelling at a constant speed of 100 kilometres per hour. From the moment when the front of the sedan is level with the back of the truck to the moment  when the front of the truck is level with the back of the sedan, how many seconds would it take?

5. Consider all six-digit numbers consisting of each of the digits ‘0’, ‘1’, ‘2’, ‘3’, ‘4’ and ‘5’ exactly once in some order. If they are arranged in ascending order, what is the 502^nd number?

6. How many seven-digit numbers are there in which every digit is ‘2’ or ‘3’, and
no two ‘3’s are adjacent?

7. How many five-digit multiples of 3 have at least one of its digits equal to ‘3’?

8. ABCD is a parallelogram. M is a point on AD such that AM=2MD, N is a point on AB such that AN=2NB. The segments BM and DN intersect at O. If the area of
ABCD is 60 cm2, what is the total area of triangles BON and DOM?

9. ABCD is a square of side length 4 cm. E is the midpoint of AD and F is the midpoint of BC. An arc with centre C and radius 4 cm cuts EF at G, and an arc with centre F and radius 2 cm cuts EF at H. The difference between the areas of the region bounded by GH and the arcs BG and BH and the region bounded by EG, DE and the arc DG is of the form m? ?n cm², where m and n are integers. What is the value of m+n?

10.  In a chess tournament, the number of boy participants is double the number of girl participants. Every two participants play exactly one game against each other. At the end of the tournament, no games were drawn. The ratio between the number of wins by the girls and the number of wins by the boys is 7:5. How many boys were there in the tournament?

11. In the puzzle every different symbol stands for a different digit.

What is the answer of this expression which is a five-digit number?

12. In the figure below, the positive numbers are arranged in the grid follow by the arrows’ direction.

For example,
“8”is placed in Row 2, Column 3.
“9” is placed in Row 3, Column 2.

Which Row and which Column that “2008” is placed?

13. As I arrived at home in the afternoon. The 24-hour digital clock shows the time as below (HH:MM:SS). I noticed instantly that the first three digits on the platform clock were the same as the last three, and in the same order. How many times in twenty four hours does this happen?

Note: The clock shows time from 00:00:00 to 23:59:59.

SOAL LATIHAN OLIMPIADE

1. Untuk setiap m, kurva y=(m−1)x+n+2 selalu melalui titik A. Koordinat titik A adalah ...
2. Angka satuan dalam 31001.71002.131003 adalah ...
3. Banyaknya digit dalam 416 x 525 jika dituliskan dalam bilangan tidak berpangkat adalah ....
4. Sebuah toko menetapkan harga suatu barang dalam dollar dan cents demikian hingga jika ditambahkan pajak penjualan 4%, tidak perlu diadakan pembulatan sebab hasilnya tepat n dollar. Nilai terkecil untuk  n adalah ...
5. Andaikan x = 0,123456789101112...998999 yang angkanya diperoleh dengan menuliskan berurutan dari 1 hingga 999. Angka ke 1983 di sebelah kanan desimal adalah...
6. Jika a dan b adalah bilangan real positif dan masing-masing persamaan x2+ax+2b=0 dan x2+2bx+a=0 mempunyai akar real. Nilai minimun a + b = ....
7. Luas permukaan suatu balok adalah 22 cm2 dan panjang semua rusuk-rusuknya 24 cm. Panjang diagonal ruangnya adalah ....
8. 12+44+98+1616+2532+...= ...
9. Jika diketahui bahwa 14y2−20y+48−−−−−−−−−−−−−√+14y2−20y−15−−−−−−−−−−−−−√=9, maka tentukan nilai dari 14y2−20y+48−−−−−−−−−−−−−√−14y2−20y−15−−−−−−−−−−−−−√.
10. 15−10+1+15−9+1+...+15−1+1+150+1+...+1510+1=
11. Tiga orang pelari A, B, dan C berlomba dengan menempuh lintasan yang sama. Ketika pelari A mencapai finish, pelari B tertinggal 12 yard dan pelari C tertinggal 18 yard. Sedangkan pelari B mencapai finish, pelari C masih tertinggal 8 yard. Berapakah panjang lintasan yang mereka tempuh jika diasumsikan mereka berlari dengan kecepatan konstan?(dalam yard)
12. 5x+1−51−x=11. Tentukan nilai x!
13. Nilai dari 1+28.29.30.31−−−−−−−−−−−−√ adalah ...
14. Dari gambar di samping, ∠A+∠B+∠C+∠D+∠E= ...
15. Koordinat titik pada garis y = 2x-15 yang terdekat dengan titik asal adalah ...
16. Diketahui seperti tergambar di atas. Titik A, B, dan C terletak pada lingkaran yang berpusat di O. Jika ∠AOC=60∘, maka ∠OCB−∠OAB= ...

Jawaban :

1. y=(m−1)x+m+2 
Kita tidak dapat mengetahui nilai m-nya karena nilai dari m tersebut sebarang. 
Yang dicari di soal ini adalah untuk sebarang nilai m kita harus mencari koordinat yang selalu dilewati grafik tersebut. 
Perhatikan untuk x=−1 nilai dari y tidak tergantung nilai dari m, 
maka subtitusi x=−1 kita dapat y=3. 
Berarti koordinat yang selalu dilalui adalah titik (−1,3)

2. Angka satuan sama dengan sisa pembagian dengan 10.

31001⋅71002⋅131003(mod10)=====(3⋅7)1001⋅7⋅31003(mod10)1⋅7⋅3⋅31002(mod10)31002(mod10)34k+2(mod10)9(mod10).

3. Jawabannya 28 digit.

416⋅525====232⋅525(2⋅5)25⋅27128⋅1025128000…028 digit.

4. x + 0.04x = n, 1,04x = n, x > 10, 0,4x = 1, x = 2.5, x = 10 + 2.5 = 12.5
diperoleh n terkecil  13 dengan harga sebelum pajak 12,50 dollar

5. Angka dua digit habis pada 9+90.2=189 digit pertama. 

Sehingga masih dibutuhkan 1983−189=1794 

Angka itu diperoleh 17943=598, jadi angka ke-1983 adalah 8

6. x^2 + ax + 2b = x^2 + 2bx + a = 0. Karena a dan b positif, maka x harus negatif.
Misalkan x = -1
1 - a + 2b = 0, a = 1 + 2b
1 - 2b + a = 0
1 - 2b + 1 + 2b <> 0
Misalkan x = -2
4 - 2a + 2b = 0, a = 2 + b
4 - 4b + 2 + b = 0
6 - 3b = 0, b = 2, a = 5
a + b = 5 + 2 = 7
7. Perhatikan bahwa diagonal ruang suatu balok adalah p2+l2+t2−−−−−−−−−√. Dari luas permukaan balok didapat persamaan 

2(pl+lt+tp)pl+lt+tp==2211.

Dari panjang sisi-sisi balok didapat

4(p+l+t)p+l+t==246.

 Maka diagonal ruangnya didapat dengan cara

p2+l2+t2p2+l2+t2−−−−−−−−−√====(p+l+t)2−2(pl+lt+tp)36−221414−−√.

8. Un = n^2 / 2^n,
Sn = 1/2n (a + Un) = 1/2n (1/2 + n^2 / 2^n) = n/4 + n^3 / 2^(n+1)

9. Idenya sederhana.

x9x9x9xx=====14y2−20y+48−−−−−−−−−−−−−√−14y2−20y−15−−−−−−−−−−−−−√(14y2−20y+48−−−−−−−−−−−−−√+14y2−20y−15−−−−−−−−−−−−−√)(14y2−20y+48−−−−−−−−−−−−−√−14y2−20y−15−−−−−−−−−−−−−√)14y2−20y+48−(14y2−20y−15)637.

10. Perhatikan bahwa 15−a+1+15a+1=1. Maka

===15−10+1+15−9+1+…+15−1+1+150+1+…+1510+110+150+11012212.

11. S = Va x t1, S = Vb x t1 + 12, S = Vc x t1 + 18, S = Vb x t2, S = Vc x t2 + 8
t1 = S / Va, S = VbS/Va + 12, S = VcS/Va + 18, t2 = S / Vb, S = VcS/Vb +8
VbS/Va + 12 = VcS/Va + 18, VbS = VcS + 8Vb
(Vb - Vc)S = 6Va, (Vb - Vc)S = 8Vb
6Va = 8Vb
Va = 4/3Vb
S = VbS/Va + 12
S = 3S/4 + 12
S/4 = 12
S = 48 yard
 

12.

13. Bentuk tersebut merupakan penjabaran dari bentuk (n(n+1) – 1)²
Karena (n(n+1) – 1)² = (n-1)n(n+1)(n+2) + 1
Akar kuadrat dari (n(n+1) – 1)² = n(n+1) – 1
Jadi, akar kuadrat dari 1+28.29.30.31= 29.30-1= 869.

14.

15. Titik A (7.5, 0), titik B (0, -15)
 Luas Segitiga OAB = 1/2 x 15 x 7.5 = 56.25
Garis AB = akar(7.5^2 + 15^2) = 16.77
1/2 x 16.77 x t = 56.25, t = 6.7

16.Misalkan BOC = X.
Maka OCB = (180 - X) / 2 = 90 - X/2
Dan OAB = (180 - (60 + X)) / 2 = 60 - X/2
OCB - OAB = 90 - X/2 - (60 - X/2) = 30

Buktinya :
OA = OB = OC, AOC = OAC = OCA = 60.
Jika OA // BC, maka OCB = 60, dan BOC = 60
Sehingga BOA = 60 + 60 = 120, OAB = OBA = 30
OCB - OAB = 60 - 30 = 30

Jika BO dan OA segaris (OAB = OBA = 0), maka BOC = 120
Sehingga OCB = OBC = 30
OCB - OAB = 30 - 0 = 30