Soal Menarik dari Math Camp dan AMC year 9 2011

Soal 1 :
a + b = 100, a x b = 25, 1/a + 1/b = ?

Penyelesaian 1 :
1/a + 1/b = b/ab + a/ab = (b + a)/ab = 100/25 = 4

Soal 2 :
(100 – 1^2) x (100 – 2^2) x …….. x (100 – 2011^2) = ?

Penyelesaian 2 :
(100 – 1^2) x (100 – 2^2) x …….. x (100 – 2011^2) =
100 x (1 - 1^2) x (1 - 2^2) x ... x (1 - 2011^2) =
Karena 1 - 1^2 = 0, maka hasilnya = 0

Soal 3 :
Selisih jari-jari 2 lingkaran = 7. Berapa selisih kelilingnya ?

Penyelesaian 3 :
rA - rB = 7
(2 x 22/7 x rA)  - (2 x 22/7 x rB) =
2 x 22/7 x (rA - rB) =
2 x 22/7 x 7 = 44

Problem 1 :
How many 3-digit numbers can be written as the sum of three (not necessarily different) 2-digit numbers ?

Answer 1 :
Nilai maksimal bilangan 2-digit adalah 99 
Nilai maksimal bilangan 3-digit yang bisa dicapai adalah 99 + 99 + 99 = 297

Nilai minimal bilangan 3-digit adalah 100 yang bisa di dapat dari 33 + 33 + 34
maka ada 297 - 100 + 1 = 198 bilangan 3-digit yang dapat ditulis sebagai penjumlahan bilangan 2-digit.

Problem 2 :
A 36 cm by 24 cm rectangle is drawn on 1 cm grd paper such that the 36 cm side contains 37 grid points and the 24 cm side contain 25 grid points. A diagonal of the rectangle is drawn. how many grid points lie on that diagonal ?


Answer 2 :
See that the diagonal gradient will be 23 which means for every 3 grid on the 36 cm direction it will got 1 point. hence it will have 363+1=13points lying on it.

Problem 3 :
I drive a distance of 200 km around the city and my car's average speed is 25 km/h. How far do I then need to drive at an average speed of 100 km/h to raise my car's average speed for the whole time to 40 km/h ?

Answer 3 :
s = v x t.
200 = 25 x t
t = 8

200+100t8+t=40
200+100t=320+40t
60t=120
t=2

s = 100 x 2 = 200 km

Problem 4 :
Mary has 64 square blue tiles and a number of square red tiles. All tiles are the same size. She makes a rectangle with red tiles inside and blue tiles on the perimeter. What is the largest number of red tiles she could have used ?

Answer 4 :
Assume that the rectangle has side of x and y. 
So the number of square blue tiles needed will be 2x+2y+4(4 corners)
We can build the equation to find x and y which must be integer:
2x+2y+4=64
2x+2y=60
x+y=30
The maximum amount will be reached if x=yso, 2x=30
x=15
x2=225


Problem 5 :
A 40 X 40 white square is divided into 1 X 1 square by lines parallel to its sides. Some of these 1 X 1 squares are coloured red so that each of the 1 X 1 squares, regardless of whether it is coloured red or not, shares a side with at most one red square. (not counting itself). What is the largest possible number of red squares ?

Answer 5 :
40 x 40/2 = 800, pola zig zag