Elementary Mathematics International Contest Solutions

Solution 1 :
(A + B + C + C + D + E) / 5 = X
(3 + 3 + 4.5 + 4.5 + E) = 5X
X + 1 = 5X - 15
4X = 16
X = 4
E = 4 + 1 = 5 tons

Solution 3 :
A = 100 x 3 = 300 (not possible reds)
B = 105 x 3 = 315 (not possible reds)
C = 110 x 3 = 330 (not possible reds)
D = 115 x 3 = 345
E = 130 x 3 = 390 (not possible reds)

The only possible to be reds is D = 115
Green = 115 x 3 = 345.
B + C + E = 105 + 110 + 130 = 345.
A is taken away.

Solution 4 :
(10000A + 2004) divisible by 123, A = 9
92004 / 123 = 748

Solution 5 :
2 moves :
I. Take out 800g = 2000 - (1000 + 200)
II. Take in 200g from 800g in the first move.
800 - 200 = 600 g

Solution 6 :
6 x 3 = 18 minutes.
First 6 minutes 4 fishes side A
Next 6 minutes 3 fishes side B + 1 fish side A
Next 6 minutes 2 fishes side B

Solution 7 :
1 + 3 + 5 + ... <= 1012, a = 1, b = 2
1/2n x (2a + (n - 1) x b) <= 1012
1/2n x (2 + (n - 1) x 2) <= 1012
1/2n x (2 + 2n - 2) <= 1012
1/2n x 2n <= 1012
n^2 <= 1012
n <= 31

2 + 4 + 6 + ... =
1/2 x 31 x (2 x 2 + 30 x 2) =
1/2 x 31 x 64 = 992

The original number of candies in the bag = 1012 + 992 = 2004

Solution 8 :
A B C D E, A + E = 13

B = 1/3 x (B + C + D + 13)
3B = B + C + D + 13
2B = C + D + 13
B = 1/2 x (C + D + 13)

C = 1/4 x (B + C + D + 13)
4C = 1/2 x (C + D + 13) + C + D + 13
8C = C + D + 13 + 2C + 2D + 26
5C = 3D + 39
C = 1/5 x (3D + 39)

D = 1/5 x (B + C + D + 13)
5D = 1/2 x (C + D + 13) + C + D + 13
10D = C + D + 13 + 2C + 2D + 26
7D = 3C + 39
7D = 3/5 x (3D + 39) + 39
35D = 9D + 117 + 195
26D = 312
D = 12

C = 1/5 x (36 + 39) = 15

B = 1/2 x (15 + 12 + 13) = 20, is the largest number.

Solution 9 :
Basketball = 80%, Non Basketball = 20%
Football = 85%, Non Football = 15%
Baseball = 74%, Non Baseball = 26%
Volleyball = 68%, Non Volleyball = 32%
Minimum percent of the students who participated in all the four sports events =
100% - (20% + 15% + 26% + 32%) = 7%

Solution 10 :
321, 421, 431, 432, 521, 531, 532, 541, 542, 543, ---, 987
1 + (1 + 2) + (1 + 2 + 3) + (1 + 2 + 3 + 4) + (1 + 2 + 3 + 4 + 5) + (1+ 2 + 3 + 4 + 5 + 6) + (1 + 2 + 3 + 4 + 5 + 6 + 7) = 1 + 3 + 6 + 10 + 15 + 21 + 28 = 84

Solution 11 :
a^2 = b^2 +9^2
c^2 = b^2 +16^2
a^2 + c^2 = (9 + 16)^2
b^2 + 81 + b^2 + 256 = 625
2b^2 = 288
b^2 = 144
b = 12
a^2 = 144 + 81 = 225
a = 15
c^2 = 144 + 256 = 400
c = 20
a + b + c = 15 + 12 + 20 = 47