Gold Panning Wa State

how to solve 38SB+8VB=698? i worked to this step and got stucked?

For the Pan Australian Games the three states at the top of the medals table (Victoria, S.A. and W.A) had won a total of 186 medals. Victoria has won the most gold medals and SA had as many gold medals as bronze medals. Victoria and SA won the same number of silver medals. WA had two more silver medals than bronze medals and their gold medals numbered one more than Victoria’s bronze medals. Victoria had as many gold medals as the bronze medals of SA and WA combined and this number was also three quarters the total number of medals won by SA. The number of gold medals won by the three states was one less than the total number of medals won by Victoria. How many of each medal did each state win?

Let Victoria win x, y, z number of gold, silver and bronze medals respectively, similarily, SA wins p, q and r and WA wins a, b and c.
Then the various conditions can be written as
1. x + y + z + p + q + r + a + b + c = 186
2. p = r
3. y =q
4. b = c + 2
5.a= z +1
6. x = r + c
7. x = (3/4)*(p + q + r)
8. x + p + a = x + y + z -1

5 and 8 give :

p = y – 2 ——(9)

9, 2, 3,7 give

x = (3/4)*(p+ y +p)
=> x = (3/4)*(2p + y)
=> x = (3/4)*(2y – 4 + y)
=> x = (3/4)*(3y – 4)——(10)

6, 9 and 10 give:
(3/4)*(3y – 4) = y – 2 + c
=> c = 2 – 3 + (9/4)y – y
=> c = (5/4)y -1 ————(11)

4 and 11 give:

b = (5/4)y -1 +2
=> b = (5/4)y +1 ————–(12)

Using 2, 3,5, 9, 10, 11,12 in 1

(3/4)*(3y – 4) + y + z + (y – 2) + y + (y – 2) + a + (5/4)*y + 1 +(5/4)*y – 1 = 186

=> a + z + (35/4) * y = 193
=> 2*a + (35/4) * y = 194
=> a = (194 – (35/4)*y) / 2 —– (13)

Now it has been mentioned that Victoria has won the most gold medals i.e., x > p and x > a.

Its evident from equation 13 that “y” cannot exceed the value of 22.17 ( “a” will be negative otherwise)

Now by trial and error method,

Case 1: for y = 4,
a = 79.5 , should be whole number, not possible.

Case 2: for y = 8,
a = 62 , x = 15, not possible (x >a).

Case 3: for y = 12,
a = 44.5, should be a whole number, not possible.

Case 4: for y = 16,
a = 27 , x = 33, possible.
p = 14

Case 5: for y = 20,
a = 9.5 , should be a whole number, not possible.

So case 4 solves our problem:

Victoria :
no of gold metals: x = 33
no of silver medals: y = 16
no of bronze medals:z = 26

SA:
no of gold metals: p = 14
no of silver medals: q = 16
no of bronze medals:r = 14

WA:
no of gold metals: a = 27
no of silver medals: b = 21
no of bronze medals:r = 19

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