To solve this problem, we can set up a system of equations.
Let's use the variables x, y, and z to represent the number of gold, silver, and bronze medals, respectively.
From the problem, we can set up the following equations:
30x + 20y + 10z = 410 (Equation 1)
40x + 25y + 15z = 545 (Equation 2)
To solve this system of equations, we can use substitution or elimination method. Let's use substitution method here.
Solving Equation 1 for x, we get:
x = (410 - 20y - 10z) / 30 (Equation 3)
Substitute Equation 3 into Equation 2:
40((410 - 20y - 10z) / 30) + 25y + 15z = 545
Simplifying this equation gives us:
(4100 - 200y - 100z + 1500y + 900z) / 30 = 545
Combine like terms:
(5800 + 1300y + 800z) / 30 = 545
Multiply both sides by 30:
5800 + 1300y + 800z = 545 * 30
5800 + 1300y + 800z = 16350
Rearrange the equation:
1300y + 800z = 10550 (Equation 4)
Now we have a system of two equations (Equation 3 and Equation 4) with two variables (y and z). We can solve this system to find the possible sets of values for x, y, and z.
(a) One possible solution:
From Equation 4, we can see that z must be a multiple of 50 to get an integer value for y. Let's try z = 50:
1300y + 800(50) = 10550
1300y + 40000 = 10550
1300y = -29450
y = -22.65
Since y cannot be a negative value, this solution is not possible.
Another possible solution:
Let's try z = 0:
1300y + 800(0) = 10550
1300y = 10550
y = 8.12
Since y cannot be a decimal value, this solution is not possible either.
Therefore, there are no possible sets of values for x, y, and z that satisfy both equations.
(b) Based on the calculations above, we can conclude that the solutions obtained are the only possible sets of values.
A club has to buy x gold medals, y silver medals and z bronze medals for an event. When the prices of gold, silver and bronze medals are $30, $20 and $10 each respectively, the total cost is $410. When the prices are $40, $25 and $15 each respectively, the total cost is $545.
(a) Find two possible sets of values of x, y and z.
(b) Do you think the solutions you have obtained are the only possible sets? Give your reasons.
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