Question
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.
(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.
Answers
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GPT 3.5
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.
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.
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