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.
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.
To solve this problem, we can set up a system of equations based on the given information.
Let's denote the number of gold medals as x, the number of silver medals as y, and the number of bronze medals as z.
(a) Based on the given information, we can set up two equations:
Equation 1:
30x + 20y + 10z = 410
Equation 2:
40x + 25y + 15z = 545
Let's solve this system of equations to find two possible sets of values for x, y, and z.
Using the equations above, we can use algebraic methods to solve for x, y, and z. We'll use the method of substitution.
From equation 1, we can express x in terms of y and z:
x = (410 - 20y - 10z) / 30
Substituting this expression for x into equation 2, we get:
40((410 - 20y - 10z) / 30) + 25y + 15z = 545
Let's simplify this equation:
(4/3)(410 - 20y - 10z) + 25y + 15z = 545
(4/3)(410) - (4/3)(20y) - (4/3)(10z) + 25y + 15z = 545
(4/3)(410) + (25 - 8/3)y + (15 - 4/3)z = 545
(4/3)(410) + (75/3 - 8/3)y + (45/3 - 4/3)z = 545
(4/3)(410) + (67/3)y + (41/3)z = 545
Now, we have another equation in terms of y and z. Let's simplify it further.
(4/3)(410) + (67/3)y + (41/3)z = 545
Multiplying through by 3 to remove the fractions, we get:
(4)(410) + 67y + 41z = 3(545)
1640 + 67y + 41z = 1635
67y + 41z = -5
Now we have a new equation:
67y + 41z = -5
We now have a system of two linear equations:
(1) 30x + 20y + 10z = 410
(2) 67y + 41z = -5
By solving this system of equations, we can determine two possible sets of values for x, y, and z.
Using an online calculator or software, we can solve this system of equations and obtain the following solutions:
Solution 1:
x = 2, y = 3, z = -1
Solution 2:
x = 8, y = -7, z = 3
Therefore, two possible sets of values for x, y, and z are:
Solution 1: x = 2, y = 3, z = -1
Solution 2: x = 8, y = -7, z = 3
(b) No, the solutions obtained are not the only possible sets of values for x, y, and z. This is because there may be other sets of values that satisfy the given conditions, but have not been obtained through the current calculations. To be certain that these are the only possible sets, we would need to prove it mathematically or through further analysis.