a math test is worth 100 points and gas 38 problems. Each problem is worth either 5 points or 2 points. How many problems of each point value are on the test?
To solve this problem, let's assume that there are x problems worth 5 points and y problems worth 2 points on the test.
Given that there are a total of 38 problems on the test, we can write the equation:
x + y = 38 ----(Equation 1)
The total points earned from problems worth 5 points would be 5x and from problems worth 2 points would be 2y.
We know that the test is worth 100 points, so we can write another equation:
5x + 2y = 100 ----(Equation 2)
Now we have a system of two equations (Equation 1 and Equation 2) with two variables (x and y). We can solve this system of equations using various methods like substitution or elimination.
Let's solve it using the elimination method:
Multiply Equation 1 by 5 and Equation 2 by 2 to make the coefficients of x in both equations the same:
5(x + y) = 5(38)
2(5x + 2y) = 2(100)
Simplifying these equations, we have:
5x + 5y = 190 ----(Equation 3)
10x + 4y = 200 ----(Equation 4)
Now, we can eliminate the variable y by subtracting Equation 3 from Equation 4:
10x + 4y - (5x + 5y) = 200 - 190
10x - 5x + 4y - 5y = 10
5x - y = 10 ----(Equation 5)
We now have a new equation (Equation 5) with only one variable, x. We can solve this equation to find the value of x:
5x - y = 10
5x = 10 + y
x = (10 + y)/5
Since the number of problems cannot be in fractions or decimals, we need to find positive integer solutions for x and y.
To find the values of x and y, we can substitute different values for y (starting with y = 0) and calculate the corresponding values of x:
When y = 0, x = (10 + 0)/5 = 2
When y = 1, x = (10 + 1)/5 = 2.2 (decimal value)
When y = 2, x = (10 + 2)/5 = 2.4 (decimal value)
...
We can observe that as y increases, x also increases in decimal values. However, we need positive integer values for x and y. So, let's continue substituting values until we find integer solutions.
When y = 3, x = (10 + 3)/5 = 2.6 (decimal value)
When y = 4, x = (10 + 4)/5 = 2.8 (decimal value)
When y = 5, x = (10 + 5)/5 = 3
Now we have found an integer solution where x = 3 and y = 5. This means there are 3 problems worth 5 points and 5 problems worth 2 points on the test.
just write down what you know.
If there are x at 5 points and y at 3 points, then
x+y = 38
5x+2y = 100