A caterer is making cookie trays for upcoming holiday parties. This morning, she made 3 small trays and 2 large trays, which contain a total of 133 cookies. In the afternoon, she made 5 small trays and 4 large trays, which contain a total of 290 cookies. How many cookies do the different sized trays contain?
Let's use algebra to solve this problem step by step.
Let's assume the number of cookies in a small tray is represented by 's' and the number of cookies in a large tray is represented by 'l'.
According to the problem, in the morning, the caterer made 3 small trays and 2 large trays, which contain a total of 133 cookies. So we can form the equation:
3s + 2l = 133 -- Equation 1
In the afternoon, the caterer made 5 small trays and 4 large trays, which contain a total of 290 cookies. So we can form the equation:
5s + 4l = 290 -- Equation 2
Now we have a system of equations with two unknowns (s and l). We can solve this system of equations simultaneously to find the values of s and l.
Step 1: Multiply Equation 1 by 4 and Equation 2 by 2 to eliminate the variable 'l'.
4(3s + 2l) = 4(133)
2(5s + 4l) = 2(290)
Simplifying these equations gives us:
12s + 8l = 532 -- Equation 3
10s + 8l = 580 -- Equation 4
Step 2: Subtract Equation 3 from Equation 4 to eliminate the variable 'l'.
(10s + 8l) - (12s + 8l) = 580 - 532
Simplifying this equation gives us:
-2s = 48
Step 3: Divide both sides of the equation by -2 to solve for the value of 's'.
s = 48 / -2
s = -24
Step 4: Substitute the value of 's' into either Equation 3 or Equation 4 to solve for the value of 'l'. Let's use Equation 3.
12s + 8l = 532
Substituting the value of 's' gives us:
12(-24) + 8l = 532
-288 + 8l = 532
Step 5: Add 288 to both sides of the equation to isolate 'l'.
-288 + 288 + 8l = 532 + 288
Simplifying this equation gives us:
8l = 820
Step 6: Divide both sides of the equation by 8 to solve for the value of 'l'.
l = 820 / 8
l = 102.5
Since the number of cookies cannot be a fraction, we need to check our calculations.
In Equation 1: 3s + 2l = 133,
Substituting the values s = -24 and l = 102.5 gives us:
3(-24) + 2(102.5) = 133
-72 + 205 = 133
133 = 133
The calculations check out, so our solution is valid.
Therefore, a small tray contains 24 cookies and a large tray contains 102.5 cookies.
To solve this problem, we need to set up a system of equations. Let's denote the number of cookies in a small tray as "s" and the number of cookies in a large tray as "l".
From the given information, we can set up the following equations:
3s + 2l = 133 (equation 1)
5s + 4l = 290 (equation 2)
To solve this system of equations, we can use the method of substitution or elimination. In this case, let's use the method of substitution.
Solve equation 1 for "s":
3s + 2l = 133
3s = 133 - 2l
s = (133 - 2l)/3 (equation 3)
Substitute equation 3 into equation 2:
5s + 4l = 290
5((133 - 2l)/3) + 4l = 290
Multiply both sides by 3 to get rid of the fraction:
5(133 - 2l) + 12l = 870
665 - 10l + 12l = 870
2l = 205
l = 205/2
l = 102.5
Substitute the value of l into equation 3 to find s:
s = (133 - 2(102.5))/3
s = 133 - 205/3
s = 133 - 68.333
s = 64.667
So, the small tray contains approximately 64.667 cookies, and the large tray contains approximately 102.5 cookies.
3S+2L = 133
5S+4L = 290
Now solve as usual