Lee and Fred are elementary school teachers. Fred works for a charter school in Pacific Palisades, California, where class size reduction is a goal for 2013. Lee works for a noncharter school where funds do not allow for class size reduction policies. Lee’s fifth-grade class has 1.2 times as many students as Fred’s.
a. If there are a total of 55 students, how many students does Fred’s class have? (Round your answers to the nearest whole number. )
b. If there are a total of 55 students, how many students does Lee’s class have? (Round your answers to the nearest whole number. )
I would appreciate some help and learning the formula! I have finals in the morning and I can't seem to solve this!
After cutting through all that unnecessary verbiage ....
"Lee’s fifth-grade class has 1.2 times as many students as Fred’s."
Fred's class-size --- x
Lee's class-size --- 1.2x
Now simply solve:
x + 1.2x = 55
23
32
Sure! Let's break down the problem step by step and solve it.
Given information:
- Total number of students: 55
- Lee's class has 1.2 times as many students as Fred's class.
Step 1: Assign variables for the number of students in Fred's class and Lee's class.
Let's say the number of students in Fred's class is 'x'. So, Fred's class has x students.
Since Lee's class has 1.2 times as many students as Fred's class, the number of students in Lee's class would be 1.2x.
Step 2: Write down the equation based on the given information.
The total number of students is equal to the sum of the number of students in Fred's class and Lee's class.
55 = x + 1.2x
Step 3: Solve the equation to find the number of students in Fred's class.
Combine like terms:
55 = 2.2x
Divide both sides of the equation by 2.2 to isolate x:
55/2.2 = x
25 = x
So, Fred's class has approximately 25 students (rounded to the nearest whole number).
Step 4: Find the number of students in Lee's class.
We found that Fred's class has 25 students, and Lee's class has 1.2 times as many students as Fred's class:
Lee's class = 1.2 * Fred's class
Lee's class = 1.2 * 25
Multiply:
Lee's class = 30
So, Lee's class has approximately 30 students (rounded to the nearest whole number).
To summarize:
a. Fred's class has approximately 25 students.
b. Lee's class has approximately 30 students.
I hope this helps! Good luck with your finals!
To solve this question, we can set up a system of equations and use algebraic methods to find the answers.
Let's assign variables to the number of students in Fred's class and Lee's class. We'll call the number of students in Fred's class "F" and the number of students in Lee's class "L".
According to the information given, we know that Lee's class has 1.2 times as many students as Fred's class. This can be written as the equation L = 1.2F.
We also know that the total number of students in both classes is 55. This can be written as the equation F + L = 55.
Now we have a system of two equations with two variables:
L = 1.2F (Equation 1)
F + L = 55 (Equation 2)
To solve this system, we can use the substitution method:
Step 1: Rearrange Equation 1 to solve for L in terms of F:
L = 1.2F
Step 2: Substitute the expression for L from Equation 1 into Equation 2:
F + 1.2F = 55
Step 3: Combine like terms:
2.2F = 55
Step 4: Divide both sides by 2.2 to solve for F:
F = 55 / 2.2
Step 5: Calculate F using a calculator:
F ≈ 25
Now that we have the value of F, we can substitute it back into Equation 1 to solve for L:
L = 1.2F
L = 1.2 * 25
L ≈ 30
Therefore, we have found the answers to the questions:
a. Fred's class has approximately 25 students.
b. Lee's class has approximately 30 students.
Remember to round the answers to the nearest whole number, as stated in the question. Good luck with your finals!