There was a surge in math club membership after the fun Halloween party. When the principal asked Mrs. Germain how many students were currently in the club, she replied, "Three times our number plus a third of our number plus a fourth of our number plus you and me is two hundred sixty."
The principal smiled and nodded, then added, "If you can pick up another eight students before your Christmas party you will have exactly 10% of the student body involved in the club."
How many students are enrolled in the math club? How many students are in the school?
Be sure to explain how you found your answer.
Please help
a. 260 - 2 = 258 Students in club.
b. 258 + 8 = 0.1n.
n = 2660 Students in the school.
So, if there are x in the club, and y students in all, then we know that
3x + x/3 + x/4 + 2 = 260
x+8 = y/10
so, just find x, and then use that to find y.
Let's solve this step by step.
Step 1: Set up equations based on the given information.
From Mrs. Germain's response:
3x + (1/3)x + (1/4)x + 2 = 260
From the principal's statement:
3x + (1/3)x + (1/4)x + 2 + 8 = 0.1s
Where:
x = the number of students currently in the math club
s = the total number of students in the school
Step 2: Simplify the equations.
3x + (1/3)x + (1/4)x + 2 = 260
Multiplying through by 12 to remove fractions:
36x + 4x + 3x + 24 = 3120
43x + 24 = 3120
43x = 3096
x = 72
3x + (1/3)x + (1/4)x + 2 + 8 = 0.1s
Substituting x = 72:
3(72) + (1/3)(72) + (1/4)(72) + 2 + 8 = 0.1s
216 + 24 + 18 + 10 = 0.1s
268 = 0.1s
s = 2680
Step 3: Solve for the number of students in the math club and the total number of students in the school.
From the equations:
x = 72 (number of students in the math club)
s = 2680 (total number of students in the school)
Therefore, there are 72 students enrolled in the math club and 2680 students in the school.
To find the number of students currently enrolled in the math club, we need to solve two equations using two variables. Let's call the number of students currently in the club "x" and the total number of students in the school "y."
The first equation is given by Mrs. Germain's statement:
3x + (1/3)x + (1/4)x + 2 = 260
The second equation is given by the principal's statement:
x + 8 = 0.1y
Let's solve these equations step by step:
Equation 1:
3x + (1/3)x + (1/4)x + 2 = 260
Multiplying through by 12 to remove the fractions:
36x + 4x + 3x + 24 = 3120
43x + 24 = 3120
Subtract 24 from both sides:
43x = 3096
Divide both sides by 43:
x = 72
Equation 2:
x + 8 = 0.1y
Substituting x = 72:
72 + 8 = 0.1y
80 = 0.1y
Divide both sides by 0.1:
y = 800
So, there are currently 72 students in the math club and the total number of students in the school is 800.
To find the answer, we solved the simultaneous equations provided by Mrs. Germain's and the principal's statements. Equation 1 was obtained by interpreting Mrs. Germain's statement, where the sum of three times the number of students, a third of the number of students, a fourth of the number of students, and the counts of Mrs. Germain and the principal equals 260. Equation 2 was obtained by interpreting the principal's statement, where the sum of the number of students and 8 equals 10% of the total student body. By solving these equations, we found the values of x and y, which represent the number of students in the math club and the total number of students in the school, respectively.