One half of the boys and one-third of the girls of Freemont High attended the homecoming game, wheras one-third of the boys and one-half of the girls attended the homecoming dance. If there were 570 students at the game and 580 at the dance, then how many students are there at Freemont High?
b/2 + g/3 = 570 --->3b+2g = 3420 (#1)
b/3 + g/2 = 580 --->2b+3g = 3480 (#2)
#1 times 2 --> 6b+4g = 6840
#2 times 3 --> 6b+12g = 10440
subtract them:
8g = 3600
g = 450
back in #1:
3b + 900 = 3420
3b=2520
b=840
450 boys and 840 girls
Let's assume the total number of boys at Freemont High is B and the total number of girls is G.
Given that one-half of the boys and one-third of the girls attended the homecoming game, we can write the equation:
(1/2) * B + (1/3) * G = 570
Similarly, for the homecoming dance, we can write the equation:
(1/3) * B + (1/2) * G = 580
To solve this system of equations, we can multiply both equations by 6 to eliminate the fractions:
3B + 2G = 3420 (Equation 1)
2B + 3G = 3480 (Equation 2)
Now, let's solve these equations using the method of substitution or elimination.
Multiplying Equation 1 by 2 and Equation 2 by 3, we get:
6B + 4G = 6840 (Equation 3)
6B + 9G = 10440 (Equation 4)
Subtracting Equation 3 from Equation 4, we have:
5G = 3600
Dividing both sides by 5, we get:
G = 720
Substituting the value of G back into Equation 1, we have:
3B + 2(720) = 3420
3B + 1440 = 3420
3B = 1980
Dividing both sides by 3, we get:
B = 660
Therefore, the total number of boys is 660 and the total number of girls is 720.
To find the total number of students at Fremont High, we add the number of boys and girls:
Total number of students = B + G
Total number of students = 660 + 720
Total number of students = 1380
So, there are 1380 students at Freemont High.
To find the total number of students at Freemont High, we need to set up a system of equations based on the given information.
Let's assume:
- The total number of boys at Freemont High is B.
- The total number of girls at Freemont High is G.
According to the given information, one-half of the boys and one-third of the girls attended the homecoming game, which means:
- The number of boys at the game is (1/2)B.
- The number of girls at the game is (1/3)G.
We are also given that 570 students attended the game, so we can write the first equation:
(1/2)B + (1/3)G = 570 ----(Equation 1)
Similarly, one-third of the boys and one-half of the girls attended the homecoming dance, which means:
- The number of boys at the dance is (1/3)B.
- The number of girls at the dance is (1/2)G.
We are given that 580 students attended the dance, so we can write the second equation:
(1/3)B + (1/2)G = 580 ----(Equation 2)
To solve this system of equations, we can use substitution or elimination method. In this case, let's use the elimination method to find the values of B and G.
First, multiply Equation 1 by 6 and Equation 2 by 6 to eliminate fractions:
3B + 2G = 3420 ----(Equation 3) [Multiply Equation 1 by 6]
2B + 3G = 3480 ----(Equation 4) [Multiply Equation 2 by 6]
Now, subtract Equation 4 from Equation 3:
(3B + 2G) - (2B + 3G) = 3420 - 3480
B - G = -60
Rewriting the equation, we get:
B = G - 60 ----(Equation 5)
We need to find the total number of students, which is the sum of boys and girls:
Total students = B + G
Substituting the value of B from Equation 5:
Total students = (G - 60) + G
Total students = 2G - 60 ----(Equation 6)
Now, we can substitute Equation 6 into either Equation 1 or Equation 2 to find the value of G. Let's use Equation 1:
(1/2)B + (1/3)G = 570
Substituting the value of B from Equation 5:
(1/2)(G - 60) + (1/3)G = 570
(3/6)(G - 60) + (2/6)G = 570
(3/6)G - 30 + (2/6)G = 570
(5/6)G - 30 = 570
Add 30 to both sides:
(5/6)G = 600
Multiply both sides by 6/5:
G = (600 * 6) / 5
G = 720
Substituting the value of G into Equation 5:
B = G - 60
B = 720 - 60
B = 660
Therefore, the total number of students at Freemont High is the sum of boys and girls:
Total students = B + G
Total students = 660 + 720
Total students = 1380
So, there are 1380 students at Freemont High.