In a senior secondary school,80 students play football or hockey.The number that play football is 5 more than twice the number that play hockey.If 15 students play both games and every student in the school plays at least one game,find (i)the number of students that play football (ii)the number of students that play football but not hockey (iii)the number of students that play hockey but not football.
Let's solve this step-by-step:
Step 1: Let's represent the unknowns in this problem with variables.
Let's assume the number of students playing hockey is "x."
Then, the number of students playing football would be "2x + 5" (as stated in the problem).
Step 2: We know that the total number of students playing football or hockey is 80.
So, we can write the equation:
x + (2x + 5) - 15 = 80
Explanation: The first term "x" represents the number of students playing only hockey, the second term "2x + 5" represents the number of students playing both hockey and football, and the "-15" represents the number of students playing both games.
Step 3: Solve the equation.
x + 2x + 5 - 15 = 80
3x - 10 = 80
3x = 80 + 10
3x = 90
x = 90 / 3
x = 30
Step 4: Substitute the value of x into the expressions to find the other unknowns.
(i) The number of students playing football:
2x + 5 = 2(30) + 5 = 60 + 5 = 65
(ii) The number of students playing football but not hockey:
2x + 5 - 15 = 65 - 15 = 50
(iii) The number of students playing hockey but not football:
x = 30
So, the answers are:
(i) The number of students playing football is 65.
(ii) The number of students playing football but not hockey is 50.
(iii) The number of students playing hockey but not football is 30.
To solve this problem, we'll use a method called the principle of inclusion-exclusion.
Let's assign variables to the unknowns:
Let F be the number of students that play football.
Let H be the number of students that play hockey.
According to the problem, we know the following information:
(i) F + H = 80 (the total number of students playing football or hockey is 80).
(ii) F = 2H + 5 (the number of students playing football is 5 more than twice the number of students playing hockey).
(iii) 15 students play both games.
To find the answer to the individual questions, we'll solve them one by one:
(i) The number of students that play football:
From equation (i), F + H = 80, we substitute the value of F from equation (ii):
(2H + 5) + H = 80
3H + 5 = 80
3H = 75
H = 25
Substituting H = 25 back into equation (ii):
F = 2H + 5 = 2(25) + 5 = 50 + 5 = 55
Therefore, the number of students that play football is 55.
(ii) The number of students that play football but not hockey:
To find this, we subtract the number of students playing both games from the total number of students playing football:
Number of students playing football but not hockey = F - (students playing both games) = 55 - 15 = 40
Therefore, the number of students that play football but not hockey is 40.
(iii) The number of students that play hockey but not football:
To find this, we subtract the number of students playing both games from the total number of students playing hockey:
Number of students playing hockey but not football = H - (students playing both games) = 25 - 15 = 10
Therefore, the number of students that play hockey but not football is 10.
To summarize:
(i) The number of students that play football is 55.
(ii) The number of students that play football but not hockey is 40.
(iii) The number of students that play hockey but not football is 10.
f = only football , h = only hockey
f + h = 80 - 15 = 65
f + 15 = 2(h + 15) + 5 ... f + 15 = 2 h + 35 ... f = 2 h + 20
solve the system for f and h