How many different ways can teams of four members be formed from a class of 20?
1 answer:
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Answer:
Number of restaurant-purchased meals eaten in a restaurant = 70
Number of restaurant-purchased meals eaten in a car = 22
Number of restaurant-purchased meals eaten in a home = 60
Step-by-step explanation:
Let the number of people that will eat in a restaurant be R.
Let the number of people that will eat in a car be C.
Let the number of people that will eat in a home be H.
From the information given in the problem we have,
R+C+H = 152 ____equation (1)
C+H-R = 12 ____equation (2)
R = H+10 ____equation (3)
1) Plugging in R=H+10 from equation 3 into the equation 2, we get
C+H-R=12
=> C+H-(H+10)=12
=> C+H-H-10=12
2) Cancelling out +H and -H, we get
C-10=12
3) Add 10 to both sides
C-10+10=12+10
4) Cancelling out -10 and +10, we get
C=22
5) Plugin C=22 in equation 1, we have
R+C+H = 152
=> R+22+H=152
Subtract 22 from both sides,
R+22+H-22=152-22
Cancelling out +22 and -22 from the left side, we get
R+H=130 ____let it be equation (4)
6) Plugin C=22 in equation 2, we have
C+H-R = 12
22+H-R = 12
Subtracting 22 from both the sides, we get
22+H-R-22 = 12-22
Cancelling out +22 and -22 from the left side,
H-R = -10 ____let it be equation (5)
7) Adding equation 4 and equation 5, we get
(R+H)+(H-R) = 130 + (-10)
=> R+H+H-R = 130-10
8) Cancelling out R and -R from the left side, we get
2H = 120
9) Dividing both sides by 2, we get
10) Cancelling out the 2's from the left side, we have
H=60
11) Plugging in C=22 and H=60 in the equation 1, we have
R+C+H = 152
=> R+22+60=152
=> R + 82 = 152
12) Subtracting 82 from both the sides, we get
R + 82 -82 = 152 -82
13) Cancelling out +82 and -82 from the left side, we get
R = 70
<em>So, C=22, H=60, R=70</em>