In a apartment block there are 20 apartments.
5 apartments house 4 tenants each.
6 apartments house 3 tenants each.
The rest of the apartments house 5 or 7 tenants.
On average each apartment housesy−2 tenants.
How many apartments are there where 5 tenants live?
To solve this problem, we first note the setup: 5 apartments with 4 tenants and 6 apartments with 3 tenants are explicitly mentioned. That accounts for:
- 5×4=20 tenants from the 4-tenant apartments.
- 6×3=18 tenants from the 3-tenant apartments.
Now, the remaining 9 apartments (since 20−(5+6)=9) can house either 5 or 7 tenants. Let x1 be the number of 5-tenant apartments and x2 be the number of 7-tenant apartments:
- x1+x2=9
- Total number of tenants = 20+18+5x1+7x2
The average number of tenants per apartment is given by y−2. We express the total tenant number equation:
2020+18+5x1+7x2=y−2
Substitute x2=9−x1 into the tenant equation:
Total number of tenants = 38+5x1+7(9−x1)
= 38+5x1+63−7x1
= 101−2x1
Average tenants = 20101−2x1=y−2
Multiplying throughout by 20:
101 - 2x_1 = 20(y - 2)
101 - 2x_1 = 20y - 40
Solving for x1:
−2x1=20y−40−101
−2x1=20y−141
x1=2141−20y
Therefore, x1=70.5−10y.
Thus, the correct answer is 70.5−10y.
70.5−10y