Calculate Number of Apartments with 5 Tenants: Solving y-2 Average Equation

Question

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 housesy2 y-2 tenants.

How many apartments are there where 5 tenants live?

Video Solution

Step-by-Step Solution

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 5 \times 4 = 20 tenants from the 4-tenant apartments.
  • 6×3=18 6 \times 3 = 18 tenants from the 3-tenant apartments.

Now, the remaining 9 9 apartments (since 20(5+6)=9 20 - (5 + 6) = 9 ) can house either 5 or 7 tenants. Let x1 x_1 be the number of 5-tenant apartments and x2 x_2 be the number of 7-tenant apartments:

  • x1+x2=9 x_1 + x_2 = 9
  • Total number of tenants = 20+18+5x1+7x2 20 + 18 + 5x_1 + 7x_2

The average number of tenants per apartment is given by y2 y - 2 . We express the total tenant number equation:

20+18+5x1+7x220=y2\frac{20 + 18 + 5x_1 + 7x_2}{20} = y - 2

Substitute x2=9x1 x_2 = 9 - x_1 into the tenant equation:

Total number of tenants = 38+5x1+7(9x1) 38 + 5x_1 + 7(9 - x_1)

= 38+5x1+637x1 38 + 5x_1 + 63 - 7x_1

= 1012x1 101 - 2x_1

Average tenants = 1012x120=y2\frac{101 - 2x_1}{20} = y - 2

Multiplying throughout by 20:

101 - 2x_1 = 20(y - 2)

101 - 2x_1 = 20y - 40

Solving for x1 x_1 :

2x1=20y40101-2x_1 = 20y - 40 - 101

2x1=20y141-2x_1 = 20y - 141

x1=14120y2 x_1 = \frac{141 - 20y}{2}

Therefore, x1=70.510y x_1 = 70.5 - 10y .

Thus, the correct answer is 70.510y\boxed{70.5 - 10y}.

Answer

70.510y 70.5-10y