Three fire trucks and 4 ambulances can fit into a parking lane 152 ft long. Two fire trucks and 5 ambulances can fit into a lane 136 ft long. How many feet long must a parking lane be for 1 fire truck and 5 ambulances? Assume

there is 1 ft of space between each vehicle.

A.) 75 ft

B.) 90 ft

C.) 100 ft

D.) 105 ft

E.) 110 ft

there is 1 ft of space between each vehicle.

A.) 75 ft

B.) 90 ft

C.) 100 ft

D.) 105 ft

E.) 110 ft

F = length of one fire truck

A = length of one ambulance

F = length of one fire truck

A = length of one ambulance

F + _ + F + _ + F + _ + A + _ + A + _ + A + _ + A = 152 ------>

**3F + 4A + 6 = 152**

F + _ + F + _ + A + _ + A + _ + A + _ + A + _ + A = 136 ------>

**2F + 5A + 6 = 136**

3F + 4A = 146 {subtracted 6 from each side in 1st equation}

2F + 5A = 130 {subtracted 6 from each side in 2nd equation}

-6F – 8A = -292 {multiplied 1st equation by -2}

__6F + 15A = 390__{multiplied 2nd equation by 3}

7A = 98 {added the two equations}

A = 14 {divided each side by 7}

**Ambulance is 14 ft long**

6F + 15A = 390

6F + 15(14) = 390 {substituted 14 for A}

6F + 210 = 390 {multiplied 15 by 14}

6F = 180 {subtracted 210 from each side}

F = 30 {divided each side by 6}

**Fire truck is 30 ft long**

**How many feet must a parking lane be for 1 fire truck and 5 ambulances?**

1F + _ + A + _ + A + _ + A + _ + A + _ + A ----->

**F + 5A + 5 = ?**

= 30 + 5(14) + 5 {substituted 30 for F and 14 for A}

= 30 + 70 + 5 {multiplied 5 by 14}

=

**105 feet of parking space**for 1 fire truck and 5 ambulances

Answer:

**D.)**

*- Algebra House*