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# can you solve this word problem simply?

During a 4th of July weekend, 32 vehicles became trapped on the Sunshine Skyway Bridge while it was being repaved. A recent city ordinance decreed that only cars with 4 wheels and trucks with six wheels could be on the bridge at any given time. If there were 148 tires that needed to be replaced to due to damage, how many cars and trucks were involved in the incident?

### 6 Answers

- PranilLv 79 years agoFavorite Answer
Let there be x cars involved

then (32 – x) trucks were also involved

4x + 6(32 – x) = 148

4x + 192 – 6x = 148

2x = 44

x = 22 cars and 10 trucks

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- Damon LyonLv 79 years ago
If you want to do it with one variable, you can say that

x = the number of cars

That makes (32 - x) the number of trucks

When you multiply the number of tires on the cars and trucks and add them together, you get 148.

4x + 6(32 - x) = 148

Now you can solve

4x + 192 - 6x = 148

-2x = -44

x = 22

So with 22 cars, there must have been 10 trucks.

- VamanLv 79 years ago
Let A be the number of cars and B be the number of trucks. The total number of vehicles A+B=32. Each car has 4 tires and each truck has 6 tires. Therefore the total number of tires 4A+6B=148. Solving these two equations, we get the number of trucks to be 10 and number of cars will be 22.

- 9 years ago
C + T = 32 (C=no of car, T=no of trucks) (1)

4C + 6T = 148 (2)

Multiply (1) by 4 :

4C+4T = 128 (3)

(2) - (3) :

2T = 20, so T=10 and C=22

10 trucks, 22 cars

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- Anonymous9 years ago
so 10 trucks and 22 cars were involved

- FIRST BLv 79 years ago
Let # of cars = x, # of trucks = y

1) x+y=32

2) 4x+6y=148

from 1), x=32-y

substituting into 2), 4(32-y)+6y=148

128-4y+6y=148

2y=20

y=10

Substituting y =10 into 1), x=22

So 22 cars, 10 trucks

/p