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?

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  • Pranil
    Lv 7
    9 years ago
    Favorite 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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  • 9 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.

  • Vaman
    Lv 7
    9 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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  • Anonymous
    9 years ago

    so 10 trucks and 22 cars were involved

  • 9 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

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