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Zorluk: Çok zorBasic Algebraic Expressions and One-Step Equations

A logistics company distributes TT tons of cargo among a fleet of trucks. A standard truck has a maximum capacity of cc tons, and a heavy-duty truck has a maximum capacity of twice that of a standard truck. The fleet consists of xx standard trucks and yy heavy-duty trucks, all loaded to their maximum capacities. If the ratio of standard trucks to heavy-duty trucks in the fleet is exactly 3:13:1, which of the following equations correctly expresses the capacity of a standard truck, cc, in terms of TT and yy?

  1. A
    c=T4yc = \frac{T}{4y}
  2. B
    c=3T7yc = \frac{3T}{7y}
  3. c=T5yc = \frac{T}{5y}Cevap
  4. D
    c=T2y+3c = \frac{T}{2y + 3}
  5. E
    c=T5yc = T - 5y

Cevap

The equation c=T5yc = \frac{T}{5y} correctly expresses the capacity of a standard truck.
The total cargo TT is the sum of the capacities of all trucks. Since there are xx standard trucks carrying cc tons each and yy heavy-duty trucks carrying 2c2c tons each, the equation is T=cx+2cyT = cx + 2cy. The 3:13:1 ratio of standard trucks to heavy-duty trucks means x=3yx = 3y. Substituting this relationship into the cargo equation gives T=c(3y)+2cy=5cyT = c(3y) + 2cy = 5cy. To isolate the capacity of a standard truck, cc, divide both sides by 5y5y to get c=T5yc = \frac{T}{5y}.

Adım Adım Çözüm

1
Write the equation for the total cargo weight by summing the capacities of all trucks in the fleet.
T=cx+2cyT = cx + 2cy
The total cargo TT is distributed among xx standard trucks, each carrying cc tons, and yy heavy-duty trucks, each carrying 2c2c tons.
2
Translate the ratio of standard trucks to heavy-duty trucks into an algebraic relationship.
x=3yx = 3y
The ratio of standard trucks (xx) to heavy-duty trucks (yy) is 3:13:1, meaning there are 3 times as many standard trucks as heavy-duty trucks.
3
Substitute the ratio relationship x=3yx = 3y into the total cargo equation.
T=c(3y)+2cyT = c(3y) + 2cy
Replacing xx with 3y3y eliminates the variable xx and expresses TT in terms of cc and yy.
4
Combine like terms to simplify the expression.
T=5cyT = 5cy
Adding 3cy3cy and 2cy2cy yields 5cy5cy.
5
Isolate the variable cc by dividing both sides of the equation by 5y5y.
c=T5yc = \frac{T}{5y}
Division is the inverse operation of multiplication, allowing us to solve the one-step equation T=(5y)cT = (5y)c for cc.

Anahtar Kavram

Translating verbal descriptions into algebraic equations and solving one-step literal equations
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