Question

On a plane, 2 men had 135 kilograms of luggage added together. The first paid $1.35 for his excess luggage and the second paid $2.70 for his excess luggage. The charge was at a flat rate. If all of the luggage belonged to one person, the ecess luggage charge would have been $8.10. At most, how many kilograms of luggag is each person permitted to bring onto the plane free of an additional charge?A. 55 kilogramsB. 50 kilogramsC. 45 kilogramsD. 40 kilograms

Answer 1

Let's call the amount of lugagge the first men had as x, and the amount of lugagge the second men had as y.

Since both had to pay for excess of luggage, both had the permitted amount of luggage plus an extra amount. Let's call the permitted amount as a, the extra the first men had as z, and the extra the second man had as w. Then, our total amounts of luggage can be rewritten as

`\begin{gathered} x=a+z \\ y=a+w \end{gathered}`

Since their total amount of luggage added together was 135 kg, this gives to us the following equation

`x+y=135`

The charge for the extra luggage is a flat rate, let's call it r, then, the price you pay for the extra luggage is given by the product between this rate and the amount of extar luggage. The first paid $1.35 for his excess luggage and the second paid $2.70 for his excess luggage, therefore

`\begin{gathered} zr=1.35 \\ wr=2.70 \end{gathered}`

If we divide one equation by the other, we can write the excess of luggage of one as a function of the other.

`\begin{gathered} (zr)/(wr)=(1.35)/(2.70) \\ (z)/(w)=(1)/(2) \\ w=2z \end{gathered}`

If all of the luggage belonged to one person, this person would have to pay $8.10 for the excess. If we call the excess of this person as p, the amount of luggage this person had would be

`a+p=135`

Using the "rate formula" we created before, we can also write p as a function of z.

`pr=8.10\Rightarrow(pr)/(zr)=(8.10)/(1.35)\Rightarrow p=6z`

Rewritting all the total amounts, we have

`\begin{gathered} a+6z=135 \\ a+2z=y \\ a+z=x \end{gathered}`

If we add the equations for x and y together, we have

`\begin{gathered} (a+z)+(a+2z)=(x)+(y) \\ a+z+a+2z=x+y \\ 2a+3z=x+y \end{gathered}`

From the first equation, we know that the sum between both total amounts of luggage is equal to 135.

`\begin{cases}x+y=135 \\ 2a+3z=x+y\end{cases}\Rightarrow2a+3z=135`

Now, we have two equations for our two variables.

`\begin{cases}2a+3z=135 \\ a+6z=135\end{cases}`

If we multiply the first equation by 2, and subtract the second equation, we're going to get a new equation only for a(the amount in kilograms of permitted luggage).

`\begin{gathered} 2\cdot(2a+3z)-(a+6z)=2\cdot(135)-135 \\ 4a+6z-a-6z=270-135 \\ 3a=135 \\ a=45 \end{gathered}`

The amount of luggage permitted to bring onto the plane for free is 45 kilograms.