Question

At the first stops on his route, a driver unloaded 2 /5 of the packages in his van. After he unloaded another three packages at his next stop, 1/2 of the original number of packages remained. How many packages were in the van before the first delivery?​

Answer 1

Let the number of packages that were in the van before the first delivery be x.

Given:

• number of packages unloaded at the first stop = 2/5 × x = 2x/5
• number of packages unloaded at the second stop = 3
• number of packages remaining = 1/2 × x = x/2

To find:

• the value of x

Solution:

We know that:

(the number of packages unloaded at the first stop) + (the number of packages unloaded at the second stop) = (the original number of packages) - (number of packages remaining)

which when putting the values would mean:

$\sf{\dfrac{2x}{5}+3=x-\dfrac{x}{2}}$

or,

$\sf{\dfrac{2x}{5}+3=\dfrac{x}{2}}$

Solving the equation:

• Transposing 3 to the RHS,

$\sf{\Rightarrow \dfrac{2x}{5}=\dfrac{x}{2}-3}$

• Transposing x/2 to the LHS

$\sf{\Rightarrow \dfrac{2x}{5}-\dfrac{x}{2}=-3}$

• Making the terms on the RHS like fractions,

$\sf{\Rightarrow \dfrac{2x\times2}{5\times2}-\dfrac{x\times5}{2\times5}=-3}$

$\sf{\Rightarrow \dfrac{4x}{10}-\dfrac{5x}{10}=-3}$

• Subtracting,

$\sf{\Rightarrow \dfrac{-x}{10}=-3}$

• Transposing 10 to the RHS

$\sf{\Rightarrow -x=-3\times10}$

• Solving further,

$\sf{\Rightarrow -x=-30}$

• Removing the minus sign from both the sides,

$\boxed{\bf{\Rightarrow x=30}}$

Therefore, thirty packages were in the van before the first delivery.