# A man sold one sort of oranges at 5 for $2, and another sort at 16 for$12. If he had sold them all at $0.5 each, he would have received$2 less. If he had sold all at 3 for $2, he would have received$18 more. How many of each sort did he sell?

Oct 15, 2015

Let's call the amounts $x \mathmr{and} y$ and the money he got $z$

#### Explanation:

What he did can be translated into:
Equation (1): $z = \frac{2}{5} x + \frac{12}{16} y = \frac{2}{5} x + \frac{3}{4} y$

The second option can be translated into:
Equation (2): $z - 2 = 0.5 \cdot \left(x + y\right) = \frac{1}{2} \left(x + y\right)$

The last option would be:
Equation (3): $z + 18 = \frac{2}{3} \left(x + y\right)$

If we look at the last two equations, we see that the difference between selling 2 for $1 or 3 for$2 is $20, or: $\frac{2}{3} \left(x + y\right) - \frac{1}{2} \left(x + y\right) = 20 \to$$\left(\frac{2}{3} - \frac{1}{2}\right) \left(x + y\right) = \left(\frac{4}{6} - \frac{3}{6}\right) \left(x + y\right) = \frac{1}{6} \left(x + y\right) = 20$$\to x + y = 120 \to z - 2 = \frac{1}{2} \cdot 120 = 60 \to z = 62$We now have the total number of oranges and the price he got according to the original scheme. We can write $y = 120 - x$Putting these into the first equation, we get: $62 = \frac{2}{5} x + \frac{3}{4} \left(120 - x\right) = \frac{2}{5} x + 90 - \frac{3}{4} x \to$$62 = \frac{8}{20} x + 90 - \frac{15}{20} x = 90 - \frac{7}{20} x \to \frac{7}{20} x = 28 \to$$x = \frac{560}{7} = 80 \to y = 120 - 80 = 40$Answer : 80 of the first sort, 40 of the other. Note : Not quite satisfactory, as he could not have sold the second sort by the 16's, but 4 for$3 would do.