1. ## Schillings

Herr Vitami buys three kinds of frut for 40 schillings,10schillings and 1 schilling each. He pays 259 schilling for 100 peices of fruit. How many of the cheapest type of fruit did he buy?

2. Hello, Rimas!

Herr Vitami buys three kinds of frut for 40 shillings,10 shillings and 1 shilling each.
He pays 259 shillings for 100 peices of fruit.
How many of the cheapest type of fruit did he buy?

Let $A$ = number of 40-shilling fruit.
Let $B$ = number of 10-shilling fruit.
Let $C$ = number of 1-shilling fruit.
. . Note: $A,\,B,\,C$ are positive integers.

He bought 100 pieces of fruit: . . $(1)\;\;A + B + C \:=\:100$

He paid 259 shillings for them: . $(2)\;\;40A + 10B + C \:=\:259$

Subtract (1) from (2): . $39A + 9B\:=\:159\quad\Rightarrow\quad 13A + 3B \:=\:53$

Solve for $B\!:\;\;B \:=\:\frac{53-13A}{3}$

Since $B$ is positive, there are only 4 choices for $A\!:\:1,\,2,\,3,\,4$

. . $\begin{array}{cccc}\text{If }A = 1,\;B = \frac{40}{3} \\
\text{If }A = 2,\:B = \;9 \\ \text{If }A = 3,\:B = \frac{14}{3} \\ \text{If }A = 4,\:B = \frac{1}{3}\end{array}$

Hence: . $A = 2,\:B = 9$

Therefore, from (1): . $\boxed{C = 89}$