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?

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- Jan 31st 2007, 03:37 PM #1

- Jan 31st 2007, 05:44 PM #2

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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 $\displaystyle A$ = number of 40-shilling fruit.

Let $\displaystyle B$ = number of 10-shilling fruit.

Let $\displaystyle C$ = number of 1-shilling fruit.

. . Note: $\displaystyle A,\,B,\,C$ are positive integers.

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

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

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

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

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

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

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