## Task 5 - 190735

Cathrin goes shopping. She has exactly 18 coins, and they are only two-mark and fifty-pfennig coins. She spends exactly half of the total amount of this money. After shopping, she realizes that she now has only two-mark and fifty-pfennig coins again, and she has as many two-mark coins as she had fifty-pfennig coins before, and as many fifty-pfennig coins as she had two-mark coins before. What amount of money did Cathrin have left after shopping?
If the number of two-mark coins that Cathrin had before shopping is denoted by $x$, then at the same time she had $(18-x)$ fifty-pfennig coins. The amount of money she had before shopping was thus

$$
(2 x+(18-x) \cdot 0.5) \text { marks }=(1.5 x+9) \text { marks }
$$

Since she spent exactly half of this amount, she had $(0.75 x+4.5)$ marks left after shopping. According to the problem, this amount consisted of $(18-x)$ two-mark coins and $x$ fifty-pfennig coins. Therefore,

$$
0.75 x+4.5=2(18-x)+0.5 x=36-1.5 x
$$

from which we get $2.25 x=31.5$, hence $x=14$.

Thus, Cathrin had exactly 14 two-mark coins and exactly 4 fifty-pfennig coins, which is a total of 30 marks, before shopping. After shopping, she had exactly 4 two-mark coins and exactly 14 fifty-pfennig coins, which is a total of 15 marks.

2. Alternative solution:

If Cathrin had exactly $p$ marks after shopping, then she had exactly $2p$ marks before. If we were to lay out the same number of coins (each of the same value) as she had before, and add the same number of coins as she had after, this would amount to 18 two-mark coins and 18 fifty-pfennig coins, which is $(36+9)$ marks $=45$ marks; on the other hand, this would also be $(2p+p)$ marks $=3p$ marks. Therefore, $3p=45$, so $p=15$.

Thus, Cathrin had exactly 15 marks after shopping.