6. How many kilograms of cherries priced at 18 kn per kilogram should be mixed with $2012 \mathrm{~kg}$ of cherries priced at $25 \mathrm{kn}$ per kilogram so that the resulting mixture of cherries can be sold at 20 kn per kilogram without a difference in profit?
6. Let $\mathrm{s} x$ be the number of kilograms of cherries at $18 \mathrm{kn}$.

The total value of cherries at $18 \mathrm{kn}$ is $x \cdot 18 \mathrm{kn}$. 1 POINT

The total value of cherries at $25 \mathrm{kn}$ is $2012 \cdot 25=50300 \mathrm{kn}$. 1 POINT

The total mass of cherries is $x+2012 \mathrm{~kg}$. 1 POINT

The total mass of cherries is sold at 20 kn per kilogram, so its value is $(x+2012) \cdot 20 \mathrm{kn}$.

1 POINT

Since the value of the total mass of cherries must be equal to the sum of the values of the cheaper and

more expensive cherries, we have

$(x+2012) \cdot 20=x \cdot 18+50300 \quad 2$ POINTS

$20 x+40240=18 x+50300$

2 POINTS

$2 x=10060$

1 POINT

$x=5030 \mathrm{~kg}$

You need to mix $5030 \mathrm{~kg}$ of cherries whose price is $18 \mathrm{kn} / \mathrm{kg}$.

1 POINT

TOTAL 10 POINTS