Stew

To prepare stew for 10 loggers we need: 2.5 kg of potatoes, 0.8 kg of meat, 18 dag of flour, 30 dag onion and 1 bread. What is the required amount of raw materials for 50 participants camps where loggers eats 1.5 times more than a participant of the camp?

Correct result:

a =  0 kg
b =  2.6667 kg
c =  0.6 kg
d =  1 kg
e =  3.3333

Solution:

$$\begin{gathered} k=50\mathrm{/}10\mathrm{/}1.5={\displaystyle \frac{10}{3}}\doteq 3.3333 \\ a=k\cdot 2.5=3.3333\cdot 2.5={\displaystyle \frac{25}{3}}\doteq 8.3333=0\text{ kg} \end{gathered}$$

$b=k\cdot 0.8=2.6667\text{ kg}$

$c=k\cdot 18\mathrm{/}100={\displaystyle \frac{3}{5}}\text{ kg}=0.6\text{ kg}$

$d=k\cdot 30\mathrm{/}100=1\text{ kg}$

$e=k\cdot 1=3.3333$

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