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?

Result

a =  0 kg
b =  2.667 kg
c =  0.6 kg
d =  1 kg
e =  3.333

Solution:

k=50/10/1.5=103≐3.3333 a=k⋅ 2.5=3.3333⋅ 2.5≐253≐8.3333≐0 kgk=50/10/1.5=\dfrac{ 10 }{ 3 } \doteq 3.3333 \ \\ a=k \cdot \ 2.5=3.3333 \cdot \ 2.5 \doteq \dfrac{ 25 }{ 3 } \doteq 8.3333 \doteq 0 \ \text{kg}
b=k⋅ 0.8=2.667 kgb=k \cdot \ 0.8=2.667 \ \text{kg}
c=k⋅ 18/100=35=0.6 kgc=k \cdot \ 18/100=\dfrac{ 3 }{ 5 }=0.6 \ \text{kg}
d=k⋅ 30/100=1 kgd=k \cdot \ 30/100=1 \ \text{kg}
e=k⋅ 1=3.333e=k \cdot \ 1=3.333

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