Question

an iron contains 18%of copper. how much kg of the ore are required to get 63 kg of copper?

Answer 1

Let X kg ore is required.

As ore contains 18% copper.

Copper needed = 63 kg

A/c to question

18 % of X = 63 kg

18X/100 =63

X = (63 ×100)÷18

X = 350 kg

Answer 2

hey buddy here is your answer

given that Iron contain Copper in it = 18% (18/100)

total copper required is 63 kg

in 1 kg we get copper is = 1 * 18/100 = 0.18 kg ...

so we can do it this way

for 180 g(0.18 kg ) copper we need = 1000g (1kg)

for 1 g copper we need = 1000/180 g

so for 63000 g (63 kg) copper we need

# ==> 1000/180*63000 = 350000 Gram

350000 g = 350 kg

thus, for 63 kg copper we needs 350 kg iron

hope it helps