Question

an ore contains 12% copper how many kilograms of those are required to get 69 kg of copper

Answer 1

Concept we will be using:
$\text{i) Whole}= \frac{\text{Part}\times100}{\text{Part percentage}}$
-------------------------------------------------------------------------------------------
Solution:
Part = mass of copper =69 kg
Part percentage = percentage of copper in the ore=12%

Whole= mass of ore required

Plug in the above values in the formula:

$\text{Whole}= \frac{\text{Part}\times100}{\text{Part percentage}}= \frac{69\times100}{12}= \frac{6900}{12}=575kg$

Answer: Required ore is 575 kg

Answer 2

Solution:-

Let the total quantity of the ore be 'x' kg
So, 12 % of that ore is copper.
According to the question.

(x*12)/100 = 69
12x/100 = 69
3x/25 = 69
3x = 69*25
3x = 1725
x = 1725/3 
x = 575 kg

So, 575 kg ore is required to get 69 kg of copper.

Answer.