Question

An ore contains 15% of iron. How much ore will be required to get 36kg of iron

Answer 1

$<b>Let the required ore in kilograms be x.<b>$
So according to question,
15% of x = 36 kg

$\bold{ \frac{15}{100} \times x = 36 }\\\bold{ \frac{3}{20} \times x = 36} \\\bold{ x = 36 \times \frac{20}{3}} \\ \bold{x = 12 \times 20} \\ \underline\bold{x = 240}$

Hence,
240 kg of the ore will be required to get 36 kg of iron.

Thanking You!

Answer 2

Answer:

Let the required ore in kilograms be x.

So according to question,

15% of x = 36 kg

\begin{lgathered}\bold{ \frac{15}{100} \times x = 36 }\\\bold{ \frac{3}{20} \times x = 36} \\\bold{ x = 36 \times \frac{20}{3}} \\ \bold{x = 12 \times 20} \\ \underline\bold{x = 240}\end{lgathered}

100

15

×x=36

20

3

×x=36

x=36×

3

20

x=12×20

x=240

Hence,

240 kg of the ore will be required to get 36 kg of iron.