Question

1kg of mixture of sand and iron,20% is iron. How much sand should be added to that so the proportion of iron become 10% select one: a. 200gms b. 1.8gms c. 800gms d. 1kg

Answer 1

Answer: The correct answer is (d).

Explanation:

Initial amount of the mixture of sand and iron = 1 kg

Amount of iron in 1 kg mixture =20% of 1 kg= $\frac{1 kg\times 20}{100}=0.2kg$

Let the total mass of the mixture after the addition be x

$\text{mass \%}=\frac{\text{mass of the component in mixture}\times 100}{\text{total amount of the mixture}}$

Since 10% of iron in present in new mixture of x kg:

$10=\frac{0.2 kg\times 100}{x}$

$x= 2kg$

Sand added to 1 kg mixture =total amount of the mixture - Initial amount of the mixture =2 - 1 kg = 1 kg

Hence,the correct option is (d).