In 1 kg. mixture of sand and iron, 20% is iron. How much sand should be added so that the proportion of iron becomes 10% ?

Question

Question

This question was previously asked in

CGPSC Civil Service 2019 Official Paper 2 (CSAT)

Answer 1

Given:

The percentage of iron in 1 kg of mixture is 20%

Calculation:

The amount of iron = 20/100 × 1000 = 200gm (1kg = 1000 gm)

The amount of sand = 1000 - 200 = 800 gm

Let the add amount of sand be x

According to the question;

⇒ 200/(1000 + x) = 10/100

⇒ 2000 = 1000 + x

⇒ x = 1000 gm = 1 kg

∴ The added amount of sand is 1 kg.

Shortcut Trick1 kg = 1000 gram

On mixing 1 kg of

Amount of sand = 1800 - 800 = 1000 gram = 1 kg

∴ On adding 1 kg of sand, the ratio of iron becomes 10%.