In 1 kg mixture of sand and iron, 23% is iron. How much sand should be added so that the proportion of iron becomes 13%

A 769.2 gms

B 800 gms

C 760.92 gms

D Data insufficient

Correct Answer: Option A

Thus initially amount of iron in the mixture = 23% if 1000

∴ initially amount of iron = 230 gm

∴ initially amount of sand = 770 gm

After adding ‘x’ gms of sand the new ratio of iron : sand will be 13:87

\(\therefore \;\frac{{230}}{{770 + x}} = \;\frac{{13}}{{87}}\)

⇒ 10010 + 13x = 20010

⇒ x = 769.2 gms