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%
1). 769.2 gms
2). 800 gms
3). 760.92 gms
4). Data insufficient
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