1.

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 gms2). 800 gms3). 760.92 gms4). Data insufficient

Answer»

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


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