Question

In 1 kg mixture of sand and iron, 20 % is iron. How much sand should be added, so that the proportion of iron becomes 5%?

Answer 1

Amount of iron in 1 kg mixture = 20% of 1000 gms  

= 20×1000100 = 200 gms  

Therefore, Amount of sand in mixture = (1000 - 200) gms = 800 gms  

Now, let the total mixture is x kg in which iron is 20%  

Therefore, According to question,  

50% of x = 200 gm  

5×x100=200gm

x=200×1005 gms

x=200005gms=4000gms

Therfore, Required answer  

= 4000 gms - 1000 gms = 3000 gms = 3 kg.

Answer 2

Answer = 3kg

Explanation:

1 kg = 1000g

iron = 20%

= 1000 × 20/100

= 200 g

if iron = 5 % =200g then

total mixture = 100%

= 5% × 20

= 200 × 20

= 4000 g

.°.4000 g - 1000 g = 3000 g

3kg sand should be added to 1 kg mixture of iron and sand so that proportion of iron becomes 5 %