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In 1 kg of a mixture of sand and iron, 20% is iron .How such sand should be added so that the proportion of iron becomes 10%
- A1 kg
- B200gm
- C800 gm
- D1.8 kg
Answer not set
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0
CCommunity
1 Kg
0
CCommunity
Solution
Total mixture of sand and iron = 1 kg
Quantity of iron = $\frac{20}{100} \times 1 = 0.2$ kg
Let $x$ kg of sand should be added, thus total iron in the mixture
=> $0.2=\frac{10}{100} \times (x+1)$
=> $2=x+1$
=> $x=2-1=1$ kg
=> Ans - (A)
Total mixture of sand and iron = 1 kg
Quantity of iron = $\frac{20}{100} \times 1 = 0.2$ kg
Let $x$ kg of sand should be added, thus total iron in the mixture
=> $0.2=\frac{10}{100} \times (x+1)$
=> $2=x+1$
=> $x=2-1=1$ kg
=> Ans - (A)
0
CCommunity
Solution
Total mixture of sand and iron = 1 kg
Quantity of iron = $\frac{20}{100} \times 1 = 0.2$ kg
Let $x$ kg of sand should be added, thus total iron in the mixture
=> $0.2=\frac{10}{100} \times (x+1)$
=> $2=x+1$
=> $x=2-1=1$ kg
=> Ans - (A)
Total mixture of sand and iron = 1 kg
Quantity of iron = $\frac{20}{100} \times 1 = 0.2$ kg
Let $x$ kg of sand should be added, thus total iron in the mixture
=> $0.2=\frac{10}{100} \times (x+1)$
=> $2=x+1$
=> $x=2-1=1$ kg
=> Ans - (A)