Alloy Composition Calculation
The problem involves calculating the final ratio of Iron (Fe) to Nickel (Ni) in a new alloy (C) formed by melting equal quantities of two existing alloys (A and B) with different Fe:Ni ratios.
Alloy A Details
Alloy A is made of Fe and Ni in the ratio 3:4.
- Fraction of Fe in Alloy A: $\frac{3}{3+4} = \frac{3}{7}$
- Fraction of Ni in Alloy A: $\frac{4}{3+4} = \frac{4}{7}$
Alloy B Details
Alloy B is made of Fe and Ni in the ratio 9:5.
- Fraction of Fe in Alloy B: $\frac{9}{9+5} = \frac{9}{14}$
- Fraction of Ni in Alloy B: $\frac{5}{9+5} = \frac{5}{14}$
Mixing Equal Quantities
Assume we take an equal quantity, say 'Q' units, of both Alloy A and Alloy B.
Amount of Fe from Alloy A = $Q \times \frac{3}{7}$
Amount of Ni from Alloy A = $Q \times \frac{4}{7}$
Amount of Fe from Alloy B = $Q \times \frac{9}{14}$
Amount of Ni from Alloy B = $Q \times \frac{5}{14}$
New Alloy C Calculation
Alloy C is formed by melting these equal quantities together.
Total Fe in Alloy C = (Fe from A) + (Fe from B)
Total Fe = $Q \times \frac{3}{7} + Q \times \frac{9}{14} = Q \times (\frac{6}{14} + \frac{9}{14}) = Q \times \frac{15}{14}$
Total Ni in Alloy C = (Ni from A) + (Ni from B)
Total Ni = $Q \times \frac{4}{7} + Q \times \frac{5}{14} = Q \times (\frac{8}{14} + \frac{5}{14}) = Q \times \frac{13}{14}$
Final Ratio in Alloy C
The ratio of Fe to Ni in Alloy C is:
Ratio = (Total Fe) : (Total Ni)
Ratio = $(Q \times \frac{15}{14})$ : $(Q \times \frac{13}{14})$
Simplifying by canceling Q and multiplying by 14, we get:
Ratio = 15 : 13
Important Questions from Mixture Problems (Notes)
- In 80 litres mixture of milk and water, the ratio of amount of milk to that of amount of water is 7 : 3. In order to make this ratio 2 : 1, how many litres of water should be added ?
- A $595$ litre of mixture contains milk and water in the ratio $17:18$. How much milk must be added to the mixture so that it contains milk and water in the proportion of $3:2$?
- How many liters of water should be added to a 30-liter mixture containing milk and water in the ratio 7:3 such that the resultant mixture has 40% water in it?
Two solutions $X$ and $Y$ containing ingredients $A, B$ and $C$, in proportions $a:b:c$ and $c:b:a$, respectively, are mixed. For the resultant mixture to have $A, B$ and $C$ in $1:1:1$ proportion, it is necessary that $a:b:c$ is
- A 360 ml aqueous solution contains 40% alcohol. How much will be the approximate percentage of alcohol if 3600 ml of water is added to the solution?
