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 is made of Fe and Ni in the ratio 3:4.
Alloy B is made of Fe and Ni in the ratio 9:5.
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 = $Q \times \frac{5}{14}$
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 = $Q \times \frac{4}{7} + Q \times \frac{5}{14} = Q \times (\frac{8}{14} + \frac{5}{14}) = Q \times \frac{13}{14}$
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
Suppose a tap mixes hot water and cold water in a ratio that depends linearly on the proportion of opening. Water out of the tap has temperature 40°C when the tap is half-open, and 30°C when it is three-fourths open. To get water at 50°C, the tap should be_______________.