N an alloy 80% is copper and the remaining tin. In an another alloy copper is 85% and tin is 12%. In what ratio should the two alloys be mixed so that the new mixture must have 15% tin. Also find the percentage of copper in the new mixture
Answers
N an alloy 80% is copper and the remaining tin. In an another alloy copper is 85% and tin is 12%. In what ratio should the two alloys be mixed so that the new mixture must have 15% tin. Also find the percentage of copper in the new mixture
Let say First Alloy is taken = x
and second alloy = y
in First Alloy
copper = 80 % = (80/100)x = 0.8x
Tin = x - 0.8x = 0.2x
in Second Alloy
copper = 85% = (85/100)y = 0.85y
Tin = 12% = (12/100)y = 0.12y
undefined = y - 0.85y - 0.12y = 0.03y
Total mixture = x + y
Total Tin = 0.2x + 0.12y
Tin % = Total Tin / Total Mixture
=> 15/100 = (0.2x + 0.12y) / (x + y)
=> 15x + 15y = 20x + 12 y
=> 3y = 5x
=> x/y = 3/5
x:y :: 3:5
x = 3y/5
Total mixture = x + y = 3y/5 + y = (3y + 5y )/5 = 8y/5 = 1.6y
Total copper = 0.8x + 0.85y = 0.8(3y/5) + 0.85y = 0.48y + 0.85y = 1.33y
Copper % = (1.33y/1.6y) * 100 = 83.125 %
Answer:83.125
Explanation:
alloy1. alloy 2.
(tin) 20% 12% (tin)
15% -----> tin in mixture
3 : 5
now in mixture alloy 1 and alloy 2 are in ratio 3:5
Cu in mixture = (3×80/100) + (5×85/100) =133/20 =66.5
%of Cu in mix. = ( 66.5×100)/(3+5) = 83.125