Two cubes of bronze have their total weight equivalent to 60 kg. The first piece contains 10 kg of pure zinc and the second piece contains 8 kg of pure z what is the percentage of zinc in the first piece of bronze if the second piece contains 15 percent more zinc than the first ?
Answers
Answer:
Total weight of two cubes = 60 kg
Let the weight of the two cubes of bronze be “x” kg and “(60 - x)” kg respectively.
Weight of zinc in the first piece = 10 kg
∴ % of pure zinc in the first piece = [10/x] * 100 = 1000/x %
And,
Weight of zinc in the second piece = 8 kg
∴ % of pure zinc in the first piece = [8/(60 – x)] * 100 = 800/(60 – x) %
According to the question, we get
[800/(60 – x)] – [1000/x] = 15
⇒ [160/(60 – x)] – [200/x] = 3
⇒ 160x - 12000 + 200x = 180x - 3x²
⇒ 3x² + 180x - 12000 = 0
⇒ x² – 60x + 4000 = 0
⇒ (x+100)(x-40) = 0
⇒ x = 40 kg
Thus,
The percentage of zinc in the first piece of bronze is
= 1000/x %
= 1000/40 %
= 25 %
Answer:
25%
Explanation:
suppose weight of 1st piece is x and that of 2nd piece is y
x+y = 60
and
(10/x)*100 + 15 = (8/y)*100
x=40 and y = 20
so weight of 1st piece is 40kg
percentage = (10/40)*100 = 25%