An alloy A contains 35% of zinc by weight.
Some alloy A and pure zinc are melted to form a new alloy that weights 400 kg and contains 66又16分之1 % of zinc.
What are the weights of alloy A and zinc needed respectively to form the new alloy?
(give the answers correct to the nearest kg)
Answer: alloy A: 209 kg, zinc: 191 kg
Show me the steps please. THX!!!
F3 Percentage question唔該幫幫手 快入來答!!!
2008-01-01 1:13 am
回答 (2)
2008-01-01 1:28 am
✔ 最佳答案
Let weight of alloy A neded be xThen weight of zinc needed = 400 - x
Weight of Zinc in alloy A is 0.35x
The % of zinc in New alloy is than
(400 - x + 0.35x) / 400 = [66 + 1 / 16] %
=> x = 208.85 or 209
and weight of zinc = 400 - 209 = 191
參考: me
2008-01-01 1:32 am
Let the weight of alloy A and pure zinc used by A kg and Z kg respectively
Consider the new alloy,
weight of zinc
= 400 x (66 又16分1 %)kg
=264.25kg
Consider A,
weight of zinc
=A x 35% kg
=0.35A kg
Therefore,
0.35A + Z = 264.25 ...... (1)
Besides, the sum of weights of Alloy A and Pure zinc should be equal to the weight of the new alloy
A + Z = 400 ...... (2)
(2) - (1) gives
A - 0.35A = 400 - 264.25
0.65A = 135.75
A = 208.84615 (to 5 decimal place)
i.e. weight of alloy A is 209kg (to the nearest kg)
substitute A=209 into (2)
208.84615 + Z = 400
Z = 191.15 (to 2 decimal place)
i.e. weight of pure zinc is 191kg (to the nearest kg)
Consider the new alloy,
weight of zinc
= 400 x (66 又16分1 %)kg
=264.25kg
Consider A,
weight of zinc
=A x 35% kg
=0.35A kg
Therefore,
0.35A + Z = 264.25 ...... (1)
Besides, the sum of weights of Alloy A and Pure zinc should be equal to the weight of the new alloy
A + Z = 400 ...... (2)
(2) - (1) gives
A - 0.35A = 400 - 264.25
0.65A = 135.75
A = 208.84615 (to 5 decimal place)
i.e. weight of alloy A is 209kg (to the nearest kg)
substitute A=209 into (2)
208.84615 + Z = 400
Z = 191.15 (to 2 decimal place)
i.e. weight of pure zinc is 191kg (to the nearest kg)
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