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02 Aug 2023, 06:00
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
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Question Stats:
89% (01:30) correct
11%
(02:07)
wrong
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A certain amount of 8% copper ore was mixed with another amount of 3% copper ore to produce 300 tons of 6% copper ore. How many tons of the 8% copper ore were used to make this mixture?
A. \(81\frac{9}{11}\)
B. \(112\frac{1}{2}\)
C. 120
D. \(163\frac{7}{11}\)
E. 180
A. \(81\frac{9}{11}\)
B. \(112\frac{1}{2}\)
C. 120
D. \(163\frac{7}{11}\)
E. 180
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02 Aug 2023, 06:23
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Bunuel
We can use teeter totter method (concept of weighted averages) to solve this question.
Attachment:
Screenshot 2023-08-02 185121.jpg [ 10.33 KiB | Viewed 39025 times ]
To obtain the 6% copper mixture, the 8% copper and the 3% copper were added in a ratio of 3 : 2
So for every 5 parts of 6% copper, 3 parts in that mixture are 8% copper.
Amount of copper user = \(\frac{3}{5} * 300 = 180\)
Option E
04 Aug 2023, 20:05
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Given: A certain amount of 8% copper ore was mixed with another amount of 3% coper ore to produce 300 tons of 6% coper ore.
Asked: How many tons of the 8% coper ore were used to make this mixture?
Let us assume that x tons of 8% copper ore was used to make the mixture.
8%x + 3%(300-x) = 300*6%
9 + 5%x = 18
5%x = 9
x = 9/5% = 900/5 = 180 tons
IMO E
Asked: How many tons of the 8% coper ore were used to make this mixture?
Let us assume that x tons of 8% copper ore was used to make the mixture.
8%x + 3%(300-x) = 300*6%
9 + 5%x = 18
5%x = 9
x = 9/5% = 900/5 = 180 tons
IMO E
