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RPWM # 1
A can complete a piece of work in 4 days. B takes double the time taken by A. C takes double that of B, and D takes double that of C to complete the same task. They are paired in groups of two each. One pair takes two-thirds the time needed by the second pair to complete the work. How are the four paired?
It appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. - Niels Henrik Abel.
Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.
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RPWM # 2
A, B and C went to buy things from a wholesale market. They had a combined sum of $ 900. A spent 80%, B spent 70% and C spent 75% of their respective amounts. Now the ratio of amounts left with them is 4 : 9 : 10. Find the ratio of amounts they had in the beginning.
It appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. - Niels Henrik Abel.
Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.
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#2
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Excellent, Ashwil!
It appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. - Niels Henrik Abel.
Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.
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RPWM # 3
Three vessels contain mixtures of milk and water in the ratio 3:1, 4:1, and 5:1. The contents of the three vessels are mixed in the ratio x:y:1. If the resulting mixture contains milk and water in the ratio 7:1, find y in terms of x.
It appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. - Niels Henrik Abel.
Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.
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Last edited by All_Is_Number (2006-07-12 09:10:13)
You can shear a sheep many times but skin him only once.
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RPWM # 3
Three vessels contain mixtures of milk and water in the ratio 3:1, 4:1, and 5:1. The contents of the three vessels are mixed in the ratio x:y:1. If the resulting mixture contains milk and water in the ratio 7:1, find y in terms of x.
I'm not sure there is a valid solution to this problem. If 3:1 implies 3 parts milk to one part water, and 4:1 implies 4 parts milk to one part water, etc; there is no way to produce a solution of seven parts milk to one part water with the three solutions available. They will all have too much water. On the other hand, if 3:1 means three parts water to one part milk, etc; all possible solutions will have too much milk. One of the initial solutions must have a ratio higher than 7:1 in order to end up with a solution of 7:1. As the problem currently stands, I believe there are no real solutions.
You can shear a sheep many times but skin him only once.
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I think that's right, AllIs.
"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman
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A & C , B&d
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#3 isn't a valid problem because the ratio of the final mixture Should be between 3:1 & 5:1.
Still I calculated this.
Last edited by G-man (2011-02-28 20:13:36)
Maths!......
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