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**ganesh****Moderator**- Registered: 2005-06-28
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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?

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**ganesh****Moderator**- Registered: 2005-06-28
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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.

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**ashwil****Member**- Registered: 2006-02-27
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#2

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**ganesh****Moderator**- Registered: 2005-06-28
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**Excellent, Ashwil! **

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**ganesh****Moderator**- Registered: 2005-06-28
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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.

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**All_Is_Number****Member**- Registered: 2006-07-10
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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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**All_Is_Number****Member**- Registered: 2006-07-10
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ganesh wrote:

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.

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**MathsIsFun****Administrator**- Registered: 2005-01-21
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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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**tan****Member**- Registered: 2010-02-17
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A & C , B&d

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**G-man****Member**- Registered: 2011-02-28
- Posts: 16

#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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