Bob

]]>Easy one now :

*Question 7*

*There are 3 numbers. *

*The second is greater than the first by the amount the third is greater than the second. *

*The product of the two smaller numbers is 85. *

*The product of the two larger numbers is 115. *

*If the numbers are x, y, z with x<y<z then the value of (2x + y+ 8z) is? *

* Question 6: How many children do they have?*

They have 3 children.

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Here’s the next one.

* Question 6*

*The combined age of a man and his wife is six times the combined ages of their children. *

*Two years ago their united ages were ten times the combined ages of their children. *

*Six years hence their united ages will be three times the combined ages of the children. *

*How many children do they have?*

That also works in the event of there being joint winners, who would then be named as such.

]]>Yes, I like that method.

Just a comment re the puzzle wording:

I know what you meant with it, but, in a strict interpretation of the wording, answering the second question could have been a bit awkward if there was a tie...and there are 3 scenarios with at least one winner in which a tie occurs.

]]>That's the correct answer!

I solved it like this. It's quite similar.

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** Question 5: The Woodcutting Competition**:

I think that there was no tie, and Third won, the felling ratio being Third:Second:First = 5:4:3.

*EDIT: Included my solution method...*

Here's another question. It's a bit different from the others. Flavours are the essence of life!

**Question 5**

**Three teams of woodcutters have decided to organize a competition. The winner is the team who fells the maximum number of trees in the given time.**

**The first and third team together felled twice the number of trees felled by the second team.**

**The second and third team together felled three fold of the number of trees felled by the first team. **

**Who won? Or in the event of there being joint winners, who would then be named as such?**

Here's a way to do this.

The x^3 term is zero and will give you a, and then the x term will give you b.

From there you can work out p and q.

Bob

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*Question 4*

*If x²+ 2x + 5 is a factor of x^4 + px² + q. *

*p, q =?*

Bob

]]>I was trying using this method:

I think it is almost the same method, just a little more direct.

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