1. ## Maths puzzle

I'm not setting a puzzle here, I'm sure it's very easy when you know how - but I don't! Would one of you maths whizzes out there tell me how to go about solving this, please? It's on a desk calendar for today. I might add that the names are their names, not mine!

<font color=blue>"If eight winkles and nine wonkles cost \$118, and nine winkles and eight wonkles cost \$120, how much will five winkles and five wonkles cost?"</font color=blue>

I'm not often stumped by this sort of thing, but for some reason I just can't work out how to go about this one!

Many thanks

2. ## Re: Maths puzzle

This one is not too hard:

<span style="background-color: #FFFF00; color: #000000; font-weight: bold"><font color=yellow>Add the two statements together. You get: seventeen winkles and seventeen wonkles cost \$118+\$120 = \$238.

Divide by 17 to get: one winkle and one wonkle cost \$238/17 = \$14.

Finally, multiply by 5 to get: five winkles and five wonkles cost 5*\$14 = \$70.</font color=yellow></span hi>

3. ## Re: Maths puzzle

Hans' solution is a rather clever one, since in this particular problem, <span style="background-color: #FFFF00; color: #000000; font-weight: bold"><font color=yellow>you only need the aggregate winkle + wonkle price</font color=yellow></span hi> to answer the question asked. More generally, you might need to derive the individual prices - for instance, if asked how much is 3 winkles and 5 wonkles.

<span style="background-color: #FFFF00; color: #000000; font-weight: bold"><font color=yellow>In such a case, you can use one piece of information to express the winkle price in terms of the wonkle price. e.g.
8wi + 9 wo = 118 means that wi = (118 - 9wo)/8

You then substitute this into the second bit of information: 9wi + 8 wo = 120 and get
9(118 - 9wo)/8 + 8wo = 120 which can be easily solved to give
wo = 428/17

and putting this result back into the substituted equation gives wi = (118 - 9 * 428/17) /8

From there, of course, the price of any winkle/wonkle combination can be calculated.</font color=yellow></span hi>

Alan

4. ## Re: Maths puzzle

Thanks for the reply, Alan, but I have to take issue with two of your comments I'm afraid -

"<span style="background-color: #FFFF00; color: #000000; font-weight: bold"><font color=yellow>... which can be easily solved ...</font color=yellow></span hi>" and
"<span style="background-color: #FFFF00; color: #000000; font-weight: bold"><font color=yellow>From there, of course, the price of any winkle/wonkle combination can be calculated.</font color=yellow></span hi>"

I'm sure it seems easy to you, but I didn't get any further than Maths 'O' level and that was 25 years ago! Any chance of a teensy, weensy little bit more detail ... pretty please?! <img src=/S/grovel.gif border=0 alt=grovel width=31 height=23>

many thanks!

5. ## Re: Maths puzzle

Okay, since both winkles and wonkles start with a w, let's say that winkles are X and wonkles are Y. From the sentence, you can make 2 equations:

8x+9y=118

And

9x+8y=120

So, the first thing you need to do, is solve for X from either equation. Let's take the top one: (8x+9y=118)
Subtract 9y from both sides and you get: 8x=118-9y
Divide both sides by 8, and you get x=14.75-1.125y

We now take the new 'definition' of x, and plug it into the OTHER equation (because plugging it into the first isn't going to do anything for us.)
the second equation is 9x+8y=120
When we replace x, with 14.75-1.125y, we get 9(14.75-1.125y)+8y=120
Multiply 9 through the parenthesis, and we get 132.75-10.125y+8y=120
(I like to keep positive numbers, so let's add 2.125y to both sides):132.75=120+2.125y
Subtract 120 from both sides:12.75=2.125y
Divide both sides by 2.125 and we get: y=6

Now, take y=6, and plug that 6 back into the definition of X (x=14.75-1,125y) and you get x=8.

So, winkles are 8 dollars a peice and wonkles are 6 dollars a peice. Therefore, 5 of each is 8*5 + 6*5, which is 40+30, or 70.

Did I break that down enough?

6. ## Re: Maths puzzle

Hi Beryl

The particular puzzle you have is not a good one for illustrative purposes, simply because the numbers are "cumbersome". More generally, there are two standard methods for solving these kinds of problems - substitution and elimination. These links give better examples of how to use each method.

It might be easier (with the numbers in your problem) to demonstrate the elimination method instead:
8wi + 9wo = 118
9wi + 8wo = 120

What we'd like to do is arrive at an equation in which either wi or wo is not present (eliminated). To eliminate wi, multiply the first equation by 9 and the second by 8:
72wi + 81wo = 1062
72wi + 64wo = 960

The second equation can be subtracted from the first and the winkle term disappears:
(81 - 64)wo = 1062 -960 or
17wo = 102 or
wo = 102/17 = 6

The key to this is that equations can be multiplied by any number and remain true. So by choosing appropriate numbers, one of the unknowns can be "eliminated" when the equations are subtracted. To solve for wo, the choice would be to multiply the first equation by 8 and the second by 9:
64wi + 72wo = 944
81wi + 72wo = 1080

Subtract the first from the second to give:
(81 - 64)wi = 1080 - 944 or
17wi = 136 or
wi = 136/17 = 8

Hans' clever solution your particular puzzle used the methods of the elimination technique, and arrived at the required answer without the need to solve for the individual prices.

Alan

7. ## Re: Maths puzzle

Yes, Drew, thanks very much, I've got my poor head round it now! I knew the principal was to multiply both side up until you could balance them, but I just couldn't see (or remember) how to do it!

Muchos gracias! <img src=/S/thankyou.gif border=0 alt=thankyou width=40 height=15>

8. ## Re: Maths puzzle

Thanks, Alan, your answer and Drew's between make both methods much clearer to me! As I said to Drew, I knew the solution was to multiply both sides up until you got them balanced, but I just couldn't see (or remember) how to do it!

Vielen Danke! <img src=/S/thankyou.gif border=0 alt=thankyou width=40 height=15>

9. ## Re: Maths puzzle

Now that you know the "Algebra" behind it, you can be lazy and have a spreadsheet do it directly:

<table border=1><td></td><td align=center>A</td><td align=center>B</td><td align=center>C</td><td align=center valign=bottom>1</td><td valign=bottom>Winkles</td><td valign=bottom>Wonkles</td><td valign=bottom>Total</td><td align=center valign=bottom>2</td><td align=right valign=bottom>8</td><td align=right valign=bottom>9</td><td align=right valign=bottom>118</td><td align=center valign=bottom>3</td><td align=right valign=bottom>9</td><td align=right valign=bottom>8</td><td align=right valign=bottom>120</td></table>

Create the "results table"
<table border=1><td></td><td align=center>E</td><td align=center>F</td><td align=center valign=bottom>1</td><td align=right valign=bottom></td><td valign=bottom>Individual</td><td align=center valign=bottom>2</td><td valign=bottom>Winkles</td><td valign=bottom> </td><td align=center valign=bottom>3</td><td valign=bottom>Wonkles</td><td valign=bottom> </td></table>

Then in select BOTH F2 and F3
Enter the ARRAY formula in the formula bar
=MMULT(MINVERSE(A2:B3),C2:C3)
Confirm the formula with ctrl-shift-enter
You will then get the results:
<table border=1><td></td><td align=center>E</td><td align=center>F</td><td align=center valign=bottom>1</td><td align=right valign=bottom></td><td valign=bottom>Individual</td><td align=center valign=bottom>2</td><td valign=bottom>Winkles</td><td align=right valign=bottom>8</td><td align=center valign=bottom>3</td><td valign=bottom>Wonkles</td><td align=right valign=bottom>6</td></table>

Without having to do any algera. The results are "Live" so you can change anything in the original "problem" and get new answers

Steve

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