Results 1 to 9 of 9

Thread: Maths puzzle

  1. #1
    Silver Lounger
    Join Date
    Jun 2001
    Location
    Morden, Surrey, United Kingdom
    Posts
    1,838
    Thanks
    3
    Thanked 0 Times in 0 Posts

    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
    Beryl M


  2. #2
    Plutonium Lounger
    Join Date
    Mar 2002
    Posts
    84,353
    Thanks
    0
    Thanked 29 Times in 29 Posts

    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. #3
    Platinum Lounger
    Join Date
    Nov 2001
    Location
    Melbourne, Victoria, Australia
    Posts
    5,016
    Thanks
    0
    Thanked 0 Times in 0 Posts

    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. #4
    Silver Lounger
    Join Date
    Jun 2001
    Location
    Morden, Surrey, United Kingdom
    Posts
    1,838
    Thanks
    3
    Thanked 0 Times in 0 Posts

    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!
    Beryl M


  5. #5
    4 Star Lounger
    Join Date
    Aug 2002
    Location
    Dallas, Texas, USA
    Posts
    594
    Thanks
    0
    Thanked 0 Times in 0 Posts

    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
    add the y components: 132.75-2.125y=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. #6
    Platinum Lounger
    Join Date
    Nov 2001
    Location
    Melbourne, Victoria, Australia
    Posts
    5,016
    Thanks
    0
    Thanked 0 Times in 0 Posts

    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. #7
    Silver Lounger
    Join Date
    Jun 2001
    Location
    Morden, Surrey, United Kingdom
    Posts
    1,838
    Thanks
    3
    Thanked 0 Times in 0 Posts

    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>
    Beryl M


  8. #8
    Silver Lounger
    Join Date
    Jun 2001
    Location
    Morden, Surrey, United Kingdom
    Posts
    1,838
    Thanks
    3
    Thanked 0 Times in 0 Posts

    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>
    Beryl M


  9. #9
    WS Lounge VIP sdckapr's Avatar
    Join Date
    Jul 2002
    Location
    Pittsburgh, Pennsylvania, USA
    Posts
    11,225
    Thanks
    14
    Thanked 342 Times in 335 Posts

    Re: Maths puzzle

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

    Enter your data in Excel:

    <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

Posting Permissions

  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts
  •