On Friday I set this puzzle….

Imagine a test with 26 questions. When you get a question wrong you are deducted 5 points. When you get a question right you are given 8 points. If you answered all of the questions, and had a final score of 0 points, how many questions did you answer correctly?

If you have not tried to solve it, have a go now. For everyone else the answer is after the break.

You answered 10 correctly – but can anyone explain why?

I have produced an ebook containing 101 of the previous Friday Puzzles! It is called **PUZZLED** and is available for the **Kindle **(UK here and USA here) and on the **iBookstore** (UK here in the USA here). You can try 101 of the puzzles for free here.

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10×8 =(26-10)x5

10×8=16×5

80=80

Wrong answer – X

right answer – y

x+y=26

-5x+8y=0

-5x+8(26-x)=0

-5x+208-8x=0

-13x==-208

x=16

y=10

My answer is 22/2 = 11 but then again I am a little bit behind on the Friday Puzzles. Does anyone have the answer to the ‘flagpole’ puzzle posted on 16th September 2011?

If I may repeat the simplest possible explaination:

To have a score of ZERO I must get FIVE questions RIGHT for every EIGHT that I get WRONG.

Yet if I only answered THIRTEEN questions (FIVE+EIGHT) I would have only completed half the test.

Thus I must have got TEN RIGHT and SIXTEEN WRONG.

I know some have claimed that cannot be the right answer as it uses only one equation while they claim the correct solution is only possible by using a second equation.

But it is the right answer so I am vindicated.

YOU are also a TWAT.

@eddie

You perhaps lack language skills to match your lack of maths skills.This was a friendly discussion until you booted your head into it.

Nethertheless I will remain kindly and address myself beyond you to those who deserve a deeper explaination.

None of the solutions put forward need “similteneous” equations. Not Richard’s. Not Roy Gillett’s. Not Darek Olak’s. Not kroketje’s.

It is a travesty of mathematics to say that there is only one way to solve the puzzle and that way is by the use of “similteneous” equations. The world has proved that to be inaccurate. Please pay attention and you may learn something of benefit.

No, no. Eddie nailed it first time.

I agree with you when you say there is no need for simultaneaous equations to solve this problem but you ARE using simultaneous equations albeit the second one is hidden by your reasonning.

When you say that “if I only answered THIRTEEN questions (FIVE+EIGHT) I would have only completed half the test.” you’re actually saying that “number of wrong answers plus number of right answers must be equal to the number of questions on the test” and that’s your second equation.

@Eddie

Thank you for your contribution this morning. Whilst discussing the intricacies of the Friday puzzle, it is good to have people submitting alternative inputs, giving us fresh approaches to solving the problems. It is particularly pleasing to see solutions which are clear, succinct, and lead us quickly to the answer. I always look forward to your insightful contributions to the debate, and today’s post adds much to the discussion. Thank you.

@grizzly – You are BG AICMFP 🙂

Whether or not you solve this puzzle through the direct or indirect use of simultaneous equations, or another method (such as quessing or drawing graphs etc) the fact that the answer is a discrete number is key here.

Under different conditions (a different puzzle or a real-life scenario possibly involving saving money at the supermarket) this might not be so, in which case solving simultaneous equations will always lead to the right answer (although admittedly a supermarket aisle is not the ideal place to be doing this sort of thing!)

The judgement therefore is whether you spend time looking for a short-cut which will allow you to get to the answer quickly but if it isn’t there or isn’t easily seen will have wasted you time, or whether you adopt a tried and tested method which may take a little longer but guarantee the answer (if you do it right of course)

That’s what I thought too Chris R. You can take the man out of BG but you can’t take the BG out of the man (mmmm…. maybe that aphorism needs a bit of work).

I also had to look up what AICMFP meant. I thought the MF bit might be something to do with familial love…..

@ChrisR

Sorry Chris but you don’t win five pounds. I can assure you that I am not BR. I am just some bloke who has grown tired of children calling each other names on the internet. Whatever you think of our Barry, hurling abuse at him is not going to win the argument.

May I thank those who have supported me for their support and kind words. It is well received in a world where harshness seems to dominate.

I am comforted too that 100 years ago Albert Einstein’s special and general theories of relativity did away forever with the concept of similteneity.

Forever that is except in the comments to this blog where ancient maths techniques live on among those who have not moved with the times.

I am sorry that people are rude to you but you are not seeing the truth and aare behaving obnoxiously. You are using 2 equations even if you can’t see it yourself and you are giving away spoilers even if you can’t see that either. I think a lot of people would prefer for you to refrain from the second because they are sick to death of your selfrighteousness.

Hear hear, Cathy! Well said.

I agree that the name-calling is unfortunate but it is understandable when someone behaves with so much arrogant disregard for the wishes of others by repeatedly spoiling the puzzles for everyone else. That their conceit extends to denying blatant truth and proclaiming themselves to be the sole voice of reason while deriding everyone else just adds insult to injury.

Perhaps we should have a vote to see how the community as a whole would like to resolve this problem?

@Cathy

Thank you for your apology on behalf of the not so well behaved elements on this forum. Politeness is a virtue that has yet to find it’s full expression on the internet.

One of the founding principles of the internet is that “information wants to be free”. Delaying that freedom of information for even a weekend for commercial or intellectual gain is undermining the very basis of the internet.

Think of it like astrology. The stars are in the sky for all of us to see. Any of us can interpret the motion of the stars to find relevant facts about our lives and world. For others to restrict our ability to freely do that is not a kind or helpful gesture.

I choose freedom. What do you choose?

It’s funny isn’t it, Barry? You arrogantly state that, despite this community’s expressed wishes, you think that you can behave however you like and spoil everyone’s fun because the “information wants to be free”, and yet you complain when other people behave how

theylike when they call you out on your unpleasantness.I think that Gerry nailed you in his comment and, since you haven’t disagreed with him, you presumably concur. If we were to have a vote, I know what mine would be.

For someone who is supposedly so in touch with their feelings, you display an alarming lack of introspection.

@Barry Goddard

My intuitive answer was 10, which is correct and it used one less equation than you claimed to need to solve the problem. So, no, you are not vindicated.

Oh! In order to demonstrate to that my intuitive solution was correct for the numbers used in the question, and to provide a derivation that would remain correct for different numbers, required the use of two equations in order to prove that my answer wasn’t just a lucky guess and that there wasn’t more than one correct answer to the question and similar questions.

Caveat: There are other mathematically correct answers, but they are irrelevant in the context of the implicit real-world constrains of this particular practical question.

That’s exactly how I reasoned out the answer, Barry.

Lowest common multiplication of both 5 and 8 is 40 – 5 right answers and 8 wrong, but this is too low in total.

The next is 80, which gives 10 correct and 16 error answers – total answers is correct. That’s it!

Exactly how I did it. Nice one.

if you have all the answers wrong you’ll have -130 points and for every wright answer you’ll get 13 points (8 for right answer and – -5 for one less answer wrong) so you need 10 right answers to reach 0 points

Nice bit of reasoning and not a single simil-bloody- taneous equation to be seen.

Using Wolfram Alpha almost certainly counts as cheating, but it adopts the graphical approach: draw a graph of the two equations (8x-5y=0 and x+y=26) and the solutions of x and y are where the two lines intersect (x=10, y = 16).

I used Excel

Five points lost for every wrong answer means answering all twenty six wrong brings you to a hundred and thirty points lost. Then, with thirteen points at stake per right answer (the five for getting it wrong reimbursed pus the eight for getting it right) you only need ten right answers to get to zero points.

It’s very basic maths that a kid could do without the aid of a calculator.

I agree with Barry that there are often very different ways of solving problems.

However, it seems to me that for this particular problem, Barry’s sentences 2&3 are doing it the same way as others (two simultaneous equations) but putting the two equations into words instead of using x and y.

And Barry/Eddie, where is our politeness in the discussions?

To be fair, Barry was being perfectly polite.

The equations are:

8x-5y=0 (1)

and

x+y=0 (2)

Therefore using (2)

y=26-x

Insert this in (1)

8x-5(26-x)=0

13x-5*26x=0

13x=5*26

divide by 13:

x = 5*2 = 10

oops equation (2) should be:

x+y=26

Now I see why my mum always told me that linear algebra would be the key to my future success…

Wonderful article! We will be linking to this particularly great post on our website.

Keep up the great writing.

10*8=16*5

8x – 5(26 – x) = 0

8x + 5x – 130 = 0

13x = 130

X = 10

Everyone solved this with only 2 variables, I instinctively used 3.

x = right answers

y = wrong answers

z = unanswered questions

x,y,z are all non-negative integers

x+y+z=26

8x-5y=0

y = 8x/5

13x/5+z=26

13x=130-5z

x=10-5z/13

==> 5z/13 is a non-negative integer and must therefore be a multiple of 13

x= 10 where z=0

x = 5 where z = 13

x = 0 where z = 26