20 Mathematical Problems Suitable For Higher Tier GCSE .
FMSP GCSE Problem Solving ResourcesFurther Mathematics Support Programme20 Mathematical Problems suitablefor Higher Tier GCSE StudentsA collection of 20 mathematical problems toencourage the development of problem-solvingskills at KS4.Each includes suggested questions to askstudents to help them to think about theproblem and a full worked solution.The problem sheets are available for freedownload separately fromwww.furthermaths.org.ukGCSE Problem Solving bookletCM 12/08/15Version 1
FMSP GCSE Problem Solving ResourcesGCSE Problem Solving bookletCM 12/08/15Version 1
FMSP GCSE Problem Solving ResourcesProblem 1Can you arrange the numbers 1 – 15 in a trianglewhere the number below each pair of numbers isthe difference between those two numbers?Here’s an example using the numbers 1 – 10:61510948273Can you complete the triangle below?6?12?4?2?5GCSE Problem Solving bookletCM 12/08/15Version 1
FMSP GCSE Problem Solving ResourcesSuggested Questions to ask studentsabout Problem 1Note: It’s important that students fully understand the term ‘difference’. A differencebetween two numbers is never negative: it’s just the distance between the numbers ifyou were to mark them on the number line. So the difference between 5 and 2 is 3,as is the difference between 2 and 5.Ask questions like‘What is the difference between 2 and 5?’. Then ‘What is the difference between 5and 2?’.‘If the difference between x and 5 is 2, what could x be? (there are two possibilities)‘If the difference between x and y is 10 what could x and y be?’.Now relate this to the triangle problem, if necessary asking questions like ‘what couldthe missing numbers be in each of the following diagrams?’.?510113?2Getting into Problem 1Now look at the triangle in the problem. Ask if there are any numbers that candefinitely be filled in straight away from the information given.Once this is done, look at parts of the diagram where there is some informationabout the missing numbers but not enough to work them out exactly straight away.Ask what the possibilities are and lots of ‘what if?’ type questions.Remember that only the numbers from 1 – 15 inclusive can be used and that nonumber should appear twice. Keep asking question like ‘what if this number is a 6?’.Like when doing a Sudoku problem, it might be useful to make rough notes of whichnumbers could go where before committing to them.GCSE Problem Solving bookletCM 12/08/15Version 1
FMSP GCSE Problem Solving ResourcesProblem 1 – Solution1. Here is the initial grid2. Some numbers can be entered from thebottom of the diagram, 9 then 11 then 1.These three are easy to see:3. It’s worth noting that the numbers in thetwo positions highlighted below are notimmediately obvious based on the differenceproperty of the table alone. There are twopossibilities for each of them, if you onlyconsider that property.4. However the position marked 15 or 7actually has to be 7 because 15 itself cannotbe a difference and so it needs to be in thetop row. 8 in the second row follows fromthis.The position marked 14 or 10 has to be 10because, of the available numbers, 14 canonly be the difference between 15 and 1 but1 is already used.5. The remaining numbers to be put in are 3, 13, 14 and 15. With a bit of thought it’s possibleto see that this is the complete solution.GCSE Problem Solving bookletCM 12/08/15Version 1
FMSP GCSE Problem Solving ResourcesProblem 2What number, when multiplied by itself,is equal to 27 x 147?GCSE Problem Solving bookletCM 12/08/15Version 1
FMSP GCSE Problem Solving ResourcesSuggested Questions to ask students about Problem 2It’s important that students realise they are looking for the square root of 27 x 147 inthis question.It’s worth asking, as an example to make this point,‘What is another name for the number, which when multiplied by itself, equals 57?’.Ask students to approximate 27 x 147.Ask whether they can suggest some numbers whose square is this big? They may ormay not be able to do this but it’s a good thing for them to think about. The idea inthis question is that there is a much better way to approach this.Ask students a good way to find the divisors of a number?Ask students the following ‘here is a number expressed as a product of primenumbers, what is its square root?’3 x 3 x 3 x 3 x 5 x 5 x 11 x 11 x 17 x 17 x 17 x 17Getting into problem 2Remember that you don’t have a calculator to do this.The first thing to think about is the phrase “number, when multiplied by itself,gives ”. This means that you’re looking for the square root of a certain number.Here that number is 27 x 147. If you actually do that multiplication by hand you’ll geta pretty big number (about 4,000 or so) and it will be difficult to spot the square root.Therefore you are looking for a better way to do this.This problem is about knowing divisors of 27 x 147 and picking out the one that isthe square root.Finding prime factors is the best way to find out about all the divisors.An important realisation here is that the prime factors of 27 x 147 are the primefactors of 27 together with the prime factors of 147.When you write 27 x 147 as a product of prime factors you then need to think abouthow this tells you about its square root.GCSE Problem Solving bookletCM 12/08/15Version 1
FMSP GCSE Problem Solving ResourcesProblem 2 – Solution1. The number which, when multiplied by itself, is equal to 27 x 147 isthe square root of 27 x 147. To get an idea of what this might be it’suseful to investigate the prime factor of 27 x 147. These will be theprime factors of 27 together with the prime factors of 147.279333x147349772. So 27 x 147 3 x 3 x 3 x 3 x 7 x 7. Because there are an evennumber of 3s and an even number of 7s we can write this as a wholenumber squared:27 x 147 (3 x 3 x 7) x (3 x 3 x 7) (3 x 3 x 7)2 6323. Think about why this would not have been possible if there had notbeen an even number of each prime factor.GCSE Problem Solving bookletCM 12/08/15Version 1
FMSP GCSE Problem Solving ResourcesProblem 3If A is divided by B the result is 2/3.If B is divided by C the result is 4/7.What is the result if A is divided by C?GCSE Problem Solving bookletCM 12/08/15Version 1
FMSP GCSE Problem Solving ResourcesSuggested Questions to ask students about Problem 3The key to this question is getting students to be able to write down a piece ofinformation like ‘if A is divided by B the answer is 2/3’ as a mathematical equation.Ask questions like‘If A is divided by B and the answer is 2/3 which is the bigger A or B?”‘Give me some examples of number A and B where if A is divided by B the answer is2/3’‘Can you write down an equation is satisfied by A and B if A divided by B is 2/3?’To get students thinking about A divided by C you could ask questions like ‘If A istwice as big as B as B is four times as big as C how many times bigger is A than C’?Getting into Problem 3Think about what ‘if A is divided by B the answer is 2/3’ tells you about A and B?There are lots of different ways to write this. One isA 2B 3another is that the ratio A : B is 2 : 3. You could even rearrange the above to get3A 2B.Now think about what you need to do. You have a statement about A and B andanother statement about B and C but you want to end up with a statement about Aand C.Somehow you need to combine the two pieces of information so that B is no longerinvolved. What does this remind you of?Sometimes when you are dealing with more than one equation you try to eliminatesome of the letters.GCSE Problem Solving bookletCM 12/08/15Version 1
FMSP GCSE Problem Solving ResourcesProblem 3 – Solution1. First of all write down what you know. This is thatA 2B 4and thatB 3C 72. The key thing to realise in this question is thatA B AB C CThis is just the usual cancellation law when you multiply fractions.3. This means thatA A B 2 4C B C 3 7 21GCSE Problem Solving bookletCM 12/08/15Version 1
FMSP GCSE Problem Solving ResourcesProblem 4Each shape contained within the largest square is also asquare.The number in each square gives the length of its sides.What are the values of A, B, C and D?1015CB121012A4D1348104GCSE Problem Solving bookletCM 12/08/15Version 1
FMSP GCSE Problem Solving ResourcesSuggested Questions to ask students about Problem 4What is the length of the sides of the big square?So how long is any horizontal line going from side to side or vertical line going fromtop to bottom of the big square? (This is a slightly strange question which possiblyovercomplicates a rather trivial point so you might not want to use this.)Write down the equations you get using this fact for some randomly selectedhorizontal or vertical lines? Are some of the equations easier than others?Getting into Problem 4Look at the diagram, think about what you are trying to work out and whatinformation you can get from the diagram.You should be able to work out how long the sides of the big square are.Now remember that any horizontal line going from side to side or vertical line goingfrom top to bottom of the big square will have that length.Such horizontal or vertical lines will give you different equations involving A, B, C andD.Can you find any equations this way that just involve one of these letters?GCSE Problem Solving bookletCM 12/08/15Version 1
FMSP GCSE Problem Solving ResourcesProblem 4 – SolutionD1. Looking at the bottom edge of the whole square, it’s clear that its sidelength is 4 8 10 13 35.2. Now, by looking at Dotted Line 1, it can be seen that 12 A 4 12 35. Solving this equation gives A 7.3. By looking at Dotted Line 2, and remembering that A 7 it can beseen that 15 B 7 10 35. This means that B 3.4. By looking at Dotted Line 3 and remembering that B 3, it can beseen that 12 3 C 12 35 which means that C 8.5. Finally, by using either the value of A or the value of C we can get avalue of D 3.GCSE Problem Solving bookletCM 12/08/15Version 1
FMSP GCSE Problem Solving ResourcesProblem 5Over the course ofnumbering everypage in a book, amechanical stampprinted 2,929individual digits.How many pages does the book have?GCSE Problem Solving bookletCM 12/08/15Version 1
FMSP GCSE Problem Solving ResourcesSuggested Questions to ask students about Problem 5If a book had nine pages how many printed digits would there be for the pagenumbers?If a book has ninety-nine pages how many printed digits would there be?How about one hundred and ninety-nine pages?Using this, have a guess as to how many pages this book might have? Checkwhether you answer is correct?If it’s not correct do you need to increase or decrease the number of pages? Howcan you work out how many you need to increase it or decrease it by?Getting into Problem 5To do this problem you need to start thinking about the number of digits on pagenumbers of a book starting from the start.The first nine pages are pages 1, 2, 3, 4, 5, 6, 7, 8, 9. The total number of digits hereis just 9.After that you get pages which have two digits in them. These are pages 10 to 99inclusive. How many pages like this are there (be careful)? What is the total numberof digits that appear on these pages. So how many digits would there be on the first99 pages? If the book in the problem has more than this number of digits then it musthave more than 99 pages.Keep going using this approach.Once you think you have a rough idea of how many pages there are you will need tothink carefully about how to get the exact number of pages.GCSE Problem Solving bookletCM 12/08/15Version 1
FMSP GCSE Problem Solving ResourcesProblem 5 – Solution1. Our book needed 2,929 digits to be stamped. Start by getting a feelfor how many pages this might be. To do this it’s good to think aboutthe pages in sequence, from the beginning, thinking about how manydigits there are in each page number.2. Pages 1-9 have one digit each, so the total number of digits to bestamped for these pages is 9. This means our book has more than 9pages.3. Pages 10-99 all need two digits to be stamped. From pages 10 to 99inclusive there are 90 pages. So, for these 90 pages 90 x 2 180digits need to be stamped. This means that from page 1 – 99, 189digits need to be stamped. So our book has more than 99 pages.4. Pages 100-999 all need three digits to be stamped. There are 900pages from page 100 to page 999 inclusive and so 900 x 3 2700digits need to be stamped. This means that from page 1 – 999, 2889digits need to be stamped. So our book has more than 999 pages.5. Our book has 2,929 digits stamped. How many more than 2,889 isthis? This is just 2,929 – 2,889 40.6. Pages 1000, 1001, 1002 now all have 4 digits. This must mean thatthere are 40/4 10 more pages after page 999.7. This means our book has 1009 pages.GCSE Problem Solving bookletCM 12/08/15Version 1
FMSP GCSE Problem Solving ResourcesProblem 6Can you make the two columns ofnumbers add up to the same total byswapping just two cards between thecolumns?GCSE Problem Solving booklet13247598CM 12/08/15Version 1
FMSP GCSE Problem Solving ResourcesSuggested Questions to ask students about Problem 6What do the two columns add up to at the moment?So how many bigger does the smaller one need to get (or equivalently how manysmaller does the bigger one need to get)?Think about possible swaps. If you swap the 2 in the left hand column with the 5 inthe right hand column how much will the left hand column go up by and how muchwill the right hand column go down by? So would that swap make the two columnsequal?What about other swaps? Is there any pattern in what happens with the swaps?Getting into Problem 6Remember that when you swap a number in the left hand column with a number inthe right hand column, both totals will change.Think about examples of this. If you swap the 2 in the left hand column with the 5 inthe right hand column, the left hand column total will go up by 3 and the left handcolumn total will go down by 3.You may now start to spot that this is going to be tricky.You might be able to prove that any swap you try will not do what you need.You might be able to find another reason that you can’t do this. Ask yourself what theimplications would be if you were able to do this?It’s useful to think about odd and even numbers to really see what is going on in thisproblem.GCSE Problem Solving bookletCM 12/08/15Version 1
FMSP GCSE Problem Solving ResourcesProblem 6 – Solution1. This is a tough question. You may have noticed that whatever youseem to try you can’t get the two columns to add up to the same total.2. In fact it’s impossible and here is why:Suppose you could make the two columns add up to the samenumber, call this n. That would mean that the total on all the cardswas n n 2n.Now n is a whole number and so 2n is an even number.But if you add up all the cards you get 1 2 3 4 5 7 8 9 39, not an even number.3. This is actually an example of ‘proof by contradiction’. The thing wedid wrong, which led to the contradiction in the above is to supposethat we could make the columns add to the same number. Thereforeit must be impossible to make the columns add to the same number.GCSE Problem Solving bookletCM 12/08/15Version 1
FMSP GCSE Problem Solving ResourcesProblem 7What is the smallest number divisible by1, 2, 3, 4, 5, 6, 7, 8 and 9?GCSE Problem Solving bookletCM 12/08/15Version 1
FMSP GCSE Problem Solving ResourcesSuggested Questions to ask students about Problem 7Ask students whether the smallest number divisible by both 6 and 10 is 6 x 10 60?What is the smallest number divisible by both 6 and 10? How do prime factorisationshelp with this?So will the answer to this problem be 1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 x 9?How can prime factorisations help with this?Getting into Problem 7Think about how you would do this problem if there are only two numbers.It might be useful to think about the smallest number that is divisible by both 6 and10.Take the prime factorisations of 6 and 10. These are 6 2 x 3 and 10 2 x 5.The smallest number that is divisible by both 6 and 10 must have a primefactorisation which includes all of these primes. It needs to include each prime themaximum number of times it appears in any of the factorisations of 6 and 10.So the smallest number divisible by 6 and 10 is 2 x 3 x 5 30.GCSE Problem Solving bookletCM 12/08/15Version 1
FMSP GCSE Problem Solving ResourcesProblem 7 – Solution1. To do this it’s useful to think of the prime factorisations of all thenumbers from 2 to 9. These are listed below.Number Primefactorisation223342x25562x37782x2x293x3Primes in the prime factorisation, withrepeats232, 252, 372, 2, 23, 32. Now think about the prime factorisation of the smallest number whichis a multiple of all the numbers from 1 to 9. It needs to include all theprimes which are listed in the right hand column above. The numberof times we need each prime is given by the highest number ofoccurences of it in the right hand column above. The primesappearing are 2, 3, 5, 7. The highest number of occurences of 2 is 3(in 8 2 x 2 x 2). If each prime appears that many times we will havethe smallest number that is divisible by all of the above.3. This gives us the number 2 x 2 x 2 x 3 x 3 x 5 x 7 2520. You cansee from the prime factorisation that it is divisible by all the numbersfrom 1 to 9.Number23456789Appearance as a 2x2x3x3x5x7GCSE Problem Solving bookletCM 12/08/15Version 1
FMSP GCSE Problem Solving ResourcesProblem 8This problem is about dividing with aremainder.As an example of this, when 29 is divided by 6the answer is 4 and the remainder is 5.When a whole number x is divided by 5, theanswer is y and the remainder is 4.When x is divided by 4, the answer is v andthe remainder is 1.Can you write an equation expressing y interms v?GCSE Problem Solving bookletCM 12/08/15Version 1
FMSP GCSE Problem Solving ResourcesSuggested Questions to ask students about Problem 8Ask students how they can write a statement like “When x is divided by 5 the answeris y and the remainder is z” as a mathematical expression.Ask students how the fact that 29 7 x 4 1 shows that when 29 is divided by 4, theanswer is 7 and the remainder is 1.Ask them about the significance of 1 4 in the example above (remainders must beless than the number you are dividing by).Getting into Problem 8Think about an example like ‘When 29 is divided by 3 the answer is 9 and theremainder is 2’.Another way of saying that ‘When 29 is divided by 9 the answer is 3 and theremainder is 2’ is to state that 29 3 x 9 2.RemainderAnswerThis idea should give you two algebraic expressions and you can go on from there.GCSE Problem Solving bookletCM 12/08/15Version 1
FMSP GCSE Problem Solving ResourcesProblem 8 – Solution1. You need to find a way of writing these ideas using algebra2. If 29 divided by 4 is 6 with remainder 5, we can write this as:29 (4 6) 5Similarly, if x divided by 5 is y with remainder 4, we can write this as:x (5 y) 4x 5y 4And finally, if x divided by 4 is v with remainder 1, we can write thisas:x (4 v) 1x 4v 13. Now we have 2 formulae for x, which we can equate. This gives:5y 4 4v 1Now rearrange this to make y the subject5y 4v – 3y 4/5v - 3/5GCSE Problem Solving bookletCM 12/08/15Version 1
FMSP GCSE Problem Solving ResourcesProblem 9In the diagram below what is the sum ofthe four shaded angles?GCSE Problem Solving bookletCM 12/08/15Version 1
FMSP GCSE Problem Solving ResourcesSuggested Questions to ask students about Problem 9Ask students about how they can find a way to talk about the individual angles in thediagram without having to point at them. (You are trying to steer them towardslabelling them here.)Ask students what facts about angles they can remember and what might berelevant to this problem.Once the angles are labelled ask the students to form any equation
FMSP GCSE Problem Solving Resources GCSE Problem Solving booklet CM 12/08/15 Version 1 Problem 1 – Solution 1. Here is the initial grid 2. Some numbers can be entered from the bottom of the diagram, 9 then 11 then 1. These three are easy to see: 3.
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