# If 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?

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1 Problem 3 If 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?

3 Problem 3 Solution 1. First of all write down what you know. This is that A B 2 3 and that B C 2. The key thing to realise in this question is that A B B C A C This is just the usual cancellation law when you multiply fractions. 3. This means that A C A B B C 2 3 2

4 Problem 4 Each shape contained within the largest square is also a square. The number in each square gives the length of its sides. What are the values of A, B, C and D? B C A 4 D

5 Suggested Questions to ask students about Problem 4 What 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 from top to bottom of the big square? (This is a slightly strange question which possibly overcomplicates a rather trivial point so you might not want to use this.) Write down the equations you get using this fact for some randomly selected horizontal or vertical lines? Are some of the equations easier than others? Getting into Problem 4 Look at the diagram, think about what you are trying to work out and what information 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 going from 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 and D. Can you find any equations this way that just involve one of these letters?

6 Problem 4 Solution 1. Looking at the bottom edge of the whole square, it s clear that its side length is = Now, by looking at Dotted Line 1, it can be seen that 12 + A = 35. Solving this equation gives A = By looking at Dotted Line 2, and remembering that A = 7 it can be seen that 15 + B = 35. This means that B = Finally, by looking at Dotted Line 3 and remembering that B = 3, it can be seen that C + 12 = 35 which means that C = 8.

7 Problem 5 Over the course of numbering every page in a book, a mechanical stamp printed 2,929 individual digits. How many pages the book must have?

8 Suggested Questions to ask students about Problem 5 If a book had nine pages how many printed digits would there be for the page numbers? 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? Check whether you answer is correct? If it s not correct do you need to increase or decrease the number of pages? How can you work out how many you need to increase it or decrease it by? Getting into Problem 5 To do this problem you need to start thinking about the number of digits on page numbers 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 here is just 9. After that you get pages which have two digits in them. These are pages 10 to 99 inclusive. How many pages like this are there (be careful)? What is the total number of digits that appear on these pages. So how many digits would there be on the first 99 pages? If the book in the problem has more than this number of digits then it must have 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 to think carefully about how to get the exact number of pages.

9 Problem 5 Solution 1. Our book needed 2,929 digits to be stamped. Start by getting a feel for how many pages this might be. To do this it s good to think about the pages in sequence, from the beginning, thinking about how many digits there are in each page number. 2. Pages 1-9 have one digit each, so the total number of digits to be stamped for these pages is 9. This means our book has more than 9 pages. 3. Pages all need two digits to be stamped. From pages 10 to 99 inclusive there are 90 pages. So, for these 90 pages 90 x 2 = 180 digits need to be stamped. This means that from page 1 99, 189 digits need to be stamped. So our book has more than 99 pages. 4. Pages all need three digits to be stamped. There are 900 pages from page 100 to page 999 inclusive and so 900 x 3 = 2700 digits need to be stamped. This means that from page 1 999, 2889 digits 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 is this? This is just 2,929 2,889 = Pages 1000, 1001, 1002 now all have 4 digits. This must mean that there are 40/4 = 10 more pages after page This means our book has 1009 pages.

10 Problem 6 Can you make the two columns of numbers add up to the same total by swapping just two cards?

11 Suggested Questions to ask students about Problem 6 What do the two columns add up to at the moment? So how many bigger does the smaller one need to get (or equivalently how many smaller does the bigger one need to get)? Think about possible swaps. If you swap the 2 in the left hand column with the 5 in the right hand column how much will the left hand column go up by and how much will the right hand column go down by? So would that swap make the two columns equal? What about other swaps? Is there any pattern in what happens with the swaps? Getting into Problem 6 Remember that when you swap a number in the left hand column with a number in the 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 in the right hand column, the left hand column total will go up by 3 and the left hand column 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 the implications 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 this problem.

12 Problem 6 Solution 1. This is a tough question. You may have noticed that whatever you seem 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 same number, call this n. That would mean that the total on all the cards was 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 = 39, not an even number. 3. This is actually an example of proof by contradiction. The thing we did wrong, which led to the contradiction in the above is to suppose that we could make the columns add to the same number. Therefore it must be impossible to make the columns add to the same number.

13 Problem 7 What is the smallest number divisible by 1, 2, 3, 4, 5, 6, 7, 8 and 9?

14 Suggested Questions to ask students about Problem 7 Ask 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 factorisations help 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 7 Think 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 and 10. 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 prime factorisation which includes all of these primes. It needs to include each prime the maximum 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.

15 Problem 7 Solution 1. To do this it s useful to think of the prime factorisations of all the numbers from 2 to 9. These are listed below. Number Prime factorisation Primes in the prime factorisation, with repeats x 2 2, x 3 2, x 2 x 2 2, 2, x 3 3, 3 2. Now think about the prime factorisation of the smallest number which is a multiple of all the numbers from 1 to 9. It needs to include all the primes which are listed in the right hand column above. The number of times we need each prime is given by the highest number of occurences of it in the right hand column above. The primes appearing 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 have the 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 = You can see from the prime factorisation that it is divisible by all the numbers from 1 to 9. Number Appearance as a divisor 2 2 x 2 x 2 x 3 x 3 x 5 x x 2 x 2 x 3 x 3 x 5 x x 2 x 2 x 3 x 3 x 5 x x 2 x 2 x 3 x 3 x 5 x x 2 x 2 x 3 x 3 x 5 x x 2 x 2 x 3 x 3 x 5 x x 2 x 2 x 3 x 3 x 5 x x 2 x 2 x 3 x 3 x 5 x 7

16 Problem 8 This problem is about dividing with a remainder. As an example of this, when 29 is divided by 6 the answer is 4 and the remainder is 5. When a whole number x is divided by 5, the answer is y and the remainder is 4. When x is divided by 4, the answer is v and the remainder is 1. Can you write an equation expressing y in terms v?

18 Problem 8 Solution 1. You need to find a way of writing these ideas using algebra 2. If 29 divided by 4 is 6 with remainder 5, we can write this as: 29 = (4 6) + 5 Similarly, if x divided by 5 is y with remainder 4, we can write this as: x = (5 y) + 4 x = 5y + 4 And finally, if x divided by 4 is v with remainder 1, we can write this as: x = (4 v) + 1 x = 4v Now we have 2 formulae for x, which we can equate. This gives: 5y + 4 = 4v + 1 Now rearrange this to make y the subject 5y = 4v 3 y = 4/5v - 3/5

19 Problem 9 In the diagram below what is the sum of the four shaded angles?

21 1. Look at triangle WXY first. Problem 9 Solution Its angles should total 0, so we know that b + (a + 30) + 40 = 180 Rearranging a + b = a + b 0 2. Now let s examine triangle XYZ. Its angles should also total 0, so we know that 30 + (40 + d) + c = 180 Rearranging c + d = c + d 0 3. Combining this, we see that the sum of the angles is a + b + c + d = (a + b) + (c + d) =

22 Problem 10 Five numbers are arranged in order from least to greatest: x, x 3, x 4, x 2, x 0 Where does x 1 belong in the list above?

23 Suggested Questions to ask students about Problem 10 One of the numbers in the list given doesn t depend on x, which one? What does this tell you about x? When you square a number which is between 0 and 1 is the answer bigger or smaller than the number you started with? Combined with the information you ve been given what does this tell you about x? What happens if you cube a number between -1 and 0? Is the answer to the right or left of the original number on the number line? What happens if you cube a number which is less than -1? Is the answer to the left or to the right on the number line? Getting into Problem 10 To do this question aim to try to work out whether x is positive or negative and what its magnitude is compared to 1 by thinking about the answers to the questions above. To answer these questions, it s useful to consider examples in each case. Once you know this you should be able to work out where -1/x should go.

24 We are told that What does this tell us about x? Problem 10 Solution x < x 3 < x 4 < x 2 < x 0 First of all x 0 = 1 so we know that x < 1. If x = 0 then x 4 = x 2 which is not the case. If 0 < x < 1 then we would have x 2 < x (think about x = 0.5 for example). So it must be the case that x < 0. If x < -1 then we would have x 3 < x (then about x = -2 for example). If x = -1 then x 3 = x which is not the case. From all the above we can deduce that -1 < x < 0. Now think about x -1. If you take a number between -1 and 0 then the reciprocal is less than -1. So the negative of the reciprocal is greater than 1.

25 Problem 11 Explain why the 1st of March is always on the same day of the week as the 1st of November Make a deduction about the day of the week that 31st May falls on compared to the 1st August.

26 Suggested questions to ask students about Problem 11 Ask questions like What do you need to think about when you are doing the first part of this question? How could you get caught out? How does what you have found in the first part of the question help with the second part? Getting into Problem 11 First of all work out how many days there are between 1st March and 1st November. It s important to define what is meant by between, which dates are counted and which are not. One this is done then a division should show the reason why the two dates are on the same day of the week. Think about 31st May and 1st August, it should be possible able to use the same techniques to see the relationship between the days on which these two dates fall.

27 Problem 11 Solution 1. Let s look at the cycle of days through the week. Take any start date. We know that 7 days later, it will be the same day of the week as the start date. In addition, 14 days, 21 days, 28 days etc. later will also be the same day of the week. 2. So let s count the number of days between st March and st November in each year. We add up 31 days for March, 30 for April, 31 for May, 30 for June, 31 for July, 31 for August, 30 for September, and 31 for October. This gives 245 in total. 3. Now let s look at 3 st May and st August. There are = 62 days between these dates, which is not a multiple of 7. However, 62 is equal to 8 lots of 7 with a remainder of 6. So we can say for sure that if 1st August will be the day of the week previous to whatever day of the week the 31st May was. For example, in 2011, 31st May was a Tuesday, and 1st August was the day of the week before Tuesday a Monday!

28 Problem 12 A ball is dropped and bounces up to a height that is 75% of the height from which it was dropped. It then bounces again to a height that is 75% of the previous height and so on. How many bounces does it make before it bounces up to less than 20% of the original height from which it was dropped?

30 Problem 12 Solution 1. Let s call the original height the ball was dropped from h. After the first bounce we know it reached 75% of this height, so we can write this as 0.75h. After the next bounce it reached a height of: 0.75 x 0.75h = h 2. The question is asking to find the number of bounces it took for the height of the ball to be less than 20% of the original height. In other words less than 0.2h. 3. After 3 bounces, the ball reached 0.75 x h = h. This is only 42.19% of the original height, so it is not small enough yet. 4. After 4 bounces, the ball reached 0.75 x h = h. This is still not small enough. 5. After 5 bounces, the ball reached 0.75 x h = h. Still not small enough! 6. Finally, after 6 bounces, the ball s height was 0. 5 x h h, or 17.8% of its original height, which is less than 20%! So the answer is 6 bounces.

31 Problem 13 Adam has three coins in his pocket, and they are all different from each other. Ben has three coins in his pocket and they are all the same as each other. Adam has half as much money as Ben. What are the coins they each have? What happens if Adam has twice as much money as Ben?

33 Problem 13 Solution 1. Let s look at the possible types of coins that Ben could have in his pocket, and the total amount of money this would be. 2. We can also use this to see how much Adam would have in each case, because we know it is half as much as Ben. 3. Then we can ask ourselves whether it s possible to make Adam s amount with three different coins. Ben Adam Make with 3 different coins? 3 x 1p 1.5p NO can t have 0.5p! 3 x 2p 3p NO the smallest 3 coins make 8p! 3 x 5p 7.5p NO can t have 0.5p! 3 x 10p 15p NO 3 x 20p 30p NO 3 x 50p 75p YES 3 x NO 3 x NO 4. The only way it would work is if Ben had three 50p coins in his pocket, totalling This would mean that Adam had 75p in total, made up of a 50p coin, a 20p coin, and a 5p coin. 5. If Adam has twice as much money as Ben then Ben Adam Make with 3 different coins? 3 x 1p 6p NO 3 x 2p 12p NO (1 x 10p and 2 x 1p) 3 x 5p 30p NO (1 x 20p and 2 x 5p) 3 x 10p 60p NO (1 x 50p and 2 x 5p) 3 x 20p 1.20 NO (1 x 1 and 2 x 20p) 3 x 50p 3.00 NO (1 x 2 and 2 x 50p or 3 x 1) 3 x NO 3 x NO In this case there is no solution if all the coins must be different but there would be several solutions if two coins could be the same.

34 Problem 14 p and q are two numbers each greater than zero. (p 2 + 5q) = 8 (p 2 3q) = 6 Find the values of p and q.

36 Problem 14 Solution To solve this problem we need to manipulate some algebra. We are starting with 2 equations: (p 2 + 5q) = 8 (p 2-3q) = 6 First of all, let s square both sides of each equation to get rid of the square root p 2 + 5q = 64 (1) p 2-3q = 36 (2) We can solve this pair of simultaneous equations by taking equation (2) away from equation (1) p 2 p 2 + 5q (-3q) = q = 28 q = 2 Now we have to substitute our value of q into one of the equations to get p; it doesn t matter which one we use Using equation (1) gives p x 2 = 64 p = 64 2p = 128 2p 2 = 93 p 2 = So p = or 93 2, but we know p > 0, so p =

37 Problem 15 A set of five numbers has: a mode of 24 a median of 21 a mean of 20 What could the numbers be?

38 Suggested questions to ask students about Problem 15 Students will probably already have a good understanding of the terms: mode, median and mean. Questions to ask could be Think of any group of five numbers with a mode of 24. What does this fact alone tell you about the number of 24s in the list? Think of any group of five numbers with a median of 2. It s probably best to think of the numbers in ascending or descending order. What exactly does this fact tell you if you were to write the numbers like this? How you would calculate the mean of a group of five numbers? If the mean is 20 what must the total be? Getting into Problem 15 If the mode is 24, then 24 must appear more times than the other numbers in the set. What is the best number of 2 s to include? If the median number is 21, then 21 must be in the middle of the set, What does this tell you about the position of the two 24s if the list is written in order? The mean of the numbers must be 20 so use this fact to work out the total of the five numbers and some possible values for the two other numbers in the list. Think about whether there could be more than one solution to this problem and, if there is, how many soutions are there?

39 Problem 15 Solution 1. If the mode is 24, we need 24 to appear more times than the other numbers in the set. Let s have two 2 s in our set, and the other three numbers different to each other. 2. If the median number is 21, we need to have 21 in our set, and also two numbers lower, and two numbers higher. Our two 2 s are higher, so we now just need two different numbers lower than The mean of the numbers must be 20. Let s call the last two unknown numbers a and b. To have a mean of 20, we need: a + b = 20 5 a + b a + b a + b 3 4. We can now pick a and b to be any two numbers that add up to 31, e.g. a = 17 and b = 14 would give the solution 14, 17, 21, 24 and 24 But there are several other correct solutions.

40 Problem 16 Can you work out the shaded area in the diagram (the line shown just touches the smaller circle)?

42 Problem 16 Solution 1. Start by calling the radius of the small circle r and the radius of the big circle R. The shaded area will have an area of = X 2. Now, using the 10cm line given, draw a triangle in the circle with vertices O (centre of circle), X (one end of the 10cm line) and Y (midpoint of the line). Y 3. We now have a right angled triangle, with sides XY = 5cm, OX = R and OY = r. (It is right angled because a radius of a circle passing through the midpoint of a chord will create a right angle.) O 4. Using Pythagoras theorem, R 2 = r 2 R 2 r 2 = Substitute this into our formula for the shaded area that we got in step 1 gives

43 Problem 17 You have eight coins that all look identical but only one is solid gold. The solid gold coin weighs slightly more than the fakes. You can use the balance only twice. How can you work out which is the real gold coin?

44 Suggested questions to ask students about Problem 17 The key to this problem is that if you put some coins on the scales and it balances, the heavier coin is not one of the ones on the scale. Ask students to think about all possible things that could happen on the second weighing if, for the first weighing they put four on each side put three on each side put two on each side put one on each side Getting into Problem 17 Now you need to think about what the first weigh will be. You can only use the balance twice so after the first weighing there is only one more weigh so you need to think about what will happen next for all the different things you could do on the first weighing. It s helpful to split the coins into groups, but how many groups, two, three, four, and how many coins in each group? Listing the possibilities and all their outcomes is a sensible approach to this problem.

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