Algebra

First note that Carl ran (35 − 22) m = 13 m while Bronwen ran 15 m. Let the length of the race be \(x\) m. Since their speeds were constant, the ratio of the distances they ran in any time is also constant. Therefore, since Carl ran \((x - 35)\) m while Bronwen ran \((x - 15)\) m, we have \(\dfrac{x - 35}{x - 15} = \dfrac{13}{15}\). Therefore \(15x - 15 \times 35 = 13x - 13 \times 15\) and hence \(2x = 15 \times 35 - 13 \times 15\). Therefore \(2x = 22 \times 15\) and hence \(x = 11 \times 15\). Therefore the distance they ran is 165 m.

We are given that \(10 \times 2^m = 2^n + 2^{n+2}\). Therefore \(5 \times 2 \times 2^m = 2^n(1 + 2^2)\).
Hence \(5 \times 2^{m+1} = 2^n \times 5\). So \(m + 1 = n\). Therefore the difference between \(m\) and \(n\) is 1.

One face has total 7 which can only be obtained from the given numbers by adding 1, 2 and 4. Therefore it is a triangular face, and so the label of the top vertex, \(x\) say, is one of these three values. Hence the square face has 5 at one vertex. So the smallest possible face total for the square is 5 + 1 + 2 + 3 = 11. Therefore the four face totals given, 7, 8, 9 and 10, must be the face totals of the triangular faces and their sum is 34. Note that each vertex except the top belongs to two triangles; and the top belongs to all four. So the sum 34 is twice the sum of all the labels plus an extra \(2x\); that is \(34 = 2(1 + 2 + 3 + 4 + 5) + 2x = 30 + 2x\). Hence \(x = 2\) and the face total for the square face is 1 + 3 + 4 + 5 = 13.

The first four expressions expand to \(x^2 - 1\);  \(x^2 - \tfrac{1}{4}\);  \(x^2 - \tfrac{1}{9}\);  \(x^2 - \tfrac{1}{16}\) respectively.
Note that \(\tfrac{1}{16} < \tfrac{1}{9} < \tfrac{1}{4} < 1\). Therefore the least value is \(x^2 - \tfrac{1}{16}\), that is \((x + \tfrac{1}{4})(x - \tfrac{1}{4})\).
This result is irrespective of the value of \(x\).

Suppose that at a particular stage there are \(m\) tarts available for a Knave to eat and that there are \(n\) left after he has finished eating.
Then \(n = m - (\tfrac{1}{2}m + \tfrac{1}{2}) = \tfrac{1}{2}m - \tfrac{1}{2}\).
Therefore, \(m = 2n + 1\).
As the Knave of Spades received one tart, then the number of tarts which the Knave of Clubs was given was 2 × 1 + 1 = 3.
Similarly, the number of tarts which the Knave of Diamonds was given was 2 × 3 + 1 = 7.
Finally, the number of tarts which the Knave of Hearts stole was 2 × 7 + 1 = 15.

By dividing \(17x + 51y = 102\) by 17 we get \(x + 3y = 6\). Multiplying by 9 gives \(9x + 27y = 54\).

Let the number of chickens and goats sold be \(c\) and \(g\) respectively.
Then, since 80 animals are sold, we have \(c + g = 80\).
Also, \(2c + 4g = 200\), since the chickens have two legs and the goats have four legs.
Dividing the second equation by 2, we obtain \(c + 2g = 100\).
Then, subtracting the first equation gives \(g = 100 - 80 = 20\).
So 20 goats were sold.

Each diameter has the same sum and contains the central cell, so the pair at the end of each diameter must have the same sum. These four pairs have sum 40, so each pair must have sum 10. Since each diameter has sum 13, the central number must be 3.

Since \(ab = 2cd\), the product \(abcd = 2cd \times cd = 2(cd)^2\), hence it must be twice a perfect square. This is true for all the options, except 100 since \(100 = 2 \times 50\) but 50 is not a perfect square. [\(50 = 2 \times 5^2\); \(200 = 2 \times 10^2\); \(450 = 2 \times 15^2\); and \(800 = 2 \times 20^2\).]

The sum of the row totals is the sum of all the nine numbers in the 3 × 3 square. Likewise, the sum of the column totals is the sum of these nine numbers. Therefore \(24 + 26 + 40 = 27 + 20 + x\), where \(x\) is the sum of the numbers in the third column. Therefore \(90 = 47 + x\) and hence \(x = 43\). Therefore the sum of the numbers in the third column is 43.

Let \(s\) be the number of squares, and \(t\) the number of triangles that Mary started with. The number of vertices was \(4s + 3t\). Also, there were 10 pieces of paper, so
\[s + t = 10. \quad [1]\]
When she cuts a square diagonally to create two triangles, she increases the number of vertices by 2 (from 4 to 6). Hence before she cut the three squares, she had \(42 - 3 \times 2 = 36\) vertices. Therefore
\[4s + 3t = 36. \quad [2]\]
Subtracting three times equation [1] from equation [2] gives \(s = 6\). Hence \(t = 4\).

The four integers cannot all be non-positive since the total is positive. Also, they cannot all be non-negative since the smallest possible total would then be 0 + 1 + 2 + 3 = 6. So, they must have at least one negative integer, and at least one positive integer, and hence will include -1, 0 and 1. Since these have a total of 0, the fourth number must be 2. Thus the least integer is -1.

The numbers on the left use the digits M and K in the Units column, and the digits J and L in the Tens column. The numbers on the right use the digits M and K twice in the Units columns, and the digits J and L twice in the Tens column, so the total is exactly twice that of the numbers on the left. Twice 79 is 158, answer B.

Let \(n\) be the total number of passengers in the middle two carriages. These are the ninth and tenth carriages and so, since there are 199 passengers in any block of five adjacent carriages, the total number of passengers in the sixth, seventh and eighth carriages is \(199 - n\). There are 199 passengers in total in carriages 1 to 5 inclusive, in carriages 9 to 13 inclusive and in carriages 14 to 18 inclusive. Since there are 700 passengers in total on the train, we have \(199 + 199 - n + 199 + 199 = 700\). Therefore \(796 - n = 700\) and hence \(n = 96\). Therefore there are 96 passengers in total in the middle two carriages of the train.

Let the largest number of teams that can enter the tournament be \(n\). Hence there would be \(3n\) players in total. Each player in a team will play a game against every player from all other teams and hence the total number of games played is \(\tfrac{3n \times (3n - 3)}{2} = \tfrac{9n(n - 1)}{2}\). Since no more than 250 games can be played, we have \(\tfrac{9n(n - 1)}{2} \leqslant 250\). Therefore \(n(n - 1) \leqslant \tfrac{500}{9} = 55\tfrac{5}{9}\). Since \(8 \times 7 = 56 > 55\tfrac{5}{9}\), we have \(n < 8\). Also, since \(7 \times 6 = 42 < 55\tfrac{5}{9}\), \(n = 7\) is a possible solution. Therefore the largest number of teams that can enter the tournament is 7.

The triangle is isosceles when one of the following three equations is true:

When equation (1) is true, we have \(n^2 - n - 12 = 0\), so that \((n - 4)(n + 3) = 0\).
Hence either \(n = 4\) or \(n = -3\). However, when \(n = -3\) then \(3n + 3 < 0\), so that no triangle can be formed. There is, though, an isosceles triangle when \(n = 4\), as the sides of the triangle are then 20, 20 and 15.
When equation (2) is true,we have \(n^2 - 2n - 3 = 0\), so that \((n - 3)(n + 1) = 0\).
Hence either \(n = 3\) or \(n = -1\). However, when \(n = -1\) then \(3n + 3 = 0\), so that no triangle can be formed. There is, though, an isosceles triangle when \(n = 3\), as the sides of the triangle are then 12, 18 and 12.
When equation (3) is true, we have \(n = 9\).
But then \(n^2 + n = 90\) and \(3n + 3 = 30 = 2n + 12\), so that no triangle can be formed at all in this case, because 90 > 30 + 30.
Hence the required sum is 3 + 4 = 7.

Let the digits of \(a\) be ‘\(pqp\)’, so \(a = 101p + 10q\).Also \(c = 2b + 1 = 2(2a + 1) + 1 = 4a + 3\) which is \(404p + 40q + 3\). This is less than 1000, so \(p = 1\) or 2.

If \(p = 2\), then \(c = 808 + 40q + 3\). This ends with digit 1 (8 + 3 = 11) but \(c\) is a 3-digit number greater than 808 so can’t begin with 1.

If \(p = 1\), then \(c = 404 + 40q + 3\). This ends with digit 4 + 3 = 7, so must also begin with 7, hence \(700 \leq 404 + 40q + 3 < 800\) and thus \(293 \leq 40q < 393\). Therefore \(q = 8\) or 9.

When \(q = 8\), this gives \(a = 181\), \(b = 363\), \(c = 727\). When \(q = 9\), this gives \(a = 191\), \(b = 383\), \(c = 767\).

Hence there are two possibilities for \(a\), 181 and 191.

From the equation \(\dfrac{x}{y} = \dfrac{2}{3}\) we deduce that \(3x = 2y\). We are given that \(x + y = \dfrac{2}{3}\), so \(3x + 3y = 2\).

Substituting for \(3x\) gives \(2y + 3y = 2\), so \(y = \dfrac{2}{5}\).

Therefore \(x - y = x + y - 2y = \dfrac{2}{3} - \dfrac{4}{5} = \dfrac{10}{15} - \dfrac{12}{15} = -\dfrac{2}{15}\).

We are given that \(xy = x + y\). So \(xy - x = y\). Therefore \(x(y - 1) = y\) and hence \(x = \dfrac{y}{y - 1}\).

(Note that division by \(y - 1\) is allowed because if \(y = 1\), then we have \(x = x + 1\) which is impossible.)

The value of ‘\(ab\)’ is \(10a + b\) so \(3a + 4b = 10a + b\) which gives \(3b = 7a\). Hence \(a\) is a multiple of 3 and \(b\) is a multiple of 7. Also \(a\) is non-zero (as it is the leading digit), and hence \(b\) also cannot be zero. The only single-digit solution of \(3b = 7a\) is \(a = 3\) and \(b = 7\), so \(a + b = 10\).

Let the number of apples Andrew had be \(6n\). When Boris divided the same number of apples into five piles, each pile contained two more apples than each of Andrew’s piles. Therefore \(6n = 5(n + 2)\) and hence \(6n = 5n + 10\). This has solution \(n = 10\). Therefore the number of apples Andrew had was 6 × 10 = 60.

Clearly, there are no values of \(x\) for which \(x = x + 6\). So the possible values of \(x\) will be solutions of the equation \(x = x^2\) or the equation \(x + 6 = x^2\).
Consider the first equation: \(x = x^2\). This simplifies to \(x(1 - x) = 0\). So \(x = 0\) or \(x = 1\).
The equation \(x + 6 = x^2\) simplifies to \(x^2 - x - 6 = 0\). So \((x - 3)(x + 2) = 0\).
Therefore \(x = 3\) or \(x = -2\). So there are four different values of \(x\).

Let the points \(O\), \(P\) etc, be labelled as \(A_1\), \(A_2\), etc, in the order in which they are drawn. Without loss of generality, we can draw the line horizontally with \(A_2\) on the right of \(A_1\). Since \(A_1\) is the midpoint of \(A_2\) and \(A_3\), we then put \(A_3\) to the left of \(A_1\) at a distance of 1 from \(A_1\). Since \(A_2\) is the midpoint of \(A_3\) and \(A_4\), we put \(A_4\) on the right of \(A_2\) at a distance of 2 from \(A_2\). Proceeding in this fashion, the odd points end up further to the left of \(A_1\), and the even points further to the right, with the distances doubling at each stage. The diagram shows the points and the distances between neighbouring points, though not to scale! The distance \(A_1A_{12}\) is \(1 + 2 + 8 + 32 + 128 + 512 = 683\).

Let the width of the pool be \(x\) m. Therefore the total distance Ruth runs is \(5(2 \times 50 + 2x)\) m = \((500 + 10x)\) m. The total distance Sarah swims is 6 × 50 m = 300 m. Since Ruth runs three times as fast as Sarah swims, \(500 + 10x = 3 \times 300\). Therefore \(10x = 400\) and hence \(x = 40\).

Let the four numbers be \(a\), \(b\), \(c\), \(d\) with \(a \leqslant b \leqslant c \leqslant d\). Abdul chooses the numbers in turn, giving
\[\begin{array}{lll} a + \tfrac{1}{3}(b + c + d) = 17 \quad (1) & \qquad & b + \tfrac{1}{3}(a + c + d) = 21 \quad (2) \\ c + \tfrac{1}{3}(a + b + d) = 23 \quad (3) & \qquad & d + \tfrac{1}{3}(a + b + c) = 29 \quad (4) \end{array}\]
Adding these equations gives:
\[\begin{array}{lr} & a + b + c + d + \tfrac{1}{3}(3a + 3b + 3c + 3d) = 90 \\ \text{that is} & 2a + 2b + 2c + 2d = 90. \end{array}\]
Hence \(a + b + c + d = 45\). Subtracting a third of this from equation (4), we get
\[d + \tfrac{1}{3}(a + b + c) - \tfrac{1}{3}(a + b + c + d) = 29 - 15.\]
Which gives \(\tfrac{2}{3}d = 14\), so \(d = 21\). We can continue in a similar way to find \(a = 3\), \(b = 9\) and \(c = 12\).

Let the amount Chloe spent be £\(x\). Therefore Amy spent £\(1.6x\) and Becky spent £\(0.15x\). The total amount spent is £55 and hence \(x + 1.6x + 0.15x = 55\). Therefore \(2.75x = 55\), which has solution \(x = 20\). Hence Amy spent £(1.6 × 20) = £32.

As the three rectangles have equal areas: \(xy = (x + 4)(y - 3) = (x + 8)(y - 4)\).
So \(xy = xy - 3x + 4y - 12 = xy - 4x + 8y - 32\).
Therefore \(-3x + 4y - 12 = 0\) and \(-4x + 8y - 32 = 0\).
Rearranging: \(3x - 4y = -12\) … (1) and \(4x - 8y = -32\) … (2).
2 × (1) − (2) gives: \(6x - 4x = -24 - (-32) = 8\). So \(x = 4\).
Substituting for \(x\) in (1): \(12 - 4y = -12\). So, \(4y = 24\), that is \(y = 6\).
So \(x + y = 4 + 6 = 10\).

The number of votes left to be cast is 130 – (24 + 29 + 37) = 40. Let the number of these votes Alan receives be \(x\). Since Katie is the closest challenger to Alan, for Alan to be certain of having the most votes, \(37 + x > 29 + 40 - x\). Therefore \(2x > 32\) and hence \(x > 16\). Therefore Alan needs at least 17 more votes to be certain to finish with the most votes.

The value of ‘abc’ equals \(100a + 10b + c\). We need to find integers of the form ‘abc’ such that ‘cba’ = ‘abc’ + 99.
So \(100c + 10b + a = 100a + 10b + c + 99\), that is \(99c = 99a + 99\). This condition simplifies to \(c = a + 1\). So there are 10 such integers of the form ‘1b2’, 10 of the form ‘2b3’, 10 of the form ‘3b4’, …, and 10 of the form ‘8b9’. Therefore the required number is 8 × 10 = 80.

Let the distance between the first two points be \(x\) cm. Consider the other nine points. The sum of their distances from the first point is \((2018 - x)\) cm and the sum of their distances from the second point is \((2000 - x)\) cm. Also, for each of these nine points, its distance from the second point is \(x\) cm less than its distance from the first point. Therefore, when we total the distances over the nine points, we obtain \(9x = (2018 - x) - (2000 - x)\) and hence \(9x = 18\). Therefore \(x = 2\) and so the distance between the first and second points is 2 cm.

Let the three digits be \(a\), \(b\), \(c\). Then the value of the 3-digit number is \(100a + 10b + c\) and the 2-digit number (made by deleting the middle digit) is \(10a + c\).
Hence \(100a + 10b + c = 9(10a + c)\), which gives \(10(a + b) = 8c\), or \(5(a + b) = 4c\). Hence \(c\) is divisible by 5. But \(c\) cannot be zero (because this would mean \(a + b = 0\), and the digits cannot all be zero). So \(c = 5\). Then \(a + b = 4\), giving four possibilities: \(a = 1\), \(b = 3\) or \(a = 2\), \(b = 2\) or \(a = 3\), \(b = 1\) or \(a = 4\), \(b = 0\). These all give valid solutions (135 = 9 × 15, 225 = 9 × 25, 315 = 9 × 35, 405 = 9 × 45).

Let \(D\) be the amount that Peter's dad gave, and \(B\) the amount his elder brother gave.
Then \(D = \tfrac{1}{2}(B + 10) \ldots\) (1) and \(B = \tfrac{1}{3}(D + 10) \ldots\) (2).
Substitute (1) into (2) to get \(B = \tfrac{1}{3}\left(\tfrac{1}{2}(B + 10) + 10\right) = \tfrac{1}{6}B + 5\).
Hence \(\tfrac{5}{6}B = 5\), so \(B = 6\) and \(D = \tfrac{1}{2}(6 + 10) = 8\). This gives a total of 6 + 8 + 10 = 24. So the book cost 24 euros.

The difference of two squares \(a^2 - b^2\) factorises to \((a + b)(a - b)\). Therefore for it to be possible to express an integer as the difference of two squares, it must be possible to factorise the integer in the form \((a + b)(a - b)\), where \(a\) and \(b\) are both integers.
Note that \(5 = (3 + 2)(3 - 2)\); \(7 = (4 + 3)(4 - 3)\); \(8 = (3 + 1)(3 - 1)\); \(9 = (5 + 4)(5 - 4)\).
So 5, 7, 8, 9 can all be written as the difference of two squares.
(It is left to the reader to show that 10 cannot be the difference of two squares as it cannot be written in the form \((a + b)(a - b)\) where a and b are both integers.)

All distances in this solution are in metres. Let \(x\) be the distance of the bus stop from the second building. Then the total distance walked by the 150 students in this building is \(150x\). The distance travelled by each of the students in the first building is \(250 - x\), so the total of their distances is \(100(250 - x)\). Adding these to get the total distance of all students gives \(150x + 100(250 - x) = 50x + 25000\). This is minimised by setting \(x = 0\), so the bus stop should be in front of the second building!

The largest possible value of \(a - b\) comes from subtracting the smallest possible value of \(b\) from the largest possible value of \(a\). Hence the largest possible value is 10 − (−5) = 10 + 5 = 15.

Let \(n\) be the middle number. Then the five numbers are \(n - 2\), \(n - 1\), \(n\), \(n + 1\), \(n + 2\) and have sum \(5n = 10^{2018}\).
Therefore \(n = \dfrac{10^{2018}}{5} = \dfrac{10 \times 10^{2017}}{5} = 2 \times 10^{2017}\).

Let \(t\) be the height of the table, \(b\) the height of the bottle and \(c\) the height of the can (all measured in cm). The first diagram shows \(t + b = c + 150\); the second diagram shows \(t + c = b + 110\). Adding these equations gives \(2t + b + c = 260 + b + c\) so \(2t = 260\). So the table has height 130 cm.

The product may be written \(\sqrt{\frac{3}{2}} \times \sqrt{\frac{4}{3}} \times \sqrt{\frac{5}{4}} \times \sqrt{\frac{6}{5}} \times \ldots = \frac{\sqrt{3}}{\sqrt{2}} \times \frac{\sqrt{4}}{\sqrt{3}} \times \frac{\sqrt{5}}{\sqrt{4}} \times \frac{\sqrt{6}}{\sqrt{5}} \times \ldots\).
Notice that the numerator of each fraction is cancelled out by the denominator of the following fraction and the only terms which are not cancelled are the denominator of the first fraction and the numerator of the last fraction. So the \(n\)th term of the sequence is \(\frac{\sqrt{n + 2}}{\sqrt{2}} = \sqrt{\frac{n + 2}{2}}\). As \(n \geqslant 1\) the product is not equal to 1.
The product increases with \(n\). So the next possible integer value to consider is 2 and this does occur when \(n = 6\) as \(\sqrt{\frac{6 + 2}{2}} = \sqrt{4} = 2\).
So the smallest number of terms required for the product to be an integer is 6.

Let the number of Brachycephalus frogs and common frogs in the bucket be \(b\) and \(c\) respectively. Note that each Brachycephalus frog has 6 toes and 4 fingers, while a common frog has 10 toes and 8 fingers.
Therefore, \(6b + 10c = 122\) (1); \(4b + 8c = 92\) (2). Subtracting (2) from (1) gives \(2b + 2c = 30\), so \(b + c = 15\).

Let the capacity of the tank be \(x\) litres. Then \(30 = \frac{5x}{6} - \frac{4x}{5} = \frac{25x - 24x}{30} = \frac{x}{30}\). So \(x = 30 \times 30 = 900\).

Each number \(a\), \(b\), \(c\), \(d\), \(e\), \(f\), \(g\) differs from its neighbour by one, so they alternate odd and even. To obtain an odd total, we must have an odd number of odd numbers in the list. Hence \(b\), \(d\), \(f\) are odd and cannot be equal to 286.
If \(c = 286\), then the biggest total possible is 288 + 287 + 286 + 287 + 288 + 289 + 290 = 2015 which is too small. By reversing this list, we can also rule out \(e = 286\).
We can obtain the total 2017 if we start with \(a = 286\) since 286 + 287 + 288 + 289 + 290 + 289 + 288 = 2017. By reversing this, we could also end with \(g = 286\).

Label the numbers Ella writes down as shown in the diagram.

Since the sum of the numbers in any two adjacent cells is the same, \(2 + a = a + b\) and hence \(b = 2\). Therefore \(b + 3 = 2 + 3 = 5\). Hence the sum of the numbers in any two adjacent cells is 5. It is now straightforward to see that \(a = c = f = 3\) and that \(b = d = e = g = 2\). Therefore the sum of all the numbers in the grid is 5 × 2 + 4 × 3 = 22.

Let the two positive integers be \(m\) and \(n\). Then \(mn = 2(m + n) = 6(m - n)\).
So \(2m + 2n = 6m - 6n\), that is \(8n = 4m\). Therefore \(m = 2n\). Substituting for \(m\) gives: \((2n)n = 2(2n + n)\). So \(2n^2 = 6n\), that is \(2n(n - 3) = 0\).
Therefore \(n = 0\) or 3. However, \(n\) is positive so the only solution is \(n = 3\).
Therefore \(m = 2 \times 3 = 6\) and \(m + n = 6 + 3 = 9\).

Note that \(64 = 2^6\) and \(512 = 2^9\). Therefore \((2^6)^x = (2^9)^5\). So \(2^{6x} = 2^{45}\).
Hence \(6x = 45\), that is \(x = 7.5\).

Let \(k\) cm be the difference between the heights.
Then the heights in cm are: Tobias 184, Victor \(184 + k\), Peter \(184 - k\), and Oscar \(184 - 2k\). The mean is 178 so \(\tfrac{1}{4}\left(184 + 184 + k + 184 - k + 184 - 2k\right) = 178\). Hence \(4 \times 184 - 2k = 4 \times 178\), giving \(2k = 4 \times 184 - 4 \times 178 = 4 \times 6 = 24\). Hence \(k = 12\).
Therefore Oscar's height in cm is \(184 - 2 \times 12 = 160\) cm.

Since the sum of the numbers in the first three boxes is to be 22, the sum of the numbers in the last three boxes is to be 25 and the sum of the numbers in all five boxes is to be 35, Ellie will write (22 + 25 − 35) = 12 in the middle box. Therefore she will write (22 − 3 − 12) = 7 in the second box and (25 − 12 − 4) = 9 in the fourth box. Hence the product of the numbers in the shaded boxes is 7 × 9 = 63.

Let the ages of Alice, Bob, Clare, Dan and Eve be \(a\), \(b\), \(c\), \(d\), \(e\) respectively.
So, for example, \(b + c = 40\) and \(d + e = 44\). Adding these gives \(b + c + d + e = 84\). We are also told that \(a + b + c + d + e = 105\). The difference between these equations shows that \(a = 105 - 84 = 21\). Hence \(b = 39 - 21 = 18\) and, similarly, \(c = 40 - 18 = 22\), \(d = 38 - 22 = 16\) and \(e = 44 - 16 = 28\). So Dan is the youngest.

The three squares will be approximately a third of 770, so roughly 250. Adding up \(15^2 = 225\), \(16^2 = 256\) and \(17^2 = 289\) gives the total 770.
Alternatively, using algebra we could let \(n\) be the middle integer and then add the squares \((n - 1)^2 + n^2 + (n + 1)^2 = n^2 - 2n + 1 + n^2 + n^2 + 2n + 1 = 3n^2 + 2 = 770\).
This gives \(n^2 = 256\) and hence \(n = 16\), so the largest integer is 17.

First note that as ‘ab’ and ‘ba’ are two digit numbers, neither \(a\) nor \(b\) is equal to zero. Now ‘ab’ − ‘ba’ = \((10a + b) - (10b + a) = 9a - 9b = 9(a - b)\). So for the difference to be as large as possible, \(a\) must be as large as possible, that is 9, and \(b\) must be as small as possible, that is 1.
So the required difference is 91 − 19 = 72.

Let the numbers of passengers in the five carriages be \(p\), \(q\), \(r\), \(s\) and \(t\) respectively with \(p\), \(q\), \(r\), \(s\) and \(t\) all at least 1. Consider the neighbours of the passengers in the first and second carriages. Since each passenger has 5 or 10 neighbours, we have \(p - 1 + q = 5\) or 10 and \(p + q - 1 + r = 5\) or 10. Therefore \(p + q = 6\) or 11 and \(p + q + r = 6\) or 11. However, we know that \(r \geqslant 1\) and hence \(p + q \leqslant 10\) and therefore \(p + q = 6\) and \(r = 5\). Similarly, considering the neighbours of the passengers in the fourth and fifth carriages, we obtain \(s + t = 6\) and (again) \(r = 5\). Therefore, the total number of passengers in the train is 6 + 5 + 6 = 17. (Note that while the total number of passengers in the train is uniquely determined, the arrangement of these passengers in all but the centre carriage is not unique.)

Let the integers written on the small cubes in the bottom layer be arranged as shown.

Hence, since the integers written on the cubes in the second and third layers are the sum of the integers on the four cubes underneath, the following is written on the cubes in the second layer.

Therefore the integer written on the top cube is
\[\begin{aligned}&(a + b + d + e) + (b + c + e + f) + (d + e + g + h) + (e + f + h + i)\\ &\quad = (a + b + c + d + e + f + g + h + i) + (b + d + f + h) + 3e.\end{aligned}\]
Since the sum of the integers on the bottom layer is 50, the integer written on the top cube is equal to \(50 + (b + d + f + h) + 3e\). To maximise this, we first require \(e\) to be as large as possible which will be obtained when the other eight integers are as small as possible. Therefore \(e = 50 - (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8) = 14\). Secondly, \((b + d + f + h)\) should now be made as large as possible and hence \(b\), \(d\), \(f\) and \(h\) are 5, 6, 7 and 8 in any order. Therefore \((b + d + f + h) = 5 + 6 + 7 + 8 = 26\). Hence the greatest possible integer she can write on the top cube is 50 + 26 + 3 × 14 = 118.

Let the radius of the circle with centre \(A\) be \(x\) cm. Therefore, since the circles drawn on each side of the pentagon touch, the radius of the circle with centre \(B\) is \((16 - x)\) cm. Similarly, the radius of the circle with centre \(C\) is \((14 - (16 - x))\) cm = \((x - 2)\) cm, the radius of the circle with centre \(D\) is \((17 - (x - 2))\) cm = \((19 - x)\) cm and the radius of the circle with centre \(E\) is \((13 - (19 - x))\) cm = \((x - 6)\) cm. However, the radius of the circle with centre \(E\) is also equal to \((14 - x)\) cm since the circle with centre \(A\) has radius \(x\) cm. Therefore \(14 - x = x - 6\), which has solution \(x = 10\). Hence the radii of the circles centres \(A\), \(B\), \(C\), \(D\) and \(E\) are 10 cm, 6 cm, 8 cm, 9 cm and 4 cm respectively. Therefore point \(A\) is the centre of the largest circle Sephideh draws.

Let \(D\) be the distance from \(X\) to \(Y\), and let \(u\) be the speed of the boat and \(v\) be the speed of the current. When travelling downstream the overall speed of the boat in the current is \(u + v\), and travelling upstream against the current it is \(u - v\). Then using time \(= \dfrac{\text{distance}}{\text{speed}}\), we get \(\dfrac{D}{u + v} = 4\) for the journey downstream and \(\dfrac{D}{u - v} = 6\) for the journey upstream. Inverting the equations gives \(\dfrac{u + v}{D} = \dfrac{1}{4}\) or \(4u + 4v = D \ldots\) (1) and \(\dfrac{u - v}{D} = \dfrac{1}{6}\) or \(6u - 6v = D \ldots\) (2). Multiplying (1) by 3 and (2) by 2, we get \(12u + 12v = 3D \ldots\) (3) and \(12u - 12v = 2D \ldots\) (4).
Subtracting, (3) − (4) gives \(24v = D\) which rearranges to \(\dfrac{D}{v} = 24\) (hours), which is the time taken for the log to float downstream at the speed of the current alone.

Let \(x\) be the number of waffles Little Red Riding Hood delivered to each of the grannies. She delivered \(x\) waffles to the third granny and so, since the Big Bad Wolf eats half of the waffles in her basket just before she enters each granny's house, she must have arrived with \(2x\) waffles. Therefore she had \(3x\) waffles before giving the second granny her waffles and hence had \(2 \times 3x = 6x\) waffles when she arrived. Therefore she had \(6x + x = 7x\) waffles before giving the first granny her waffles and hence had \(2 \times 7x = 14x\) waffles in her basket when she arrived. Hence, since we do not know the value of \(x\), the only numbers that definitely divide the number of waffles she started with are the numbers that divide 14, namely 1, 2, 7 and 14. Therefore, of the options given, only 7 definitely divides the number of waffles she started with.

Let \(G\) be the number of grey kangaroos, and \(P\) the number of pink kangaroos, so \(G + P = 2016\). For each grey kangaroo we calculate the fraction as \(\dfrac{P}{G}\); and there are \(G\) of these, so the total of the fractions for grey kangaroos is \(G \times \dfrac{P}{G} = P\).
Similarly the total of the fractions calculated for the pink kangaroos is \(G\). Thus the total of all the fractions is \(P + G = 2016\).

Let the number of boys in the class be \(x\) and let the number of girls be \(y\). Since one third of the boys sit with a girl and one half of the girls sit with a boy, we have \(\tfrac{1}{3}x = \tfrac{1}{2}y\) and hence \(\tfrac{2}{3}x = y\). The total number of students in the class is 20. Therefore \(x + \tfrac{2}{3}x = 20\). Hence \(\tfrac{5}{3}x = 20\), which has solution \(x = 12\). Therefore there are 12 boys in the class.

Let there be \(g\) girls and \(b\) boys in Tegwen's family. Then, as she has the same number of brothers as she does sisters, \(b = g - 1\). Also, each of her brothers has 50% more sisters than brothers. Therefore \(g = \frac{3}{2}(b - 1)\). So \(b + 1 = \frac{3}{2}(b - 1)\) and hence \(2b + 2 = 3b - 3\). Rearranging this equation gives \(b = 5\). So \(g = 5 + 1 = 6\). Therefore there are 5 + 6 = 11 children in Tegwen's family.

In the seven lines each of the integers from 1 to 7 is used twice and each of the integers from 8 to 14 is used once. So the sum of the numbers in the seven lines is (1 + 2 + … + 14) + (1 + 2 + … + 7) = 105 + 28 = 133. Therefore the total of the numbers in each line is 133 ÷ 7 = 19.

It is left as an exercise for the reader to show that it is possible to complete the diagram so that the total of the three numbers in each line is indeed 19.

Since the two kangaroos jump at the same time and in the same direction, Ing will catch Bo when they have jumped the same distance. After \(n\) jumps, Bo has jumped \(6n\) metres whereas Ing has jumped \((1 + 2 + 3 + \ldots + n)\) metres. The sum of the whole numbers from 1 to \(n\) is given by the formula \(\tfrac{1}{2}n(n + 1)\) (it is left as an exercise for the reader to prove this) and hence \(\tfrac{1}{2}n(n + 1) = 6n\). Hence \(n + 1 = 12\), which has solution \(n = 11\). Therefore Ing will catch Bo after 11 jumps.

Assume that \(a > b\). Then \(a + b = 7\) and \(a - b = 2\). Adding these two equations together gives \(2a = 9\). So \(a = \frac{9}{2}\) and hence \(b = 7 - \frac{9}{2} = \frac{14 - 9}{2} = \frac{5}{2}\).
Therefore \(a \times b = \frac{9}{2} \times \frac{5}{2} = \frac{45}{4} = 11\frac{1}{4}\).

Let \(x\) be the age of each of Tim, Tom and Tam. Therefore the age of both Jon and Jim is \(x - 3\). Therefore the sum of all their ages is \(3x + 2(x - 3) = 5x - 6\) and \(5x - 6\) can also be written as \(5(x - 1) - 1\). Hence the sum of their ages is always one less than a multiple of 5. Of the options given, the only number for which this is true is 89 (when Tim, Tom and Tam are 19 and Jon and Jim are 16).

Note that \(d = c^2 + 7\), so \(d > c\); \(d = b^2 + 3\), so \(d > b\); and \(d = a + 9\), so \(d > a\). Hence \(d\) is the largest.

Let \(g\) and \(s\) be the number of goats and sheep respectively. Then \(s = 2g\) and \(12 = s - g = 2g - g = g\). Hence the number of animals is \(s + g = 3g = 36\).

By train, the distance in miles of the second sign from Edinburgh is \(200 - 3\frac{1}{2}\). This sign is halfway between London and Edinburgh, so the distance in miles between the two cities is \(2(200 - 3\frac{1}{2}) = 400 - 7 = 393\).

It is certainly possible to underline two numbers. In the list below both 1 and −1 are equal to the product of the other nine numbers \(1, -1, -2, \tfrac{1}{2}, 3, \tfrac{1}{3}, 4, \tfrac{1}{4}, 5, \tfrac{1}{5}\).
However, it is not possible to underline three numbers. For supposing it was possible to underline the numbers \(a\), \(b\), \(c\). Let \(N\) be the product of the other seven numbers. Then we have
\[a = b \times c \times N \ldots (1) \qquad b = a \times c \times N \ldots (2) \qquad c = a \times b \times N \ldots (3)\]
Substituting (1) into (2) gives \(b = (b \times c \times N) \times c \times N = b \times c^2 \times N^2\) so \(c^2 \times N^2 = 1\). Thus \(c \times N = 1\) or \(c \times N = -1\). But if \(c \times N = 1\), then (1) becomes \(a = b\), contradicting that \(a\), \(b\) are distinct. Hence \(c \times N = -1\), and (1) becomes \(a = -b\). Substituting this into (3) gives \(c = (-b) \times b \times N\). Multiplying both sides by \(c\) gives \(c^2 = -b^2 \times c \times N = -b^2 \times (-1) = b^2\). Hence either \(c = b\) (contradicting that they are distinct) or \(c = -b = a\) (contradicting that \(a\), \(c\) are distinct).
Hence there is no way that \(a\), \(b\), \(c\) can be distinct numbers and also underlined. (If the numbers are allowed to be the same, it is possible to underline them all by choosing the numbers to be all equal to one).

The sum of 1, 2, … \(n\) is \(\tfrac{1}{2}n(n + 1)\). So the smallest the sum could be if one number were deleted would be \(\tfrac{1}{2}n(n + 1) - n = \tfrac{1}{2}(n^2 + n - 2n) = \tfrac{1}{2}n(n - 1)\). Thus the mean of any \(n - 1\) of these numbers is at least \(\tfrac{1}{2}n\). So for the mean to be 4.75, we must have \(n \leqslant 9\).
Similarly, the largest the sum could be with one number deleted would be \(\tfrac{1}{2}n(n + 1) - 1 = \tfrac{1}{2}(n^2 + n - 2) = \tfrac{1}{2}(n - 1)(n + 2)\). Thus the mean of any \(n - 1\) of the numbers is at most \(\tfrac{1}{2}(n + 2)\). So we must have \(\tfrac{1}{2}(n + 2) \geqslant 4.75\) and so \(n \geqslant 8\).
Since \(4.75 = 19/4\) and the mean is obtained by dividing by \(n - 1\), we require \(n - 1\) to be a multiple of 4. So \(n = 9\). The sum of 1, … , 9 is 45 and we need the 8 numbers used to have a mean of 4.75. So their total is 38 and hence 7 is the number to be eliminated.

Let the five points be \(P\), \(Q\), \(R\), \(S\) and \(T\) with individual distances between them of \(w\), \(x\), \(y\) and \(z\) as shown.

The maximum distance between any two points is 22 so \(PT = 22\). The next largest distance is 20 and, since no distance is 1, this is either \(PS\) or \(QT\). Assume \(QT = 20\) so that \(PQ = w = 2\). The next largest distance is 17 and, since no distance is 3 this cannot be \(QS\) or \(RT\) and so is \(PS\). Hence \(ST = z = 5\) and \(QS = 15\). The remaining distances are 6, 8, 9 and \(k\) which represent the lengths of \(PR\), \(QR\), \(RS\) and \(RT\) in some order with \(QR + RS = 15\). Since \(k > 9\), the only possible pair of distances adding to 15 is 6 and 9 and so \(QR = x = 6\) and \(RS = y = 9\) (since \(QR = 9\) and \(RS = 6\) would mean two distances are 11) leaving \(PR = 8\) and \(RT = k = 14\). Hence the value of \(k\) is 14 with the distances between the points taking the values shown below.

(Note: The same distances would result but in the reverse order if we assumed \(PS = 20\).)

Let the five integers be \(a\), \(b\), \(c\), \(d\) and \(e\) with \(a \leqslant b \leqslant c \leqslant d \leqslant e\). The smallest total is 57, which is an odd number so \(b \neq a\). Similarly, the largest total is 83, which is also an odd number so \(d \neq e\). Hence we now have \(a < b \leqslant c \leqslant d < e\) and \(a + b = 57\) and \(d + e = 83\). Only one possible total remains and so \(b = c = d\) with \(c + d = 70\). This gives \(c = d\ (= b) = 35\) and therefore \(e\), the largest integer, is 83 − 35 = 48 (whilst \(a = 22\)).

Let the weights in kg of the box, 1 plate and 1 cup be \(b\), \(p\) and \(c\) respectively.
Then: \(b + 20p + 30c = 4.8\) (i); \(b + 40p + 50c = 8.4\) (ii).
Subtracting (i) from (ii): \(20p + 20c = 3.6\) (iii). So \(10p + 10c = 1.8\) (iv).
Subtracting (iv) from (i): \(b + 10p + 20c = 3\). So the required weight is 3 kg.

Suppose the quadratic has roots \(p\) and \(q\). So it factorises as \((x - p)(x - q)\). This expands to give \(x^2 - (p + q)x + pq\). Comparing this with \(x^2 - 85x + c\) shows that \(p + q = 85\) and \(pq = c\). It follows that one of \(p\), \(q\) is even and one is odd. The only even prime is 2 so the primes \(p\), \(q\) are 2 and 83. Therefore \(c = 2 \times 83 = 166\), and therefore has digit sum 13.

Let the length of the rectangle be \(x\) cm and let the width be \(y\) cm. The information in the question tells us that \(2x + y = 44\) and \(x + 2y = 40\). Add these two equations to obtain \(3x + 3y = 84\). Hence \(x + y = 28\) and so the perimeter, which is equal to \(2(x + y)\) cm, is 56 cm.

Suppose the first three terms of the sequence are \(a\), \(b\), \(c\). Then \(c = \tfrac{1}{2}(a + b)\) and so \(a + b = 2c\). The mean of the first three terms is then \(\tfrac{1}{3}(a + b + c) = \tfrac{1}{3}(2c + c) = c\), so the fourth term is \(c\). Similarly, the following terms are all equal to \(c\). Since one of these terms is 26 and \(a = 8\) then \(b = 2c - a = 52 - 8 = 44\).

From the first picture, we can see:
\[\begin{array}{lrcll} \text{From the right scale:} & Z & > & Y & \quad (1) \\ \text{From the left scale:} & X & > & W & \quad (2) \\ \text{From the large scale:} & Y + Z & > & W + X & \quad (3) \end{array}\]
It follows from (1), (2) and (3) that \(Z + Z > Y + Z > W + X > W + W\), so \(2Z > 2W\), and hence \(Z > W\) … (4).
We can show that most swaps give a contradiction of these inequalities:
Firstly, suppose that weight \(Z\) doesn’t move. Then there are three possible swaps:
\(X\) and \(Y\) swap: then in the second picture we must have \(Z < X\) and \(Y < W\), which add to give \(Y + Z < W + X\), contradicting (3).
\(Y\) and \(W\) swap: then we would have \(Z < W\), contradicting (4).
\(X\) and \(W\) swap: then we would have \(X < W\), contradicting (2).
Hence \(Z\) must swap, which can happen in three ways:
\(Z\) and \(Y\) swap: then we would have \(Z < Y\), contradicting (1).
\(Z\) and \(X\) swap: then we would have \(Z < W\), contradicting (4).
\(Z\) and \(W\) swap: This is the only possibility left, and it can work when the weights \(W\), \(X\), \(Y\), \(Z\) are 1, 2, 3, 4. The swap would give 4, 2, 3, 1 which matches the picture on the right. However, this does depend on the values of \(W\), \(X\), \(Y\), \(Z\); it would not work for 1, 5, 3, 4.

The maximum number of leaves the bush could have is 10 × 5 = 50. Each branch that has two leaves and a flower instead of five leaves reduces the number of leaves the bush has by three. Therefore the total number of leaves the bush has is of the form \(50 - 3n\) where \(n\) is the number of branches with two leaves and a flower. It is straightforward to check that none of options A to D has that form and so the answer is E.

Firstly suppose that any two knights X and Y win \(x\) and \(y\) bouts respectively and that \(x\) is at least as large as \(y\). The difference between their total scores would be the same as if X had won \(x - y\) bouts and Y had won none, since each of the separate totals would have been reduced by the same amount, namely 20\(y\). A similar procedure applies to losses. For example, if X won 3 and lost 6, while Y won 8 and lost 2, the difference between their total scores is the same as if X won 0 and lost 4, while Y won 5 and lost 0. In each case the difference is 32.
This argument shows that, in the case of the Black Knight, B, and the Red Knight, R, the smallest number of bouts will be achieved when one of B, R wins all his bouts and the other loses all his bouts. Also B has to score one more point than R. The possible scores for the knight who wins all his bouts are 20, 40, 60, 80, 100, 120, … . while the possible scores for the knight who loses all his bouts are 17, 34, 51, 68, 85, 102, 119, 136, … .
The first two numbers to differ by 1 are 119 and 120. Thus the Black Knight has a total of 120 corresponding to winning all of his 6 bouts and the Red Knight has a total of 119 corresponding to losing all of his 7 bouts.

Note that \(8^m = (2^3)^m = 2^{3m} = (2^m)^3\) and \(27 = 3^3\); so \(2^m = 3\). Therefore \(4^m = 2^m \times 2^m = 9\).

Let Zac's number be \(x\). Then \(\tfrac{1}{2}x + 8 = 2x - 8\). So \(x + 16 = 4x - 16\).
Therefore \(32 = 3x\), that is \(x = 10\tfrac{2}{3}\).

Gill's car uses \(p/100\) litres of petrol for every one kilometre travelled. So for a journey of length \(d\) km, \(pd/100\) litres of petrol are required.

Since \(m\) is even, \(m = 2k\) for some integer \(k\). So \(3m + 4n = 2(3k + 2n)\); \(5mn = 2(5kn)\); \(m^3n^3 = 8k^3n^3\) and \(5m + 6n = 2(5k + 3n)\), which are all even. As \(n\) is odd, \(3n\) is also odd. So \(m + 3n\) is an even integer plus an odd integer and is therefore odd. The square of an odd integer is odd so \((m + 3n)^2\) is odd.

The image of a straight line under a rotation is also a straight line. The centre of rotation, the point (1, 1), lies on the given line and so also lies on the image. The given line has slope 1 and so its image will have slope −1. Hence graph A shows the image.

Let the number of scores of 1 be \(n\). Then the product of the scores is 1^\(n\) × 2 × 3 × 5 = 30. Therefore 1 × \(n\) + 2 + 3 + 5 = 30, i.e. \(n\) = 20. So Jim threw 23 dice.

The key to the solution is to note that the squares of the odd numbers occur on the diagonal leading downwards and to the right from the cell which contains the number 1, and the squares of the even numbers occur on the diagonal which leads upwards and to the left of the cell which contains the number 4.
The squares of the even numbers have the form \((2n)^2\), that is, \(4n^2\).We see that the number \(4n^2 + 1\) occurs to the left of the cell containing \(4n^2\). Below \(4n^2 + 1\) there occur the numbers \(4n^2 + 2, 4n^2 + 3, \ldots, 4n^2 + 2n + 1\), and then in the cells to the right of the cell containing \(4n^2 + 2n + 1\), there occur the numbers \(4n^2 + 2n + 2, 4n^2 + 2n + 3, \ldots, 4n^2 + 4n + 1 = (2n + 1)^2\).
Now \(44^2 = 1936\) and \(45^2 = 2025\). Thus 2011 is in the same row as 2025 and to the left of it, in the sequence 1981,….,2012,…,2025, and below these occur the numbers 2163,…., \(2209 = 47^2\), with 2208 below 2025 as shown below. It follows that 2195 is the number below 2012. In the diagram below the square numbers are shown in bold.

The terms of the sequence are 6, 3, 14, 7, 34, 17, 84, 42, 21, 104, 52, 26, 13, 64, 32, 16, 8, 4, 2, 1, 4, 2, 1, … . As can be seen, there will now be no other terms in the sequence other than 4, 2 and 1. It can also be seen that the only values of \(n\) for which the \(n\) th term = \(n\) are 13 and 16.

The mean of the first three numbers is \(\tfrac{1}{3}(20 + x)\); the mean of the last four numbers is \(\tfrac{1}{4}(33 + x)\). Therefore \(4(20 + x) = 3(33 + x)\), that is \(80 + 4x = 99 + 3x\), so \(x = 99 - 80 = 19\).

\(5^p = 9\) and \(9^q = 12\). Therefore \((5^p)^q = 12\), that is \(5^{pq} = 12\). Similarly, as \(12^r = 16\) then \(5^{pqr} = 16\), as \(16^s = 20\) then \(5^{pqrs} = 20\) and, finally, as \(20^t = 25\) then \(5^{pqrst} = 25\). Therefore \(pqrst = 2\).

Let Max and Molly meet after the latter has travelled \(x\) miles. Then Max has travelled \((6 - x)\) miles. So \(6 - x = 2x\), thus \(x = 2\). Therefore Molly walks a total of 4 miles.

\(3 = \dfrac{2 + 2^2}{2}\); \(10 = \dfrac{4 + 4^2}{2}\); \(15 = \dfrac{5 + 5^2}{2}\); \(21 = \dfrac{6 + 6^2}{2}\). However \(\dfrac{7 + 7^2}{2} = 28\) and \(\dfrac{8 + 8^2}{2} = 36\), so 30 is not exactly halfway between a positive integer and its square.
(Note that every number which is exactly halfway between a positive integer and its square is a triangle number. Can you explain why this is so?)

\(19\tfrac{1}{2} \times 20\tfrac{1}{2} = (20 - \tfrac{1}{2}) \times (20 + \tfrac{1}{2}) = 20^2 - (\tfrac{1}{2})^2 = 400 - \tfrac{1}{4} = 399\tfrac{3}{4}\).

The base of the open box is a square. Let its side be of length \(x\) cm. Then the total surface area of the box in cm² is \(x^2 + 4 \times 2x = x^2 + 8x\). Hence \(x^2 + 8x = 180\), that is \(x^2 + 8x - 180 = 0\). Therefore \((x + 18)(x - 10) = 0\), which gives \(x = -18\) or \(x = 10\). As \(x\) is positive, it may be deduced that the open box has dimensions 10 cm × 10 cm × 2 cm. So its volume is 200 cm³.

The diameter is (5 × 0.3 + 2.1) mm = (1.5 + 2.1) mm = 3.6 mm.

Let the distance from the chalet to the top of the mountain be \(x\) miles. Then, at 6 mph Supergran would take \(\dfrac{x}{6}\) hours, whereas at 10 mph she would take \(\dfrac{x}{10}\) hours. So \(\dfrac{x}{6} - \dfrac{x}{10} = 2\), that is \(5x - 3x = 60\), so \(x = 30\). Hence Supergran's departure time is 8 am and to arrive at 12 noon she should walk at \(\dfrac{30}{4}\) mph, that is \(7\tfrac{1}{2}\) mph.

After \(x\) hours, the first clock will have gone forward \(2x\) hours and the second clock will have gone back \(x\) hours. So the next time they agree is when \(2x + x = 24\), that is when \(x = 8\). The correct time then is 21:00.

The line \(y = 2x + 6\) intersects the \(y\)-axis when \(x = 0\) and \(y = 6\). It intersects the \(x\)-axis when \(x = -3\) and \(y = 0\). So E is the correct line.
(Alternatively: \(y = 2x + 6\) may be rearranged to give \(x = \tfrac{1}{2}y - 3\). So the required line looks the same as the line \(y = \tfrac{1}{2}x - 3\) when the axes are drawn in the traditional way.)

Adding the three equations gives \(3x + 3y + 3z = 30\), so \(x + y + z = 10\).
(The equations may be solved to obtain \(x = 2\), \(y = 3\), \(z = 5\). However, as the above method shows, this is not necessary in order to find the value of \(x + y + z\).)

Let the prices of a fork handle and a candle be £\(x\) and £\(y\) respectively.
Then \(x + y = 6.1\) and \(x - y = 4.6\). Adding these two equations gives \(2x = 10.7\).
So a fork handle costs £5.35 and a candle costs £0.75.
Therefore the required total is £10.70 + £3.00 = £13.70.

Reading from the left, we number the statements I, II, III, IV and V.
Statement I is true if and only if \(-1 < x < 1\); statement II is true if \(x > 1\) or if \(x < -1\).
By considering the graph of \(y = x - x^2\), which intersects the \(x\)-axis at (0, 0) and (1, 0) and has a maximum at \((\tfrac{1}{2}, \tfrac{1}{4})\), it may be seen that statement V is true if and only if \(0 < x < 1\).
We see from the table below that a maximum of three statements may be true at any one time.

Let the time for which Mary drove at 70 mph be \(t\) hours. Then the total distance covered was (55 × 2 + 70 × \(t\)) miles. Also, as her average speed over (2 + \(t\)) hours was 60 mph, the total distance travelled was 60(2 + \(t\)) miles.

Therefore 110 + 70\(t\) = 120 + 60\(t\), that is 10\(t\) = 10, that is \(t\) = 1.

So, in total, Mary's journey took 3 hours.

As \(\dfrac{1}{x} = 3.5 = \dfrac{7}{2}\), \(x = \dfrac{2}{7}\). So \(x + 2 = \dfrac{16}{7}\). Hence \(\dfrac{1}{x + 2} = \dfrac{7}{16}\).

Note that 2004 is a multiple of 3 (since its digit sum is a multiple of 3) and also a multiple of 4 (since its last two digits form a multiple of 4). So 2004 is a multiple of 12 and hence the part of the pattern between 2007 and 2011 is the same as the part of the pattern between 3 and 7.

Let the number be \(x\). Then \(x^2 = 2x^3\), that is \(x^2(1 - 2x) = 0\). So \(x = 0\) or \(x = \tfrac{1}{2}\). However, \(x\) is positive, so the only solution is \(x = \tfrac{1}{2}\).

The two acute angles in the quadrilateral in the centre of the diagram are both \((180 - 2x)\)° and the two obtuse angles are both \(y\)°, so \(360 - 4x + 2y = 360\). So \(y = 2x\).

Using ‘the difference of two squares’:
\((1 + x + y)^2 - (1 - x - y)^2 = (1 + x + y + 1 - x - y)(1 + x + y - 1 + x + y) = 2(2x + 2y) = 4(x + y)\).

The problem may be solved by firstly calculating the third and fourth terms of the sequence, but an algebraic method does reduce the amount of calculation involved. Let the first two terms be \(x\) and \(y\) respectively. Then the third term is \(\tfrac{1}{2}(x + y)\), whilst the fourth term is \(\tfrac{1}{4}(x + 3y)\). So the fifth term is \(\tfrac{1}{8}(3x + 5y)\). Putting \(x = \dfrac{2}{3}\) and \(y = \dfrac{4}{5}\), we obtain \(\dfrac{2 + 4}{8} = \dfrac{3}{4}\).

Let the original number be \(x\). Then \(x^2 = 0{\cdot}7x\), that is \(x(x - 0{\cdot}7) = 0\). So \(x = 0\) or \(x = 0{\cdot}7\), but as \(x\) is non-zero it is 0·7.

Note that

So when \(n > 12\), the remainder when \(n^2\) is divided by \(n + 4\) is always 16. For \(1 \leqslant n \leqslant 12\), the remainder when \(n^2\) is divided by \(n + 4\) is shown in the table below.

So there are 9 different remainders, namely 0, 1, 2, 3, 4, 5, 6, 7, 16.

Both \(p\) and \(-q\) are positive numbers. Hence \(p + (-q)\) is the largest of the alternatives.

The formula which gives \(C\) in terms of \(d\) is \(C = \pi d\), so the graph of \(C\) versus \(d\) is a straight line which passes through the origin.

If there are \(f\) females and \(m\) males then \(86 = 8f + 9m = 8(f + m) + m\). Now 86 = 8 × 10 + 6 = 8 × 9 + 14 = 8 × 8 + 22…. Only the first of these is feasible (because \(f + m \geqslant m\)). So \(m = 6\) and \(f = 4\).

The sum of the three numbers is \(3x\), so \(y + z = 2x\). Hence the mean of \(y\) and \(z\) is \(x\).

As \(n\) is clearly odd, the series may be written as \(1 + [-2 + 3] + [-4 + 5] + \ldots + [-(n - 1) + n]\). So \(1 - 2 + 3 - 4 + 5 - 6 + \ldots + (n - 2) - (n - 1) + n\) \(= 1 + 1 + \ldots + 1\) \(= \tfrac{1}{2}(n + 1)\). So \(\tfrac{1}{2}(n + 1) = 2006\), that is \(n = 4011\).

Note that \(4^x + 4^x + 4^x + 4^x = 4 \times 4^x = 4^{x+1}\). So \(x + 1 = 16\).

Al and Bertie have £55 between them, so Chris and Di have £150 − £55, that is £95, between them. As Al and Chris have £65 between them, the difference between the amounts Al and Di have is £95 − £65.

The equations of the three lines must be considered in pairs to find the coordinates of their points of intersection, i.e. the coordinates of the vertices of the triangle. It is left to the reader to show that these are (−15, −9), (0, 6) and (5, 1). The area of the triangle may now be found by enclosing it in a rectangle measuring 20 × 15 and subtracting the areas of the three surrounding triangles from that of the rectangle. This gives 300 − (112\(\tfrac{1}{2}\) + 12\(\tfrac{1}{2}\) + 100) = 75.

The mean of the nine consecutive positive integers is the fifth of the numbers, so their sum is nine times the fifth number. As nine is itself a perfect square, the sum will be a perfect square if and only if the fifth number is a perfect square. For the options given, the fifth numbers are 114, 124, 134, 144 and 154 respectively.

The area of the rectangle is 48 cm², so the unshaded area is 12 cm². Therefore \(\tfrac{1}{2} \times x \times 2 + \tfrac{1}{2} \times (6 - x) \times 8 = 12\), that is \(x + 24 - 4x = 12\), so \(x = 4\).

Let the weights in kg of the head and body of the fish be \(h\) and \(b\) respectively. Then \(h = 9 + \tfrac{1}{3}b\) and \(b = h + 9\). So \(b = 9 + \tfrac{1}{3}b + 9\), that is \(\tfrac{2}{3}b = 18\), which gives \(b = 27\). Hence \(h = 18\), so the whole fish weighed 54 kg.

We may deduce that \(p > q\), so, as \(p\) and \(q\) are both positive, \(\dfrac{p^2}{q^2} > \dfrac{\sqrt{p}}{\sqrt{q}} > 1\). We may also deduce that \(0 < \dfrac{q^2}{p^2} < \dfrac{q}{p} < \dfrac{\sqrt{q}}{\sqrt{p}} < 1\) and hence \(\dfrac{q^2}{p^2}\) is the least of the five numbers.

If you answer all questions correctly, you receive \(m\) marks for each of the \(N\) questions plus an extra 2 marks for the last \((N - q)\) questions. So the maximum possible score \(= m \times N + 2 \times (N - q) = mN + 2N - 2q = (m + 2)N - 2q\).

Note that the number at the end of the \(n\) th row is \(n^2\), so 400 will lie at the end of the 20 th row. The row below will end in 21², i.e. 441, so the number directly below 400 will be 440.

We note first that \(y = 5\) since that is the only non-zero digit which, when it is multiplied by 3, has itself as the units digit. So there is a carry of 1 into the tens column. We note also that \(a = 1\) or \(a = 2\) as “\(fly\)” < 1000 and therefore 3 × “\(fly\)” < 3000. We now need 3 × \(l\) + 1 to end in either 1 or 2 and the only possibility is \(l = 7\), giving \(a = 2\) with a carry of 2 into the hundreds column. As \(a = 2\), \(f\) must be at least 6. However, if \(f = 6\) then \(w = 0\) which is not allowed. Also the letters represent different digits, so \(f \neq 7\) and we can also deduce that \(f \neq 9\) since \(f = 9\) would make \(w = 9\).
Hence \(f = 8\), making \(w = 6\) and the letters represent 875 × 3 = 2625.

Points (−3, −3), (−2, −1), (4, 11) and (5, 13) all lie on the line whose equation is \(y = 2x + 3\), but (2, 5) does not lie on this line.

The 49 black squares will be in a 7 × 7 formation, so the board will measure 15 × 15 squares. Hence the number of white squares = 225 − 49 = 176.

Let the numbers be \(x\) and \(y\). Then \(x + y = 2\); \(x - y = 4\). Adding these equations gives \(2x = 6\), so \(x = 3\). Hence \(y = 2 - 3 = -1\) so the numbers are 3 and −1.