Earlier today I set you ten questions form the UKMT Intermediate Mathematical Challenge, sat by more than 200,000 British schoolchildren aged 13-16 on Feb 4 this year.
More than 30,000 of you took the test today, and on the whole you did better than the students did - but then I bet you took more than your allotted 24 minutes.
The students who enter the IMC are of all abilities. Some schools enter the three years 9, 10 and 11, some enter single years and some just the top set.
Here are the answers, with explanations. I have also included the percentage of students who got each answer right, and the percentage of Guardian readers who did too.
The average mark overall is roughly speaking three questions right out of ten. So if you got four or more, gold stars all round!
Click on these links to see the other 15 questions from the test paper or read more in-depth explanations of the answers.
1. What is the value of 6102 − 2016?
3994- 4086
409641144994
Method: subtract one from the other.
Students: 91 per cent. Guardian readers: 88 per cent.
2. Which of the following fractions is closest to 1?
7/88/79/10- 10/11
11/10
Method: calculate the difference between each fraction and 1: 1/8, 1/7, 1/10, 1/11, 1/10. The smallest is 1/11.
Students: 66 per cent. Guardian readers: 83 per cent.
3. One third of the animals in Jacob’s flock are goats, the rest are sheep. There are twelve more sheep than goats.
How many animals are there altogether in Jacob’s flock?
1224- 36
4860
Method: One third of the animals in Jacob’s flock are goats. Therefore two thirds of these animals are sheep. Therefore, the 12 more sheep than goats amount to one third of the flock. Since 12 animals make up one third of the flock, there are 3 × 12 = 36 animals in Jacob’s flock.
Students: 78 per cent. Guardian readers: 82 per cent.
4. The net shown consists of squares and equilateral triangles. The net is folded to form a rhombicuboctahedron, as shown.
When the face marked P is placed face down on a table, which face will be facing up?
Solution: D
Method: When the net is folded up to form the rhombicuboctahedron, the left-hand edge of the square marked X is joined to the right-hand edge of the square marked E so that the eight squares at the centre of the net form a band around the solid. In this band, the square opposite square P is the square which is four squares away from P, that is square D. So if the square marked P is placed face down on a table, then the square marked D will be facing up.
Students: 65 per cent. Guardian readers: 88 per cent.
5. Tegwen has the same number of brothers as she has sisters. Each one of her brothers has 50% more sisters than brothers.
How many children are in Tegwen’s family?
579- 11
13
Method: 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 per cent more sisters than brothers. Therefore g = (3/2)(b –1). So (b + 1) = (3/2)(b–1) and hence 2b + 2 = 3b – 3. Rearranging this equation gives b = 5. So g = 5 + 1 = 6. Therefore there are 11 children in Tegwen’s family.
Students: 15 per cent. Guardian readers: 23 per cent.
6. How many of the following positive integers are divisible by 24?
22 x 32 x 52 x 73
22 x 32 x 53 x 72
22 x 33 x 52 x 72
23 x 32 x 52 x 72
0- 1
234
Method: The prime factorisation of 24 is 2 x 2 x 2 x 3, or 23 x 3. Therefore all multiples of 24 must include both 23 and 3 in their prime factorisation. Of the options given, only the last includes 23. As it is also a multiple of 3, it is a multiple of 24.
Students: 14 per cent. Guardian readers: 36 per cent.
7. A list of positive integers has a median of 8, a mode of 9 and a mean of 10.
What is the smallest possible number of integers in the list?
5- 6
789
Method: From the information given, there are at least two 9s in the list, since 9 is the mode, and at least one number greater than 10, since 10 is the mean. So there are at least three numbers greater than 8 in the list. Therefore the list must contain at least six numbers, as the median of the numbers is 8. Moreover, it is possible to find suitable lists of six numbers with sum 60 (as the mean is 10), for example 1, 2, 7, 9, 9, 32.
Students: 12 per cent. Guardian readers: 24 per cent.
8. In a particular group of people, some always tell the truth, the rest always lie. There are 2016 in the group. One day, the group is sitting in a circle. Each person in the group says, “Both the person on my left and the person on my right are liars.”
What is the difference between the largest and smallest number of people who could be telling the truth?
072126288- 336
Method: For brevity, let T denote a truth teller and L a liar. Clearly each T has to have an L on each side. Each L either (i) has a T on each side or (ii) has an L on one side and a T on the other side. The largest number of Ts will occur if (i) is always the case. This gives the arrangement TLTLTL… which, since 2 divides 2016, joins up correctly after going round the table. In this case the number of Ts is (1/2)x 2016 . The smallest will occur if case (ii) always is the case. This gives the arrangement LLTLLTLLT… which, since 3 divides 2016, also joins up correctly. In this case the number of Ts is (1/3)x 2016 . The difference is (1/6) x 2016 = 336.
Students: 5 per cent. Guardian readers: 23 per cent
9. Every day, Aimee goes up an escalator on her journey to work. If she stands still, it takes her 60 seconds to travel from the bottom to the top. One day the escalator was broken so she had to walk up it. This took her 90 seconds.
How many seconds would it take her to travel up the escalator if she walked up at the same speed as before while it was working?
3032- 36
4575
Method: Let the distance from the bottom of the escalator to the top be d . Then, when she stands still, Aimee travels d/60 every second. When she is walking, Aimee travels d/90 every second. So when Aimee walks up the working escalator, the distance which she travels every second is
(d/60) + (d/90) = (3d + 2d)/180 = 5d/180 = d/36
So the required number of seconds is 36.
Students: 5 per cent. Guardian readers: 35 per cent.
10. The tiling pattern shown uses two types of tile, regular hexagons and equilateral triangles, with the length of each side of the equilateral triangles equal to half the length of each side of the hexagons. A large number of tiles is used to cover a floor.
Which of the following is closest to the fraction of the floor that is shaded black?
1/81/101/12- 1/13
1/16
Method:
We see that the entire plane can be tessellated by the shape labelled S in the above diagram. This shape is made up of one tile in the shape of a regular hexagon and two tiles in the shape of equilateral triangles.
The diagram shows that, as the side length of the triangular tiles is half that of the hexagonal tiles, the area of the hexagonal tiles is 24 times the area of the triangular tiles.
[This may be seen by counting the number of small triangles, each congruent to the triangular tiles, into which the hexagonal tile is divided by the dashed lines. Alternatively, we see that the hexagonal tile may be divided into 6 equilateral triangles each with side length twice that of the triangular tiles, and hence each with an area 4 times that of the triangular tiles. It follows that the area of the hexagonal tile is 6 × 4, that is, 24, times that of the triangular tiles.]
Therefore the shape S is made up of 2 black triangles and a hexagon equal in area to 24 of the black triangles. So the total area of S is equal to that of 2 + 24 = 26 of the triangles. It follows that the fraction of the area of S that is black is 2/26 = 1/13.
A large area of the floor will be covered by a large number of tiles which make up shapes congruent to S, together with a few bits of tiles. Therefore the fraction of the floor that is shaded black will be very close to 1 13 .
Students: 3 per cent. Guardian readers: 9 per cent.
Thanks to Rachel Greenhalgh of the UKMT.
I post a puzzle here every second Monday.
My most recent book is the mathematical adult colouring book Snowflake Seashell Star. (In the US its title is Patterns of the Universe.)
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And if you know of any great puzzles that you would like me to set here, get in touch.
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