Download [PDF] 2011 UKMT UK IMC Solutions Intermediate Mathematical Challenge United Kingdom Mathematics Trust

File Information

Filename:	[PDF] 2011 UKMT UK IMC Solutions Intermediate Mathematical Challenge United Kingdom Mathematics Trust.pdf
Filesize:	307.54 KB
Uploaded:	09/08/2021 16:46:06
Keywords:	2011 ukmt uk imc solutions intermediate mathematical challenge united kingdom mathematics trust pdf
Description:	Download file or read online UKMT past exam paper intermediate mathematical challenge 2011 UK IMC solutions - United Kingdom Mathematics Trust
Downloads:	2

File Preview

Download Urls

Short Page Link

https://www.edufilestorage.com/5O9

Full Page Link

https://www.edufilestorage.com/5O9/PDF_2011_UKMT_UK_IMC_Solutions_Intermediate_Mathematical_Challenge_United_Kingdom_Mathematics_Trust.pdf

HTML Code

<a href="https://www.edufilestorage.com/5O9/PDF_2011_UKMT_UK_IMC_Solutions_Intermediate_Mathematical_Challenge_United_Kingdom_Mathematics_Trust.pdf" target="_blank" title="Download from eduFileStorage.com"><img src="https://www.edufilestorage.com/cache/plugins/filepreviewer/2329/pdf/150x190_middle_46f4e7862b1eb5bd4935adbbba5d79e8.jpg"/></a>

Forum Code

[url=https://www.edufilestorage.com/5O9/PDF_2011_UKMT_UK_IMC_Solutions_Intermediate_Mathematical_Challenge_United_Kingdom_Mathematics_Trust.pdf][img]https://www.edufilestorage.com/cache/plugins/filepreviewer/2329/pdf/150x190_middle_46f4e7862b1eb5bd4935adbbba5d79e8.jpg[/img][/url]

[PDF] 2011 UKMT UK IMC Solutions Intermediate Mathematical Challenge United Kingdom Mathematics Trust.pdf | Plain Text

1 UK INTERMEDIATE MATHEMATICAL CHALLENGE February 3rd 2011 SOLUTIONS These solutions augment the printed solutions that we send to schools. For convenience, the solutions sent to schools are confined to two sides of A4 paper and therefore in many cases are rather short. The solutions given here have been extended. In some cases we give alternative solutions, and we have included some Extension Problems for further investigations. The Intermediate Mathematical Challenge (IMC) is a multiple choice contest, in which you are presented with five alternative answers, of which just one is correct. It follows that often you can find the correct answers by working backwards from the given alternatives, or by showing that four of them are not correct. This can be a sensible thing to do in the context of the IMC, and we often give first a solution using this approach. However, this does not provide a full mathematical explanation that would be acceptable if you were just given the question without any alternative answers. So usually we have included a complete solution which does not use the fact that one of the given alternatives is correct. (A few questions do not lend themselves to such a treatment.) Thus we have aimed to give full solutions with all steps explained. We therefore hope that these solutions can be used as a model for the type of written solution that is expected in the Intermediate Mathematical Olympiad and similar competitions. We welcome comments on these solutions, and, especially, corrections or suggestions for improving them. Please send your comments, either by e-mail to enquiry@ukmt.co.uk or by post to IMC Solutions, UKMT Maths Challenges Office, School of Mathematics, University of Leeds, Leeds LS2 9JT. Quick Marking Guide 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 B A E E C B A E A E E D D E A C C D B D B B A C D  UKMT, 2011. These solutions may be used freely within your school or college. You may, without further permission, post these solutions on a website which is accessible only to staff and students of the school or college, print out and distribute copies within the school or college, and use them within the classroom. If you wish to use them in any other way, please consult us at the address given above.

2 1. What is the value of 5.45.45.55.4    ? A 36.5 B 45 C 50 D 90 E 100 Solution: B Since you are not allowed to use a calculator in the IMC, it is a good idea to look for a way to avoid having to do multiplication sums. The presence of the factor 5.4in both products provides a clue to an efficient method. If we take out this common factor, we obtain 45105.4)5.45.5(5.45.45.45.55.4         . Extension Problem. 1. Without using a calculator find the value of 6.8764.1234.1234.123    2. To find the diameter in mm of a Japanese knitting needle, you multiply the size by 0.3 and add 2.1. What is the diameter in mm of a size 5 Japanese knitting needle? A 3.6 B 7.4 C 10.8 D 12 E 17.1 Solution: A We need to work out 1.23.05   . This gives 6.31.25.1   as the answer. 3. The consecutive digits 1, 2, 3, 4 in that order can be arranged to make the correct division, 4312  . One other sequence of four consecutive digits p,q,r,smakes a correct division, srpq '' . What is the value of sin this case? A 4 B 5 C 6 D 7 E 8 Solution: E In the context of the IMC it is good enough to try out all the cases in turn until we find a correct sum. We see that 5423  , 6534  , 7645  but 8756  . To check that this really gives the only sequence that gives a correct division sum you could just check the remaining case 6p . However, it is more illuminating to tackle the question algebraically. Suppose the digits p, q ,rand sare consecutive, then 1 pq , 2 pr and 3 ps . Now ‘ pq ’ represents the number qp 10 , that is, the number 111)1(10    ppp . So the equation ‘pq ’ s r  is equivalent to 3)2()111(     ppp . We now have that, as 02 p , 3)2()111(     ppp if and only if )3)(2(111     ppp , if and only if 65111 2     ppp , if and only if 056 2    pp , if and only if 0)5)(1(   pp . This gives just the two solutions 5,1 p corresponding to the correct equations 4312  and 8756  .

3 4. The angles of a triangle are in the ratio 2:3:5. What is the difference between the largest angle and the smallest angle? A 9 B  18 C  36 D  45 E  54 Solution: E The sum of the angles in a triangle is  180 . Since the angles are in the ratio 2:3:5 we have to divide  180 in the ratio 2:3:5. The difference between the largest angle the smallest angle is therefore    54180 10 3 180 5 3 2 25        . 5. The diagram shows a rectangle placed on a grid of 1 cm 1 cm squares. What is the area of the rectangle in cm 2? A 15 B 21 22 C 30 D 36 E 45 Solution: C The total area of the 88 grid in cm 2is 648 2 . The two larger triangles in the top left and bottom right corners of the grid make up a 55 square with area 25 cm 2. The two smaller triangles in the other corners of the grid make up a 33 square with area 9 cm 2. Hence the area of the rectangle, in cm 2, is 3092564   . 6. When I glanced at my car milometer it showed 24942, a palindromic number. Two days later, I noticed that it showed the next palindromic number. How many miles did my car travel in those two days? A 100 B 110 C 200 D 220 E 1010 Solution: B The only 5-digit palindromic number that begins 249 is 24942, so the next palindromic must be greater than 24999. The only 5-digit palindromic number beginning 250 is 25052, so this is the next palindromic number after 24942. We have that 1102494225052  , and so the answer is 110. Extension problems: 1. Which is the next palindromic number after 25052? 2. How many 5-digit palindromic numbers are there? 3. Can you find a formula for the number of n-digit palindromic numbers? [Hint: you may need to consider the cases where nis even, and where nis odd, separately.]

4 7. What is the value of xin this diagram? x A 30 B 35 C 40 D 45 E 50  70  80 x x Solution: A We label the points P, Q, R, S, T, U and V as shown. P The vertically opposite angles PRS and QRU are x equal, as are PSR and TSV . We let these angles be y Q R y z S T and z, as shown in the diagram.  70 y z  80 We can find the value of xin more than one way. 1. Since UPVQUP  , the lines QU and PV are x parallel. Hence the corresponding angles RQU and U x TSV are equal. So 70z .Since the angles of a triangle add V up to 180 0, we have from triangle STV that 308070180   x . 2. From the triangles PRS ,QRU and STV we have that 180   zyx , (1) 18070  yx , (2) and 18080  zx . (3) From (1) and (2), 70z and hence, from (3), 30x . 8. A square piece of card has a square of side 2 cm cut out from each of its corners. The remaining card is then folded along the dotted lines shown to form an open box whose total internal surface area is 180 cm 2. What is the volume of the open box in cm 3? A 100 B 128 C 162 D 180 E 200 Solution: E Suppose that the dotted square corresponding to the base of the box has side length xcm. Then the base has area 2x cm, and each of the sides of the box has area 2 xcm. Hence the internal surface area of the box, in cm 2, is xx24 2   , that is, xx8 2 . Therefore 1808 2   xx . Now 1808 2   xx if and only if 01808 2    xx if and only if 0)10)(18(   xx with solutions 18x and 10x . Since xmust be positive, we deduce that 10x . The volume of the open box is the height multiplied by the area of the base, and, in cm 3, this is 2 2x , that is, 200.

5 9. In the diagram, XY is a straight line. What is the value of x?  80 A 170 B 160 C 150  40 D 140 E 130  20 x  150 X Y Solution: A Let the angles be as marked in the diagram.  80  40  20 x r q p  150 The exterior angle of a triangle is the sum of the two opposite interior angles. Therefore we have that 15080 p (1) pq  40 (2) and qr 20 (3) From (1), 70p . Hence, from (2), 30q . Therefore, from (3), 10r . Since the angles on a line add up to  180 , it follows that 170x . 10. Merlin magically transforms a 6 tonne monster into mice with the same total mass. Each mouse has a mass of 20g. How many mice does Merlin make? A 30 B 300 C 3000 D 30 000 E 300 000 Solution: E 1 tonne = 1000 kg = g10001000  , so 6 tonne = g100010006   . So the number of 20 g mice that Merlin makes is 0003000005065010006 20 100010006         . 11. What is the value of 21 21 2019 ? A 250 B 41 380 C 41 390 D 395 E 43 399 Solution: E Since in the IMC we are not allowed to use a calculator, we should look for a better method than just multiplying the two given numbers. The clue is that both numbers differ from 20 by 21, so we can make use of the standard “difference of two squares” formula, 22 ))(( bababa    . This gives 43 41 221 2 21 21 21 21 399400)(20)20)(20(2019         . Extension problem. 1. Find the values of (a) 21 21 200199 and (b) 31 32 1013986 without using a calculator .

6 12. What is the sum of the first 2011 digits when 1120 is written as a decimal? A 6013 B 7024 C 8035 D 9046 E 10057 Solution: D We see from the following long division sum 11 ...8181.1 ...0000.20 11 90 88 20 that 1120 has the recurring decimal expansion 1.8181…. . Hence the first 2011 digits consist of 1006 1s and 1005 8s. So the sum of these digits is 9046804010068100511006      . 13. The three blind mice stole a piece of cheese. In the night, the first mouse ate 31of the cheese. Later, the second mouse ate 31of the remaining cheese. Finally, the third mouse ate 31of what was then left of the cheese. Between them, what fraction of the cheese did they eat? A 27 16 B 27 17 C 3 2 D 27 19 E 27 20 Solution: D Our first method is a direct calculation: The first mice ate 31of the cheese, leaving 32 31 1   . The second mouse ate 31of this, namely 92 32 31   . This left 94 92 32   of the cheese. The third mouse ate 31of this, namely 274 94 31   . So between them, the amount of the cheese that the three mice ate was 2719 27469 274 92 31      . Our second method is to first calculate the amount of cheese that remains. Each mouse eats 31rd of the remaining cheese, and so leaves 32rd of it. Hence, after the three mice have eaten the cheese, there remains 3 32 of the initial amount of cheese. So the fraction which is eaten is  2719 278 3 32 11     . Extension problem. 1. Suppose there are nmice, and each, in turn, eats n1th of the remaining amount of cheese. Find a formula for the fraction of cheese that they eat between them. 14. The number 6 lies exactly halfway between 3 and 23 . Which of the following is not halfway between a positive integer and its square? A 3 B 10 C 15 D 21 E 30 Solution: E The number klies exactly halfway between the positive integer nand its square 2n if and only if 2 )1( 2 2     nnnn k . So the numbers of this form are, putting nsuccessively equal to 1, 2, 3, … in this formula, the numbers in the sequence 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, … We see that of the alternative answers given, the only one that does not occur in this sequence is 30.

7 Note: You will probably recognize the numbers in the sequence 1, 3, 6, … as the triangle (or triangular )numbers. They are so-called because they are the number of dots in a triangular array such as           The number of dots in this array is 4321   . In general a triangular number is a number which, for some positive integer n,is the sum of the first npositive integers. So, if we let nT be the nth triangular number, ....321nT n      Extension Problems. 1. The formula for the sum n    ...321 is given by , 2 )1(   nn Tn .Can you prove that for each positive integer n, 2 )1( ...321       nn n ? 2. Since the sequence, 1, 3, 6, … of triangular numbers is increasing it is, in principle, possible to check whether a number is triangular by seeing whether or not it occurs in the sequence. However this is not very easy for a large number such as 6 126 750. Here is an alternative method. The positive integer kis triangular if and only 2 2 nn k   for some positive integer n, that is, if and only if the quadratic equation 02 2    knn has a positive integer solution for n. Use the formula for the solutions of a quadratic equation to find a necessary and sufficient condition for kto be a triangular number. Then use this criterion to determine whether the following numbers are triangular (a) 6 975 163, (b) 76 205 685. 3. Note that the sequence of triangular numbers, 1, 3, 6, 10, 15, 21, 28, 36, 45 ,… , includes two squares, 1 and 36. Find some more triangular numbers that are also squares. Are there infinitely many triangular numbers that are squares? 15. The equilateral triangle ABC has sides of length 1 C A B C and AB lies on the line XY . The triangle is rotated clockwise around Buntil BC lies on the line XY . It is then rotated similarly around C and then X Y about Aas shown in the diagram. A B C A B What is the length of the path traced out by point C during this sequence of rotations? A 3 4 B 32 C 3 8 D 3 E 3 2 Solution: A For convenience we have used 1A , 2A etc for the C 1A 1B 2C subsequent positions taken up by A,Band C. C first moves along a circular arc with centre B. This arc has radius 1. Since 0 1 120 CBC , X Y which corresponds to one-third of a complete A B 1C 2A 2B turn, the length of this arc is 2 31 32 .When the triangle is rotated about the point 1C the vertex C does not move at all. Finally, when the triangle is rotated about the point 2A ,C again turns through an angle 120 0. and so again moves along an arc of length 32 . Therefore the total length of the path traced out by C is    34 32 32   .

8 16. The diagram shows an L-shape divided into 1 1 squares. Gwyn cuts the shape along some of the lines shown to make two pieces neither of which is a square. She then uses the pieces to form a 62 rectangle. What is the difference between the areas of the two pieces? A 0 B 1 C 2 D 3 E 4 Solution: C The L-shape needs to be divided as shown if Gwyn is to make a 62 rectangle from two pieces which are not squares. Note that one piece must be turned over. The pieces have areas 7 and 5, whose difference is 2. 17. A shop advertised “Everything half price in our sale”, but also now advertises that there is “An additional 15% off sale prices”. Overall, this is equivalent to what reduction on the original prices? A 7.5% B 35% C 57.5% D 65% E 80% Solution: C To reduce prices by a half, we multiply the price by 0.5. To reduce prices by a further 15%, we then multiply by 0.85. Therefore the final price is obtained by multiplying the original price by .425.085.05.0   So the final price is 42.5% of the original price. So the reduction is %5.57 . 18. The diagram contains six equilateral triangles with sides of length 2 and a regular hexagon with sides of length 1. What fraction of the whole shape is shaded? A 8 1 B 7 1 C 6 1 D 5 1 E 4 1 Solution: D Each of the equilateral triangles with side length 2, may be divided into 4 equilateral triangles, each with side length 1, as shown. In a similar way, the regular hexagon may be divided up into 6 equilateral triangles each with side length 1. So the whole shape is made up of 30646   of these congruent small triangles, of which 6 are shaded. So the proportion of the figure that is shaded is  30 6 5 1.

9 19. Harrogate is 23 km due north of Leeds, York is 30 km due east of Harrogate, Doncaster is 48 km due south of York, and Manchester is 70 km due west of Doncaster. To the nearest kilometre, how far is it from Leeds to Manchester, as the crow flies? A 38km B 47km C 56km D 65km E 74km Solution: B We see from the diagram that Manchester is 40 km Harrogate 30 York west of Leeds, and 25 km south of Leeds. So, by Pythagoras’ Theorem, the distance between them 23 is, in km, 8958554025 2222    . Since ,1008981   we have that 10899  , Leeds 48 and therefore 5089545  . So of the options given, 47 is the best approximation. 25 [To check that this really is the best integer approximation to 895 . Manchester 40 30 Doncaster you would need to check that, 5.9894.9   . 70 To do this you need to check, by a direct calculation, that 225.9894.9   .] 20. Max and his dog Molly set out for a walk. Max walked up the road and then back down again, completing a six mile round trip. Molly, being an old dog, walked at half Max’s speed. When Max reached the end of the road, he turned round and walked back to the starting point, at his original speed. Part way back he met Molly, who then turned around and followed Max home, still maintaining her original speed. How far did Molly walk? A 1 mile B 2 miles C 3 miles D 4 miles E 5 miles. Solution: D 3miles Max completes a 6 mile trip, so he walks 3 miles before xmiles x3 miles turning round. Suppose that he meets Molly after Molly has walked xmiles. So when they meet Max has covered x3 miles of his return journey, and so he has walked xx    6)3(3 miles while Molly has walked xmiles. Since Molly walks at half Max’s speed, )6( 21 xx  . So xx21 3  . Hence 3 23 x , and so 2x . Molly altogether walks x x miles, that is, 4 miles.

10 21. A regular octagon is placed inside a square, as shown. The shaded square connects the midpoints of the four sides of the octagon. What fraction of the outer square is shaded? A 12 B 2 1 C 4 12 D 5 22 E 4 3 Solution: B Suppose that the side length of the regular octagon is sunits. The right angled isosceles triangles in the diagram have hypotenuses of length s. Suppose that the length of each of the other sides of these triangles is t. By Pythagoras’ Theorem, 222stt  , and hence st21  . Hence the side length of the larger square is sssts)21(22      . t s t The side length of the smaller square is tstts    21 21 t s t21 t21 s t ssss)21()12( 21 21 21       . t21 t21 Hence the fraction of the outer square that is shaded is   2 1 2 1 )21(( )21( 2 2 2 21          s s . 22. You are given that 95p , 129q , 1612r , 2016s and 2520t . What is the value of pqrst ? A 1 B 2 C 3 D 4 E 5 Solution: B The key to the solution is to first calculate the value of pqrst5 . We have that 252520)16(16)12(12)9(9)5(5          ttsststrrstrstqqrstqrstppqrst . Hence 2 pqrst 23. A window frame in Salt’s Mill consists of two equal semicircles and a circle inside a large semicircle with each touching the other three as shown. The width of the frame is 4m. What is the radius of the circle, in metres? 4m A 3 2 B 2 2 C 4 3 D 122 E 1 Solution: A T Suppose that the radius of the circle is xm. We let Pbe the centre of the large semicircle, and Q and Rbe S centres of the smaller semicircles. We let Sbe the centre of the circle and Tthe point where PS meets the large semicircle. The large semicircle has radius 2 m and the smaller Q P R semicircles have radius 1 m.

11 So we have that PQ has length 1, PS has length x2 and SQ has length x1 .PS is a tangent to the smaller semicircles, and therefore 0 90 SPQ . So, applying Pythagoras’ Theorem to the triangle PQS , we have 222)1()2(1xx    that is 22 21441 xxxx      and hence x64 and therefore 3 2x . Notes: 1. It follows that in the right angled triangle QPS the side lengths are in the ratio 1: 32 321:2  , that is, 3:4:5. 2. Salt’s Mill was textile mill, built by Titus Salt in 1853 at Saltaire near Bradford. It was bought by Jonathan Silver in 1987. Today it is centre for art and commerce, with the Hockney art galleries, restaurants, shopping and office space. You will find more information about it on its website: http://www.saltsmill.org.uk/ . 24. Given any positive integer n, Paul adds together the distinct factors of n, other than nitself. Which of these numbers can never be Paul’s answer? A 1 B 3 C 5 D 7 E 9 Solution: C The only factor of 2, other than 2 itself is 1. So the sum of these factors is 1. [Note that whenever p is a prime number, the sum of the factors of p, other than p, is 1.] The factors of 4, other than 4 itself, are 1 and 2, whose sum is 3. The factors of 8, other than 8 itself, are 1, 2, 4, whose sum is 7. The factors of 15, other than 15 itself, are 1, 3, 5, whose sum is 9. So options A, B, D and E can each be Paul’s answer. In the context of the IMC this is enough for us to be able to select C as the correct option. However, to give a mathematically complete answer, we need to give a reason why the sum of the factors of n, other than nitself, cannot equal 5. Clearly, we need only consider the case where 1n . So, one of the factors of n, other than nitself, is 1. Suppose that the other factors are a,b,… . Then a, b,… are distinct, none of them is 1, and 5...1    ba . So 4...  ba . However, there is no way of expressing 4 as the sum of more than one distinct positive integer none of which is 1. So the only possibility is that 4 is the only factor of n,other than 1 and n. However, this is impossible, since if 4 is a factor of n, then so also is 2. Therefore option C is not possible. Extension Problems. 1. Investigate the sums of the factors of n, other than nitself, for all the positive integers in the range 321 n that are not prime numbers. 2. We have shown above that 5 does not occur as the sum of the factors of nother than nitself. Find another value that does not occur. 3. Prove that the sum of the factors of a positive integer n, other than nitself, is never 52.

12 25. The diagram shows a square, a diagonal and a line joining a vertex to the midpoint of a side. What is the ratio of area Pto area Q? Q A 2:1 B 3:2 C 1:2 D 2:5 E 1:3 P Solution: D Let I,J,K and Lbe the vertices of the square. Let M be the midpoint of JK , I J let Nbe the point where the diagonal IK meets LM . Let the line through N parallel to LK meet IL at Rand JK at S. Let Tbe the foot of the perpendicular from N to LK. Let the square have side length s. N M R S In the triangles INL and KNM , the opposite angles INL and KNM are equal. Also, as IL is parallel to JK , the alternate angles LIN and MKN L T K are equal. Therefore the triangles INL and KNM are similar. Hence 2 MK IL NK IN . Similarly, the triangles INR and KNS are similar. Therefore 2 NK IN NS NR . So sNR32 and sNS31 . Now, NTKS is square, because its angles are all right-angles, and  45 NKT . Therefore sNSNT31  . It follows that the area of the triangles LNK, INL and MNK are 2 61 31 21 )( sss   , 2 31 32 21 )( sss   , and 2 121 31 21 21 )( sss   , respectively. The area of the region P, is that of the triangle LNK , that is, 2 61s . The area of the region Q is obtained by subtracting the areas of the triangles LNK, INL and MNK from the area of the square. So region Q has area 2 125 2 121 2 31 2 61 2 sssss     . So the ratio of these areas is 2 125 2 61 : ss 5:2: 125 61  .