Download [PDF] 2015 UKMT Senior Kangaroo Solutions Mathematical Challenge United Kingdom Mathematics Trust

File Information

Filename:	[PDF] 2015 UKMT Senior Kangaroo Solutions Mathematical Challenge United Kingdom Mathematics Trust.pdf
Filesize:	219.48 KB
Uploaded:	09/08/2021 18:19:30
Keywords:	2015 ukmt senior kangaroo solutions mathematical challenge united kingdom mathematics trust pdf
Description:	Download file or read online UKMT past exam paper senior mathematical challenge SMC senior kangaroo 2015 solutions - United Kingdom Mathematics Trust
Downloads:	3

File Preview

Download Urls

Short Page Link

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

Full Page Link

https://www.edufilestorage.com/5PX/PDF_2015_UKMT_Senior_Kangaroo_Solutions_Mathematical_Challenge_United_Kingdom_Mathematics_Trust.pdf

HTML Code

<a href="https://www.edufilestorage.com/5PX/PDF_2015_UKMT_Senior_Kangaroo_Solutions_Mathematical_Challenge_United_Kingdom_Mathematics_Trust.pdf" target="_blank" title="Download from eduFileStorage.com"><img src="https://www.edufilestorage.com/cache/plugins/filepreviewer/2441/pdf/150x190_middle_46f4e7862b1eb5bd4935adbbba5d79e8.jpg"/></a>

Forum Code

[url=https://www.edufilestorage.com/5PX/PDF_2015_UKMT_Senior_Kangaroo_Solutions_Mathematical_Challenge_United_Kingdom_Mathematics_Trust.pdf][img]https://www.edufilestorage.com/cache/plugins/filepreviewer/2441/pdf/150x190_middle_46f4e7862b1eb5bd4935adbbba5d79e8.jpg[/img][/url]

[PDF] 2015 UKMT Senior Kangaroo Solutions Mathematical Challenge United Kingdom Mathematics Trust.pdf | Plain Text

UKMT UKMT UKMT SENIOR ‘KANGAROO’ MATHEMATICAL CHALLENGE Friday 27th November 2015 Organised by the United Kingdom Mathematics Trust SOLUTIONS 1. 180 The number of gold coins in the original pile is . These form 20% of the final pile. Therefore there are coins left. Hence the number of silver coins Simon removes is .0.02×200=4 4×5=20 200−20=180 2. 28 The expression can be simplified in stages as follows: 1+ 1 1+ 1 1+ 1 5=1+ 1 1+ 1 ( 6 5 ) =1+ 1 1+ 5 6=1+ 1 ( 11 6 ) =1+ 6 11= 17 11= a b. Hence the value of is .a+b17+11=28 3. 11 Let the three missing integers be and , as shown.x,yz Consider the ‘top’ three faces. Since the sum of the three numbers at the vertices of each face is the same, we have 1+5+x=1+x+y=1+5+y and hence . Therefore the sum of the numbers on a face is equal to . But is equal to the sum of the numbers on a face. Hence the sum of the other three numbers that Andrew will write is 11.x=y=5 5+5+1=11 x+y+z 1 5 xy z 4. 7 The first six socks Rachel takes out could consist of three different coloured socks and the three socks with holes in, in which case she would not have a pair of socks the same colour without holes in. However, whatever colour her next sock is, she must then complete a pair. Hence she must take seven socks to be certain of getting a pair of socks the same colour without holes in.

5. 50 Let and be the centres of the large and small circles respectively and label points and as shown in the diagram. Let the radius of the small circle be cm. Draw line so that is on and is parallel to . Draw in line . Since triangle is right-angled and isosceles, by Pythagoras. Hence . Similarly, since triangle is right- OP QS r PR R QS PR OS OQ OQS OQ 2 =10 2+10 2 OQ=10 2cmPQR OQ S PR angled and isosceles, . Note that angle angle so is a straight line. Therefore . This has solution PQ=r 2cmOQS=PQS=45°OPQ 10 2=10+r+r 2 r= 10 ( 2Š1 ) 2+1= 10 ( 2Š1 )( 2Š1 ) ( 2+1 )( 2Š1 ) = 10 (2+1Š2 2) 2Š1=30Š20 2. Hence the radius of the small circle is and the value of is . (30Š20 2)cma+b30+20=50 6. 7 Let the five integers be , , , and with . The median of the list is and, since the mode is one less than the median, and . The mean is one more than the median and hence the total of the five integers is . Therefore and hence . Since the smallest possible value of is , the maximum value of is . Hence the largest possible value of the range of the five integers is .pqr s t pqrst rp=q=rŠ1r

10. 288 Katherine catches James after 8 minutes when she has jogged laps. In that time, James will have jogged one lap fewer so will have jogged laps. Therefore, James jogs laps in 8 minutes which is the same as 480 seconds. Hence he will jog of a lap in 96 seconds and so he jogs a whole lap in 288 seconds. 8 3 5 3 5 3 1 3 11. 52 A solution can be obtained by reflecting the square repeatedly in the cushion the ball strikes. The path of the ball is then represented by the line in the diagram.AB The length of the path can be calculated using Pythagoras Theorem. We have . Therefore and so metres and hence the value of is 52. (AB )2 =( 3×2) 2+( 2×2) 2 (AB )2 =36+16=52AB= 52 kA B 12. 35 Chris's time for the ride when he rode at an average speed of was hours. His planned speed was when his time would have been hours. The question tells us that he completed the ride 1 hour earlier than planned, so .xkm/h 210 x (xŠ5)km/h 210 xŠ5 210 xŠ5Š 210 x=1 Therefore and hence . Thus and hence . Therefore, since is positive, . 210xŠ210(xŠ5)=x(xŠ5)1050=x 2Š5x x 2Š5xŠ1050=0(x−35)(x+30)=0xx=35 13. 13 Assume the man at the front of the queue is telling the truth and that everyone behind him always lies. However, then the person in third place in the queue would be telling the truth when he says that the person in second place always lies. This contradicts the original assumption and so the man at the front of the queue is lying. In this case, the man in second place is telling the truth, the man in third place is lying etc. Hence, every other person, starting with the first, is lying and so there are people in the queue who always lie.1+ 1 2 ×24=13 14. 30 The smallest number of contestants solving all four problems correctly occurs when the contestants who fail to solve individual problems are all distinct. In that case, the number failing to solve some question is and the number solving them all is .10+15+20+25=70 100−70=30 15. 14 First note that . Hence, for to be exactly divisible by 165, it must be exactly divisible by 3, 5 and 11. A number is divisible by 3 if and only if the sum of its digits is divisible by 3 so is divisible by 3 and hence is divisible by 3. A number is divisible by 5 if and only if its last digit is 5 or 0 so or 0. Since is divisible by 3 then . A number is divisible by 11 if and only if the sum of its digits with alternating signs is divisible by 11 so is divisible by 11. Hence is divisible by 11 and so . Hence the value of is .165=3×5×11 ‘XX4XY’ 3X+4+Y4+Y Y=5 4+YY=5 XŠX+4ŠX+Y 9ŠXX=9 X+Y9+5=14 16. 64 Since adjacent digits differ by 1, each time the number has a digit that is a 1 or a 3, there is only one choice for the next digit as it must be a 2 whereas each time the number has a digit that is a 2, there are two choices for the next digit, namely 1 or 3. Consider all 10- digit numbers starting in a 1. There is only one choice for the second digit since it must be a 2, then two choices for the third digit, then one for the fourth etc. Altogether there are such numbers. Similarly there are 16 such numbers starting in 3. However, if we consider numbers starting in 2, there are two choices for the second digit then only one choice for the third then two for the fourth etc. Altogether there are such numbers. Hence there are such numbers with the required property.1×2×1×2×1×2×1×2×1=16 2×1×2×1×2×1×2×1×2=32 16+16+32=64

17. 16 JM L KN P Q R Let points and be the points where the perpendiculars from to and meet the lines and extend line so it meets at , as shown in the diagram. Since is the bisector of angle , angle . Since angle is , triangle is isosceles and . Let the length of be cm. Hence the lengths of and are also cm. Observe that triangles and are similar since they have the same angles. Therefore and so since is positive. The length of is equal to the sum of the lengths of and . Therefore, the length of is cm. Hence, the value of is 16.PQ NMLKL PN J K R J N MJK NJR=45°JRN90°JRN JR=RN RN x J R PM x NKQ MNP 1 x= x 8x= 8xKL NP NR KL (8+ 8) a+b 18. 2 Consider the equation . Multiply each side by to get and so . Therefore . Hence or . Similarly, if we consider the equations and , then or and or respectively. Therefore, the possible values of when all three equations are satisfied simultaneously occur when , giving , or when , giving . Hence there are two possible values of . a b+c= b c+a(b+c)(c+a) a 2+ac=b 2+bc a 2Šb 2+acŠbc=0(aŠb)(a+b+c)=0 a=ba+b+c=0 b c+a= c a+b c a+b= a b+cb=ca+b+c=0c=aa+b+c=0 k a=b=ck= 1 2 a+b+c=0 k=Š1k 19. 100 Let the lengths of , and be , and centimetres respectively. Let the area of be and let the area of be . Note that is one quarter of the square which has as an edge. Hence . Next, using Pythagoras, . Hence . Therefore the area in of is .BC AB AC x y z AC D Ucm 2 ABC Vcm 2 AC D AC U= 1 4z2 z2=x 2+y 2=(x+y) 2Š2xy=20 2Š4V U= 1 4(400Š4V)=100ŠV cm 2 ABC D U+V=100 AD C B yx z (Note:Since the answer to the problem is independent of x and y, one could observe that the given properties of quadrilateral are satisfied by a square of side 10 cm which has area and conclude that this is therefore the required answer.) ABC D 100 cm 2 20. 247 First factorise twice using the difference of two squares i.e. . This shows that both 80 and 82 are divisors of in the required range. The question tells us that 193 is a divisor of N and, since 193 is prime, it must be a divisor of . Now observe that and that . Therefore or . N N=3 16 Š1= (38Š1 )(38+1 )= (34Š1 )(34+1 )(38+1 )=80×82× (38+1 ) N 3 8+1=81×81+1=6562 6562=2×3281 3281÷193=17N=80×82×2×17×193 N= (24×5 )× (2×41 )×2×17×193 Next consider the integers from 75 to 85 inclusive to see which could be divisors of . Because has no prime factors of 3 or 7, we know that 75, 77, 78, 81 and 84 are not divisors of while the initial argument established that 80 and 82 are divisors of . Both 79 and 83 are prime and so, since does not have a prime factor of 79, 83 or 19, these must also be excluded. This only leaves 85 to be considered. Note that and both 5 and 17 are prime factors of so 85 is a divisor of . Hence the divisors of in the required range are 80, 82 and 85 with sum 247.N N NN 76=4×19N 85=5×17NN N