Download [PDF] 2017 UKMT Junior Kangaroo Solutions Mathematical Challenge United Kingdom Mathematics Trust

File Information

Filename:	[PDF] 2017 UKMT Junior Kangaroo Solutions Mathematical Challenge United Kingdom Mathematics Trust.pdf
Filesize:	76.61 KB
Uploaded:	09/08/2021 18:02:29
Keywords:	2017 ukmt junior kangaroo solutions mathematical challenge united kingdom mathematics trust pdf
Description:	Download file or read online UKMT past exam paper junior mathematical challenge JMC junior kangaroo 2017 solutions - United Kingdom Mathematics Trust
Downloads:	1

File Preview

Download Urls

Short Page Link

https://www.edufilestorage.com/3dv

Full Page Link

https://www.edufilestorage.com/3dv/PDF_2017_UKMT_Junior_Kangaroo_Solutions_Mathematical_Challenge_United_Kingdom_Mathematics_Trust.pdf

HTML Code

<a href="https://www.edufilestorage.com/3dv/PDF_2017_UKMT_Junior_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/2369/pdf/150x190_middle_46f4e7862b1eb5bd4935adbbba5d79e8.jpg"/></a>

Forum Code

[url=https://www.edufilestorage.com/3dv/PDF_2017_UKMT_Junior_Kangaroo_Solutions_Mathematical_Challenge_United_Kingdom_Mathematics_Trust.pdf][img]https://www.edufilestorage.com/cache/plugins/filepreviewer/2369/pdf/150x190_middle_46f4e7862b1eb5bd4935adbbba5d79e8.jpg[/img][/url]

[PDF] 2017 UKMT Junior Kangaroo Solutions Mathematical Challenge United Kingdom Mathematics Trust.pdf | Plain Text

Tuesday 13th June 2017 Junior Kangaroo Solutions 1. D Kieran makes 4 jumps in 6 seconds so makes 2 jumps in 3 seconds. Therefore it will take him seconds to make 30 jumps.(30÷2)×3 seconds=45 2. C Label the numbers to be written in the cells of the grid as shown. da eb c f 1 21 Each row and column contains the digits 1, 2 and 3 exactly once. Hence . Therefore (and and for completeness). Hence the sum of the digits in the shaded cells is .c=d=3 b=e=2a=3f=1 2+2=4 3. B The number of small cubes along each edge of the large cube is . Therefore Ben has small cubes in total. Hence the row he forms is long. Since there are 100 cm in 1 m, his row is 400 m long.100÷5=20 20×20×20=8000 8000×5cm=40 000 cm 4. B The smallest and largest totals Beattie can obtain are and respectively. Totals of 10 and 11 can also be obtained, for example from and . Therefore, since all Beattie's totals will be integers, she can obtain four different totals.1+3+5=9 2+4+6=12 2+3+5=10 1+4+6=11 5. E When Anna is 13, Annie is and so Annie is 26 years older than Anna. When Anna is three times as old as she is now, she will be 39 and Annie will still be 26 years older. Therefore Annie will be 65.3×13=39 6. C Let Hasan's two-digit number be , which is equal to . The four-digit number he forms is therefore , which is equal to and hence to‘ab’10a+b ‘abab’ 1000a+100b+10a+b . Therefore the ratio of his four-digit number to his two-digit number is 101 : 1. 100(10a+b)+10a+b=101×(10a+b) 7. E The length of the edge of Charlie's original square is . Since he cuts his square into two rectangles, he cuts parallel to one side of the square to create two rectangles each with two sides 5 cm long as shown in the diagram. (20÷4)cm=5cm Hence the total perimeter of his two rectangles is longer than the perimeter of his square. Since the perimeter of one of the rectangles is 16 cm, the perimeter of the other rectangle is . 2×5cm=10 cm (20+10−16)cm=14 cm 5 cm 5 cm 5 cm 5 cm 8. E Let the number of birds remaining in each tree be . Therefore , which has solution . Hence the number of birds originally perched in the second tree is .x x+6+x+8+x+4=60x=14 14+8=22 9. A The two longest diagonals of an square grid each contain squares. When is an odd number, the two diagonals meet at the square in the centre of the grid and hence there are squares in total on the diagonals. Alex coloured 2017 squares and hence , which has solution . Therefore the size of the square grid is .n×nnn 2nŠ1 2nŠ1=2017n=1009 1009×1009

10. A The sequence KANGAROOKANGAROOKANG... repeats every 8 letters. Since , the 2017th letter in the sequence is the first of the repeating sequence and hence is K. 2017=8×252+1 11. D On each net, label the four vertices of the right-hand square 1, 2, 3 and 4 as shown. Also label any vertex on any of the other squares that will meet vertices 1, 2, 3 or 4 when the net of the cube is assembled into a cube with the corresponding value. A 1 1 2 22 3 3 34 4 B 1 1 2 22 3 3 34 4 C 1 1 2 22 3 3 34 4 D 1 1 2 22 3 3 34 4 Since there are three vertices of the original cube at which two diagonals meet, to be a suitable net for the cube shown, any diagonal drawn meets another diagonal at a vertex with the same label. As can be seen, only in net D are the ends of the diagonals at vertices with the same label. Therefore Usman could only use net D to make the cube shown. 12. A Let the lengths of the four rectangles be cm, cm, cm and cm with . The lines Maddie draws join the centres of two pairs of rectangles and hence have total length . Therefore the sum of the lengths of the lines she draws is .pqr s p+q+r+s=36 (1 2p+ 1 2q)cm+ (1 2r+ 1 2s)cm= 1 2(p+q+r+s )cm 1 2 ×36 cm=18 cm 13. D Let the size in degrees of and of be and . Therefore the size of and of are and respectively. Since the angles between parallel lines (sometimes called co-interior or allied angles) add to , we have . This has solution . Similarly and hence . Therefore the value of is 5.PQRQRS x kxSPQ RSP2x2×2x=4x 180°2x+4x=180 x=30x+kx=180 30k=150 k 14. C Let Taran's original number be . When he multiplied it, he obtained either or . When Krishna added 5 or 6, his answer was one of , , or . Finally, when Eshan subtracted 5 or 6, his answer was one of , , , , , , or . Since the final result was 73 and since 73 is neither a multiple of 5 or 6, nor 1 less than a multiple of 5 or 6, nor 1 more than a multiple of 5, the only suitable expression for the answer is . The equation has solution . Hence the number Taran chose is 12.x5x6x 5x+55x+66x+56x+6 5x5x+16x6x+1 5xŠ15x6xŠ16x 6x+16x+1=73 x=12 15. E Consider the two unshaded triangles. Each has height equal to 12 cm and hence their total area is . Therefore the shaded area is . (1 2×PQ×12+ 1 2×QR×12 )cm 2=6×(PQ+QR)cm 2=6×20 cm 2=120 cm 2 (20×12Š120)cm 2 =120 cm 2 16. D The path indicated follows three sides of each of the squares shown. The sum of the lengths of one side of each square is equal to the length of , which is 24 cm. Therefore the length of the path is .PQ 3×24 cm=72 cm 17. B Let the length of the shortest ribbon be . Therefore the lengths of the other ribbons are , and . The perimeter of the first shape (starting from the lower left corner and working clockwise) is while the perimeter of the second shape (again starting from the lower left corner) is . Hence the difference between the two perimeters is .xcm (x+25)cm(x+50)cm(x+75)cm (x+10+25+10+25+10+25+10+x+75+40)cm=(2x+230)cm (x+50+10+25+10+50+10+75+10+x+40)cm=(2x+280)cm (2x+280)cmŠ(2x+230)cm =50 cm

18. E Draw in lines and as shown. Since angles in a triangle add to and we are given and , we obtain . Therefore is isosceles and hence . Therefore, since as it is a side of the square, is also isosceles. Since and PT T S 180°SPT=75°TSP=30° PT S=75°PT S TS=PS=10 cmRS=10 cm RS TRSP=90° P QRS T 10 cm , we have . Therefore is isosceles with one angle equal to . Hence is equilateral and therefore the length of is 10 cm. TSP=30°RST=60°RS T 60°RS T T R 19. A Let the length of a side of and of be . Consider quadrilateral .PQRS W X Y Z xcmQXRW P Q R S W X YZ The diagonals and are perpendicular and of length . Therefore the area of is half the area of a rectangle with sides equal in length to and and hence is equal to . Similarly, the area of quadrilateral is also . Therefore the total shaded area is . However, the question tells us that the shaded area is equal to . Therefore . Hence the area of is .QR W X xcm QXRW QR W X 1 2 ×QR×WX= 1 2x2cm 2 SW RZ 1 2x2cm 2 x2cm 2 1cm 2 x2 =1PQRS1cm 2 20. C Note first that . Therefore either the two-digit number or the three-digit number is a multiple of 53. Since the multiplication uses each of the digits 1 to 9 once and 7632 contains a 3, the option is not allowable. Hence we need to find a three-digit multiple of 53 that does not share any digits with 7632 and divides into 7632 leaving an answer that also does not share any digits with 7632. We can reject since it contains a 6 but is a possibility. The value of is which does not have any digits in common with 7632 nor with 159. We can also check that no other multiple of 53 will work. Therefore the required multiplication is and hence the value of is 5.7632=2×2×2×2×3×3×53 de=53abc de=53 2×53=106 3×53=159 7632÷159 2×2×2×2×3=48 159×48=7632 b 21. B The information in the question tells us that the numbers on touching faces of the solid are the same and that numbers on opposite faces of a die add to 7. Since the number 4 is visible on the rear of the right-hand side of the solid, there is a 3 on the left-hand face of the rear right die and hence a 3 and a 4 on the right- and left- 1234 ? 6 hand faces of the rear left die. Similarly, since the number 1 is visible on the left-hand side of the front of the solid, there is a 6 and a 1 on the front and back faces of the rear left die. Therefore the top and bottom faces of the rear left die have a 2 and a 5 written on them. Since the four dice are identical, comparison with the front right die of the solid tells us that a die with a 6 on its front face and a 3 on its right-hand face has a 2 on its lower face and hence a 5 on its upper face. 22. C The possible groups of three integers with product 36 are (1, 1, 36), (1, 2, 18), (1, 3, 12), (1, 4, 9), (1, 6, 6), (2, 2, 9), (2, 3, 6) and (3, 3, 4) with sums 38, 21, 16, 14, 13, 13, 11 and 10 respectively. The only value for the sum that occurs twice is 13. Hence, since Topaz does not know what the three integers chosen are, the sum of Harriet's three integers is 13.

23. E Since the triangle formed when the trapeziums are put together is equilateral, the smaller angles in the isosceles trapeziums are both . Consider one trapezium split into a parallelogram and a triangle as shown.60° a x b bŠa Since the original trapezium contains two base angles of , the triangle also contains two base angles of . Hence the triangle is equilateral and has side length . Now consider the large equilateral triangle with the hole. The perimeter of the hole is where is the length of the shortest sides of the trapezium. Therefore the perimeter of the hole is .60° 60° (bŠa ) 3(aŠx ) x 3 (aŠ(bŠa) ) =3(2aŠb)=6aŠ3b 24. A Let the number of pencils Zain takes on Monday and Tuesday be and respectively. Therefore . Hence, when we multiply the equation through by 6 to eliminate the fractions and simplify, we obtain . Since and are both positive integers and since the units digit of is 0, the units digit of is 6 and hence . Therefore and hence the number of pencils Zain takes is . Therefore the number of pencils Jacob takes is .xy x+ 2 3x+y+ 1 2y=21 10x+9y=126xy 10x9y y=4x=9 9+4=13 21Š13=8 25. E Let the three-digit number be . Since each suitable number is 34 times the sum of its digits, we have . Therefore . Since the left-hand side of this equation is a multiple of 11, the right- hand side is also a multiple of 11 and hence . Therefore and hence . Therefore the three-digit numbers with the required property are 102, 204, 306 and 408 and hence there are four three-digit numbers with the required property.100a+10b+c 100a+10b+c=34(a+b+c) 66aŠ33c=24b b=066aŠ33c=0 c=2a