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Tuesday 14th June 2016 Junior Kangaroo Solutions 1. D At all the times given, the minute hand is pointing to 12. When the minute hand is pointing to 12 and the angle between the hands is , the hour hand has turned of a complete turn. Therefore the hour hand will point at 5 and the time will be 5 pm. (There are other times when the angle between the hands is but, of these, only at 7 pm does the minute hand point to 12 and 7 pm is not one of the times given.)150° 150 360 = 5 12 150° 2. B The number of people who can sit on each side of the square table is . When eight of these tables are arranged to make a long rectangular table, there will be room for people on each long side and for three extra people at each end. Hence, the number of people that can sit round the long table is .12÷4=3 8×3=24 2×24+2×3=48+6=54 3. E Since one ball and one bat cost £90, two balls and two bats cost . Now, since three balls and two bats cost £210, one ball costs . Therefore a bat costs .2×£90=£180 £210Š£180=£30 £90Š£30=£60 4. A The surface areas of the two solids are the same. Hence the same amount of paint is required to cover them. Therefore it would take 9 litres of paint to cover the surface of the second solid. 5. B The calculation is equivalent to . 1 10× 3 10× 5 10×7000=1×3×5×7=105 6. E Let the number of pupils in the class be . The information in the question tells us that , which has solution . Hence the number of sheets of paper Miss Spelling has is .x 3x+31=4x+8x=23 3×23+31=100 7. C Nets A and D would produce cubes with holes on two edges of the same face. Net E would produce a cube with a hole in the centre of two opposite faces while net B would produce a cube with one hole on an edge and two small holes. The given partial cube has holes on two opposite edges and therefore its net will have a hole on the edge of four different faces. Hence only net C can be used to build the required shape. 8. E Since one angle of the isosceles triangle is , there are two possibilities. Either the other two angles are equal, in which case the difference between them is , or one of the other angles is . In this case, since angles in a triangle add to , the second missing angle is and hence the difference between the two missing angles is .30° 0° 30°180° 120° 120°Š30°=90° 9. E Label vertices , , , , , and as shown. Since the hexagon is regular, it can be divided into six equilateral triangles as shown. Therefore quadrilateral is a rhombus and hence its diagonal is a line of symmetry. Therefore, if vertex is folded onto , the fold will be along . Similarly, if vertices and are folded onto , the ABC DEE F OABC AC BO AC D F O O A BCD E F folds will be along and respectively. Hence the figure that is formed will be a triangle and, since all three of the rhombuses , and are made out of two congruent equilateral triangles, the lengths of their diagonals and will be equal. Hence the shape that is formed is an equilateral triangle.CE EA OABC OCDE OEFA AC,CE EA AC E

10. D The radius of each of the circles is 5 cm and hence the diameter of each is 10 cm. The length of the side of the square is equal to the sum of the diameters of two circles and hence is equal to 20 cm. The length of each side of the equilateral triangle is equal to the length of the side of the square. Hence the perimeter of the star, which is made up of eight sides of congruent equilateral triangles, is . 8×20 cm=160 cm 11. B Joey's two numbers are 99 and 12 and hence his sum is 111. Zoë's two numbers are 98 and 10 and hence her sum is 108. Therefore the difference between their answers is . 111Š108=3 12. C A BCD E F G H Since is the midpoint of and is the midpoint of , the length of is . Similarly, since is the midpoint of and is the midpoint of , the length of is . Therefore the area of the shaded rectangle is .EADFAEFE1 2 × 1 2 ×4cm=1cmGABH AG H G 1 2 × 1 2 ×1cm= 1 4cm (1× 1 4)cm 2 = 1 4cm 2 13. B Let the units digit of the number be . Hence the tens digit of the number is and the sum of the digits of the number is . The information in the question tells us that . Hence which has solution . Therefore the sum of the digits of the two-digit number is .xx+3 2x+3 10(x+3)+x=7(2x+3)+311x+30=14x+24 x=22×2+3=7 14. B Consider the case where two opposite faces are coloured red. Whichever of the four remaining faces is also coloured red, the resulting arrangement is equivalent under rotation to a cube with top, bottom and front faces coloured red. Hence, there is only one distinct colouring of a cube consisting of three red and three blue faces with two opposite faces coloured red. Now consider the case where no two opposite faces are coloured red. This is only possible when the three red faces share a common vertex and, however these faces are arranged, the resulting arrangement is equivalent under rotation to a cube with top, front and right-hand faces coloured red. Hence there is also only one distinct colouring of a cube consisting of three red and three blue faces in which no two opposite faces are coloured red. Therefore there are exactly two different colourings of the cube as described in the question. 15. C The diagonals of a rectangle bisect each other at the midpoint of the rectangle. Hence, the midpoint of a rectangle is equidistant from all four vertices and is the centre of a circle through its vertices. In this case, the diameter of the circle is 10 cm. This is equal to the sum of the lengths of the diagonals of four of the smaller rectangles. Hence the diagonal of each small rectangle has length 2.5 cm. The perimeter of the marked shape is made up of eight diagonals of the small rectangles and hence has length .8×2.5cm=20 cm

16. D Since Leonhard's walk always goes over bridge 1 first, it must conclude by going over bridge 5 to enable him to reach to B. Note also that bridges 2 and 6 must be crossed consecutively, in some order, as they are the only way to get to and from the opposite bank to the one from which he started and is to finish and so can be considered together. 1 23 45 6 A B Hence the number of days he can walk without repeating the order in which he crosses the bridges is the same as the number of ways of choosing ordered crossings of bridges 3, 4 and the pair 2 and 6. These can be chosen in six different ways (three choices for the first bridge, two for the second and then only one choice for the third). Hence Leonhard can walk for six days without repeating the order in which he crosses the bridges. (The six orders are 126345, 126435, 134265, 136245, 143265 and 146235.) 17. A Let the length of each of the rectangles be cm and the width be cm. The perimeter of each of the rectangles is 40 cm and hence . Therefore . From the diagram we can see that the length of each side of the square is . Therefore the square has side length 20 cm. Hence the area of is .xy 2x+2y=40x+y=20 ABC D(x+y)cm ABC D ABC D (20×20 )cm 2 =400 cm 2 18. D Let the cost of a can of cola be pence and the cost of a croissant be pence. The information in the question tells us that and that both sides of the equation represent the total amount of money Ellen has. Hence . Therefore the total amount of money she has is . Hence she could buy 16 croissants if she bought only croissants.xy 6x+7y=8x+4y 3y=2x 3×3y+7ypence=16ypence 19. C If each person paid their fair share, each would have paid five times. Therefore Adam has paid on an extra three occasions and Bill has paid on an extra two occasions. Hence the £30 Chris owes should be divided in the ratio 3:2. Therefore Adam should get and Bill should get . 3 5×£30=£18 2 5 ×£30=£12 20. D Consider one of the right-angled isosceles triangles as shown. The longest side is . The triangle can be divided into two identical right-angled isosceles triangles with base 3 cm and hence with height 3 cm. Therefore the area of each of the original triangles is . Hence the total shaded area is .(30 / 5)cm=6cm (1 2 ×6×3 )cm 2 =9cm 2 5×9cm 2 =45 cm 2 45°45° 3cm 3cm3cm 21. C The information that in any group of four pencils, at least two have the same colour, tells us that there at most three different coloured pencils in Carl's pencil case. The information that in any group of five pencils, at most three have the same colour, tells us that there are at most three pencils of any single colour in the pencil case. Hence there are three pencils of each of the three different colours and so Carl's pencil case contains three blue pencils.

22. C Let the distance from London to Brighton be miles. Since time = distance/speed, the times Lewis spent on the two parts of his journey are hours and hours. Hence the total time in hours that he travelled isd d 60 d 40 d 60+ d 40= 2d+3d 120= 5d 120= d 24. Therefore his average speed for the whole journey is .2d÷ (d 24)mph=48 mph 23. E a b c +a c b c 4 a. Since is the digit in the hundreds column of the answer, we can deduce that . Therefore, there must be a carry from the units column to the tens column and hence . Since there will also be a carry from the tens column to the hundreds column, we have . Therefore, and hence . Therefore the value of is .cc>a a=4Š1=3 c=a+a+1=77+b=13b=6 a+b+c3+6+7=16 24. A Consider the quadrilateral as shown with , and .PQRS PQ=QR=RSRQP=60° SRQ=100° PQ R S60° 100° Draw line . Since and , triangle is equilateral and hence and . Since , . Since , triangle is isosceles and hence . Therefore the largest angle of the quadrilateral is .PR PQ=QRPQR=60°PQR PR=PQ=QR=RSPRQ=60°SRQ=100° SRP=100°Š60°=40°PR=RS PRS RPS=PSR= 1 2(180°Š40° ) =70° QPS=70°+60°=130° 25. C There is no number that is both a multiple of three and a multiple of four without also being a multiple of two. Hence, the numbers underlined exactly twice are those that are a multiple of two and of three but not of four and those that are a multiple of two and four but not of three. The first set of numbers consists of the set of odd multiples of six. Since , there are 336 multiples of 6 in the list of numbers and hence odd multiples of six that would be underlined in red and blue but not green. The second set of numbers consists of two out of every three multiples of four and, since , there are numbers that would be underlined in red and green but not blue. Hence there are numbers that Moritz would underline exactly twice. 2016÷6=336 336÷2=168 2016÷4=504 2 3 ×504=336 168+336=504