UK JUNIOR MATHEMATICAL CHALLENGE TUESDAY 27TH APRIL 1999 Organised by the United Kingdom Mathematics Trust from the School of Mathematics, University of Leeds On S@qa% SOLUTIONS LEAFLET This is the first solutions leaflet which has been provided for a JMC. It is sent in the hope that it might provide all concerned with some alternative solutions to the ones they have obtained. It is not intended to be definitive. The organisers would be very pleased to receive alternatives created by candidates.10. 11. 12. 13. 14, 15. 16. 17. Thirty-four hundredths = 0.34. 200ml is approximately one third of a pint or just under two thirds of the volume of a typical can of soft drink (330ml). The hour hand will be half way between the ‘one’ and the ‘two’ whilst the minute hand will still point to the ‘six’. 7000000 is exactly divisible by 7, so we need only to calculate the remainder when 10 is divided by 7. ‘There must be seven 26p stamps to give a total ending in 2. 7 x 26p = £1.82 and therefore there is only one 20p stamp. 19 +99 + 19x99 =99 + 19x 1+ 19x 99 = 99 + 19 x 100 = 99 + 1900 or 19 +99 + 19 x 99 = 19+99 x 1499 x 19 = 19+ 99 x 20 = 19 + 1980 There are eight people, each of whom buys seven eggs each. Total debt = £55 x 55 = £3025. £50 x 50 + £5 x 5 = £2525. I have £500 too little. Only 21 of the 22 slices may be used and these will make seven ‘double- decker’ sandwiches. Each sandwich requires four sides of bread to be buttered. Laa-laa's luggage is 15kg overweight. The limit above which a charge is made is therefore 35kg. Po's luggage weighs less than this and therefore she will not be charged. 2434547411 413417419 +23 = 100, (By coincidence, the number of prime numbers less than 100 is 25.) There are 3 routes from S to U and 2 from U to V. The number of different routes from S to V is therefore 3x 2 = 6. Each of these may be followed by any one of three different routes from V to 7, making a total of 6 x 3 = 18 routes in all. ZPMR = 110° (adjacent angles on a straight line); ZPRM = 50° (angle sum of a triangle) ZPRS = 130° (adjacent angles on a straight line), (A shorter method uses the exterior angle —_—theorem: ZPRS = ZMPR+ ZPMR.) Let Ross drink x ml. Then Rachel drinks 3x ml. $x +.x = 750 => $x = 750 => x = 300. Rachel, therefore, drinks 450ml. The distance the snail moves East = 2—4+6-8+10=6, The distance the snail moves North= 1-3+5-—7+9=5. -14+4=3; 2+2=4, 14+6=5, 24+6=8 lity by 11 is to add alternate 1+34*+7=11+*%; 24+44+6+8=20, If the original number is a multiple of 11 then these two totals will be the same or will differ by a multiple of 11. In this case, 11 +*=20 > *=9, Or, you can solve it without knowing a rule as follows:— 1234*678 = 12340678 + 1000* = (11 x 1121879 +9) +11 x 90* + 10* and hence is divisible by 11 <=> 10* + 9 is divisible by 11. So* = 9,18. 19. 20. 21. 22, 25, 6729 6729 x 2 = 13458 and therefore 13458 digits from 1 to 9 inclusive. The required fold lines are shown (A folds to A’). is equal to one half and uses all the x Yas Zz a , At / | ue m ’ f ’ A - A With the front face as shown, the cubes A ~ D would appear as shown: Lf 2H) GE —_ A (e B c D iy A | N i § of Granny's age is 84. Her age, therefore, is 84 x 3 = 98. The hands are at right angles to each other twice in every hour except that the only time between 2.00 and 3.00 is just after 2.27 and the only time between 8.00 and 9.00 is just after 8.27. In the diagram the triangles labelled X can be reached from ¢ in one ‘move’; those labelled Y are triangles which are one ‘move’ away from ©. This shows that it is not possible to go from * to © in one ‘move’, nor in two ‘moves’ since none of the triangles are labelled both X and Y. However, some of the triangles labelled Y may be reached from a triangle labelled X in one ‘move’ and therefore it is possible to complete the required task in three ‘moves’ Let the time which Boris takes to run up five steps be T. Therefore: Boris takes 27 to run up the 99 steps; Spike takes #7 to run up the (99 — 21) steps; Percival takes $7 to run up the (99 — 38) steps. 2% < ® < 4Land therefore the order is Spike, Boris, Percival. The multiplication sum is 43 x 29. There cannot be a carry from the units column into the tens column in the final addition and this tells us we have 860 on the fourth line. This must come from 43 multiplied by a 43 number between 20 and 29 inclusive. The only single digit x 2.9 which multiplies 43 to give a three digit number with 8 as 387 the middle digit (third row) is 9 and therefore we have 860 v * a