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1 SASMO 2014 Round 1 Primary 6 Solutions 1. A clock takes 3 seconds to make 4 chimes. Assuming that the rate of chiming is constant, and the duration of each chime is negligible, how long does the clock take to make 8 chimes? Solution Time taken by the clock between 2 chimes = 3 s / 3 gaps = 1 s  time taken by the clock to make 8 chimes = 1 s  7 gaps = 7 s 2. If n is a whole number, for what values of n is also a whole number? Solution is a whole number if n is a factor of 36. 36 = 1  36 = 2  18 = 3  12 = 4  9 = 6  6  is a whole number if n = 1, 2, 3, 4, 6, 9, 12, 18, 36 . 3. Find the smallest whole number between 1 and 100 that is divisible by 12 and by 30. Solution Smallest whole number between 1 and 100 that is divisible by 12 and by 30 = LCM( 12, 30) = 60 4. A shop sells sweets where every 3 sweet wrappers can be exchanged for one more sweet. Catherine has enough money to buy only 26 sweets. What is the biggest number of sweets that she can get from the shop? Solution Method 1 26 sweets  26 wrappers  8 sweets and 2 wrappers  10 wrappers  3 sweets and 1 wrapper  4 wrappers  1 sweet and 2 wrappers  biggest no. of sweets = 26 + 8 + 3 + 1 = 3 8 [Common mistakes: 34, 37] Method 2 26 sweets  8 sweets and 2 wrappers  2 sweets and 4 wrappers  2 sweets  biggest no. of sweets = 26 + 8 + 2 + 2 = 3 8 [Common mistakes: 34, 36] Свалено от Klasirane.Com



3 Method 2 (Model Method) Duck Chicken The idea of average is to “even out”. This means that the 4-kg mark will divide 1.2 kg into 2 equal parts.  mass of duck = 4 kg + 0.6 kg = 4.6 kg Method 3 (Algebra) Let the mass of the chicken be x kg. Then the mass of the duck is ( x + 1.2) kg. Average mass of the duck and the chicken = = 4 2x + 1.2 = 8 2x = 6.8 x = 3.4  mass of duck = 3.4 + 1.2 = 4.6 kg 8. A whole number is between 50 and 100. When it is divided by 5, the remainder is 3. When it is divided by 7, the remainder is 5. Find the number. Solution Since the number leaves a remainder of 3 when divided by 5, then it must end with 3 or 8, i.e. the possible values of the number are 53, 58, 63, 68, …, 98. Since the number leaves a remainder of 3 when divided by 5, we can either check each of the above possible values one by one, or we can start with 7  7 + 5 = 54, and then add 7 until we get a number ending in 3 or 8, i.e. 54, 61, 68, …  the number is 68 . 9. A man travelled at 120 km/h for the first half of a 12 -km journey. Then he travelled at 6 0 km/h for the rest of his journey. What is his average speed for the whole journey? Solution Time taken for first half of the 12-km journey = 6 km / 120 km/h = 0.05 h Time taken for second half of the 12-km journey = 6 km / 60 km/h = 0.1 h  average speed for whole journey = 12 km / 0.15 h = 80 km/h Note: Common mistake is 90 km/h. Since speed = distance / time, if you want to average speeds, then the base (in this case, the time taken) must be the same. But instead the distance travelled for both parts of the journey is the same, so you cannot average speeds. 1.2 4 kg Свалено от Klasirane.Com

4 10. Amy buys an item for a 2 0% discount during a sale, but she still needs to pay a 5% GST (Goods and Services Tax). She is given two options. Option A: 20% discount first, then add 5% GST Option B: Add 5% GST first, then 20% discount Which option is cheaper for Amy? Or does it not matter? Solution Option A: 0.8  1.05  Sales Price Option B: 1.05  0.8  Sales Price  both options are the same, so it does not matter. Note: Common mistake is Option A. If you pay 5% GST first (Option B), although this GST is bigger than the GST in Option A, this GST is also discounted 20%, so it ends up the same as if you pay the GST after the discount (Option A). Notice that the base is the same (in this case, the selling price). 11. The digits 3, 4, 5, 6 and 7 are all used to write a five-digit number ABCDE where the three-digit number ABC is divisible by 4, BCD is divisible by 5 and CDE is divisible by 3. Find all the possible values of the five-digit number ABCDE. Solution Since BCD divisible by 5, then D = 5. Since ABC divisible by 4, then BC is also divisible by 4, i.e. BC = 36 or 64. Since CDE is divisible by 3, then C + D + E = C + 5 + E is also divisible by 3. If BC = 36, then C + 5 + E = 6 + 5 + E = 11 + E is also divisible by 3, i.e. E = 4 or 7, so A = 7 or 4 respectively. If BC = 64, then C + 5 + E = 4 + 5 + E = 9 + E is also divisible by 3, i.e. E = 3, so A = 7.  the possible values of ABCDE are 43657, 73654 and 76453. 12. At a workshop, there are 20 participants. Each of them shakes hand once with one another. How many handshakes are there? Solution The first participant will shake hand with 19 other participants; the second participant will shake hand with 18 other participants; the third participant will shake hand with 17 other participants; etc. Thus total no. of handshakes = 19 + 18 + 17 + … + 3 + 2 + 1 Method 1 1 + 19 = 20 2 + 18 = 20 3 + 17 = 20 9 pairs Свалено от Klasirane.Com

5 9 + 11 = 20 10  total no. of handshakes = 20  9 + 10 = 190 Method 2 Since 1 + 2 + 3 + … + n = , then total no. of handshakes = = 190 13. In the following cryptarithm, different letters represent different digits and an asterisk * may represent any digit. The product MATH is a four-digit number less than 5000. What number does MATH represent? E H  E H * * 6 + * * 8 M A T H Solution In the ones column, H  H = _6 implies H = 6. In the tens column, E  H = E  6 = _8 implies E = 3 or 8. If E = 8, then 86  86 > 5000. So E = 3.  MATH = 36  36 = 1296 . 14. There are 100 buns to be shared among 100 monks (consisting of senior and junior monks). The senior monks get 3 buns each and every 3 junior monks share 1 bun. How many senior monks are there? Solution Method 1 (By Observation) Observe that if you group one senior monk and 3 junior monks together, they will get a total of 4 buns. And 25 such groups will give a total of 100 monks and 100 buns.  there are 25 senior monks . Method 2 (Algebra) Let the no. of senior monks be x. Then the no. of junior monks is 100  x. Total no. of buns = 3 x + = 100 9x + (100  x) = 300 Свалено от Klasirane.Com

6 8x = 200 x = 25  there are 25 senior monks . 15. What is the least number of cuts required to cut 5 identical sausages so that they can be shared equally among 9 people? Solution Fraction of sausage each person will get = This means that there must be at least 5 cuts since no one will get one whole sausage. Cut each of the 5 sausages at the -mark. Then 5 people will get one sausage each. There are 5 times sausages left. But each person must get of a sausage. This means that one of the sausages must be cut into 4 equal parts each (i.e. 3 cuts each) so that the remaining 4 people will get one sausage and one sausage each.  least no. of cuts = 5 + 3 = 8 16. How many digits are there before the 50 th ‘8’ of the following number? 85855855585555855555… Solution No. of ‘8’s before the 50th ‘8’ of the number = 49 No. of ‘5’s before the 50th ‘8’ of the number = 1 + 2 + 3 + … + 49 Method 1 1 + 49 = 50 2 + 48 = 50 3 + 47 = 50 24 + 26 = 50 25  total no. of digits before the 50 th ‘8’ = 49 + 50  24 + 25 = 1274 Method 2 Since 1 + 2 + 3 + … + n = , then total no. of digits before the 50 th ‘8’ of the number = 49 + = 1274 24 pairs Свалено от Klasirane.Com

7 17. Find the total number of triangles in the diagram. Solution Label the vertices as shown in the diagram. These are the vertices that form a triangle: 123, 124, 125, 126, 134, 135, 137, 145, 156, 157, 158, 167, 168, 236, 256, 347, 356, 357, 367, 457, 458, 568, 578, 678  there is a total of 2 4 triangles in the diagram. 18. A teacher has a bag of sweets to treat her class. If she gave 5 sweets to each student, then she would have 13 sweets left. If she gave 8 sweets to each student, then she would have 20 sweets short. How many students and how many sweets are there? Solution If the teacher gave 5 sweets to each student, then she would have 13 sweets left. But 13 sweets left are not enough to give 3 more sweets to each student since she would have 20 sweets short. So if she had 20 more sweets, then she could give 20 + 13 = 33 sweets to the students so that each student has 3 more sweets, i.e. 8 sweets each. Since each student receives only 3 more sweets (i.e. the 6 th , 7 th and 8 th sweets), then there are 33  3 = 11 students .  there are 11  5 + 13 = 68 sweets. 19. Find the value of . Solution Method 1 Observe the following pattern:  . 1 2 3 4 5 6 7 8 Свалено от Klasirane.Com

8 Method 2 Since for all n > 1, then 20. Susie has 23 coins. She divides them into 5 piles so that each pile has a different number of coins. Find the smallest possible number of coins in the biggest pile. Solution For each pile to have a different number of coins, and the biggest pile to have the smallest possible number of coins, put 1 coin in the 1 st pile, 2 coins in the 2 nd pile, 3 coins in the 3 rd pile, 4 coins in the 4 th pile and 5 coins in the 5 th pile. So the biggest pile is the 5 th pile, but there are only 1 + 2 + 3 + 4 + 5 = 15 coins. The 16 th coin will have to go to the 5 th pile so that each pile will have a different number of coins. The 17 th coin cannot go to the 5 th pile because we want to find the smallest possible number of coins in the biggest pile, so the 17 th coin will have to go to the 4 th pile. Similarly, the 18 th , 19 th and 20 th coins will go to the 3 rd , 2 nd and 1 st piles respectively. The 21 st coin will go to the 5 th pile again, the 22 nd coin to the 4 th pile, and the 23 rd coin to the 3 rd pile.  the largest pile (which is the 5 th pile) will contain 5 + 1 + 1 = 7 coins. 21. Find the next term of the following sequence: 1, 4, 9, 7, 7, 9, 4, 1, 9, 1, … Solution Observe that the first 3 terms are perfect squares: 1 2 , 2 2 and 3 2 . But subsequent terms are not perfect squares. However, if you compare the given sequence with perfect squares (1, 4, 9, 16, 25, 36, 49, 64, 81, 100, …), you will observe that the n -th term can be found by squaring n and then adding the digits continuously until a single-digit number is obtained.  the next term, which is the 11 th term, is obtained by: 11 2 = 121  1 + 2 + 1 = 4 . 22. Given that 5! means 5  4  3  2  1, find the last two digits of 2014!. Solution Since 2014! contains the factor 100, and any number multiplied by 100 will have 00 as the last two digits, then the last two digits of 2014! are 0 0. Свалено от Klasirane.Com

9 23. There were 210 students in the hall. of the boys and of the girls wore T-shirts. If there were 78 students in the hall who wore T-shirts, how many boys were in the hall? Solution Method 1 (Model Method) The first model shows of the boys and of the girls weari ng T-shirts (indicated by solid boxes, not the dotted ones). We will call an unshaded box from the boys as B- box, and a shaded box from the girls as G-box. Boys Girls The 2 solid B-boxes and the one solid G -box (i.e. B 1 + B 1 + G 1) represent a total of 78 students. If you take 2 dotted B-boxes and 1 dotted G-box (i.e. B 2 + B 2 + G 2), the 3 dotted boxes also represent a total of 78 students. But you can’t do anything with the remaining 2 dotted boxes. So we double all the boxes as shown in the second model, where the total no. of students is 210  2 = 420. Boys Girls From the second model, we are able to form 5 groups of 2 B-boxes and 1 G-box (i.e. from B 1 and G 1 to B 5 and G 5), leaving behind only one G-box, where each group of 2 B-boxes and 1 G-box represent a total of 78 students. This means that the last G-box represents 420  5  78 = 420  390 = 30 students.  there are 30  3 = 90 girls and 210  90 = 120 boys in the hall. Method 2 (Algebra) Let the no. of boys be x. Then the no. of girls is 210  x. Total no. of students who wore T-shirts = x + (210  x) = 78 6 x + 5(210  x) = 78  15 6 x + 1050  5x = 1170 x = 120  there are 120 boys. 210 420 B 1 B 1 B 2 B 2 G 1 G 2 B 1 B 1 B 2 B 2 B 3 B 3 B 4 B 4 B 5 B 5 G 1 G 2 G 3 G 4 G 5 Свалено от Klasirane.Com 52 3 1 52 3 1 52 3 1

10 24. A farmer wants to plant 10 trees in 5 rows such that there are exactly 4 trees in each row. Draw a diagram to show how the trees should be planted. Solution If you start with 10 trees and then try fitting 5 lines onto them, you will realise that the lines must overlap because there are not enough trees, and it’s not easy to fit 5 lines onto 10 trees.  try to dr aw 5 overlapping lines first, and a common figure with 5 overlapping lines is the following star: Then put in the trees and yes, it works. 25. Frank knows 5 women: Amy, Betty, Cheryl, Doris and Elaine. a . 3 women are under 30 and the other 2 women are over 30. b . 3 women are nurses and the other 2 women are teachers. c . Amy and Cheryl are in the same age bracket. d . Doris and Elaine are in different age brackets. e . Betty and Elaine have the same occupation. f . Cheryl and Doris have different occupations. g . Of the 5 women, Frank will marry the teacher over 30. Who will Frank marry? Solution Conditions a , c and d  Amy and Cheryl are under 30, and Betty is over 30. Conditions b , e and f  Betty and Elaine are nurses, and Amy is a teacher. 3 women under 30 2 women over 30 3 Nurses 2 Teachers Amy Cheryl ? Betty ? Betty Elaine ? Amy ? So Amy is a teacher under 30. Conditions b and g  the other teacher must be over 30. Condition f  Cheryl or Doris is the other teacher over 30. But Cheryl is under 30, so Doris is the other teacher over 30.  F rank will marry Doris. Свалено от Klasirane.Com