[PDF] SASMO 2014 Round 1 Primary 5 Solutions.pdf | Plain Text

1 SASMO 2014 Round 1 Primary 5 Solutions 1. 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 24. 24 = 1  24 = 2  12 = 3  8 = 4  6  is a whole number if n = 1, 2, 3, 4, 6, 8, 12, 24 . 2. A textbook is opened at random. To what pages is it opened if the product of the facing pages is 600? Solution Since 25  25 = 625, try 24  25 = 600.  the pages are 2 4 and 25. 3. The diagram shows a quadrant OAB of a circle with centre O. OPQR is a rectangle. Given that PR = 7 cm, find the length of OA. Solution OA = OQ (radii of quadrant) = PR (diagonal of rectangle) = 7 cm 4. Find an even number between 300 and 400 that is divisible by 5 and by 7. Solution Since 2, 5 and 7 are relatively prime, if a number is divisible by 2, by 5 and by 7, then the number is divisible by 2  5  7 = 70, i.e. the number is a multiple of 70.  an even number between 300 and 400 that is divisible by 5 and by 7 is 350. A B O P Q R Свалено от Klasirane.Com

2 5. A shop sells sweets where every 3 sweet wrappers can be exchanged for one more sweet. Navin has enough money to buy only 29 sweets. What is the biggest number of sweets that he can get from the shop? Solution 29 sweets  29 wrappers  9 sweets and 2 wrappers  11 wrappers  3 sweets and 2 wrappers  5 wrappers  1 sweet and 2 wrappers  3 wrappers  1 sweet  biggest no. of sweets = 29 + 9 + 3 + 1 + 1 = 43 6. Find the next term of the following sequence: 2, 3, 4, 10, 38, … Solution From the third term onwards, the next term is obtained by multiplying the previous two terms and then subtracting 2.  the next term is 10  38  2 = 378 . 7. The percentage passes in an exam for two classes are 80% and 60%. The numbers of students in the two classes are 20 and 30 respectively. Find the overall percentage pass for the two classes. Solution No. of students in the first class who pass the exam = 80%  20 = 16 No. of students in the second class who pass the exam = 60%  30 = 18 Total no. of students in both classes who pass the exam = 16 + 18 = 34  overall percentage pass for the two classes = 34 / 50  100% = 68% Note: Common mistake is 70%. But class size is different: the weaker class will pull down the percentage because it has more students than the better class. So you can’t average percentages unless the base (in this case, the number of students in each class) is the same. 8. A clock takes 9 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 3 chimes? Solution Time taken by the clock between 2 chimes = 9 s / 3 gaps = 3 s  time taken by the clock to make 3 chimes = 3 s  2 gaps = 6 s Свалено от Klasirane.Com

3 9. Evaluate . Solution = = 10. A frog fell into a drain that was 50 cm deep. After one hour, it mastered enough energy to make a jump of 6 cm but it then slid down 4 cm. If it continued in this manner after every one hour, how many hours will it take to get out of the drain? Solution After 22 hou rs, the frog would have jumped 22  2 = 44 cm. In the next hour, it would have jumped the last 6 cm and out of the drain.  it takes the frog 22 + 1 = 23 hours to get out of the drain. 11. A farmer’s chickens produced 4028 eggs one day. Was he able to pack all the eggs in full cartons of one dozen eggs each? Solution A dozen is equal to 12. If a number is divisible by 12 (= 3  4), then it is also divisible by 3 and 4, since 3 and 4 are relatively prime. Using the divisibility test for 4, 4028 is divisible by 4 since the last two digits 28 is divisible by 4. Using the divisibility test for 3, 4028 is not divisible by 3 since 4 + 0 + 2 + 8 = 14 is not divisible by 3, i.e. 4028 is not divisible by 12.  the farmer was not able to pack all the eggs in full cartons of one dozen eggs. 12. A farmer wants to find out the number of sheep and ducks that he has. He count ed a total of 40 heads and 124 legs. How many sheep and how many ducks does he have? Solution Method 1 (Supposition) Suppose all the animals are ducks. Then there would be 40  2 = 80 legs. So the remaining 124  80 = 44 legs must belong to the “extra 2 legs” of the sheep.  there are 44  2 = 22 sheep and 40  22 = 18 ducks . Свалено от Klasirane.Com                             20141 1 4 1 1 3 1 1 2 1 1                              20141 1 4 1 1 3 1 1 2 1 1                         20142013 4 3 3 2 2 1  2014 1

4 Method 2 (Systematic Guess and Check) Make a systematic list, starting with a random guess: 20 sheep and 20 ducks. No. of S heep No. of Ducks No. of Sheep Legs No. of Duck Legs Total No. of Legs 20 20 80 40 120 21 19 84 38 122 22 18 88 36 124  there are 22 sheep and 18 ducks . Method 3 (Algebra) Let the no. of sheep be x. Then there are 40  x ducks. Total no. of legs = 4 x + 2(40  x) = 124 4 x + 80  2x = 124 2x = 44 x = 22  there are 22 sheep and 40  22 = 18 ducks . 13. Jaime puts some blue and red cubes in a box. The ratio of the number of blue cubes to the number of red cubes is 2 : 1. She adds 12 more red cubes in the box and the ratio becomes 4 : 5. How many blue cub es are there in the box? Solution Method 1 (Model Method) Before Blue Red After Blue Red From the diagram, 3 units = 12 1 unit = 4 4 units = 16  there are 16 blue cubes . 12 Свалено от Klasirane.Com

5 Method 2 (Algebraic Method) Let the no. of red cubes at the start be x. Then the no. of blue cubes will be 2 x. After adding 12 more red cubes, there are x + 12 red cubes. Ratio of blue cubes to red cubes is now 4 : 5, i.e. . Then 10 x = 4 x + 48 6 x = 48 x = 8 2 x = 16  there are 16 blue cubes . 14. The diagram shows a rectangle with its two diagonals. What percentage of the rectangle is shaded? Solution Let the length and breadth of a rectangle be l and b respectively. Then area of rectangle = lb and area of shaded triangle =  base  height =  b  = .  percentage of the rectangle that is shaded =  100% = 25% 15. Given that a  b = 2014, and a and b are whole numbers such that a < b , how many possible pairs ( a , b) are there? Solution 2014 = 2  19  53, where 2, 19 and 53 are prime numbers.  2014 = 1  2014 = 2  1007 = 19  106 = 38  53  there are 4 possible pairs ( a , b). Свалено от Klasirane.Com

6 16. Amy had 3 times as much money as Betty. After they had spent $60 each, Amy had 4 times as much money as Betty. How much money did Amy have at first? Solution Method 1 (Model Method) After Amy Betty Before Amy Betty From the Before diagram, where Amy had 3 times as much money as Betty, 1 unit = $60 + $60 = $120 4 units = $480  Amy had $480 + $60 = $54 0 at first. Method 2 (Algebraic Method) Let the amount of money Betty had at first be $ x. Then the amount of money Alice had at first was $3 x. After they had spent $60 each, Betty had $( x  60) and Alice had $(3 x  60). Then $(3 x  60) = 4  $( x  60) 3x  60 = 4 x  240 x = 180 3x = $540  Amy had $54 0 at first. $60 $60 $60 $60 Свалено от Klasirane.Com

7 17. Billy uses identical square tiles to make the following figures. If he continues using the same pattern, in which figure will there be 6044 tiles? Figure 1 Figure 2 Figure 3 Figure 4 Solution Method 1 The two corner tiles are the same for all figures.  figure number with 6044 tiles = (6044  2)  3 = 2014 Method 2 The tiles in the top row have this pattern: 3, 4, 5, 6, … Thus the nth figure will have (n + 2) + 2  n = 3 n + 2 tiles.  figure number with 6044 tiles = (6044  2)  3 = 2014 Method 3 The tiles in each of the vertical column have this pattern: 2, 3, 4, 5, … Thus the nth figure will have (n + 1)  2 + n = 3 n + 2 tiles .  figure number with 6044 tiles = (6044  2)  3 = 2014 Method 4 The no. of tiles in each figure is equal to the “area of the rectangle” minus the “area of the hole in the middle”. The pattern for the “area of the rectangle” is 2  3, 3  4, 4  5, 5  6, … The pattern for the “area of the hole in the middle” is 1  1, 2  2, 3  3, 4  4, … Thus the nth figure will have (n + 1)  (n + 2)  n  n = 3 n + 2 tiles .  figure number with 6044 tiles = (6044  2)  3 = 2014 18. What is the least number of cuts required to cut 12 identical sausages so that they can be shared equally among 20 people? Solution Fraction of sausage each person will get = = This means that there must be at least 12 cuts since no one will get one whole sausage. Cut each of the 12 sausages at the -mark. Then 12 people will get one sausage each. Свалено от Klasirane.Com

8 There are 12 times sausages left . But each person must get of a sausage. This means that 4 of the sausa ges must be cut into half each so that the remaining 8 people will get one sausage and one sausage each.  least no. of cuts = 12 + 4 = 16 19. In the following alphametic, all the different letters stand for different digits. Find the two -digit sum PI. I S I S I S + I S P I Solution If I  3, then the sum will be a 3-digit number. So I = 1 or 2. In the ones column, S + S + S + S = 4  S = _I. Since 4  S is even, then I must be even, so I = 2. But 4  S = _2 implies that S = 3 or 8. If S = 8, then 28  4 is a 3-digit number, so S = 3. Thus 23 + 23 + 23 + 23 = 92 , i.e. P = 9 .  the two-digit sum PI is 91 . 20. A teacher has a bag of sweets to treat her class. If she gave 6 sweets to each student, then she would have 5 sweets left. If she gave 7 sweets to each student, then she would have 30 sweets short. How many students and how many sweets are there? Solution If the teacher gave 6 sweets to each student, then she would have 5 sweets left. But 5 sweets left are not enough to give one more sweet to each student since she would have 30 sweets short. So if she had 30 more sweets, then she could give 30 + 5 = 35 sweets to the students so that each student has one more sweet, i.e. 6 sweets each. Since each student receives only 1 more sweet (i.e. the 6 th sweet), then there are 35 students .  there are 35  6 + 5 = 215 sweets. 21. A box contains 80 coloured pens: 36 black, 24 blue, 12 red and 8 green. Alice takes some pens from the box without looking at the colours of the pens. What is the least number of pens she must take so that she has at least 20 pens of the same colour? Свалено от Klasirane.Com

9 Solution Worst case scenario for all the pens taken out by Alice to be of different colours: 19 black, 19 blue, 12 red and 8 green, i.e. a total of 58 pens. Then the next pen taken out has to be either black or blue, i.e. Alice will have at least 20 pens of the same colour, either 20 black or 20 blue pens.  least no. of pens Alice must take so that she has at least 20 pens of the same colour = 58 + 1 = 59 22. Find the value of . Solution Observe the following pattern:  . 23. The diagram shows a square being divided into four rectangles. If the sum of the perimeter of the four rectangles is 40 cm, find the area of the square. Solution Perimeter of the four rectangles = 8  length of square = 40 cm So length of square = 5 cm Area of square = 25 cm2 24. Given that 5! means 5  4  3  2  1, find the last digit of 2014!. Solution Свалено от Klasirane.Com

10 Since 2014! contains the factor 10, and any number multiplied by 10 will have 0 as the last digit, then the last digit of 2014! is 0 . 25. Two women, Ann and Carol, and two men, Bob and David, are athletes. One is a swimmer, a second is a skater, a third is a gymnast, and a fourth is a tennis player. On a day they were seated around a square table: a . The swimmer sat on Ann’s left. b . The gymnast sat across from Bob. c . Carol and David sat next to each other. d . A woman sat on the skater’s left. Who is the tennis player? Solution There are only two possibilities to satisfy the first two conditions: Only Possibility 2 satisfies Condition c. Condition d says that Ann cannot be the skater since Bob is on her left.  Ann must be the tennis player. Ann s wimm er Bob gym nast Possibility 1 Ann swimmer Bob gymnast Possibility 2 Свалено от Klasirane.Com