[PDF] SASMO 2014 Round 1 Primary 2 Solutions.pdf | Plain Text

1 SASMO 2014 Round 1 Primary 2 Solutions Section A [1 mark for each question] 1. What is 2014 + 2  0  1  4 equal to? (a) 2014 (b) 2016 (c) 2021 (d) 2022 (e) None of the above Solution 2014 + 2  0  1  4 = 2014 + 0 = 2014 (a) 2. Ten lampposts are equally spaced along a straight line. The distance between two consecutive lampposts is 40 m. What is the distance between the first and the last lampposts? (a) 360 m (b) 380 m (c) 400 m (d) 420 m (e) None of the above Solution Distance between the first and the last lampposts = 40 m  9 gaps = 360 m (a) 3. Find the next term of the following sequence: 1, 1, 2, 3, 5, … (a) 6 (b) 7 (c) 8 (d) 9 (e) 10 Solution From the third term onwards, the next term is obtained by adding the previous two terms.  the next term is 3 + 5 = 8 (c) 4. Jane wrote the word STUDENTS thrice. How many times did she write the letter S? (a) 2 (b) 4 (c) 6 (d) 8 (e) 10 Свалено от Klasirane.Com

2 Solution ‘Thrice’ means 3 times.  Jane wrote the letter S a total of 2  3 = 6 times (c) 5. What number between 37 and 47 is exactly divisible by both 2 and 3? (a) 38 (b) 39 (c) 42 (d) 44 (e) 45 Solution Method 1 Numbers between 37 and 47 that are exactly divisible by 2 are: 38, 40, 42, 44 and 46. Of these 5 numbers, only 42 is exactly divisible by 3.  the number between 37 and 47 that is exactly divisible by both 2 and 3 is 42 (c). Method 2 A number that is exactly divisible by both 2 and 3 must also be exactly divisible by 6.  he only number between 37 and 47 that is exactly divisible by 6 is 42.  the number between 37 and 47 that is exactly divisible by both 2 and 3 is 42 (c). 6. A shop sells sweets where every 3 sweet wrappers can be exchanged for one more sweet. Ali has enough money to buy only 7 sweets. What is the biggest number of sweets that he can get from the shop? (a) 7 (b) 8 (c) 9 (d) 10 (e) 11 Solution 7 sweets  7 wrappers  2 sweets and 1 wrapper  3 wrappers  1 sweet  biggest no. of sweets = 7 + 2 + 1 = 10 (d) 7. There are 14 children playing “The eagle catches the chicks.” One of them is the ‘eagle’ while another child is the ‘mother hen’ whose job is to protect the ‘chicks’. The rest of the children are the ‘chicks’. After a while, the ‘eagle’ has caught 5 ‘chicks’. How many ‘chicks’ are still running around? (a) 6 (b) 7 (c) 8 (d) 9 (e) 10 Свалено от Klasirane.Com

3 Solution No. of ‘chicks’ still ru nning around = 14  1 (eagle)  1 (mother hen)  5 = 7 (b) 8. Find the number A such that the following statement is true: 7  A = 3  8 + 4  8. (a) 3 (b) 4 (c) 5 (d) 7 (e) 8 [Ans] Solution Method 1 7  A = 3  8 + 4  8 = 24 + 32 = 56  A = 56  7 = 8 (e) Method 2 7  A = 3  8 + 4  8 = (3 + 4)  8 = 7  8  A = 8 (e) 9. Two $1 coins and ten 50¢ coins are randomly distributed among 4 children such that each child receives the same number of coins. What is the difference between the biggest amount and the smallest amount a child can receive? (a) 50¢ (b) $1 (c) $1.50 (d) $2 (e) None of the above Solution There are a total of 2 + 10 = 12 coins. So each child receives 12  4 = 3 coins. Biggest amount a child can receive = $1 + $1 + 50¢ = $2.50 Smallest amount a child can receive = 50¢ + 50¢ + 50¢ = $1.50  difference between biggest amount and smallest amount = $2.50  $1.50 = $1 (b) 10. Tim is 8 years old and Sally is 4 years old. How old will Sally be when Tim is 14 years old? (a) 7 (b) 8 (c) 9 (d) 10 (e) None of the above Свалено от Klasirane.Com

4 Solution Method 1 Tim will be 14 years old in 14  8 = 6 years’ time.  Sally will be 6 + 4 = 10 years old (d). Method 2 Difference in age between Tim and Sally = 8  4 = 4 years  when Tim is 14 years old, Sally will be 14  4 = 10 years old (d) . Section B [3 marks for each question] 11. In the following alphametic, all the different letters stand for different digits. Find P and I. I I + I P I Solution I  3 = _I By guess and check, the only possible solution for I is 5.  5 + 5 + 5 = 15 , i.e. P = 1 and I = 5 . 12. A box contains 4 balls of different colours (red, green, yellow and blue) lying in a row. The green ball is not the second ball. The red ball is neither the first nor the last ball. The yellow ball is neither next to the red ball nor next to the blue ball. What is the order of the balls in the box from first to last? Solution The red ball is neither the first nor the last ball. Suppose the red ball is the third ball: _____, _____, __R __ , _____ The yellow ball is neither next to the red ball nor next to the blue ball. This means that the yellow ball is the first ball, and the blue ball is the last ball: __ Y __, _____, __R __ , __ B __ So the green ball is the second ball, which is a contradiction. Свалено от Klasirane.Com



6 Method 3 The pattern for the no. of white squares is 3  3  1, 3  4  2, 3  5  3, … (total no. of squares minus no. of black squares in each figure)  no. of white squares that will surround one row of 50 black squares = 3  52  50 = 106 (e) Method 4 The pattern for the no. of white squares is 8, 10, 12, …, which is equal to 2  4, 2  5, 2  6, …  no. of white squares that will surround one row of 50 black squares = 2  53 = 106 (e) 15. The diagram shows 9 points. Draw 4 consecutive line segments (i.e. the start point of the next segment must coincide with the endpoint of the previous segment) to pass through all the 9 points. Solution If you try to draw the line segments within the region bounded by the dots, you will realise that you need at least 5 consecutive line segments.  you must draw some of the line segments outside the region as shown: Свалено от Klasirane.Com