Đề thi Toán học quốc tế SASMO 2014 Vòng 1 Lớp 3 - Solutions

 SASMO 2014 Round 1 Primary 3 Solutions 
Section A [1 mark for each question] 
1. Jane is 9 years old and John is 5 years old. How old will John be when Jane is 15
 years old?
 Solution 
 Method 1 
 Jane will be 15 years old in 15 9 = 6 years’ time. 
  John will be 6 + 5 = 11 years old. 
 Method 2 
 Difference in age between Jane and John = 9 5 = 4 years 
  when Jane is 15 years old, John will be 15 4 = 11 years old. 
2. A textbook is opened at random. To what pages is it opened if the product of the
 facing pages is 110?
 Solution 
 Since 10 10 = 100, try 10 11 = 110. 
  the pages are 10 and 11. 
3. Find the number B such that the following statement is true: 8 B = 3 9 + 5 9.
 Solution 
 Method 1 
 8 B = 3 9 + 5 9 = 27 + 45 = 72 
  A = 72  8 = 9 
 Method 2 
 8 B = 3 9 + 5 9 = (3 + 5) 9 = 8 9 
  A = 9 
4. It is given that a  b = a b + a b. For example, 2  3 = 2 3 + 2 3 = 5. Find the
 value of 4  3 3  4.
 Solution 
 Method 1 
 4  3 = 4 3 + 4 3 = 13 
 1 
 Свалено от Klasirane.Com 11 sweets 11 wrappers 3 sweets and 2 wrappers 5 wrappers 1 sweet and 2 
 wrappers 3 wrappers 1 sweet 
  biggest no. of sweets = 11 + 3 + 1 + 1 = 16 [Common mistakes: 14, 15] 
9. At a workshop, there are 10 participants. Each of them shakes hand once with one 
 another. How many handshakes are there? 
 The first participant will shake hand with 9 other participants; 
 the second participant will shake hand with 8 other participants; 
 the third participant will shake hand with 7 other participants; etc. 
 Thus total no. of handshakes = 9 + 8 + 7 +  + 3 + 2 + 1 
 1 + 9 = 10 
 2 + 8 = 10 
 3 + 7 = 10 4 pairs 
 4 + 6 = 10 
 5 
  total no. of handshakes = 10 4 + 5 = 45 
10. Ali uses identical square tiles to make the following figures. If he continues using the 
 same pattern, how many tiles will there be in the 15th figure? 
 Solution 
 Method 1 
 The two corner tiles are the same for all figures. 
  the 15th figure will have 15 3 + 2 = 47 tiles. 
 Method 2 
 The tiles in the top row have this pattern: 3, 4, 5, 6,  
  the 15th figure will have 17 + 2 15 = 47 tiles. 
 Method 3 
 The tiles in each of the vertical column have this pattern: 2, 3, 4, 5,  
  the 15th figure will have 16 2 + 15 = 47 tiles. 
 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”. 
 3 
 Свалено от Klasirane.Com  the pen costs $1.75 
 Method 2 (Model Method) 
 Pen 60¢ 
 $2.90 
 Pencil 
 2 units = $2.90 60¢ = $2.30 
 1 unit = $1.15 
  the pen costs $1.15 + 60¢ = $1.75 
 Method 3 (Algebraic Method) 
 Let the cost of the pencil be $x. 
 Then the cost of the pen is $(x + 0.6). 
 So x + (x + 0.6) = 2.9 
 2x + 0.6 = 2.9 
 2x = 2.3 
 x = 1.15 
  the pen costs $(1.15 + 0.60) = $1.75 
14. If the three-digit number 3N3 is divided by 9, the remainder is 1. Find N. 
 Solution 
 Since 3N3 gives a remainder of 1 when divided by 9, then 3N3 1 = 3N2 is divisible 
 by 9. 
 Using the divisibility test for 9, 3 + N + 2 = N + 5 is also divisible by 9. 
  N = 4. 
15. Charles has 16 marbles. He divides them into 4 piles so that each pile has a different 
 number of marbles. Find the smallest possible number of marbles in the biggest pile. 
 Solution 
 For each pile to have a different number of marbles, and the biggest pile to have the 
 smallest possible number of marbles, put 1 marble in the 1st pile, 2 marbles in the 
 2nd pile, 3 marbles in the 3rd pile and 4 marbles in the 4th pile. So the biggest pile is 
 the 4th pile, but there are only 1 + 2 + 3 + 4 = 10 marbles. 
 The 11th marble will have to go to the 4th pile so that each pile will have a different 
 number of marbles. The 12th marble cannot go to the 4th pile because we want to 
 find the smallest possible number of marbles in the biggest pile, so the 12th marble 
 will have to go to the 3rd pile. Similarly, the 13th and 14th marbles will go to the 2nd 
 and 1st piles respectively. 
 The 15th marble will go to the 4th pile again, and the 16th marble to the 3rd pile. 
  the largest pile (which is the 4th pile) will contain 4 + 1 + 1 = 6 marbles. 
 5 
 Свалено от Klasirane.Com  there are 17 7 + 6 = 125 sweets. 
19. What are the last 2 digits of the sum 1 + 11 + 111 +  + 111111? 
 50 digits 
 Solution 
 Last digit: 50 0 (and carry 5) 
 Second last digit: 49 + 5 (carry) = 54 4 
  the last 2 digits are 40. 
20. Alvin tells the truth on Monday, Tuesday, Wednesday and Thursday. He lies on all 
 other days. Doris tells the truth on Monday, Friday, Saturday and Sunday. She lies on 
 all other days. One day they both said, “Yesterday I lied.” When was that ‘one day’? 
 Solution 
 Mon Tue Wed Thur Fri Sat Sun 
 Alvin     
 Doris     
  indicates the person telling the truth 
 If Alvin tells the truth on that ‘one day’ that he lied ‘yesterday’, then that ‘one day’ 
 must be Monday. 
 If Alvin tells the lie on that ‘one day’ that he lied ‘yesterday’, then he must be telling 
 the truth ‘yesterday’ and so that ‘one day’ must be Friday. 
  based on Alvin, it has to be either Monday or Friday. 
 If Carol tells the truth on that ‘one day’ that she lied ‘yesterday’, then that ‘one day’ 
 must be Friday. 
 If Carol tells the lie on that ‘one day’ that she lied ‘yesterday’, then she must be 
 telling the truth ‘yesterday’ and so that ‘one day’ must be Tuesday. 
  based on Carol, it has to be either Tuesday or Friday. 
 Hence, that ‘one day’, when they both said that they lied ‘yesterday’, has to be Friday. 
 7 
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