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

SASMO 2014 Round 1 Secondary 1 Solutions 1. Find the next term of the following sequence: 2, 1, 3, 4, 7, …Solution From the third term onwards, the next term is obtained by adding the previous two terms.  the next term is 4 + 7 = 11 . Note: These are called Lucas numbers, which is a particular Lucas sequence. 2. Find the product of the highest common factor and the lowest common multiple of 8 and 12. Solution Method 1 For any two natural numbers a and b , HCF( a, b )  LCM( a , b ) = ab . [It does not work for 3 or more numbers.]  HCF(8, 12)  LCM(8, 12) = 8  12 = 96 Method 2 8 = 2 3 12 = 2 2  3 HCF(8, 12) = 2 2 = 4 LCM(8, 12) = 2 3  3 = 24  HCF(8, 12)  LCM(8, 12) = 4  24 = 96 3. Solve for x and y in the following equation . Solution Since and cannot be negative for any values of x and y, then the only way for to be equal to 0 is when = 0 and = 0.  x = 7 and y = 8 4. The last day of 2013 was a Tuesday. There are 365 days in 2014. In what day of the week will 2014 end? Solution 365 days = 52 weeks and 1 day Since the last day of 2013 was a Tuesday, then the last day of 2014 will be Wednesday . Свалено от Klasirane.Com

5. What is the maximum number of parts that can be obtained from cutting a circular disc u sing 3 straight cuts? Solution Maximum number of parts = 7 6. A man bought two paintings and then sold them for $300 each. He made a profit of 20% for the first painting, but a loss of 20% for the second painting. Overall, did he make a profit, a loss or break even? If he did not break even, state the amount of profit or loss. Solution Profit of 20% for first painting = $300 / 120  20 = $50. Loss of 20% for second painting = $300 / 80  20 = $75.  the man lost $25. 7. Solve = 2. Solution = 2 = 4 x + 2 = 4 x = 2 8. Given that xyz = 2014, and x, y and z are positive integers such that x < y < z, how many possible triples ( x, y, z) are there? Solution 2014 = 2  19  53, where 2, 19 and 53 are prime numbers.  2014 = 1  1  2014 = 1  2  1007 = 1  19  106 = 1  38  53 = 2  19  53  there are 5 possible pairs ( x, y, z). Свалено от Klasirane.Com

9. At a workshop, there are 27 participants. Each of them shakes hand once with one another. How many handshakes are there? Solution The first participant will shake hand with 26 other participants; the second participant will shake hand with 25 other participants; the third participant will shake hand with 24 other participants; etc. Thus total no. of handshakes = 26 + 25 + 24 + … + 3 + 2 + 1 Method 1 1 + 26 = 27 2 + 25 = 27 3 + 24 = 27 13 + 14 = 27  total no. of handshakes = 27  13 = 351 Method 2 Since 1 + 2 + 3 + … + n = , then total no. of handshakes = = 351 10. A perfect number is a positive integer that is equal to the sum of its proper positive factors. Proper positive factors of a number are positive factors that are less than the number. For example, 6 = 1 + 2 + 3 is a perfect number because 1, 2 and 3 are the only proper positive factors of 6. Find the next perfect number. Solution By systematic trial and error from 7 to 28, since 28 = 1 + 2 + 4 + 7 + 14, the next perfect number is 28. 11. The dimensions of a rectangle are x cm by y cm, where x and y are integers, such that the area and perimeter of the rectangle are numerically equal. Find all the possible values of x and y. Solution Method 1 Area = perimeter Since x  2 > 0, then x > 2. When x = 3, y = 6. When x = 4, y = 4. 13 pairs Свалено от Klasirane.Com   yx xy  2 2 2   x x y

When x = 5, y = is not an integer. When x = 6, y = 3. If x > 6, y < 3. By symmetry, , i.e. y > 2. So there are no other solutions.  the dimensions of all the rectangles with integral sides whose area and perimeter are numerically equal are 3 by 6 , 4 by 4, and 6 by 3 . Me thod 2 Area = perimeter Since x  2 > 0, then x > 2. For y to be an integer, must also be an integer. This means that 4  x – 2, i.e. x  6. So the only possible solutions are when x = 3, 4, 5 and 6. When x = 3, y = 6. When x = 4, y = 4. When x = 5, y = is not an integer. When x = 6, y = 3.  the dimensions of all the rectangles with integral sides whose area and perimeter are numerically equal are 3 by 6 , 4 by 4, and 6 by 3 . 12. If a and b are positive integers such that a < b and ab = ba , find a possible value for a and for b . Solution By guess and check, a = 2, b = 4 . In fact, this is the only solution. Свалено от Klasirane.Com 2 2   yy x   yx xy  2 2 2   x x y 2442    x x  2422    x x 2 4 2   x 2 4  x

13. Find the value of . Solution Let = x. Then = x, i.e. 2 x2 + x  1 = 0.  (2x  1) ( x + 1) = 0, i.e. x = or  1 (rejected because x > 0)  = . 14. Find the last digit of 2014 2014 . Solution Since the last digit of a product ab depends only on the last digit of a and of b, then 4 = 4 4  4 = 16 6  4 = 2 4  the last digit repeats with a period of 2. Since the index 2014 is even, then the last digit of 2014 2014 is 6 . 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             1 1 21 1 21 1             1 1 21 1 21 1 x 21 1              1 1 21 1 21 1

16. What are the last 5 digits of the sum 1 + 11 + 111 + … + 111…111? Solution Last digit: 2014  4 Second last digit: 2013 + 201 = 2214  4 Third last digit: 2012 + 221 = 2233  3 Fourth last digit: 2011 + 223 = 2234  4 Fifth last digit: 2010 + 223 = 2233  3  the last 5 digits are 34344. 17. What is the least number of cuts required to cut 10 identical sausages so that they can be shared equally among 18 people? Solution Method 1 Fraction of sausage each person will get = = This means that there must be at least 10 cuts since no one will get one whole sausage. Cut each of the 10 sausages at the -mark. Then 10 people will get one sausage each. There are 10 times sausages left . But each person must get of a sausage. This means that 2 of the sausages must be cut into 4 equal parts each (i.e. 3 cuts each) so that the remaining 8 people will get one sausage and one sausage each.  least no. of cuts = 10 + 3  2 = 16 Method 2 Least number of cuts required to cut one sausage so that it can be shared equally among n’ people = n ’  1 If m’ and n ’ are relatively prime, it can be proven by putting the m’ identical sausages end to end in one row that the least number of cuts required to cut them so that they can be shared equally among n ’ people is still n ’  1, i.e. the cuts will not coincide with the gaps between the sausages. If m and n are not relatively prime, some cuts will coincide with the gaps between the sausages, and this occurs at the end of every set of m’ identical sausages, where m = m ’  HCF( m, n) and n = n ’  HCF( m, n ), i.e. where m’ and n ’ are relatively prime. Thus the least number of cuts required to cut m identical sausages so that they can be shared equally among n people is (n ’  1)  HCF( m, n) = n  HCF(m , n). In general, this formula is true for any positive integers m and n , even if m and n are relatively prime since HCF( m, n) = 1 in the latter case.  least no. of cuts required to cut 12 identical sausages so that they can be shared equally among 20 people = 18  HCF(10 ,18) = 18  2 = 16 20 14 digits Свалено от Klasirane.Com

18. Divide the following shape into 4 identical parts. Solution The shape actually consists of three identical squares. But 3 and 4 are relatively prime, so it is not easy to divide the three squares into four identical parts. So we divide the shape into LCM(3, 4) = 12 parts. From the 12 parts, we then try to regroup into 4 identical parts as shown below: 19. Solve the following equation: x5 + 2 x 3  x2  2 = 0. Solution x 5 + 2 x 3  x2  2 = 0 x 3 (x 2 + 2)  (x 2 + 2) = 0 ( x 2 + 2) ( x3  1) = 0 Since x2 + 2 > 0, then x 3  1 = 0, i.e. x3 = 1.  x = 1. Note: If participants solve by guess and check correctly, award 0 mark because they cannot exclude the possibility that there are other solutions. Свалено от Klasirane.Com

20. Find the value of . Solution Method 1 1 + 2014 = 2015 2 + 2013 = 2015 3 + 2012 = 2015 1007 + 1008 = 2015 So 1 + 2 + 3 + … + 2014 = 1007  2015  = = = = 1007.5 Method 2 Using t he formula 1 + 2 + 3 + … + n = , 1 + 2 + 3 + … + 2014 = .  = = = 1007.5 21. In the following cryptarithm, all the letters stand for different digits. Find the values of A, B, C and D. A 8 B C  3 D 9 8 2 0 1 4 Solution Method 1 In the ones column, C  8 < 4 since C  9, so need to borrow from the tens column.  10 + C  8 = 4 implies that C = 2. In the tens column, B  1  9 = B  10 < 1 since B  9, so need to borrow from the hundreds column.  10 + B  10 = 1 implies that B = 1. In the hundreds column, 8  1  D = 0 implies that D = 7. In the thousands column, A  3 = 2 implies that A = 5. Check: 5812  3798 = 2014 There are 1007 pairs of numbers that add up to 2015 Свалено от Klasirane.Com

Method 2 Rephrase as addition: 3 D 9 8 + 2 0 1 4 A 8 B C In the ones column, 8 + 4 = 12, so C = 2. There is a carryover of 1 from the ones column to the tens column, so 1 + 9 + 1 = 11, i.e. B = 1 . There is a carryover of 1 from the tens column to the hundreds column, so 1 + D + 0 = 8 or 1 + D + 0 = 18. If 1 + D + 0 = 18, then D = 17, which is not possible. So 1 + D + 0 = 8, i.e. D = 7. In the thousands column, 3 + 2 = A, so A = 5. Check: 5812  3798 = 2014 22. Find the sum of the terms in the n th pair of brackets: (1, 2), (3 , 4), (5, 6), (7, 8), … Solution The sums of the terms in each pair of brackets form the following sequence: 3, 7, 11, 15, … Method 1 Common difference between consecutive terms, m = 4 Term before the first term, c = 3  4 = 1  sum of the terms in the n th pair of brackets = mn + c = 4 n  1 Method 2 Common difference between consecutive terms, d = 4 First term, a = 3  sum of the terms in the n th pair of brackets = a + ( n  1) d = 3 + ( n  1)  4 = 4 n  1 Свалено от Klasirane.Com

23. In the diagram, PQ is parallel to RS, PA = PB and RB = RC . Given that BCA = 60  , find BAC . Solution Method 1 Let PAB = x. Then PB A = x (base s of isos. ABP) APB = 180  2x ( sum of  ABP) BRC = 180    APB (corr.  s; PQ // RS) = 180   (180   2x) = 2 x RBC = (base s of isos. RBC ) = 90   x ABC = 180    PBA   RBC (adj. s on a str. ln) = 180   x  (90   x) = 90   BAC = 180   90   60  ( sum of AB C) = 30  Method 2 Let PAB = x. Then PB A = x (base s of isos. ABP) APB = 180  2x ( sum of  ABP) BRC = 180    APB (corr.  s; PQ // RS) = 180   (180   2x) = 2 x RC B = (base s of isos. RBC ) = 90   x Now  BAC + PAB + APB + BRC + RC B + 60  = 360  ( sum of quad APRC)  BAC + x + (180   2x) + 2 x + (90   x) + 60  = 360   BAC = 30  P Q R S A B C 60   Свалено от Klasirane.Com

24. Find the remainder when 2 2014 is divided by 7. Solution Method 1 Observe the following pattern: when divided by 7, 2 1 leaves a remainder of 2, 2 2 = 4 leaves a remainder of 4, 2 3 = 8 leaves a remainder of 1, 2 4 = 16 leaves a remainder of 2, 2 5 = 32 leaves a remainder of 4, 2 6 = 64 leaves a remainder of 1, … This means that the remainder will repeat with a period of 3. Since 2014 = 671  3 + 1, then 2 2014 , when divided by 7, will leave a remainder of 2 . Method 2 When divided by 7, 2 1 leaves a remainder of 2. If 2 k leaves a remainder of 2 when divided by 7, it means 2 k = 7 p + 2 for some integer p . Then 2 k +1 = 2(7 p + 2) = 14 p + 4 = 7(2p) + 4, i.e. 2 k +1 will leave a remainder of 4 when divided by 7. Let 2 k +1 = 7 q + 4 for some integer q . Then 2k +2 = 2(7 q + 4) = 14 q + 8 = 7(2 q + 1) + 1, i.e. 2 k +2 will leave a remainder of 1 when divided by 7. Let 2 k +2 = 7 r + 1 for some integer r. Then 2 k +3 = 2(7 r + 1) = 14 r + 2 = 7(2 r) + 2, i.e. 2 k +3 will leave a remainder of 2 when divided by 7. This means that the remainder will repeat with a period of 3. Since 2 1 leaves a remainder of 2 when divided by 7, and 2014 = 671  3 + 1, then 2 2014 , when divided by 7, will leave a remainder of 2 . 25. The diagram shows a triangle ABC where AB = AC , BC = AD and BAC = 20  . Find  ADB . A B C D 20   Свалено от Klasirane.Com

Solution Draw AE perpendicular to BC. Then AE bisects BAC , i.e. BAE = 10 ----- (1) Draw the point F on AE such that BCF is an equilateral triangle [diagram not drawn to scale], i.e. FBC = 60 . Draw the line DF.  ABC = (base  of isos. ABC ) = 80   ABF = ABC   FBC = 80   60  = 20  = BAD (given) BF = BC (sides of equilateral ) = AD (given) Since ABF = BAD and BF = AD , then ABFD is an isosceles trapezium [diagram not drawn to scale]. Let G be the point of intersection of the two diagonals, AF and BD, of the isosceles trapezium ABFD. In an isosceles trapezium, AG = BG (by symmetry), i.e. ABG is an isosceles triangle.  ABD = ABG = BAG (base  of isos. ABG ) = BAE = 10 from (1)  ADB = 180    BAD   ABD = 180   20   10  = 150  A B C D 20   F E G Свалено от Klasirane.Com