MEMORANDUM 20 QUESTION () (F) 5() 5(F) 6() 6(F) 7() 7(F) VRAAG D E C A B B B A 2 B B B B A B C D 2 A B C A E C B B E C C B E E A C 5 C C C E E D A B 5 6 E B D B D C D D 6 7 D C B B D A A B 7 8 B B E A E E D A 8 9 B A B E B E A D 9 0 B E D C B B A A 0 D E D E B A B B 2 A A E A C B B E 2 D B B D E A D E E B B C C D D C 5 D B D B E B C A 5 6 C B D D E C B D 6 7 B C B D B D E B 7 8 D B D E E E A E 8 9 C A A C E B C E 9 20 A D C D A D A A 20 2 C C D A B B D A 2 22 C C C A E B A A 22 2 D A B B B A D B 2 2 A A E C D C B B 2 25 C B D A B A B B 25 QUESTION () (F) 5() 5(F) 6() 6(F) 7() 7(F) VRAAG
NOTES ON 20 MEMORANDUM These notes are necessarily brief and often formal and symbolic. Many questions could be answered using primitive methods, e.g. If today is Wednesday, what day of the week will it be 00 days from now? can be done by counting. That would be laborious, time-consuming and error-prone. The essence of a mathematical approach is to work more smartly by using appropriate representations to model the situation and to exploit the inherent structures and patterns in the situation. GRADE (). 5 + = 8, while the others all have a sum of 7 2. Half of 8 8 6. There is a pattern of +, +, + in the numbers 7. 8 = 6 8 8. How many 8 are there in 6 8? 8. B is a mirror-image in a horizontal or vertical line of symmetry, as shown. 5 000 m 5 m = 00 2. The figure can be divided into 2 equal triangles of which 6 are shaded..,, 9,... =, 22,, So 88 = 6. 5 small cubes to a side. So 55 in bottom layer, with 5 layers, so 555 5. 27 26 + = 29 6. 8 cubes on each of the 6 sides. But then they are all counted twice! So 68 2 7. Bottom level: = 9 blocks, Second level has less: 8 blocks, Top level has 5 blocks 8. x + 5 6 =, so x 5 = so x = 8 9. Two people at the end + 2 people per table: 20 2 + 2 = 2 people 20. 58 2 = 56; 56 2 = 28 tables 2. 20 km in 60 min, so 20 km in 0 min, so 200 km in 00 min, so the time is :0 22. 7, 7, 27, 7, 77 (two!), 87, 97 is, plus 70, 7, 72, 77, 78, 79 is another 9, so 20 2. M + M + 0 =, so 2 M = 8, so Monde weighs 2 kg 2. 50 2 = 99 25. = GRADE (F). The numbers are halved: ; 2 ; ; ; 8 6 2. 8 9. pizza for children 5 pizzas children/pizza = 5 children. 58 = 9 5. These are multiples of 6. Only 82 = 6 697 is a multiple of 6 6. 09:7 to 0:8 = minutes 2:0 = min. = :0 7. Jason has 2/ of the stamps and Mary has / of the stamps 96 = 2 stamps 8. 8 257 = 8 km 9. 8 + 69 = 607 km 0. Thabo takes out of 2; /2 = / He has to pay / of R0 R0. The tower is on your left if you look at the object from the back 2. 6,8 2, 2,7 2 = 0,85. 2 2 = 0; 7 = 2 marbles 5. R5,60c 0 = R0,89c R0,89c 5 = R,5c 7. 8 2 + 2 = 22 m 20 5 8. 6 9 2. Invent some notation and system and count systematically, e.g.: 2 One area,, 2,,, 5 and 6 each form a triangle (6) Two areas - and -6 each form a triangle (2) 6 5 Three areas --2, 2--6, -6-5 and 5--each form a triangle ()
22. Do not count or calculate look for structure, e.g. For Pattern : For Pattern 2: 2 For Pattern 00: 00 For Pattern : For Pattern : Pattern 2 Pattern Pattern 2. Arrange them: O S (R) T (R) E (Ram can be between Siva and Temba or between Temba and Eby) Oscar is the shortest 2. 20c + 0c + 5c 0c + 5c 25. Number of blocks = + 2+ + +. + 8 + 9 + 50 = (+50) + (2 + 9) + = 25 5 = 275 GRADE 5(). There are 5 tiles in every metre because 000 cm 200 cm = 5. So 5 0 = 50 tiles 2. The numbers inside the square and the circle are 2 and. 2 is not inside the triangle. Try and test each possible answer!. C a rotation to the right through 90 5. reds 0 greens purples. So 2 () reds 9 () purples 6. n th row has 2n dots, so 7 th row has dots 7. n th row has 2n dots, so 70 th row has 270 = 9 dots 8. 00 2 = rem, i.e. full days bringing us to 0:00, plus more hours, i.e., 2,, :00 Or 0 + 00 = 0, 0 2 = 6 rem 9. Height = 2 cm +,5 cm/day days. So Height after 0 days = 2 +,50 = 57 cm 0. (50 cm 2 cm),5 cm/day = 92 days. One more than a multiple of 6, so it is odd, so it cannot be A or B. Test the others: 826 = 697 2. If a sack weighs S kg, then S = S + 0, so 2S = 0, so S = 5. So S = 5 kg., 6, 9, is the -times table. So 50 = 50. Mathematics is of his time, and this is 2 hours. So of his time is 2 hours = 8 hours 5. In the bottom layer there are 8 = 2 blocks, so in two layers there are 6 blocks 6. All the blocks of the bottom layer (2) and all the blocks round the side of the top layer (20) 7. 2 + 2 =8; 72 + 2 = 6; so for rectangle with length 20: 202 + 2 = 2 8. 9. 75c more per week, so 2 75c =R9 20. There are days in January, of which 5 are even (2,, 0). There are 28 or 29 days in February of which are even. All other months have 5 even days. So the total in a year is 5 + 2. The number must start and end with so list them systematically: 0 2 5 6 7 8 9 22. Share 0 litres in ratio 5 to, i.e. 25 to 5 2. The ones digit of the product of the four numbers is equal to the product of the last digits of the numbers, i.e. 2 6 2 9 which is 6. So the remainder is. 2. List them systematically: 000 00, 00, 00 2200, 2020, 2002 20, 20, 20 2020, 2002. 20 00, 00, 00 20, 20 20, 02 00, 00 02, 02 25. If a small pizza costs S rands and a large pizza costs L rands: 2S + L = 5S, so L = S, so the cost is L = R,50 = R,50
GRADE 5(F). There are nine squares, four 2 2 squares, and one square. Total = 9 + +. Build a mental picture! B, E & F. The trip is minutes long, so 2:0 + minutes = : 5. Use trial and error, i.e. try each of the given answers one by one 6. Sipho has 2 marbles, so Landi has 20 marbles. So together they have 2+20 = 52 marbles 7. (2 ) + ( 2) + ( ) + + (00 99) + (0 00) = + + + +... 00 times = 00 8. If 2 5 is 2 leaners, then 5 is 2 2 = 6 learners, and 5 5 (the whole class) is 5 6 = 0 learners 9. 2. So there is 2 left for Oscar 0. 6 + 2 = 26 cm. 000 g 725 g = 0275 g = 0,275 kg 2. Divide 20 into 7 equal parts:20 7 = 60. of these parts are dresses, i.e. 60 = 80. If Kim has stamps, then Jack has + stamps. So + + + 0 = 220 So + + = 80 So = 60, so Kim has 60 stamps. Consider the possible choices from the top row: If I choose, then the options are, 5, 9 or, 6, 8 giving products 5 or 8 respectively. 2 If I choose 2, the options are 2,, 9 or 2, 6, 7 with products 72 or 8 respectively. If I choose, the options are,, 8 or, 5, 7 with products 96 or 05. 5 6 So 05 is the maximum possible product. 7 8 9 5. T n = n +. So T 50 = 50 + = 5 6. P n = n + or (n + ), so P 50 = 5 = 5 7. The first three may be blue, red and brown. Then the next one must match one of these colours 8. You can maybe take out, e.g. 0 red, then 0 brown, then blue, then the next one is also blue 9. Look at the structure in the pictures! T : = T 2 : + = = 2 2 T : + + 5 = 9 =. T 0 : + + 5 + 7 +... to 0 numbers = 0 0 triangles 20. Look at the structure in the pictures! Count the number of triangles: # triangles in T = # triangles in T 2 = + 2 # triangles in T = + 2 + # triangles in T 0 = + 2 + + +... + 9 + 0 = ( + 0) 0/2 = 55 So # matches = 55 2. Because 65 = 52 7 +, the day of the week moves one day later each year (but remember leap years!) 20 Wed 20 Thu 205 Fri 206 Sun - 206 is a leap year! 207 Mon 208 Tue 209 Wed 22. Draw it! Fill in the information as you read. Re-read, bit by bit! 2. Be systematic, e.g. 2 2 2 2 2 2 2. There are 28 days in February of which are odd (,, 27) There are 0 days in Apr, Jun, Sep and Nov, of which 5 are odd (,, 29) The other 7 months have days of which 6 are odd days So the total odd dates in a year is + 5 + 7 6 = 86 25. Debbie is first, Peter is second, Tom is third and Robert is fourth.
GRADE 6(). Make equal parts. Each small square is half of the next bigger square. So half of half of the big square is a quarter of the big square 2. There are 8 columns, each with 2 + + 6 cubes. So 8 2 = 96 cubes. 7 = 5 5 and 5 = 7 5 so 6 5 is exactly in between them. Or ( 5 + 7 ) 2 = ( 7 5 + 5 5 ) 2 = 6 5 5. Use trial and error, i.e. try each of the given answers one by one 6. In middle row the missing number is 8 (+6) =, so in right column A = 8 (+0) = 7 8. Continue the patterns: 7, 22, 27, 2, 7, 2, 7, 52, and 7, 2,, 8, 5, 52, 9. For n dice, the number of visible faces is n + 2. So for 75 dice, 75 + 2 = 227 0. Imagine or draw the cube! If the side is times as long, the big cube contains 27 of the small cubes. So its mass is 27 times as large!. 020 + 0 + 5 20 + 0 + 5 2. 02 7 = rem, so adding, we have 05 7 = 5. B C M In the middle row, N cannot be 2, so N is or A 2 N Suppose N =. Then A = which is impossible (already a in left column). D So N =, A =. In left column B = 2. Then C = (D ), so M =, so M+N = 5. List the triangles systematically - notation and a system will help! 6. Count equal parts! There are 8 equal parts and 9 of them are shaded. 7. Vary the numbers systematically and note the behaviour of the product of the numbers: + 7 = 8 and 7 = 7 6 + 2 = 8 and 6 2 = 72 2 + 6 = 8 and 2 6 = 2 7 + = 8 and 7 = 77 + 5 = 8 and 5 = 5 8 + 0 = 8 and 8 0 = 80 + = 8 and = 56 9 + 9 = 8 and 9 9 = 8 5 + = 8 and 5 = 65 0 + 8 = 8 and 0 8 = 80 8.? = 000 = 0 + 2 (0000) = 6 + from first two balances 9. If the numbers are x and y: 6x + y = 7. So 7 y must be a multiple of 6, i.e. 2, so y = 5 Note: If 7 y = 6, y =, which is not a one-digit number! 20. and 22 (the sum of any two sides must be greater than the third side why?) 2. 6 out of 2 marbles are not blue, so the probability of choosing a not-blue marble is 6 2 = 2. 22. If the empty glass has a mass of g gram and the milk has a mass of m gram, then g + m = 70 g + 2 m = 290 So 2 m = 70 290 = 80 gram, so m = 60 gram and g = 70 60 = 20 gram 2. Each number is the sum of the two numbers above it, e.g. 6 = + 5, 5 = 5 + 0 2. If a bubble gum cost B cents and a chocolate costs C cents: B + C = 90 and 0B + 5C = 70, so 5B + 5(B + C) = 70, so 5B + 590 = 70, so B =, so C = R0,86 25.,, 9, =, 22,, 2020
GRADE 6(F). Try trial and improvement, e.g. 50 + 5 + 52 7;. But 57 + 58 + 59 = 7 Or test each of the given numbers Or, if the smallest is x, then x + (x + ) + (x + 2) = x + = 7, so x = 57 2. The average of the two numbers: (7,8 + 7,85) 2 = 5,65 2 = 7,825 7 7. ( + ) 2 = 2 = 2 2. There are different sizes of triangles as shown. Total = 8 + 8 + 2 + 2: 8 8 2 2 5. Full lorry = 65 kg; empty lorry = 258 kg; 65 258 = 2070; 2070 kg 90 kg/bag = 2 bags 6. (2000 999) + (998 997) + + (2 ) = + + + + + (000 times) 7. 8. ½ of cm 2 + cm 2 + ½ of cm 2 =, 5 cm 2 + 9 cm 2 +,5 cm 2 = 5 cm 2 9. Check each of them, e.g. for (A): 60 2 + 0 = 60 does not give 0 legs. But (E) does: 0 2 + 20 = 60 0. 00 20 = 5; 000 20 = 50. Or 5% 00. 20 202 20 202 2. 2 + 8 + 8 = ; R5 is ; R60 =. If the jersey costs Rx, the coat costs Rx + 50. Together they cost x + x + 50 = 650 So x = (650 50) 2 = R250. There is a general structure here: The denominators is twice the numerator +, i.e. 2 2 5 6 7 We can therefore investigate a general pattern, 5, 7, 9,,, 5, 2 2 Check with your calculator: = 0,..., 5 = 0,, So < 5 < 7 5 < 9 < 6 < 7 < 5 < Conclusion: the larger the denominator, the larger this kind of fraction, so 2 is the largest 5. If the number is, then + / of = 52, so / of = 52. So / of =, so / of = 9 6. Jane eats 2 2 sweets in 5 minutes; she eats 2 2 sweets in 0 minutes. Jane eats 8 sweets in 0 minutes 7. Make a representation of the situation (draw it): 8. There are nine -digit numbers, ninety 2-digit numbers giving us 9 +90 2 = 89 digits. So we need 852 89 = 66 more digits. 66 = 22 so we need 22 -digit numbers, thus the numbers from 00 to 20. So there are 20 pages 9. For 9 wheels we can have: tricycle and 8 bicycles = total of 9 (too much) tricycles and 0 bicycles = total of (too much) 5 tricycles and 2 bicycles = total of 7 (just right) 20. Look at the structure in the pictures! P = + = 5 P 2 = 2 + = 9 P = + = P 50 = 50 + = 20
2. Work systematically! 0,, 2,,, 5, 6, 7, 8, 9 this is 0 202, 22, 222, 22, 22, 252, 262, 272, 282, 292 this is 0. 909, 999, 929, 99, 99, 959, 969, 979, 989, 999 this is 0 So 9 0 = 90 22. Make a list, varying the persons systematically. If the persons are a, b, c and d: abcd, abdc, acbd, acdb, adbc, adcb and similarly if the first person is b, c, and d. So 6 = 2. Or 2 = 2 2. Let the children be A, B, C, D and E. List all the possibilities systematic, note patterns and structure: A vs B B vs C C vs D D vs E A vs C B vs D C vs E A vs D B vs E A vs E 2. A vs B B vs C. H vs I I vs J A vs C B vs D. H vs J A vs D B vs E. A vs E B vs F. A vs F B vs G. A vs G B vs H. A vs H B vs I. A vs I B vs J A vs J The structure is: 9 + 8 +... + 2 + = 5 25. Look at the structure in the pictures! P : = P 2 : + = = 2 2 P : + + 5 = 9 = P : + + 5 + 7 = 6 =. P 50 : + + 5 + 7 +... to 50 numbers = 50 50 = 2500
GRADE 7() 2. + = 9 + = 6 + = 9. nth number = 2n, so 8 rd number = 2 8 = 65. & 5. Sum of numbers 6. We know: 8, so Sum of numbers = 8 = 88 88 x If the new number is x, then. So x = 2 88 = 2 7. 2 5 5 8. First fit the tiles in the width: If one tile is used in the width, it has a side of 2,5 m, and cannot cover the length exactly. If 2 tiles are used in the width the tiles have a length of,25 m, and then of them can fit into the length. So 2 = 6 tiles are the minimum number of tiles. 9. 5 Ʊ 6: 5 6 = 7 res, dus is 5 Ʊ 6 =. Dan vir 2 Ʊ : 2 = res 0, dus is 2 Ʊ = 0 0. The largest, by guess-and-improvement is = 96. So there are squares smaller than 000. # Triangles = 2squares + 2, or 2(squares + ). So Triangles (6) = 26 + 2 = 2. Triangles (60) = 260 + 2 = 22. 2x + 2 = 60, so x = 29. Make a list, varying the numbers systematically. If the digits are a, b, c and d: abcd, abdc, acbd, acdb, adbc, adcb and similarly if the first digit is b, c, and d. So 6 = 2 5. 2 (7 + 8 + 9) = 2 2 6. Using a representation like this, Area D = b d We know a c = 2, b c = 2, a d = 20 D Multiply them all together: a 2 c 2 bd= 2 20 2 But ac = 2, so a 2 c 2 =, so bd = 220 2 = 5 7. Volume = area of base length = 7 cm 2 2 cm = 8 cm Or think of cutting out a rectangular prism: Volume = 2 2 = 7 2 8. The first digit can be 2,, 6 or 8. The second digit can be 0, 2,, 6 or 8, which gives 5 = 20 possible combinations 9. The 6 th column are multiples of 6, with formula 6 row n. So the last number in row 80 is 6 80 = 80. Then row 8 is 8, 82, 8, 20. T+F = R+7 So 7+F=5+7 So F=5 2. Fill in numbers in the calendar, and test each statement with the numbers. 22. Test specific cases, e.g. if a = 5, then b = 6, c = 2 and d =, the a+b+c+d = 6, which is not correct. Choose a better value for a Or: We know a + d = c + b, so a + b + c + d = a + d + c + b = 2(a + d) = 52. So a + d = 26, so a + (a + 8) = 26, so a = 9 2. Choose different consecutive numbers, and test each statement with the numbers. 2. lines from two corners divide the triangle in sections 0 lines from two corners will divide the triangle in sections = 2 25. 6 pencils and pens cost R62 pencils and 6 pens cost R8 Add them: 0 pencils and 0 pens cost R6 Divide by 2: 5 pencils and 5 pens cost R7 8 8 8 8
GRADE 7(F). The L-shaped region can be decomposed into a rectangle and a rectangle. So the total area is 7 cm 2 2. + + + + + = 6 cm. (2 ) + ( 2) + ( ) + + (00 99) + (0 00) = + + + +... 00 times = 00. Numbers ending with, 2, or 5 have this property. They are, 2, 5, 2, 22, 25,, 2, 5,, 2, and 5 In addition, we have 2,, 6, and 8, for a total of 7 5. n th number = 2n, so 8 rd number = 2 8 = 65 6. In middle row the missing number is 8 (+6) =, so in right column x = 8 (+0) = 7 7. Do not rush into calculation! Look for structure! 2 85 28 28 7 28 (5 7) 5 28 28 8. b and c are both less than, so b c is less than both b and c. 9. We know: Sum of numbers 8, so Sum of numbers = 8 = 88 88 x If the new number is x, then. So x = 2 88 = 2 0. 7 = 20 litres, so = 20 litres = 0 litres. So the full tank = 8 = 0 litres 8 = 20 litres 8 2 8 8 8. Look at the structure: For n dice, the number of visible faces is n + 2. So for 0 dice, 0 + 2 2. If 50 faces are visible, n + 2 = 50, so n = 6. There are 8 possibilities: GGG, GGB, GBG, GBB, BGG, BGB, BBG, BBB. At least one girl means, 2 or girls, and only in the case of BBB is there no girl. So the probability is 7/8. 2 0. 2 ( ) ( ) ( ) 85 256 5. To be divisible by 5, the last digit must be 5. But to be divisible by 2, the last digit must be 2 or. So none of these numbers can be divisible by 2 and 5, so none of them can be divisible by, 2,,, and 5. 6. Starting with the primes: 2, 29,, 7,,, 7,5, 59, 6, 67, 7, 7, 79, these are also prime, 7, 7, 7, 97, and of the five +, 7+, 7+, 7+ and 79+, only 7+ is a multiple of 7. Write as the product of factors, but do not repeat factors, e.g. do not write 6 = 2, because it is already there! 5 7 8 9 = 5 7 2 2 2 8. You must take all possible combinations of the numbers 9, 29, 59 and 79 (they are all prime). Do not calculate, simply systematically count all possible combinations. It is the same as taking all possible combinations of the letters A, B, C and D if the order does not matter: at a time: A, B, C, D so 2 at a time: AB, AC, AD, BC, BD, CD so 6 at a time: ABC, ABD, ACD, BCD so 9. If the length of a square doubles ( 2), then the area quadruples (x), as illustrated in this simple example If the dimensions of the room is a by b by c, then the area to paint is A = ab + 2ac + 2bc Double the dimensions are 2a by 2b by 2c, so the area to paint is D = (2a)(2b) + 2(2a)(2c) + 2(2b)(2c) = A 20. Add all together: 2A + 2B + 2C = 2, so A + B + C = 2 2. B + A + C = 2 and A + C = 6, so B + 6 = 2 22. Take special cases, be systematic, and notice the patterns: number: 2 numbers: numbers: 50 numbers: 2 2 2 2 6 5 9 2 6 2 5... 99 2 6... 00 50 5 Alternatively, if you know or develop some formulas: 2 5 7... 97 99 50 2 6 8... 98 00 505 2. Suppose Xolile had x marbles. After giving / to Baba, he had 2/ remaining; or 2/ of x. After giving / of the remainder to Sam, he had / of them left, or / of 2/ of x which equals 2.So /2 of x equals 2, so x = 8. This means she gave Baba / of 8 = 6 marbles 2. ++6+0+5+2 = 56 25. Do not count or calculate, investigate the structure:,, 9, =, 22,, 2020 x 50 5 x x