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| Primary Mathematics 6B 2004 printing |
This page last updated November 10, 2006
| p. 9 | 1.(c) | 6 ÷ 2/3 = 6 x 3/2 = 9 | ||
| 4.(b) | 3/4 ÷ 9/10 = 3/4 x 10/9 = 5/6 | |||
| p. 29 | example | P = 1/4 x 22/7 x 28 + (2 x 14) = 50 | ||
| p. 41 | 25 | Fraction spent on books = 1 - 1/8 - 1/4 = 5/8 5/8 -> $50 1/8 -> $50 ÷ 5 = $10 1/4 = 2/8 = $10 x 2 = $20 |
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| p. 42 | 38 | 3rd is Ali, US is Joe | ||
| p. 46 | 29 | US on top of names to the right of the drawaing should be 3d. | ||
| p. 53 | notes | delete words after 8 min -> 200/5 x 8 | ||
| p. 55 | 5.(b) | Change "Area of 186 faces=" to "Area of 18 faces =" | ||
| p. 61 | Learning Tasks | change "...in task 4 your student could find angle ZWV =..." to "...in task 4 your student could find angle VZW =..." | ||
| p. 77 | 7 | US on top of names to the right of the drawaing should be 3d. | ||
| p. 77 | 9 | 2nd to last line in solution should be $24 x 20 = $480 | ||
| p. 85 | 2 | Names for 3d edition are Devi and Gopal. | ||
| p. 89 | 9 | Diagram needs to be redrawn. The remainder is 4/5 of the total. Draw another bar under the ramainder (4 parts) and divide into 5 units. Shade two of them, the remaining 3 unshaded units is $72. Problem could be solved by setting 3 units = $72, so 5 units (the remainder) = $120, which is 4/5 of the original. So 4 parts = $120, and 5 parts = $150. Or, each of the 4/5 parts could be divided up into 5 equal little units, so the total is 25 units, and 12 of them (2/5 of the remainder) is $72, so 25 of them is $150. | ||
| p. 90 | notes | Last equation is Time = Distance/Speed | ||
| p. 95 | 19 | US on top of names to the right of the drawaing should be 3d. | ||
| p. 102 | 27 | Names for 3d edition are Gopal and Raju. | ||
| p. 107 | 4.(a) |  |
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| p. 114 | 1.(f) | answer should be 2 1/12 | ||
| p. 131 | US Exercise number is 22. | |||
