 Teacher resources and professional development across the curriculum

Teacher professional development and classroom resources across the curriculum            Solutions for Session 7, Part B

See solutions for Problems: B1 | B2 | B3 | B4    Problem B1

 a. Since this is a terminating decimal, 0.125 = 125/1,000, which can be reduced to 1/8. b. Use the method of multiplication. If F = 0.125125125..., then 1,000F = 125.125125125... Subtracting F from both sides gives you 999F = 125, so F = 125/999.   Problem B2

 a. Since this is a terminating decimal, 0.5436 = 5,436/10,000 = 1,359/2,500. b. F = 0.54365436.... Multiply by 10,000 to get 10,000F = 5,436.54365436.... Subtracting F from both sides gives you 9,999F = 5,436, so F = 5,436/9,999, which can be reduced to 604/1,111.   Problem B3

 a. 0.236 = 236/1,000 = 59/250. b. Note the difference; only the 36 is repeated! In this case, F = 0.2363636.... Multiply by 100 to get 100F = 23.6363636.... Subtracting F from both sides gives you 99F = 23.4, so F = 23.4/99, or 234/990, which can be reduced to 13/55.   Problem B4 Multiplying by 10 yields 10F = 1.111111.... Subtracting F from both sides gives you 9F = 1. Thus, F = 1/9. (It may seem counterintuitive at first that 0.11111111... = 1/9 since there are no 9s in the decimal!)     