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HANDBOOK MATH B Questions 51-100
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Question 81…how did you calculate the value 28 2/7 from 198/7 so quickly?
They probably used the fact that 7×3=21 or 7×30=210. 210 is 12 more than 198 so 7x(30-2)=7×28=210-14=196. There is 2 left over from 196 to 198. That is one way to break it up quickly in your head and get 198/7= 28 remainder 2.
I would do it differently myself. Once I had 198/7 and knew I had to subtract 25 I would convert 25 to a common denominator of 7 which is pretty easy. 25 x 7 = 175. The subtraction becomes 198/7 – 175/7 = (198-175)/7 = 23/7. Now it is much easier to calculate the result without any complex division…23/7 = 3 remainder 2 or 3 2/7. Hope that helps.
On problem 69 why did you start a prime factor tree with 9 which isn’t a prime factor?
Hi Kate. Technically it is a factor tree, but not a “prime” factor tree. It looks exactly the same. The only difference is that I began by using the sum of digits method on both numbers to determine divisibility by 9 at the start so I knew 9 was a common factor of both numbers. Breaking 9 down into 3 x 3 prime factors would just be an added step that wasn’t needed to calculate the Greatest Common Factor. Although you can always use the prime factors as the most fundamental building blocks to build up a GCF…it is a general foolproof approach…, you may find you can save a few steps if you immediately see a larger composite number is a common factor. Use the larger common factor and then factor out the rest like I did in this solution.