| Problem | Example | Equation or Rule | Solution using logic |
| Dividing by fractions | \(\huge {2{1\over 2} \over {1\over 8}}\) |
Rule:
Convert all mixed numbers to improper fractions,
invert the second fraction and multiply:
\(\Large {2{1\over 2} \div {1\over 8}} = {5\over 2} \times {8\over 1}\)
\(\Large {5\over 2} \times {8\over 1} = {40\over 2} = \normalsize 20\)
|
- How many 8ths are there in 2 whole numbers? \(16\)
- How many 8ths are there in a half: \(4\)
- \( 16+4 = 20\)
Note: If the numerator of the dividing fraction is not 1, then the answer must be divided by it. For example, if the problem was \(\large {2{1\over 2} \div {5\over 8}}\) then the answer must be divided by 5:
\(\Large {20\over 5} \normalsize = 4\)
|
| Find the \(n\)th term of an arithmetic sequence | Find the 14th term of this sequence:
\(3 \quad 6 \quad 9 ...\) |
Common difference \(d\) = 3
Equation:
\(a_n = a_1 + d(n - 1)\)
\(a_n = 3 + 3\times (14-1)\)
\(a_n = 3 + 3\times 13\)
\(a_n = 3 + 39 = 42\)
| The common difference is \(3\), so we are adding 13 of them (one less than the total number of terms) to the first one, \(3\), so the answer is \(3 + 3\times 13 = 3 + 39 = 42\)
We add one less than the total number of terms because we already have the first one! |
| Permutations | How many ways can you order 3 letters out of the word CANDY? | Equation, using factorials:
\(\large P =\Large {n!\over (n - r)!}\)
Where:
\(n\) is the number of letters and
\(r\) is the number of letters to be selected.
\(\large P = \Large {5!\over (5-3)!}\)
\(\large P = {120\over 2} = \normalsize 60\)
| Using logic:
- You select one of the 5 letters:
C, A, N, D, or Y.
- Then you select the 2nd letter from the remaining 4
- After that, you select the 3rd letter from the 3 remaining.
So, summarizing:
Ways to select the 1st letter: 5
Ways to select the 2nd letter: 4
Ways to select the 3rd letter: 3
\( 5\times 4\times 3 = 60\)
After selecting each letter, there is 1 less letter to select from. Then, just multiply! |
