| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 3.22 |
| Score | 0% | 64% |
If the ratio of home fans to visiting fans in a crowd is 4:1 and all 42,000 seats in a stadium are filled, how many home fans are in attendance?
| 33,600 | |
| 28,333 | |
| 41,667 | |
| 28,000 |
A ratio of 4:1 means that there are 4 home fans for every one visiting fan. So, of every 5 fans, 4 are home fans and \( \frac{4}{5} \) of every fan in the stadium is a home fan:
42,000 fans x \( \frac{4}{5} \) = \( \frac{168000}{5} \) = 33,600 fans.
What is the least common multiple of 8 and 10?
| 43 | |
| 40 | |
| 67 | |
| 30 |
The first few multiples of 8 are [8, 16, 24, 32, 40, 48, 56, 64, 72, 80] and the first few multiples of 10 are [10, 20, 30, 40, 50, 60, 70, 80, 90]. The first few multiples they share are [40, 80] making 40 the smallest multiple 8 and 10 have in common.
Solve for \( \frac{6!}{3!} \)
| 7 | |
| \( \frac{1}{4} \) | |
| 120 | |
| \( \frac{1}{210} \) |
A factorial is the product of an integer and all the positive integers below it. To solve a fraction featuring factorials, expand the factorials and cancel out like numbers:
\( \frac{6!}{3!} \)
\( \frac{6 \times 5 \times 4 \times 3 \times 2 \times 1}{3 \times 2 \times 1} \)
\( \frac{6 \times 5 \times 4}{1} \)
\( 6 \times 5 \times 4 \)
120
If all of a roofing company's 4 workers are required to staff 2 roofing crews, how many workers need to be added during the busy season in order to send 4 complete crews out on jobs?
| 8 | |
| 12 | |
| 4 | |
| 13 |
In order to find how many additional workers are needed to staff the extra crews you first need to calculate how many workers are on a crew. There are 4 workers at the company now and that's enough to staff 2 crews so there are \( \frac{4}{2} \) = 2 workers on a crew. 4 crews are needed for the busy season which, at 2 workers per crew, means that the roofing company will need 4 x 2 = 8 total workers to staff the crews during the busy season. The company already employs 4 workers so they need to add 8 - 4 = 4 new staff for the busy season.
Simplify \( \frac{28}{52} \).
| \( \frac{7}{18} \) | |
| \( \frac{7}{13} \) | |
| \( \frac{3}{10} \) | |
| \( \frac{5}{8} \) |
To simplify this fraction, first find the greatest common factor between them. The factors of 28 are [1, 2, 4, 7, 14, 28] and the factors of 52 are [1, 2, 4, 13, 26, 52]. They share 3 factors [1, 2, 4] making 4 their greatest common factor (GCF).
Next, divide both numerator and denominator by the GCF:
\( \frac{28}{52} \) = \( \frac{\frac{28}{4}}{\frac{52}{4}} \) = \( \frac{7}{13} \)