| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 3.00 |
| Score | 0% | 60% |
If all of a roofing company's 16 workers are required to staff 4 roofing crews, how many workers need to be added during the busy season in order to send 8 complete crews out on jobs?
| 7 | |
| 17 | |
| 16 | |
| 11 |
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 16 workers at the company now and that's enough to staff 4 crews so there are \( \frac{16}{4} \) = 4 workers on a crew. 8 crews are needed for the busy season which, at 4 workers per crew, means that the roofing company will need 8 x 4 = 32 total workers to staff the crews during the busy season. The company already employs 16 workers so they need to add 32 - 16 = 16 new staff for the busy season.
What is \( \frac{3}{6} \) ÷ \( \frac{3}{6} \)?
| 1 | |
| \(\frac{3}{16}\) | |
| \(\frac{8}{45}\) | |
| \(\frac{2}{63}\) |
To divide fractions, invert the second fraction and then multiply:
\( \frac{3}{6} \) ÷ \( \frac{3}{6} \) = \( \frac{3}{6} \) x \( \frac{6}{3} \)
To multiply fractions, multiply the numerators together and then multiply the denominators together:
\( \frac{3}{6} \) x \( \frac{6}{3} \) = \( \frac{3 x 6}{6 x 3} \) = \( \frac{18}{18} \) = 1
Convert y-2 to remove the negative exponent.
| \( \frac{-1}{y^{-2}} \) | |
| \( \frac{-2}{-y} \) | |
| \( \frac{1}{y^2} \) | |
| \( \frac{2}{y} \) |
To convert a negative exponent to a positive exponent, calculate the positive exponent then take the reciprocal.
What is \( \frac{8}{4} \) - \( \frac{5}{12} \)?
| 2 \( \frac{8}{12} \) | |
| 2 \( \frac{5}{11} \) | |
| 1\(\frac{7}{12}\) | |
| 1 \( \frac{5}{12} \) |
To subtract these fractions, first find the lowest common multiple of their denominators. The first few multiples of 4 are [4, 8, 12, 16, 20, 24, 28, 32, 36, 40] and the first few multiples of 12 are [12, 24, 36, 48, 60, 72, 84, 96]. The first few multiples they share are [12, 24, 36, 48, 60] making 12 the smallest multiple 4 and 12 share.
Next, convert the fractions so each denominator equals the lowest common multiple:
\( \frac{8 x 3}{4 x 3} \) - \( \frac{5 x 1}{12 x 1} \)
\( \frac{24}{12} \) - \( \frac{5}{12} \)
Now, because the fractions share a common denominator, you can subtract them:
\( \frac{24 - 5}{12} \) = \( \frac{19}{12} \) = 1\(\frac{7}{12}\)
If a rectangle is twice as long as it is wide and has a perimeter of 30 meters, what is the area of the rectangle?
| 2 m2 | |
| 72 m2 | |
| 50 m2 | |
| 32 m2 |
The area of a rectangle is width (w) x height (h). In this problem we know that the rectangle is twice as long as it is wide so h = 2w. The perimeter of a rectangle is 2w + 2h and we know that the perimeter of this rectangle is 30 meters so the equation becomes: 2w + 2h = 30.
Putting these two equations together and solving for width (w):
2w + 2h = 30
w + h = \( \frac{30}{2} \)
w + h = 15
w = 15 - h
From the question we know that h = 2w so substituting 2w for h gives us:
w = 15 - 2w
3w = 15
w = \( \frac{15}{3} \)
w = 5
Since h = 2w that makes h = (2 x 5) = 10 and the area = h x w = 5 x 10 = 50 m2