| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 2.82 |
| Score | 0% | 56% |
Cooks are needed to prepare for a large party. Each cook can bake either 2 large cakes or 16 small cakes per hour. The kitchen is available for 4 hours and 26 large cakes and 110 small cakes need to be baked.
How many cooks are required to bake the required number of cakes during the time the kitchen is available?
| 6 | |
| 33 | |
| 15 | |
| 13 |
If a single cook can bake 2 large cakes per hour and the kitchen is available for 4 hours, a single cook can bake 2 x 4 = 8 large cakes during that time. 26 large cakes are needed for the party so \( \frac{26}{8} \) = 3\(\frac{1}{4}\) cooks are needed to bake the required number of large cakes.
If a single cook can bake 16 small cakes per hour and the kitchen is available for 4 hours, a single cook can bake 16 x 4 = 64 small cakes during that time. 110 small cakes are needed for the party so \( \frac{110}{64} \) = 1\(\frac{23}{32}\) cooks are needed to bake the required number of small cakes.
Because you can't employ a fractional cook, round the number of cooks needed for each type of cake up to the next whole number resulting in 4 + 2 = 6 cooks.
Solve for \( \frac{3!}{5!} \)
| \( \frac{1}{20} \) | |
| \( \frac{1}{15120} \) | |
| 15120 | |
| \( \frac{1}{72} \) |
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:
\( \frac{3!}{5!} \)
\( \frac{3 \times 2 \times 1}{5 \times 4 \times 3 \times 2 \times 1} \)
\( \frac{1}{5 \times 4} \)
\( \frac{1}{20} \)
If all of a roofing company's 15 workers are required to staff 5 roofing crews, how many workers need to be added during the busy season in order to send 9 complete crews out on jobs?
| 16 | |
| 15 | |
| 14 | |
| 12 |
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 15 workers at the company now and that's enough to staff 5 crews so there are \( \frac{15}{5} \) = 3 workers on a crew. 9 crews are needed for the busy season which, at 3 workers per crew, means that the roofing company will need 9 x 3 = 27 total workers to staff the crews during the busy season. The company already employs 15 workers so they need to add 27 - 15 = 12 new staff for the busy season.
If a rectangle is twice as long as it is wide and has a perimeter of 42 meters, what is the area of the rectangle?
| 72 m2 | |
| 18 m2 | |
| 98 m2 | |
| 128 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 42 meters so the equation becomes: 2w + 2h = 42.
Putting these two equations together and solving for width (w):
2w + 2h = 42
w + h = \( \frac{42}{2} \)
w + h = 21
w = 21 - h
From the question we know that h = 2w so substituting 2w for h gives us:
w = 21 - 2w
3w = 21
w = \( \frac{21}{3} \)
w = 7
Since h = 2w that makes h = (2 x 7) = 14 and the area = h x w = 7 x 14 = 98 m2
Which of the following is an improper fraction?
\({7 \over 5} \) |
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\({2 \over 5} \) |
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\({a \over 5} \) |
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\(1 {2 \over 5} \) |
A rational number (or fraction) is represented as a ratio between two integers, a and b, and has the form \({a \over b}\) where a is the numerator and b is the denominator. An improper fraction (\({5 \over 3} \)) has a numerator with a greater absolute value than the denominator and can be converted into a mixed number (\(1 {2 \over 3} \)) which has a whole number part and a fractional part.