| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 2.66 |
| Score | 0% | 53% |
Cooks are needed to prepare for a large party. Each cook can bake either 5 large cakes or 13 small cakes per hour. The kitchen is available for 2 hours and 33 large cakes and 120 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?
| 9 | |
| 7 | |
| 8 | |
| 6 |
If a single cook can bake 5 large cakes per hour and the kitchen is available for 2 hours, a single cook can bake 5 x 2 = 10 large cakes during that time. 33 large cakes are needed for the party so \( \frac{33}{10} \) = 3\(\frac{3}{10}\) cooks are needed to bake the required number of large cakes.
If a single cook can bake 13 small cakes per hour and the kitchen is available for 2 hours, a single cook can bake 13 x 2 = 26 small cakes during that time. 120 small cakes are needed for the party so \( \frac{120}{26} \) = 4\(\frac{8}{13}\) 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 + 5 = 9 cooks.
\({b + c \over a} = {b \over a} + {c \over a}\) defines which of the following?
commutative property for multiplication |
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distributive property for division |
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distributive property for multiplication |
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commutative property for division |
The distributive property for division helps in solving expressions like \({b + c \over a}\). It specifies that the result of dividing a fraction with multiple terms in the numerator and one term in the denominator can be obtained by dividing each term individually and then totaling the results: \({b + c \over a} = {b \over a} + {c \over a}\). For example, \({a^3 + 6a^2 \over a^2} = {a^3 \over a^2} + {6a^2 \over a^2} = a + 6\).
Solve for \( \frac{5!}{3!} \)
| 20 | |
| 60480 | |
| 42 | |
| \( \frac{1}{9} \) |
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:
\( \frac{5!}{3!} \)
\( \frac{5 \times 4 \times 3 \times 2 \times 1}{3 \times 2 \times 1} \)
\( \frac{5 \times 4}{1} \)
\( 5 \times 4 \)
20
A machine in a factory has an error rate of 8 parts per 100. The machine normally runs 24 hours a day and produces 9 parts per hour. Yesterday the machine was shut down for 5 hours for maintenance.
How many error-free parts did the machine produce yesterday?
| 147.4 | |
| 157.3 | |
| 92.2 | |
| 156.4 |
The hourly error rate for this machine is the error rate in parts per 100 multiplied by the number of parts produced per hour:
\( \frac{8}{100} \) x 9 = \( \frac{8 \times 9}{100} \) = \( \frac{72}{100} \) = 0.72 errors per hour
So, in an average hour, the machine will produce 9 - 0.72 = 8.28 error free parts.
The machine ran for 24 - 5 = 19 hours yesterday so you would expect that 19 x 8.28 = 157.3 error free parts were produced yesterday.
If all of a roofing company's 10 workers are required to staff 5 roofing crews, how many workers need to be added during the busy season in order to send 8 complete crews out on jobs?
| 18 | |
| 2 | |
| 4 | |
| 6 |
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 10 workers at the company now and that's enough to staff 5 crews so there are \( \frac{10}{5} \) = 2 workers on a crew. 8 crews are needed for the busy season which, at 2 workers per crew, means that the roofing company will need 8 x 2 = 16 total workers to staff the crews during the busy season. The company already employs 10 workers so they need to add 16 - 10 = 6 new staff for the busy season.