| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 2.93 |
| Score | 0% | 59% |
In a class of 23 students, 5 are taking German and 8 are taking Spanish. Of the students studying German or Spanish, 2 are taking both courses. How many students are not enrolled in either course?
| 22 | |
| 12 | |
| 10 | |
| 11 |
The number of students taking German or Spanish is 5 + 8 = 13. Of that group of 13, 2 are taking both languages so they've been counted twice (once in the German group and once in the Spanish group). Subtracting them out leaves 13 - 2 = 11 who are taking at least one language. 23 - 11 = 12 students who are not taking either language.
Cooks are needed to prepare for a large party. Each cook can bake either 4 large cakes or 10 small cakes per hour. The kitchen is available for 2 hours and 22 large cakes and 310 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?
| 5 | |
| 15 | |
| 10 | |
| 19 |
If a single cook can bake 4 large cakes per hour and the kitchen is available for 2 hours, a single cook can bake 4 x 2 = 8 large cakes during that time. 22 large cakes are needed for the party so \( \frac{22}{8} \) = 2\(\frac{3}{4}\) cooks are needed to bake the required number of large cakes.
If a single cook can bake 10 small cakes per hour and the kitchen is available for 2 hours, a single cook can bake 10 x 2 = 20 small cakes during that time. 310 small cakes are needed for the party so \( \frac{310}{20} \) = 15\(\frac{1}{2}\) 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 3 + 16 = 19 cooks.
If all of a roofing company's 6 workers are required to staff 3 roofing crews, how many workers need to be added during the busy season in order to send 7 complete crews out on jobs?
| 15 | |
| 16 | |
| 17 | |
| 8 |
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 6 workers at the company now and that's enough to staff 3 crews so there are \( \frac{6}{3} \) = 2 workers on a crew. 7 crews are needed for the busy season which, at 2 workers per crew, means that the roofing company will need 7 x 2 = 14 total workers to staff the crews during the busy season. The company already employs 6 workers so they need to add 14 - 6 = 8 new staff for the busy season.
\({b + c \over a} = {b \over a} + {c \over a}\) defines which of the following?
distributive property for multiplication |
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commutative property for multiplication |
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commutative property for division |
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distributive 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, \({a^3 + 6a^2 \over a^2} = {a^3 \over a^2} + {6a^2 \over a^2} = a + 6\).
What is the greatest common factor of 48 and 68?
| 8 | |
| 4 | |
| 42 | |
| 40 |
The factors of 48 are [1, 2, 3, 4, 6, 8, 12, 16, 24, 48] and the factors of 68 are [1, 2, 4, 17, 34, 68]. They share 3 factors [1, 2, 4] making 4 the greatest factor 48 and 68 have in common.