| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 2.60 |
| Score | 0% | 52% |
What is \( \frac{5}{6} \) + \( \frac{6}{12} \)?
| 1 \( \frac{3}{10} \) | |
| 1 \( \frac{4}{10} \) | |
| 1\(\frac{1}{3}\) | |
| \( \frac{9}{12} \) |
To add these fractions, first find the lowest common multiple of their denominators. The first few multiples of 6 are [6, 12, 18, 24, 30, 36, 42, 48, 54, 60] 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 6 and 12 share.
Next, convert the fractions so each denominator equals the lowest common multiple:
\( \frac{5 x 2}{6 x 2} \) + \( \frac{6 x 1}{12 x 1} \)
\( \frac{10}{12} \) + \( \frac{6}{12} \)
Now, because the fractions share a common denominator, you can add them:
\( \frac{10 + 6}{12} \) = \( \frac{16}{12} \) = 1\(\frac{1}{3}\)
Solve 5 + (2 + 5) ÷ 3 x 2 - 52
| \(\frac{3}{8}\) | |
| -15\(\frac{1}{3}\) | |
| 2\(\frac{1}{3}\) | |
| 1\(\frac{1}{8}\) |
Use PEMDAS (Parentheses, Exponents, Multipy/Divide, Add/Subtract):
5 + (2 + 5) ÷ 3 x 2 - 52
P: 5 + (7) ÷ 3 x 2 - 52
E: 5 + 7 ÷ 3 x 2 - 25
MD: 5 + \( \frac{7}{3} \) x 2 - 25
MD: 5 + \( \frac{14}{3} \) - 25
AS: \( \frac{15}{3} \) + \( \frac{14}{3} \) - 25
AS: \( \frac{29}{3} \) - 25
AS: \( \frac{29 - 75}{3} \)
\( \frac{-46}{3} \)
-15\(\frac{1}{3}\)
Cooks are needed to prepare for a large party. Each cook can bake either 3 large cakes or 12 small cakes per hour. The kitchen is available for 4 hours and 21 large cakes and 230 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?
| 7 | |
| 13 | |
| 8 | |
| 5 |
If a single cook can bake 3 large cakes per hour and the kitchen is available for 4 hours, a single cook can bake 3 x 4 = 12 large cakes during that time. 21 large cakes are needed for the party so \( \frac{21}{12} \) = 1\(\frac{3}{4}\) cooks are needed to bake the required number of large cakes.
If a single cook can bake 12 small cakes per hour and the kitchen is available for 4 hours, a single cook can bake 12 x 4 = 48 small cakes during that time. 230 small cakes are needed for the party so \( \frac{230}{48} \) = 4\(\frac{19}{24}\) 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 2 + 5 = 7 cooks.
A circular logo is enlarged to fit the lid of a jar. The new diameter is 50% larger than the original. By what percentage has the area of the logo increased?
| 20% | |
| 32\(\frac{1}{2}\)% | |
| 25% | |
| 30% |
The area of a circle is given by the formula A = πr2 where r is the radius of the circle. The radius of a circle is its diameter divided by two so A = π(\( \frac{d}{2} \))2. If the diameter of the logo increases by 50% the radius (and, consequently, the total area) increases by \( \frac{50\text{%}}{2} \) = 25%
If all of a roofing company's 6 workers are required to staff 2 roofing crews, how many workers need to be added during the busy season in order to send 6 complete crews out on jobs?
| 17 | |
| 5 | |
| 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 6 workers at the company now and that's enough to staff 2 crews so there are \( \frac{6}{2} \) = 3 workers on a crew. 6 crews are needed for the busy season which, at 3 workers per crew, means that the roofing company will need 6 x 3 = 18 total workers to staff the crews during the busy season. The company already employs 6 workers so they need to add 18 - 6 = 12 new staff for the busy season.