| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 2.59 |
| Score | 0% | 52% |
On average, the center for a basketball team hits 50% of his shots while a guard on the same team hits 60% of his shots. If the guard takes 15 shots during a game, how many shots will the center have to take to score as many points as the guard assuming each shot is worth the same number of points?
| 18 | |
| 25 | |
| 19 | |
| 15 |
guard shots made = shots taken x \( \frac{\text{% made}}{100} \) = 15 x \( \frac{60}{100} \) = \( \frac{60 x 15}{100} \) = \( \frac{900}{100} \) = 9 shots
The center makes 50% of his shots so he'll have to take:
shots made = shots taken x \( \frac{\text{% made}}{100} \)
shots taken = \( \frac{\text{shots taken}}{\frac{\text{% made}}{100}} \)
to make as many shots as the guard. Plugging in values for the center gives us:
center shots taken = \( \frac{9}{\frac{50}{100}} \) = 9 x \( \frac{100}{50} \) = \( \frac{9 x 100}{50} \) = \( \frac{900}{50} \) = 18 shots
to make the same number of shots as the guard and thus score the same number of points.
In a class of 25 students, 10 are taking German and 12 are taking Spanish. Of the students studying German or Spanish, 6 are taking both courses. How many students are not enrolled in either course?
| 9 | |
| 15 | |
| 24 | |
| 21 |
The number of students taking German or Spanish is 10 + 12 = 22. Of that group of 22, 6 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 22 - 6 = 16 who are taking at least one language. 25 - 16 = 9 students who are not taking either language.
A circular logo is enlarged to fit the lid of a jar. The new diameter is 40% larger than the original. By what percentage has the area of the logo increased?
| 22\(\frac{1}{2}\)% | |
| 30% | |
| 32\(\frac{1}{2}\)% | |
| 20% |
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 40% the radius (and, consequently, the total area) increases by \( \frac{40\text{%}}{2} \) = 20%
What is \( \frac{3}{6} \) - \( \frac{5}{14} \)?
| \(\frac{1}{7}\) | |
| \( \frac{5}{10} \) | |
| \( \frac{5}{9} \) | |
| 1 \( \frac{7}{13} \) |
To subtract 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 14 are [14, 28, 42, 56, 70, 84, 98]. The first few multiples they share are [42, 84] making 42 the smallest multiple 6 and 14 share.
Next, convert the fractions so each denominator equals the lowest common multiple:
\( \frac{3 x 7}{6 x 7} \) - \( \frac{5 x 3}{14 x 3} \)
\( \frac{21}{42} \) - \( \frac{15}{42} \)
Now, because the fractions share a common denominator, you can subtract them:
\( \frac{21 - 15}{42} \) = \( \frac{6}{42} \) = \(\frac{1}{7}\)
Cooks are needed to prepare for a large party. Each cook can bake either 3 large cakes or 11 small cakes per hour. The kitchen is available for 3 hours and 24 large cakes and 440 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?
| 11 | |
| 17 | |
| 8 | |
| 9 |
If a single cook can bake 3 large cakes per hour and the kitchen is available for 3 hours, a single cook can bake 3 x 3 = 9 large cakes during that time. 24 large cakes are needed for the party so \( \frac{24}{9} \) = 2\(\frac{2}{3}\) cooks are needed to bake the required number of large cakes.
If a single cook can bake 11 small cakes per hour and the kitchen is available for 3 hours, a single cook can bake 11 x 3 = 33 small cakes during that time. 440 small cakes are needed for the party so \( \frac{440}{33} \) = 13\(\frac{1}{3}\) 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 + 14 = 17 cooks.