| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 2.64 |
| Score | 0% | 53% |
What is \( \frac{3}{3} \) + \( \frac{7}{9} \)?
| 2 \( \frac{1}{8} \) | |
| \( \frac{1}{9} \) | |
| 1\(\frac{7}{9}\) | |
| \( \frac{6}{9} \) |
To add these fractions, first find the lowest common multiple of their denominators. The first few multiples of 3 are [3, 6, 9, 12, 15, 18, 21, 24, 27, 30] and the first few multiples of 9 are [9, 18, 27, 36, 45, 54, 63, 72, 81, 90]. The first few multiples they share are [9, 18, 27, 36, 45] making 9 the smallest multiple 3 and 9 share.
Next, convert the fractions so each denominator equals the lowest common multiple:
\( \frac{3 x 3}{3 x 3} \) + \( \frac{7 x 1}{9 x 1} \)
\( \frac{9}{9} \) + \( \frac{7}{9} \)
Now, because the fractions share a common denominator, you can add them:
\( \frac{9 + 7}{9} \) = \( \frac{16}{9} \) = 1\(\frac{7}{9}\)
Cooks are needed to prepare for a large party. Each cook can bake either 3 large cakes or 17 small cakes per hour. The kitchen is available for 4 hours and 34 large cakes and 350 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 | |
| 13 | |
| 15 | |
| 12 |
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. 34 large cakes are needed for the party so \( \frac{34}{12} \) = 2\(\frac{5}{6}\) cooks are needed to bake the required number of large cakes.
If a single cook can bake 17 small cakes per hour and the kitchen is available for 4 hours, a single cook can bake 17 x 4 = 68 small cakes during that time. 350 small cakes are needed for the party so \( \frac{350}{68} \) = 5\(\frac{5}{34}\) 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 + 6 = 9 cooks.
If the ratio of home fans to visiting fans in a crowd is 5:1 and all 48,000 seats in a stadium are filled, how many home fans are in attendance?
| 28,500 | |
| 26,667 | |
| 40,000 | |
| 34,167 |
A ratio of 5:1 means that there are 5 home fans for every one visiting fan. So, of every 6 fans, 5 are home fans and \( \frac{5}{6} \) of every fan in the stadium is a home fan:
48,000 fans x \( \frac{5}{6} \) = \( \frac{240000}{6} \) = 40,000 fans.
On average, the center for a basketball team hits 30% of his shots while a guard on the same team hits 35% of his shots. If the guard takes 10 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?
| 10 | |
| 11 | |
| 8 | |
| 6 |
guard shots made = shots taken x \( \frac{\text{% made}}{100} \) = 10 x \( \frac{35}{100} \) = \( \frac{35 x 10}{100} \) = \( \frac{350}{100} \) = 3 shots
The center makes 30% 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 = \( \frac{3}{\frac{30}{100}} \) = 3 x \( \frac{100}{30} \) = \( \frac{3 x 100}{30} \) = \( \frac{300}{30} \) = 10 shots
to make the same number of shots as the guard and thus score the same number of points.
What is -3b4 - b4?
| -4b4 | |
| -2b8 | |
| -2b-8 | |
| -4b-4 |
To add or subtract terms with exponents, both the base and the exponent must be the same. In this case they are so subtract the coefficients and retain the base and exponent:
-3b4 - 1b4
(-3 - 1)b4
-4b4