| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 2.51 |
| Score | 0% | 50% |
On average, the center for a basketball team hits 30% of his shots while a guard on the same team hits 45% of his shots. If the guard takes 30 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?
| 65 | |
| 43 | |
| 39 | |
| 23 |
guard shots made = shots taken x \( \frac{\text{% made}}{100} \) = 30 x \( \frac{45}{100} \) = \( \frac{45 x 30}{100} \) = \( \frac{1350}{100} \) = 13 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 = \( \frac{13}{\frac{30}{100}} \) = 13 x \( \frac{100}{30} \) = \( \frac{13 x 100}{30} \) = \( \frac{1300}{30} \) = 43 shots
to make the same number of shots as the guard and thus score the same number of points.
If a rectangle is twice as long as it is wide and has a perimeter of 30 meters, what is the area of the rectangle?
| 72 m2 | |
| 50 m2 | |
| 162 m2 | |
| 8 m2 |
The area of a rectangle is width (w) x height (h). In this problem we know that the rectangle is twice as long as it is wide so h = 2w. The perimeter of a rectangle is 2w + 2h and we know that the perimeter of this rectangle is 30 meters so the equation becomes: 2w + 2h = 30.
Putting these two equations together and solving for width (w):
2w + 2h = 30
w + h = \( \frac{30}{2} \)
w + h = 15
w = 15 - h
From the question we know that h = 2w so substituting 2w for h gives us:
w = 15 - 2w
3w = 15
w = \( \frac{15}{3} \)
w = 5
Since h = 2w that makes h = (2 x 5) = 10 and the area = h x w = 5 x 10 = 50 m2
Solve 3 + (4 + 2) ÷ 5 x 2 - 22
| 1\(\frac{1}{6}\) | |
| 3\(\frac{1}{2}\) | |
| 1\(\frac{2}{5}\) | |
| 1 |
Use PEMDAS (Parentheses, Exponents, Multipy/Divide, Add/Subtract):
3 + (4 + 2) ÷ 5 x 2 - 22
P: 3 + (6) ÷ 5 x 2 - 22
E: 3 + 6 ÷ 5 x 2 - 4
MD: 3 + \( \frac{6}{5} \) x 2 - 4
MD: 3 + \( \frac{12}{5} \) - 4
AS: \( \frac{15}{5} \) + \( \frac{12}{5} \) - 4
AS: \( \frac{27}{5} \) - 4
AS: \( \frac{27 - 20}{5} \)
\( \frac{7}{5} \)
1\(\frac{2}{5}\)
| 1 | |
| 3.5 | |
| 1.2 | |
| 1.5 |
1
A tiger in a zoo has consumed 32 pounds of food in 4 days. If the tiger continues to eat at the same rate, in how many more days will its total food consumtion be 80 pounds?
| 1 | |
| 5 | |
| 6 | |
| 3 |
If the tiger has consumed 32 pounds of food in 4 days that's \( \frac{32}{4} \) = 8 pounds of food per day. The tiger needs to consume 80 - 32 = 48 more pounds of food to reach 80 pounds total. At 8 pounds of food per day that's \( \frac{48}{8} \) = 6 more days.