| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 2.65 |
| Score | 0% | 53% |
Cooks are needed to prepare for a large party. Each cook can bake either 2 large cakes or 18 small cakes per hour. The kitchen is available for 3 hours and 28 large cakes and 240 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?
| 8 | |
| 14 | |
| 10 | |
| 9 |
If a single cook can bake 2 large cakes per hour and the kitchen is available for 3 hours, a single cook can bake 2 x 3 = 6 large cakes during that time. 28 large cakes are needed for the party so \( \frac{28}{6} \) = 4\(\frac{2}{3}\) cooks are needed to bake the required number of large cakes.
If a single cook can bake 18 small cakes per hour and the kitchen is available for 3 hours, a single cook can bake 18 x 3 = 54 small cakes during that time. 240 small cakes are needed for the party so \( \frac{240}{54} \) = 4\(\frac{4}{9}\) 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 5 + 5 = 10 cooks.
What is \( \sqrt{\frac{49}{49}} \)?
| 1 | |
| \(\frac{3}{4}\) | |
| \(\frac{2}{3}\) | |
| \(\frac{7}{8}\) |
To take the square root of a fraction, break the fraction into two separate roots then calculate the square root of the numerator and denominator separately:
\( \sqrt{\frac{49}{49}} \)
\( \frac{\sqrt{49}}{\sqrt{49}} \)
\( \frac{\sqrt{7^2}}{\sqrt{7^2}} \)
1
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?
| 162 m2 | |
| 50 m2 | |
| 18 m2 | |
| 128 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
What is \( 7 \)\( \sqrt{48} \) - \( 8 \)\( \sqrt{3} \)
| -1\( \sqrt{16} \) | |
| 20\( \sqrt{3} \) | |
| -1\( \sqrt{3} \) | |
| -1\( \sqrt{-7} \) |
To subtract these radicals together their radicands must be the same:
7\( \sqrt{48} \) - 8\( \sqrt{3} \)
7\( \sqrt{16 \times 3} \) - 8\( \sqrt{3} \)
7\( \sqrt{4^2 \times 3} \) - 8\( \sqrt{3} \)
(7)(4)\( \sqrt{3} \) - 8\( \sqrt{3} \)
28\( \sqrt{3} \) - 8\( \sqrt{3} \)
Now that the radicands are identical, you can subtract them:
28\( \sqrt{3} \) - 8\( \sqrt{3} \)A triathlon course includes a 100m swim, a 40.5km bike ride, and a 15.4km run. What is the total length of the race course?
| 49.9km | |
| 57.1km | |
| 52.5km | |
| 56km |
To add these distances, they must share the same unit so first you need to first convert the swim distance from meters (m) to kilometers (km) before adding it to the bike and run distances which are already in km. To convert 100 meters to kilometers, divide the distance by 1000 to get 0.1km then add the remaining distances:
total distance = swim + bike + run
total distance = 0.1km + 40.5km + 15.4km
total distance = 56km