| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 2.58 |
| Score | 0% | 52% |
Simplify \( \sqrt{125} \)
| 9\( \sqrt{10} \) | |
| 5\( \sqrt{5} \) | |
| 3\( \sqrt{5} \) | |
| 9\( \sqrt{5} \) |
To simplify a radical, factor out the perfect squares:
\( \sqrt{125} \)
\( \sqrt{25 \times 5} \)
\( \sqrt{5^2 \times 5} \)
5\( \sqrt{5} \)
The total water usage for a city is 20,000 gallons each day. Of that total, 28% is for personal use and 52% is for industrial use. How many more gallons of water each day is consumed for industrial use over personal use?
| 3,300 | |
| 4,800 | |
| 2,800 | |
| 1,100 |
52% of the water consumption is industrial use and 28% is personal use so (52% - 28%) = 24% more water is used for industrial purposes. 20,000 gallons are consumed daily so industry consumes \( \frac{24}{100} \) x 20,000 gallons = 4,800 gallons.
Cooks are needed to prepare for a large party. Each cook can bake either 5 large cakes or 12 small cakes per hour. The kitchen is available for 4 hours and 25 large cakes and 100 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?
| 13 | |
| 7 | |
| 5 | |
| 8 |
If a single cook can bake 5 large cakes per hour and the kitchen is available for 4 hours, a single cook can bake 5 x 4 = 20 large cakes during that time. 25 large cakes are needed for the party so \( \frac{25}{20} \) = 1\(\frac{1}{4}\) 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. 100 small cakes are needed for the party so \( \frac{100}{48} \) = 2\(\frac{1}{12}\) 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 + 3 = 5 cooks.
What is \( 7 \)\( \sqrt{48} \) + \( 5 \)\( \sqrt{3} \)
| 12\( \sqrt{3} \) | |
| 33\( \sqrt{3} \) | |
| 12\( \sqrt{144} \) | |
| 35\( \sqrt{144} \) |
To add these radicals together their radicands must be the same:
7\( \sqrt{48} \) + 5\( \sqrt{3} \)
7\( \sqrt{16 \times 3} \) + 5\( \sqrt{3} \)
7\( \sqrt{4^2 \times 3} \) + 5\( \sqrt{3} \)
(7)(4)\( \sqrt{3} \) + 5\( \sqrt{3} \)
28\( \sqrt{3} \) + 5\( \sqrt{3} \)
Now that the radicands are identical, you can add them together:
28\( \sqrt{3} \) + 5\( \sqrt{3} \)What is \( \frac{9}{4} \) - \( \frac{3}{10} \)?
| 1 \( \frac{7}{10} \) | |
| 2 \( \frac{7}{20} \) | |
| 1\(\frac{19}{20}\) | |
| 1 \( \frac{4}{20} \) |
To subtract these fractions, first find the lowest common multiple of their denominators. The first few multiples of 4 are [4, 8, 12, 16, 20, 24, 28, 32, 36, 40] and the first few multiples of 10 are [10, 20, 30, 40, 50, 60, 70, 80, 90]. The first few multiples they share are [20, 40, 60, 80] making 20 the smallest multiple 4 and 10 share.
Next, convert the fractions so each denominator equals the lowest common multiple:
\( \frac{9 x 5}{4 x 5} \) - \( \frac{3 x 2}{10 x 2} \)
\( \frac{45}{20} \) - \( \frac{6}{20} \)
Now, because the fractions share a common denominator, you can subtract them:
\( \frac{45 - 6}{20} \) = \( \frac{39}{20} \) = 1\(\frac{19}{20}\)