| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 2.83 |
| Score | 0% | 57% |
The total water usage for a city is 25,000 gallons each day. Of that total, 35% is for personal use and 67% is for industrial use. How many more gallons of water each day is consumed for industrial use over personal use?
| 7,200 | |
| 4,550 | |
| 8,000 | |
| 9,500 |
67% of the water consumption is industrial use and 35% is personal use so (67% - 35%) = 32% more water is used for industrial purposes. 25,000 gallons are consumed daily so industry consumes \( \frac{32}{100} \) x 25,000 gallons = 8,000 gallons.
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 2 hours and 34 large cakes and 340 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?
| 19 | |
| 12 | |
| 15 | |
| 9 |
If a single cook can bake 2 large cakes per hour and the kitchen is available for 2 hours, a single cook can bake 2 x 2 = 4 large cakes during that time. 34 large cakes are needed for the party so \( \frac{34}{4} \) = 8\(\frac{1}{2}\) 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 2 hours, a single cook can bake 18 x 2 = 36 small cakes during that time. 340 small cakes are needed for the party so \( \frac{340}{36} \) = 9\(\frac{4}{9}\) 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 9 + 10 = 19 cooks.
What is \( \frac{2}{4} \) - \( \frac{2}{6} \)?
| 2 \( \frac{3}{12} \) | |
| \( \frac{2}{10} \) | |
| 2 \( \frac{3}{11} \) | |
| \(\frac{1}{6}\) |
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 6 are [6, 12, 18, 24, 30, 36, 42, 48, 54, 60]. The first few multiples they share are [12, 24, 36, 48, 60] making 12 the smallest multiple 4 and 6 share.
Next, convert the fractions so each denominator equals the lowest common multiple:
\( \frac{2 x 3}{4 x 3} \) - \( \frac{2 x 2}{6 x 2} \)
\( \frac{6}{12} \) - \( \frac{4}{12} \)
Now, because the fractions share a common denominator, you can subtract them:
\( \frac{6 - 4}{12} \) = \( \frac{2}{12} \) = \(\frac{1}{6}\)
What is \( \frac{3x^5}{8x^2} \)?
| 2\(\frac{2}{3}\)x7 | |
| \(\frac{3}{8}\)x3 | |
| \(\frac{3}{8}\)x\(\frac{2}{5}\) | |
| \(\frac{3}{8}\)x-3 |
To divide terms with exponents, the base of both exponents must be the same. In this case they are so divide the coefficients and subtract the exponents:
\( \frac{3x^5}{8x^2} \)
\( \frac{3}{8} \) x(5 - 2)
\(\frac{3}{8}\)x3
If \( \left|z + 9\right| \) + 1 = 5, which of these is a possible value for z?
| -13 | |
| 2 | |
| -12 | |
| -6 |
First, solve for \( \left|z + 9\right| \):
\( \left|z + 9\right| \) + 1 = 5
\( \left|z + 9\right| \) = 5 - 1
\( \left|z + 9\right| \) = 4
The value inside the absolute value brackets can be either positive or negative so (z + 9) must equal + 4 or -4 for \( \left|z + 9\right| \) to equal 4:
| z + 9 = 4 z = 4 - 9 z = -5 | z + 9 = -4 z = -4 - 9 z = -13 |
So, z = -13 or z = -5.