| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 2.75 |
| Score | 0% | 55% |
How many 2 gallon cans worth of fuel would you need to pour into an empty 12 gallon tank to fill it exactly halfway?
| 2 | |
| 3 | |
| 3 | |
| 6 |
To fill a 12 gallon tank exactly halfway you'll need 6 gallons of fuel. Each fuel can holds 2 gallons so:
cans = \( \frac{6 \text{ gallons}}{2 \text{ gallons}} \) = 3
Simplify \( \frac{20}{80} \).
| \( \frac{5}{8} \) | |
| \( \frac{10}{17} \) | |
| \( \frac{1}{4} \) | |
| \( \frac{3}{10} \) |
To simplify this fraction, first find the greatest common factor between them. The factors of 20 are [1, 2, 4, 5, 10, 20] and the factors of 80 are [1, 2, 4, 5, 8, 10, 16, 20, 40, 80]. They share 6 factors [1, 2, 4, 5, 10, 20] making 20 their greatest common factor (GCF).
Next, divide both numerator and denominator by the GCF:
\( \frac{20}{80} \) = \( \frac{\frac{20}{20}}{\frac{80}{20}} \) = \( \frac{1}{4} \)
The total water usage for a city is 30,000 gallons each day. Of that total, 11% is for personal use and 42% is for industrial use. How many more gallons of water each day is consumed for industrial use over personal use?
| 5,200 | |
| 9,900 | |
| 9,300 | |
| 5,000 |
42% of the water consumption is industrial use and 11% is personal use so (42% - 11%) = 31% more water is used for industrial purposes. 30,000 gallons are consumed daily so industry consumes \( \frac{31}{100} \) x 30,000 gallons = 9,300 gallons.
A circular logo is enlarged to fit the lid of a jar. The new diameter is 70% larger than the original. By what percentage has the area of the logo increased?
| 25% | |
| 35% | |
| 15% | |
| 20% |
The area of a circle is given by the formula A = πr2 where r is the radius of the circle. The radius of a circle is its diameter divided by two so A = π(\( \frac{d}{2} \))2. If the diameter of the logo increases by 70% the radius (and, consequently, the total area) increases by \( \frac{70\text{%}}{2} \) = 35%
What is \( 5 \)\( \sqrt{75} \) + \( 7 \)\( \sqrt{3} \)
| 32\( \sqrt{3} \) | |
| 35\( \sqrt{25} \) | |
| 35\( \sqrt{3} \) | |
| 12\( \sqrt{3} \) |
To add these radicals together their radicands must be the same:
5\( \sqrt{75} \) + 7\( \sqrt{3} \)
5\( \sqrt{25 \times 3} \) + 7\( \sqrt{3} \)
5\( \sqrt{5^2 \times 3} \) + 7\( \sqrt{3} \)
(5)(5)\( \sqrt{3} \) + 7\( \sqrt{3} \)
25\( \sqrt{3} \) + 7\( \sqrt{3} \)
Now that the radicands are identical, you can add them together:
25\( \sqrt{3} \) + 7\( \sqrt{3} \)