| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 3.56 |
| Score | 0% | 71% |
How many 1\(\frac{1}{2}\) gallon cans worth of fuel would you need to pour into an empty 9 gallon tank to fill it exactly halfway?
| 3 | |
| 9 | |
| 3 | |
| 2 |
To fill a 9 gallon tank exactly halfway you'll need 4\(\frac{1}{2}\) gallons of fuel. Each fuel can holds 1\(\frac{1}{2}\) gallons so:
cans = \( \frac{4\frac{1}{2} \text{ gallons}}{1\frac{1}{2} \text{ gallons}} \) = 3
What is the next number in this sequence: 1, 2, 3, 4, 5, __________ ?
| -2 | |
| 6 | |
| 2 | |
| 10 |
The equation for this sequence is:
an = an-1 + 1
where n is the term's order in the sequence, an is the value of the term, and an-1 is the value of the term before an. This makes the next number:
a6 = a5 + 1
a6 = 5 + 1
a6 = 6
How many hours does it take a car to travel 225 miles at an average speed of 45 miles per hour?
| 5 hours | |
| 2 hours | |
| 1 hour | |
| 6 hours |
Average speed in miles per hour is the number of miles traveled divided by the number of hours:
speed = \( \frac{\text{distance}}{\text{time}} \)Solving for time:
time = \( \frac{\text{distance}}{\text{speed}} \)
time = \( \frac{225mi}{45mph} \)
5 hours
A circular logo is enlarged to fit the lid of a jar. The new diameter is 75% larger than the original. By what percentage has the area of the logo increased?
| 27\(\frac{1}{2}\)% | |
| 30% | |
| 35% | |
| 37\(\frac{1}{2}\)% |
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 75% the radius (and, consequently, the total area) increases by \( \frac{75\text{%}}{2} \) = 37\(\frac{1}{2}\)%
What is the greatest common factor of 68 and 60?
| 4 | |
| 59 | |
| 27 | |
| 55 |
The factors of 68 are [1, 2, 4, 17, 34, 68] and the factors of 60 are [1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60]. They share 3 factors [1, 2, 4] making 4 the greatest factor 68 and 60 have in common.