| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 3.14 |
| Score | 0% | 63% |
What is the least common multiple of 2 and 10?
| 11 | |
| 7 | |
| 10 | |
| 17 |
The first few multiples of 2 are [2, 4, 6, 8, 10, 12, 14, 16, 18, 20] and the first few multiples of 10 are [10, 20, 30, 40, 50, 60, 70, 80, 90]. The first few multiples they share are [10, 20, 30, 40, 50] making 10 the smallest multiple 2 and 10 have in common.
A sports card collection contains football, baseball, and basketball cards. If the ratio of football to baseball cards is 7 to 2 and the ratio of baseball to basketball cards is 7 to 1, what is the ratio of football to basketball cards?
| 49:2 | |
| 3:8 | |
| 5:1 | |
| 7:1 |
The ratio of football cards to baseball cards is 7:2 and the ratio of baseball cards to basketball cards is 7:1. To solve this problem, we need the baseball card side of each ratio to be equal so we need to rewrite the ratios in terms of a common number of baseball cards. (Think of this like finding the common denominator when adding fractions.) The ratio of football to baseball cards can also be written as 49:14 and the ratio of baseball cards to basketball cards as 14:2. So, the ratio of football cards to basketball cards is football:baseball, baseball:basketball or 49:14, 14:2 which reduces to 49:2.
If a rectangle is twice as long as it is wide and has a perimeter of 48 meters, what is the area of the rectangle?
| 162 m2 | |
| 8 m2 | |
| 50 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 48 meters so the equation becomes: 2w + 2h = 48.
Putting these two equations together and solving for width (w):
2w + 2h = 48
w + h = \( \frac{48}{2} \)
w + h = 24
w = 24 - h
From the question we know that h = 2w so substituting 2w for h gives us:
w = 24 - 2w
3w = 24
w = \( \frac{24}{3} \)
w = 8
Since h = 2w that makes h = (2 x 8) = 16 and the area = h x w = 8 x 16 = 128 m2
If all of a roofing company's 20 workers are required to staff 5 roofing crews, how many workers need to be added during the busy season in order to send 9 complete crews out on jobs?
| 10 | |
| 16 | |
| 11 | |
| 4 |
In order to find how many additional workers are needed to staff the extra crews you first need to calculate how many workers are on a crew. There are 20 workers at the company now and that's enough to staff 5 crews so there are \( \frac{20}{5} \) = 4 workers on a crew. 9 crews are needed for the busy season which, at 4 workers per crew, means that the roofing company will need 9 x 4 = 36 total workers to staff the crews during the busy season. The company already employs 20 workers so they need to add 36 - 20 = 16 new staff for the busy season.
If a car travels 560 miles in 8 hours, what is the average speed?
| 45 mph | |
| 25 mph | |
| 60 mph | |
| 70 mph |
Average speed in miles per hour is the number of miles traveled divided by the number of hours:
speed = \( \frac{\text{distance}}{\text{time}} \)