| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 2.92 |
| Score | 0% | 58% |
Which of the following is not a prime number?
2 |
|
7 |
|
9 |
|
5 |
A prime number is an integer greater than 1 that has no factors other than 1 and itself. Examples of prime numbers include 2, 3, 5, 7, and 11.
10 members of a bridal party need transported to a wedding reception but there are only 2 3-passenger taxis available to take them. How many will need to find other transportation?
| 7 | |
| 9 | |
| 4 | |
| 6 |
There are 2 3-passenger taxis available so that's 2 x 3 = 6 total seats. There are 10 people needing transportation leaving 10 - 6 = 4 who will have to find other transportation.
If a rectangle is twice as long as it is wide and has a perimeter of 6 meters, what is the area of the rectangle?
| 8 m2 | |
| 32 m2 | |
| 98 m2 | |
| 2 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 6 meters so the equation becomes: 2w + 2h = 6.
Putting these two equations together and solving for width (w):
2w + 2h = 6
w + h = \( \frac{6}{2} \)
w + h = 3
w = 3 - h
From the question we know that h = 2w so substituting 2w for h gives us:
w = 3 - 2w
3w = 3
w = \( \frac{3}{3} \)
w = 1
Since h = 2w that makes h = (2 x 1) = 2 and the area = h x w = 1 x 2 = 2 m2
A machine in a factory has an error rate of 8 parts per 100. The machine normally runs 24 hours a day and produces 9 parts per hour. Yesterday the machine was shut down for 6 hours for maintenance.
How many error-free parts did the machine produce yesterday?
| 149 | |
| 85.5 | |
| 113.1 | |
| 135.4 |
The hourly error rate for this machine is the error rate in parts per 100 multiplied by the number of parts produced per hour:
\( \frac{8}{100} \) x 9 = \( \frac{8 \times 9}{100} \) = \( \frac{72}{100} \) = 0.72 errors per hour
So, in an average hour, the machine will produce 9 - 0.72 = 8.28 error free parts.
The machine ran for 24 - 6 = 18 hours yesterday so you would expect that 18 x 8.28 = 149 error free parts were produced yesterday.
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 10 complete crews out on jobs?
| 4 | |
| 20 | |
| 1 | |
| 17 |
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. 10 crews are needed for the busy season which, at 4 workers per crew, means that the roofing company will need 10 x 4 = 40 total workers to staff the crews during the busy season. The company already employs 20 workers so they need to add 40 - 20 = 20 new staff for the busy season.