| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 2.65 |
| Score | 0% | 53% |
What is \( 5 \)\( \sqrt{8} \) + \( 3 \)\( \sqrt{2} \)
| 15\( \sqrt{16} \) | |
| 15\( \sqrt{8} \) | |
| 15\( \sqrt{2} \) | |
| 13\( \sqrt{2} \) |
To add these radicals together their radicands must be the same:
5\( \sqrt{8} \) + 3\( \sqrt{2} \)
5\( \sqrt{4 \times 2} \) + 3\( \sqrt{2} \)
5\( \sqrt{2^2 \times 2} \) + 3\( \sqrt{2} \)
(5)(2)\( \sqrt{2} \) + 3\( \sqrt{2} \)
10\( \sqrt{2} \) + 3\( \sqrt{2} \)
Now that the radicands are identical, you can add them together:
10\( \sqrt{2} \) + 3\( \sqrt{2} \)A tiger in a zoo has consumed 70 pounds of food in 5 days. If the tiger continues to eat at the same rate, in how many more days will its total food consumtion be 140 pounds?
| 7 | |
| 6 | |
| 9 | |
| 5 |
If the tiger has consumed 70 pounds of food in 5 days that's \( \frac{70}{5} \) = 14 pounds of food per day. The tiger needs to consume 140 - 70 = 70 more pounds of food to reach 140 pounds total. At 14 pounds of food per day that's \( \frac{70}{14} \) = 5 more days.
A circular logo is enlarged to fit the lid of a jar. The new diameter is 55% larger than the original. By what percentage has the area of the logo increased?
| 15% | |
| 22\(\frac{1}{2}\)% | |
| 17\(\frac{1}{2}\)% | |
| 27\(\frac{1}{2}\)% |
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 55% the radius (and, consequently, the total area) increases by \( \frac{55\text{%}}{2} \) = 27\(\frac{1}{2}\)%
A machine in a factory has an error rate of 5 parts per 100. The machine normally runs 24 hours a day and produces 5 parts per hour. Yesterday the machine was shut down for 6 hours for maintenance.
How many error-free parts did the machine produce yesterday?
| 86.4 | |
| 172.9 | |
| 85.5 | |
| 97.9 |
The hourly error rate for this machine is the error rate in parts per 100 multiplied by the number of parts produced per hour:
\( \frac{5}{100} \) x 5 = \( \frac{5 \times 5}{100} \) = \( \frac{25}{100} \) = 0.25 errors per hour
So, in an average hour, the machine will produce 5 - 0.25 = 4.75 error free parts.
The machine ran for 24 - 6 = 18 hours yesterday so you would expect that 18 x 4.75 = 85.5 error free parts were produced yesterday.
Find the average of the following numbers: 7, 5, 8, 4.
| 5 | |
| 2 | |
| 9 | |
| 6 |
To find the average of these 4 numbers add them together then divide by 4:
\( \frac{7 + 5 + 8 + 4}{4} \) = \( \frac{24}{4} \) = 6