| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 2.91 |
| Score | 0% | 58% |
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?
| 20% | |
| 25% | |
| 22\(\frac{1}{2}\)% | |
| 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\(\frac{1}{2}\)%
A machine in a factory has an error rate of 9 parts per 100. The machine normally runs 24 hours a day and produces 6 parts per hour. Yesterday the machine was shut down for 8 hours for maintenance.
How many error-free parts did the machine produce yesterday?
| 186.2 | |
| 149 | |
| 87.4 | |
| 179.6 |
The hourly error rate for this machine is the error rate in parts per 100 multiplied by the number of parts produced per hour:
\( \frac{9}{100} \) x 6 = \( \frac{9 \times 6}{100} \) = \( \frac{54}{100} \) = 0.54 errors per hour
So, in an average hour, the machine will produce 6 - 0.54 = 5.46 error free parts.
The machine ran for 24 - 8 = 16 hours yesterday so you would expect that 16 x 5.46 = 87.4 error free parts were produced yesterday.
What is \( \frac{9}{5} \) - \( \frac{8}{9} \)?
| \(\frac{41}{45}\) | |
| 1 \( \frac{5}{45} \) | |
| 2 \( \frac{4}{45} \) | |
| 2 \( \frac{6}{45} \) |
To subtract these fractions, first find the lowest common multiple of their denominators. The first few multiples of 5 are [5, 10, 15, 20, 25, 30, 35, 40, 45, 50] and the first few multiples of 9 are [9, 18, 27, 36, 45, 54, 63, 72, 81, 90]. The first few multiples they share are [45, 90] making 45 the smallest multiple 5 and 9 share.
Next, convert the fractions so each denominator equals the lowest common multiple:
\( \frac{9 x 9}{5 x 9} \) - \( \frac{8 x 5}{9 x 5} \)
\( \frac{81}{45} \) - \( \frac{40}{45} \)
Now, because the fractions share a common denominator, you can subtract them:
\( \frac{81 - 40}{45} \) = \( \frac{41}{45} \) = \(\frac{41}{45}\)
What is \( \frac{-6z^6}{8z^2} \)?
| -1\(\frac{1}{3}\)z-4 | |
| -\(\frac{3}{4}\)z12 | |
| -1\(\frac{1}{3}\)z8 | |
| -\(\frac{3}{4}\)z4 |
To divide terms with exponents, the base of both exponents must be the same. In this case they are so divide the coefficients and subtract the exponents:
\( \frac{-6z^6}{8z^2} \)
\( \frac{-6}{8} \) z(6 - 2)
-\(\frac{3}{4}\)z4
What is \( \sqrt{\frac{49}{25}} \)?
| 2\(\frac{1}{2}\) | |
| 1 | |
| 1\(\frac{1}{4}\) | |
| 1\(\frac{2}{5}\) |
To take the square root of a fraction, break the fraction into two separate roots then calculate the square root of the numerator and denominator separately:
\( \sqrt{\frac{49}{25}} \)
\( \frac{\sqrt{49}}{\sqrt{25}} \)
\( \frac{\sqrt{7^2}}{\sqrt{5^2}} \)
\( \frac{7}{5} \)
1\(\frac{2}{5}\)