| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 2.95 |
| Score | 0% | 59% |
What is \( \frac{8}{3} \) - \( \frac{9}{5} \)?
| \( \frac{9}{15} \) | |
| 1 \( \frac{1}{5} \) | |
| \( \frac{2}{9} \) | |
| \(\frac{13}{15}\) |
To subtract these fractions, first find the lowest common multiple of their denominators. The first few multiples of 3 are [3, 6, 9, 12, 15, 18, 21, 24, 27, 30] and the first few multiples of 5 are [5, 10, 15, 20, 25, 30, 35, 40, 45, 50]. The first few multiples they share are [15, 30, 45, 60, 75] making 15 the smallest multiple 3 and 5 share.
Next, convert the fractions so each denominator equals the lowest common multiple:
\( \frac{8 x 5}{3 x 5} \) - \( \frac{9 x 3}{5 x 3} \)
\( \frac{40}{15} \) - \( \frac{27}{15} \)
Now, because the fractions share a common denominator, you can subtract them:
\( \frac{40 - 27}{15} \) = \( \frac{13}{15} \) = \(\frac{13}{15}\)
What is 9c3 x 9c5?
| 18c3 | |
| 81c8 | |
| 81c2 | |
| 18c8 |
To multiply terms with exponents, the base of both exponents must be the same. In this case they are so multiply the coefficients and add the exponents:
9c3 x 9c5
(9 x 9)c(3 + 5)
81c8
If a rectangle is twice as long as it is wide and has a perimeter of 30 meters, what is the area of the rectangle?
| 50 m2 | |
| 18 m2 | |
| 98 m2 | |
| 162 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 30 meters so the equation becomes: 2w + 2h = 30.
Putting these two equations together and solving for width (w):
2w + 2h = 30
w + h = \( \frac{30}{2} \)
w + h = 15
w = 15 - h
From the question we know that h = 2w so substituting 2w for h gives us:
w = 15 - 2w
3w = 15
w = \( \frac{15}{3} \)
w = 5
Since h = 2w that makes h = (2 x 5) = 10 and the area = h x w = 5 x 10 = 50 m2
A bread recipe calls for 3\(\frac{5}{8}\) cups of flour. If you only have 1\(\frac{3}{4}\) cups, how much more flour is needed?
| 3\(\frac{1}{8}\) cups | |
| 2\(\frac{1}{8}\) cups | |
| 1\(\frac{7}{8}\) cups | |
| 2\(\frac{3}{4}\) cups |
The amount of flour you need is (3\(\frac{5}{8}\) - 1\(\frac{3}{4}\)) cups. Rewrite the quantities so they share a common denominator and subtract:
(\( \frac{29}{8} \) - \( \frac{14}{8} \)) cups
\( \frac{15}{8} \) cups
1\(\frac{7}{8}\) cups
A machine in a factory has an error rate of 4 parts per 100. The machine normally runs 24 hours a day and produces 8 parts per hour. Yesterday the machine was shut down for 8 hours for maintenance.
How many error-free parts did the machine produce yesterday?
| 105.6 | |
| 122.9 | |
| 180.2 | |
| 85.5 |
The hourly error rate for this machine is the error rate in parts per 100 multiplied by the number of parts produced per hour:
\( \frac{4}{100} \) x 8 = \( \frac{4 \times 8}{100} \) = \( \frac{32}{100} \) = 0.32 errors per hour
So, in an average hour, the machine will produce 8 - 0.32 = 7.68 error free parts.
The machine ran for 24 - 8 = 16 hours yesterday so you would expect that 16 x 7.68 = 122.9 error free parts were produced yesterday.