| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 2.97 |
| Score | 0% | 59% |
What is the next number in this sequence: 1, 4, 10, 19, 31, __________ ?
| 52 | |
| 49 | |
| 46 | |
| 43 |
The equation for this sequence is:
an = an-1 + 3(n - 1)
where n is the term's order in the sequence, an is the value of the term, and an-1 is the value of the term before an. This makes the next number:
a6 = a5 + 3(6 - 1)
a6 = 31 + 3(5)
a6 = 46
A machine in a factory has an error rate of 2 parts per 100. The machine normally runs 24 hours a day and produces 9 parts per hour. Yesterday the machine was shut down for 2 hours for maintenance.
How many error-free parts did the machine produce yesterday?
| 186.2 | |
| 194 | |
| 76.8 | |
| 73.6 |
The hourly error rate for this machine is the error rate in parts per 100 multiplied by the number of parts produced per hour:
\( \frac{2}{100} \) x 9 = \( \frac{2 \times 9}{100} \) = \( \frac{18}{100} \) = 0.18 errors per hour
So, in an average hour, the machine will produce 9 - 0.18 = 8.82 error free parts.
The machine ran for 24 - 2 = 22 hours yesterday so you would expect that 22 x 8.82 = 194 error free parts were produced yesterday.
Cooks are needed to prepare for a large party. Each cook can bake either 3 large cakes or 17 small cakes per hour. The kitchen is available for 2 hours and 23 large cakes and 380 small cakes need to be baked.
How many cooks are required to bake the required number of cakes during the time the kitchen is available?
| 14 | |
| 16 | |
| 9 | |
| 6 |
If a single cook can bake 3 large cakes per hour and the kitchen is available for 2 hours, a single cook can bake 3 x 2 = 6 large cakes during that time. 23 large cakes are needed for the party so \( \frac{23}{6} \) = 3\(\frac{5}{6}\) cooks are needed to bake the required number of large cakes.
If a single cook can bake 17 small cakes per hour and the kitchen is available for 2 hours, a single cook can bake 17 x 2 = 34 small cakes during that time. 380 small cakes are needed for the party so \( \frac{380}{34} \) = 11\(\frac{3}{17}\) cooks are needed to bake the required number of small cakes.
Because you can't employ a fractional cook, round the number of cooks needed for each type of cake up to the next whole number resulting in 4 + 12 = 16 cooks.
What is \( \frac{6}{8} \) + \( \frac{8}{16} \)?
| \( \frac{1}{9} \) | |
| 2 \( \frac{3}{6} \) | |
| 1\(\frac{1}{4}\) | |
| 2 \( \frac{8}{16} \) |
To add these fractions, first find the lowest common multiple of their denominators. The first few multiples of 8 are [8, 16, 24, 32, 40, 48, 56, 64, 72, 80] and the first few multiples of 16 are [16, 32, 48, 64, 80, 96]. The first few multiples they share are [16, 32, 48, 64, 80] making 16 the smallest multiple 8 and 16 share.
Next, convert the fractions so each denominator equals the lowest common multiple:
\( \frac{6 x 2}{8 x 2} \) + \( \frac{8 x 1}{16 x 1} \)
\( \frac{12}{16} \) + \( \frac{8}{16} \)
Now, because the fractions share a common denominator, you can add them:
\( \frac{12 + 8}{16} \) = \( \frac{20}{16} \) = 1\(\frac{1}{4}\)
Simplify \( \frac{24}{56} \).
| \( \frac{9}{17} \) | |
| \( \frac{10}{13} \) | |
| \( \frac{4}{9} \) | |
| \( \frac{3}{7} \) |
To simplify this fraction, first find the greatest common factor between them. The factors of 24 are [1, 2, 3, 4, 6, 8, 12, 24] and the factors of 56 are [1, 2, 4, 7, 8, 14, 28, 56]. They share 4 factors [1, 2, 4, 8] making 8 their greatest common factor (GCF).
Next, divide both numerator and denominator by the GCF:
\( \frac{24}{56} \) = \( \frac{\frac{24}{8}}{\frac{56}{8}} \) = \( \frac{3}{7} \)