| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 2.87 |
| Score | 0% | 57% |
Solve 3 + (4 + 5) ÷ 2 x 3 - 32
| 1\(\frac{2}{7}\) | |
| \(\frac{7}{8}\) | |
| 7\(\frac{1}{2}\) | |
| \(\frac{2}{3}\) |
Use PEMDAS (Parentheses, Exponents, Multipy/Divide, Add/Subtract):
3 + (4 + 5) ÷ 2 x 3 - 32
P: 3 + (9) ÷ 2 x 3 - 32
E: 3 + 9 ÷ 2 x 3 - 9
MD: 3 + \( \frac{9}{2} \) x 3 - 9
MD: 3 + \( \frac{27}{2} \) - 9
AS: \( \frac{6}{2} \) + \( \frac{27}{2} \) - 9
AS: \( \frac{33}{2} \) - 9
AS: \( \frac{33 - 18}{2} \)
\( \frac{15}{2} \)
7\(\frac{1}{2}\)
A machine in a factory has an error rate of 8 parts per 100. The machine normally runs 24 hours a day and produces 7 parts per hour. Yesterday the machine was shut down for 7 hours for maintenance.
How many error-free parts did the machine produce yesterday?
| 165.9 | |
| 141.7 | |
| 131 | |
| 109.5 |
The hourly error rate for this machine is the error rate in parts per 100 multiplied by the number of parts produced per hour:
\( \frac{8}{100} \) x 7 = \( \frac{8 \times 7}{100} \) = \( \frac{56}{100} \) = 0.56 errors per hour
So, in an average hour, the machine will produce 7 - 0.56 = 6.4399999999999995 error free parts.
The machine ran for 24 - 7 = 17 hours yesterday so you would expect that 17 x 6.4399999999999995 = 109.5 error free parts were produced yesterday.
What is \( \frac{6}{5} \) + \( \frac{9}{9} \)?
| \( \frac{9}{12} \) | |
| \( \frac{4}{10} \) | |
| 2\(\frac{1}{5}\) | |
| 2 \( \frac{9}{45} \) |
To add 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:
\( \frac{6 x 9}{5 x 9} \) + \( \frac{9 x 5}{9 x 5} \)
\( \frac{54}{45} \) + \( \frac{45}{45} \)
Now, because the fractions share a common denominator, you can add them:
\( \frac{54 + 45}{45} \) = \( \frac{99}{45} \) = 2\(\frac{1}{5}\)
4! = ?
4 x 3 x 2 x 1 |
|
5 x 4 x 3 x 2 x 1 |
|
3 x 2 x 1 |
|
4 x 3 |
A factorial has the form n! and is the product of the integer (n) and all the positive integers below it. For example, 5! = 5 x 4 x 3 x 2 x 1 = 120.
Cooks are needed to prepare for a large party. Each cook can bake either 5 large cakes or 15 small cakes per hour. The kitchen is available for 4 hours and 40 large cakes and 220 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?
| 11 | |
| 6 | |
| 9 | |
| 8 |
If a single cook can bake 5 large cakes per hour and the kitchen is available for 4 hours, a single cook can bake 5 x 4 = 20 large cakes during that time. 40 large cakes are needed for the party so \( \frac{40}{20} \) = 2 cooks are needed to bake the required number of large cakes.
If a single cook can bake 15 small cakes per hour and the kitchen is available for 4 hours, a single cook can bake 15 x 4 = 60 small cakes during that time. 220 small cakes are needed for the party so \( \frac{220}{60} \) = 3\(\frac{2}{3}\) 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 2 + 4 = 6 cooks.