| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 2.60 |
| Score | 0% | 52% |
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?
| 72 m2 | |
| 162 m2 | |
| 50 m2 | |
| 32 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
If \( \left|c - 5\right| \) + 0 = -7, which of these is a possible value for c?
| -13 | |
| 15 | |
| -8 | |
| -2 |
First, solve for \( \left|c - 5\right| \):
\( \left|c - 5\right| \) + 0 = -7
\( \left|c - 5\right| \) = -7 + 0
\( \left|c - 5\right| \) = -7
The value inside the absolute value brackets can be either positive or negative so (c - 5) must equal - 7 or --7 for \( \left|c - 5\right| \) to equal -7:
| c - 5 = -7 c = -7 + 5 c = -2 | c - 5 = 7 c = 7 + 5 c = 12 |
So, c = 12 or c = -2.
This property states taht the order of addition or multiplication does not mater. For example, 2 + 5 and 5 + 2 are equivalent.
commutative |
|
associative |
|
PEDMAS |
|
distributive |
The commutative property states that, when adding or multiplying numbers, the order in which they're added or multiplied does not matter. For example, 3 + 4 and 4 + 3 give the same result, as do 3 x 4 and 4 x 3.
Cooks are needed to prepare for a large party. Each cook can bake either 3 large cakes or 15 small cakes per hour. The kitchen is available for 2 hours and 28 large cakes and 470 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?
| 9 | |
| 10 | |
| 21 | |
| 15 |
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. 28 large cakes are needed for the party so \( \frac{28}{6} \) = 4\(\frac{2}{3}\) 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 2 hours, a single cook can bake 15 x 2 = 30 small cakes during that time. 470 small cakes are needed for the party so \( \frac{470}{30} \) = 15\(\frac{2}{3}\) 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 5 + 16 = 21 cooks.
A machine in a factory has an error rate of 8 parts per 100. The machine normally runs 24 hours a day and produces 6 parts per hour. Yesterday the machine was shut down for 2 hours for maintenance.
How many error-free parts did the machine produce yesterday?
| 121.4 | |
| 165.6 | |
| 100 | |
| 156.2 |
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 6 = \( \frac{8 \times 6}{100} \) = \( \frac{48}{100} \) = 0.48 errors per hour
So, in an average hour, the machine will produce 6 - 0.48 = 5.52 error free parts.
The machine ran for 24 - 2 = 22 hours yesterday so you would expect that 22 x 5.52 = 121.4 error free parts were produced yesterday.