| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 3.26 |
| Score | 0% | 65% |
Cooks are needed to prepare for a large party. Each cook can bake either 4 large cakes or 12 small cakes per hour. The kitchen is available for 3 hours and 30 large cakes and 360 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 | |
| 8 | |
| 13 | |
| 6 |
If a single cook can bake 4 large cakes per hour and the kitchen is available for 3 hours, a single cook can bake 4 x 3 = 12 large cakes during that time. 30 large cakes are needed for the party so \( \frac{30}{12} \) = 2\(\frac{1}{2}\) cooks are needed to bake the required number of large cakes.
If a single cook can bake 12 small cakes per hour and the kitchen is available for 3 hours, a single cook can bake 12 x 3 = 36 small cakes during that time. 360 small cakes are needed for the party so \( \frac{360}{36} \) = 10 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 3 + 10 = 13 cooks.
Simplify \( \frac{36}{80} \).
| \( \frac{9}{20} \) | |
| \( \frac{5}{11} \) | |
| \( \frac{5}{9} \) | |
| \( \frac{5}{17} \) |
To simplify this fraction, first find the greatest common factor between them. The factors of 36 are [1, 2, 3, 4, 6, 9, 12, 18, 36] and the factors of 80 are [1, 2, 4, 5, 8, 10, 16, 20, 40, 80]. They share 3 factors [1, 2, 4] making 4 their greatest common factor (GCF).
Next, divide both numerator and denominator by the GCF:
\( \frac{36}{80} \) = \( \frac{\frac{36}{4}}{\frac{80}{4}} \) = \( \frac{9}{20} \)
Which of the following is a mixed number?
\({7 \over 5} \) |
|
\({5 \over 7} \) |
|
\({a \over 5} \) |
|
\(1 {2 \over 5} \) |
A rational number (or fraction) is represented as a ratio between two integers, a and b, and has the form \({a \over b}\) where a is the numerator and b is the denominator. An improper fraction (\({5 \over 3} \)) has a numerator with a greater absolute value than the denominator and can be converted into a mixed number (\(1 {2 \over 3} \)) which has a whole number part and a fractional part.
Solve 5 + (3 + 4) ÷ 5 x 2 - 32
| 3\(\frac{1}{2}\) | |
| 1 | |
| 1\(\frac{2}{5}\) | |
| -1\(\frac{1}{5}\) |
Use PEMDAS (Parentheses, Exponents, Multipy/Divide, Add/Subtract):
5 + (3 + 4) ÷ 5 x 2 - 32
P: 5 + (7) ÷ 5 x 2 - 32
E: 5 + 7 ÷ 5 x 2 - 9
MD: 5 + \( \frac{7}{5} \) x 2 - 9
MD: 5 + \( \frac{14}{5} \) - 9
AS: \( \frac{25}{5} \) + \( \frac{14}{5} \) - 9
AS: \( \frac{39}{5} \) - 9
AS: \( \frac{39 - 45}{5} \)
\( \frac{-6}{5} \)
-1\(\frac{1}{5}\)
What is z3 - 3z3?
| 4z9 | |
| 2z-3 | |
| -2z3 | |
| -2z-3 |
To add or subtract terms with exponents, both the base and the exponent must be the same. In this case they are so subtract the coefficients and retain the base and exponent:
1z3 - 3z3
(1 - 3)z3
-2z3