| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 3.00 |
| Score | 0% | 60% |
Cooks are needed to prepare for a large party. Each cook can bake either 3 large cakes or 19 small cakes per hour. The kitchen is available for 3 hours and 37 large cakes and 170 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?
| 10 | |
| 12 | |
| 15 | |
| 8 |
If a single cook can bake 3 large cakes per hour and the kitchen is available for 3 hours, a single cook can bake 3 x 3 = 9 large cakes during that time. 37 large cakes are needed for the party so \( \frac{37}{9} \) = 4\(\frac{1}{9}\) cooks are needed to bake the required number of large cakes.
If a single cook can bake 19 small cakes per hour and the kitchen is available for 3 hours, a single cook can bake 19 x 3 = 57 small cakes during that time. 170 small cakes are needed for the party so \( \frac{170}{57} \) = 2\(\frac{56}{57}\) 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 + 3 = 8 cooks.
A menswear store is having a sale: "Buy one shirt at full price and get another shirt for 50% off." If Charlie buys two shirts, each with a regular price of $47, how much money will he save?
| $9.40 | |
| $23.50 | |
| $18.80 | |
| $21.15 |
By buying two shirts, Charlie will save $47 x \( \frac{50}{100} \) = \( \frac{$47 x 50}{100} \) = \( \frac{$2350}{100} \) = $23.50 on the second shirt.
Solve 5 + (3 + 2) ÷ 2 x 3 - 32
| 1\(\frac{1}{4}\) | |
| 1 | |
| \(\frac{2}{3}\) | |
| 3\(\frac{1}{2}\) |
Use PEMDAS (Parentheses, Exponents, Multipy/Divide, Add/Subtract):
5 + (3 + 2) ÷ 2 x 3 - 32
P: 5 + (5) ÷ 2 x 3 - 32
E: 5 + 5 ÷ 2 x 3 - 9
MD: 5 + \( \frac{5}{2} \) x 3 - 9
MD: 5 + \( \frac{15}{2} \) - 9
AS: \( \frac{10}{2} \) + \( \frac{15}{2} \) - 9
AS: \( \frac{25}{2} \) - 9
AS: \( \frac{25 - 18}{2} \)
\( \frac{7}{2} \)
3\(\frac{1}{2}\)
What is \( \sqrt{\frac{49}{81}} \)?
| \(\frac{5}{9}\) | |
| 1 | |
| \(\frac{8}{9}\) | |
| \(\frac{7}{9}\) |
To take the square root of a fraction, break the fraction into two separate roots then calculate the square root of the numerator and denominator separately:
\( \sqrt{\frac{49}{81}} \)
\( \frac{\sqrt{49}}{\sqrt{81}} \)
\( \frac{\sqrt{7^2}}{\sqrt{9^2}} \)
\(\frac{7}{9}\)
Convert a-5 to remove the negative exponent.
| \( \frac{1}{a^5} \) | |
| \( \frac{1}{a^{-5}} \) | |
| \( \frac{-5}{a} \) | |
| \( \frac{-1}{-5a^{5}} \) |
To convert a negative exponent to a positive exponent, calculate the positive exponent then take the reciprocal.