| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 3.27 |
| Score | 0% | 65% |
A tiger in a zoo has consumed 16 pounds of food in 2 days. If the tiger continues to eat at the same rate, in how many more days will its total food consumtion be 72 pounds?
| 7 | |
| 8 | |
| 3 | |
| 5 |
If the tiger has consumed 16 pounds of food in 2 days that's \( \frac{16}{2} \) = 8 pounds of food per day. The tiger needs to consume 72 - 16 = 56 more pounds of food to reach 72 pounds total. At 8 pounds of food per day that's \( \frac{56}{8} \) = 7 more days.
How many 2 gallon cans worth of fuel would you need to pour into an empty 12 gallon tank to fill it exactly halfway?
| 3 | |
| 8 | |
| 6 | |
| 3 |
To fill a 12 gallon tank exactly halfway you'll need 6 gallons of fuel. Each fuel can holds 2 gallons so:
cans = \( \frac{6 \text{ gallons}}{2 \text{ gallons}} \) = 3
\({b + c \over a} = {b \over a} + {c \over a}\) defines which of the following?
distributive property for division |
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commutative property for multiplication |
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distributive property for multiplication |
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commutative property for division |
The distributive property for division helps in solving expressions like \({b + c \over a}\). It specifies that the result of dividing a fraction with multiple terms in the numerator and one term in the denominator can be obtained by dividing each term individually and then totaling the results: \({b + c \over a} = {b \over a} + {c \over a}\). For example, \({a^3 + 6a^2 \over a^2} = {a^3 \over a^2} + {6a^2 \over a^2} = a + 6\).
What is the greatest common factor of 24 and 64?
| 8 | |
| 22 | |
| 14 | |
| 19 |
The factors of 24 are [1, 2, 3, 4, 6, 8, 12, 24] and the factors of 64 are [1, 2, 4, 8, 16, 32, 64]. They share 4 factors [1, 2, 4, 8] making 8 the greatest factor 24 and 64 have in common.
How many hours does it take a car to travel 50 miles at an average speed of 50 miles per hour?
| 1 hour | |
| 2 hours | |
| 5 hours | |
| 7 hours |
Average speed in miles per hour is the number of miles traveled divided by the number of hours:
speed = \( \frac{\text{distance}}{\text{time}} \)Solving for time:
time = \( \frac{\text{distance}}{\text{speed}} \)
time = \( \frac{50mi}{50mph} \)
1 hour