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Question 1 of 4
A factory can produce 120 cars in 6 hours. How many cars can they produce each hour?
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A rate is a ratio of two different units.
Convert the values into fraction form
120 cars in 6 hours |
= |
120 cars6 hours |
Make the value of the denominator equal to 1 unit to get the rate.
1206 |
= |
120÷6cars6÷6hours |
Divide the values by 6 |
|
|
= |
20 carshour |
This can be read as 20 cars per hour.
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Question 2 of 4
A man bought 12 kg of tomatoes for $9. What is the cost per kilogram?
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A rate is a ratio of two different units.
Convert the values into fraction form
$9 for 12 kg |
= |
$912 kg |
Make the value of the denominator equal to 1 unit to get the rate.
912 |
= |
$9÷1212÷12kg |
Divide the values by 12 |
|
|
= |
$0.75kg |
Finally, convert the dollars into cents.
1 dollar |
= |
100 cents |
0.75×1001kg |
= |
75 centskg |
This can be read as 75 cents per kilogram.
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Question 3 of 4
Michael ran 680m in 85 seconds. Express his speed as a rate.
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A rate is a ratio of two different units.
Convert the values into fraction form
680 m in 85 seconds |
= |
680 m85 seconds |
Make the value of the denominator equal to 1 unit to get the rate.
68085 |
= |
680÷85m85÷85seconds |
Divide the values by 85 |
|
|
= |
8 msecond |
This can be read as 8 metres per second.
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Question 4 of 4
A woman walked her pet dog at a speed of 6 km/hour. How far would she travel in 3 hours?
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A rate is a ratio of two different units.
Quad boxes are used to easily find rates
(Use cross product method)
First, fill up the quad boxes with the three known values and have the unknown value be x.
Next, equate the two columns and solve for x.
6x |
= |
13 |
|
1(x) |
= |
6(3) |
Cross multiply |
x |
= |
18 |
Since x is under the km column in the quad boxes, the value will be 18 kilometres.