Rates 2

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Rates 2

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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?

(20) $\text(cars/hour)$

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A rate is a ratio of two different units.

Convert the values into fraction form

$120 \text(cars in) 6 \text(hours)$

$=$

$(120 \text(cars))/(6 \text(hours))$

Make the value of the denominator equal to

$1$ unit to get the rate.

$(120)/6$	$=$	$\frac{120\div{\color{#CC0000}{6}}\;\text{cars}}{6\div{\color{#CC0000}{6}}\;\text{hours}}$	Divide the values by $6$

	$=$	$(20 \text(cars))/(\text(hour))$

This can be read as

$20 \text(cars per hour)$ .

$20 \text(cars/hour)$

Question 2 of 4

2. Question
A man bought $12 \text(kg)$ of tomatoes for $$9$ . What is the cost per kilogram?
- (75) $\text(cents/kg)$
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A rate is a ratio of two different units.

Convert the values into fraction form

$$9 \text(for) 12 \text(kg)$ $=$ $($9)/(12 \text(kg))$

Make the value of the denominator equal to $1$ unit to get the rate.

$(9)/12$ $=$ $\frac{$9\div{\color{#CC0000}{12}}}{12\div{\color{#CC0000}{12}}\;\text{kg}}$ Divide the values by $12$

$=$ $($0.75)/(\text(kg))$

Finally, convert the dollars into cents.

$1 \text(dollar)$ $=$ $100 \text(cents)$

$(0.75xx(100)/1)/(\text(kg))$ $=$ $(75 \text(cents))/(\text(kg))$

This can be read as $75 \text(cents per kilogram)$ .

$75 \text(cents/kg)$

Question 3 of 4

Michael ran

$680\text(m)$ in

$85$ seconds. Express his speed as a rate.

(8) $\text(m/sec)$

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A rate is a ratio of two different units.

Convert the values into fraction form

$680 \text(m in) 85 \text(seconds)$

$=$

$(680 \text(m))/(85 \text(seconds))$

Make the value of the denominator equal to

$1$ unit to get the rate.

$(680)/(85)$	$=$	$\frac{680\div{\color{#CC0000}{85}}\;\text{m}}{85\div{\color{#CC0000}{85}}\;\text{seconds}}$	Divide the values by $85$

	$=$	$(8 \text(m))/(\text(second))$

This can be read as

$8 \text(metres per second)$ .

$8 \text(m/sec)$

Question 4 of 4

4. Question
A woman walked her pet dog at a speed of $6 \text(km/hour)$ . How far would she travel in $3$ hours?
- (18) $\text(km)$
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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 $x$ .

km Hour

$6$ $1$

$x$ $3$

Next, equate the two columns and solve for $x$ .

$6/x$ $=$ $1/3$

$1(x)$ $=$ $6(3)$ Cross multiply

$x$ $=$ $18$

Since $x$ is under the $\text(km)$ column in the quad boxes, the value will be $18 \text(kilometres)$ .

$18 \text(km)$

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$(9)/12$	$=$	$\frac{$9\div{\color{#CC0000}{12}}}{12\div{\color{#CC0000}{12}}\;\text{kg}}$	Divide the values by $12$

	$=$	$($0.75)/(\text(kg))$

km	Hour
$6$	$1$
$x$	$3$

$6/x$	$=$	$1/3$

$1(x)$	$=$	$6(3)$	Cross multiply
$x$	$=$	$18$