smc-4609-05-Simplifying rates

Rates, SM-Bank 004

All the water has leaked from a 10 litre bucket in 20 minutes.

Calculate the leaking rate in litres/minute. (1 mark)

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How long would it take to empty a 12 litre bucket at the same rate? (2 marks)

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a. $0.5\ \text{litres}/ \text{minute}$

b. $24\ \text{minutes}$

Show Worked Solution

a.	$10\ \text{litres}/20\ \text{minutes}$	$=\dfrac{10}{20}\ \text{litres}/\dfrac{20}{20}\ \text{minutes}$
		$=0.5\ \text{litres}/\text{minute}$

b.	$0.5\ \text{litres}/\text{hour}$	$=1\ \text{litres}/2\ \text{minutes}$
		$=12\ \text{litres}/2\times 12\ \text{minutes}$
		$=12\ \text{litres}/24\ \text{minutes}$

$\therefore\ \text{It would take}\ 24\ \text{minutes to empty a}\ 12\ \text{litre bucket.}$

Rates, SM-Bank 003

Write each of the following as a rate in simplified form.

$136 for every 8 hours worked (1 mark)

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$171 000 to purchase 3 cars (1 mark)

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112 millimetres of rain in 14 days (1 mark)

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a. $$17/ \text{hour}$

b. $$57\ 000/\text{car}$

c. $8\ \text{mm}/\text{day}$

Show Worked Solution

a.	$$136/8\ \text{hours}$	$=\dfrac{$136}{8}/\dfrac{8}{8}\ \text{hours}$
		$=$17/\text{hour}$

b.	$$171\ 000/3\ \text{cars}$	$=\dfrac{$171\ 000}{3}/\dfrac{3}{3}\ \text{cars}$
		$=$57\ 000/\text{car}$

c.	$112\ \text{mm}/14\ \text{days}$	$=\dfrac{112}{14}\ \text{mm}/\dfrac{14}{14}\ \text{days}$
		$=8\ \text{mm}/\text{day}$

Rates, SM-Bank 002

Write each of the following as a rate in simplified form.

$18 for every 5 kilograms (1 mark)

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30 animals per 6 cages (1 mark)

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63 rainy days in 7 months (1 mark)

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a. $$3.60/ \text{kilogram}$

b. $5\ \text{animals}/\text{cage}$

c. $9\ \text{rainy days}/\text{month}$

Show Worked Solution

a.	$$18/5\ \text{kg}$	$=\dfrac{$18}{5}/\dfrac{5}{5}\ \text{kg}$
		$=$3.60/\text{kg}$

b.	$30\ \text{animals}/6\ \text{cages}$	$=\dfrac{30}{6}\ \text{animals}/\dfrac{6}{6}\ \text{cages}$
		$=5\ \text{animals}/\text{cage}$

c.	$63\ \text{rainy days}/7\ \text{months}$	$=\dfrac{63}{7}\ \text{rainy days}/\dfrac{7}{7}\ \text{months}$
		$=9\ \text{rainy days}/\text{month}$

Rates, SM-Bank 001 MC

Scott owns a sheep station.

He keeps 4 sheep on every acre of the property.

How many acres would he need for 12 sheep?

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$A$

Show Worked Solution

$\text{4 sheep per acre}$

$\therefore\ \text{Acres needed for 12 sheep}$	$=\dfrac{12}{4}$
	$=3\ \text{acres}$

$\Rightarrow A$

a.	\(10\ \text{litres}/20\ \text{minutes}\)	\(=\dfrac{10}{20}\ \text{litres}/\dfrac{20}{20}\ \text{minutes}\)
		\(=0.5\ \text{litres}/\text{minute}\)

b.	\(0.5\ \text{litres}/\text{hour}\)	\(=1\ \text{litres}/2\ \text{minutes}\)
		\(=12\ \text{litres}/2\times 12\ \text{minutes}\)
		\(=12\ \text{litres}/24\ \text{minutes}\)

a.	\($136/8\ \text{hours}\)	\(=\dfrac{$136}{8}/\dfrac{8}{8}\ \text{hours}\)
		\(=$17/\text{hour}\)

b.	\($171\ 000/3\ \text{cars}\)	\(=\dfrac{$171\ 000}{3}/\dfrac{3}{3}\ \text{cars}\)
		\(=$57\ 000/\text{car}\)

c.	\(112\ \text{mm}/14\ \text{days}\)	\(=\dfrac{112}{14}\ \text{mm}/\dfrac{14}{14}\ \text{days}\)
		\(=8\ \text{mm}/\text{day}\)

a.	\($18/5\ \text{kg}\)	\(=\dfrac{$18}{5}/\dfrac{5}{5}\ \text{kg}\)
		\(=$3.60/\text{kg}\)

b.	\(30\ \text{animals}/6\ \text{cages}\)	\(=\dfrac{30}{6}\ \text{animals}/\dfrac{6}{6}\ \text{cages}\)
		\(=5\ \text{animals}/\text{cage}\)

c.	\(63\ \text{rainy days}/7\ \text{months}\)	\(=\dfrac{63}{7}\ \text{rainy days}/\dfrac{7}{7}\ \text{months}\)
		\(=9\ \text{rainy days}/\text{month}\)

\(\therefore\ \text{Acres needed for 12 sheep}\)	\(=\dfrac{12}{4}\)
	\(=3\ \text{acres}\)