- There are many times when we hear the word 'rate'.
- Till now, we have not learned the concept of rate.
- Let's learn it.

**Rate**

- Rate describes a ratio relationship between two different types of units.

**For ****example:**** **\(1.75\) miles/hour is a rate that describes a ratio relationship between hours and miles.

If an object travels at this rate then at the end of 1^{st }hour, it will cover \(1.75\) miles, after 2 hours it will cover \(3.50\) miles, after 3 hours it will cover \(5.25\) miles, and so on.

- A rate can be identified by the keyword
**'per'**.

A \(\text{30 chocolates}\)

B \(\text{3000 chocolates per 10 boxes}\)

C \(\text{10 boxes}\)

D \(\text{300 chocolates}\)

- Speed is the rate at which something moves.
- Speed is the rate of motion or the rate of change of position. It is expressed as distance moved per unit of time.

\(Speed=\dfrac{\text{Distance}}{\text{Time}}\)

**For ****example:** Carl travels \(40\) miles in \(20\) minutes. Find the distance Carl travels in \(50\) minutes.

Given statement:

Carl travels \(40\) miles in \(20 \) minutes.

Thus, the rate is \(40\) miles per \(20\) minutes.

Now, to find the unit rate, we need to make the denominator \(1\).

The GCF of \(40\) and \(20\) is \(20\).

So, divide both \(40\) and \(20\) by \(20\).

\(\Rightarrow\dfrac{40\div20}{20\div20}\)

\(\Rightarrow\dfrac{2}{1}\)

This means Carl travels \(2\) miles in \(1\)minute.

Now, to find the distance traveled in \(50\) minutes, we have to multiply both numerator and denominator by \(50\).

So,

\(\Rightarrow\dfrac{2\times50}{1\times50}\)

\(\Rightarrow\dfrac{100}{50}\)

This means Carl travels \(100\) miles in \(50\) minutes.

A \(10\; miles\)

B \(90\; miles\)

C \(70\; miles\)

D \(50\; miles\)

**Unit Rate**

- A unit rate describes a ratio relationship between two types of quantities, when one of them is \(1\).
- It describes a ratio in which the denominator is \(1\).

For example: If an airplane travels at a speed of 610 miles/hr then

\(Unit \,rate=\dfrac{610 \; miles}{1\; hour}\)

**Unit Rate Method**

- Unit Rate Method is a method in which we first find the value of one unit, and then calculate the value of the required number of units.

For example:

Casey makes \(60\) cookies in \(3\) hours.

- To write this in ratio, we need to compare \(60\) cookies with \(3\) hours. The \(60\) cookies become our numerator and \(3\) hours become our denominator.

Thus, \(\text{rate}=\dfrac{\text{60 cookies}}{\text{3 hours}}\)

Now, to find the unit rate, we have to make the denominator \(1\).

The GCF of \(60\) and \(3\) is \(3\).

\(\therefore\) Divide both \(60\) and \(3\) by \(3\).

\(\Rightarrow\dfrac{60\div3}{3\div3}\)

So, we get

\(\Rightarrow \dfrac{\text{20 cookies}}{\text{1 hour}}\)

This means Casey makes \(20\) cookies per hour.

A \(\text{50 pages in 2 hours}\)

B \(\text{60 pages in 2 hours}\)

C \(\text{60 pages in 1 hour}\)

D \(\text{80 pages in 1 hour}\)

**Rate**

- Rate describes a ratio relationship between two types of quantities.

For example: Jose can run \(3\) miles in \(2\) hours.

Here, \(3\) miles in \(2\) hours is a rate that describes a ratio relationship between hours and miles.

- Rate can be identified by the keyword '
**per**'.

**Unit Rate**

- A unit rate describes a ratio relationship between two types of quantities, when one of them is \(1\).

For example: Jose can run \(1.5\) miles per (1) hour.

Here, \(1.5\) miles per hour is a unit rate that describes a ratio relationship between miles and hour.

**Note:** Here, the unit 'miles/hour' is known as rate unit.

**Equivalent Rate**

- When two or more rates are equal, they are called equivalent rates.

For example: Jose can run \(3\) miles in \(2\) hours.

\(\text{Rate}=\dfrac{3\; miles }{2\;hours}\)

\(Unit\, Rate=\dfrac{1.5\;miles}{1\; hour}\) [By long division method]

Alex can run \(1.5\) miles per hour.

\(\text{Unit Rate}=\dfrac{1.5\;miles}{1\;hour}\)

Thus, both rates are equal.

\(\dfrac{3\; miles }{2\;hours}\) \(=\dfrac{1.5\;miles}{1\;hour}\)

These are equivalent rates.

A \(\text{6 pastries per box}\)

B \(\text{12 pastries per box}\)

C \(\text{1 pastry per box}\)

D \(\text{3 pastries per box}\)

- Cost is the amount that has to be paid or spent to buy or make something.
- If we know the cost of one unit, we can find out the cost of more units.
- Generally, the unit of cost is dollars($).

**For example:** If the cost of \(5\) chocolates is \($10\), find the cost of \(12\) chocolates.

Cost of \(5\) chocolates \(=$10\)

Thus, the cost of \(1\) chocolate = \(\dfrac{$10}{5}\)

Now, to find the unit rate, we need to make the denominator \(1\).

The GCF of \(10\) and \(5\) is \(5\).

\(\therefore\) Divide both \(10\) and \(5\) by \(5\).

\(\Rightarrow\dfrac{10\div5}{5\div5}\)

\(\Rightarrow\dfrac{2}{1}\)

This means the cost of \(1\) chocolate is \($2\).

Now, to find the cost of \(12\) chocolates, we have to multiply both numerator and denominator by \(12\).

So,

\(\text{Rate}=\dfrac{2\times12}{1\times12}\)

\(\Rightarrow\dfrac{24}{12}\)

This means the cost of \(12\) chocolates is \($24\).

A \($20\)

B \($30\)

C \($40\)

D \($60\)

- Time and work are related to each other.
- Time - work relationship shows how much work is done in a given period of time. When amount of work increases, the amount of time required to do that work also increases.

**For example:** Jacob writes \(20\) pages of notes in \(5\) hours.

Let's calculate the number of pages he can write in \(8\) hours.

The time taken by Jacob to write \(20\) pages of notes \(=5\) hours

\(\text{Rate}=\dfrac{\text{20 pages}}{\text{5 hours}}\)

To find the unit rate, we need to make the denominator \(1\).

The GCF of \(20\) and \(5\) is \(5\).

\(\therefore\) Divide both \(20\) and \(5\) by \(5\).

\(\Rightarrow\dfrac{20\div5}{5\div5}\)

\(\Rightarrow\dfrac{4}{1}\)

This means Jacob writes \(4\) pages per hour.

Now, to find the number of pages he can write in \(8\) hours, we have to multiply both numerator and denominator by \(8\).

So,

\(\Rightarrow\dfrac{4\times8}{1\times8}\)

\(\Rightarrow\dfrac{32}{8}\)

This means Jacob writes \(32\) pages in \(8\) hours.

A \(90\;kg\)

B \(48\;kg\)

C \(81\;kg\)

D \(10\;kg\)

**Rate**

- Rate describes a ratio relationship between two types of quantities.

**Unit rate**

- A unit rate describes a ratio relationship between two types of quantities, when one of them is \(1\).

**Comparison of unit rates**

The comparison of unit rates can be a good way of finding out which is the best option/best buy.

Consider the following example:

Casey prepares \(50\) kgs of dough in \(10\) hours and Jana prepares \(30\) kgs of dough in \(15\) hours. Find who prepares more dough in less time.

Given statement:

Casey prepares \(50\) kgs of dough in \(10\) hours.

Thus, we compare \(50\) kgs with \(10\) hours.

So, we can write,

\(\Rightarrow\dfrac{50\;kgs}{10\;hours}\)

Now, to find the unit rate, we need to make the denominator 1.

The GCF of \(50\) and \(10\) is \(10\).

\(\therefore\) Divide both \(50\) and \(10\) by \(10\).

\(\Rightarrow\dfrac{50\div10}{10\div10}\)

So, we get

\(\Rightarrow 5:1\)

\(\Rightarrow\dfrac{5}{1}\)

This means Casey prepares \(5\) kg dough per hour.

Given statement:

Jana prepares \(30\) kgs of dough in \(15\) hours.

Thus, we compare \(30\) kgs with \(15\) hours.

So, we can write,

\(\Rightarrow\dfrac{30\;kgs}{15\;hours}\)

Now, to find the unit rate, we need to make the denominator 1.

The GCF of \(30\) and \(15\) is \(15\).

\(\therefore\) Divide both \(30\) and \(15\) by \(15\).

\(\Rightarrow\dfrac{30\div15}{15\div15}\)

So, we get

\(\Rightarrow 2:1\)

\(\Rightarrow\dfrac{2}{1}\)

This means Jana prepares \(2\) kg dough per hour.

Thus, Casey prepares more dough in less time as compared to Jana.