Equivalent Ratios

🏆Practice ratio

To easily solve ratio problems and to gain a better understanding of the topic in general, it is convenient to know about equivalent ratios.

Equivalent ratios are, in fact, ratios that seem different, are not expressed in the same way but, by simplifying or expanding them, you arrive at exactly the same relationship.

Think of it this way,

Equivalent ratios

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Test yourself on ratio!

einstein

There are 18 balls in a box, \( \frac{2}{3} \) of which are white.

How many white balls are there in the box?

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If you have two fractions in front of you:
20004000 \frac{2000}{4000}

and

24 \frac{2}{4}

we can simplify the larger fraction and arrive at

Equivalent ratios

and even more, we can simplify the smaller fraction and arrive at:

2 - Equivalent ratios

In fact, we can say that:

40002000=24=12 \frac{4000}{2000}=\frac{2}{4}=\frac{1}{2}

All the expressions are equivalent ratios.

Do you remember we said that a ratio can be shown in the form of a fraction?

Therefore, the same rule also applies to the ratios we have learned.

We can reduce both terms of the ratio or amplify them and arrive at equivalent ratios.

To solve this type of problem easily we will always try to arrive at the smallest ratio.

We will ask ourselves by what number we can divide both terms of the ratio, in this way we will arrive at the most reduced equivalent ratio possible.


How can you tell if they are equivalent ratios?

We will ask ourselves: Will we arrive at the same ratio by reduction or by amplification?

Let's see some examples:

X3

2:5 2:5

6:16 6:16

We have managed to demonstrate that by multiplying both terms by 3 3 we arrive at the same ratio. Therefore, they are equivalent ratios!

Are these ratios equivalent?

1:3 1:3

2:6 2:6

6:18 6:18

Yes! The first ratio is equivalent to the second: we multiply both terms by 2 2 .

The first ratio is also equivalent to the third, we multiply by 6 6 .

The second ratio is equivalent to the third: multiplication of both terms by 3 3 .


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Example

José and Dani have notebooks and pencils. José has 4 4 notebooks and 8 8 pencils. 

The ratio between the notebooks and pencils that Dani has is the same as José's. Dani has 6 6 notebooks. We are asked to calculate how many pencils Dani has.

We see that the number of pencils José has is double the number of notebooks he has. Since we already know that the ratio between notebooks and pencils that José and Dani have is identical, we can deduce that Dani has 12 12 pencils (6 6 times 2 2 , so that the number of pencils is double the number of notebooks).


Examples and exercises with solutions of Equivalent Ratios

Exercise #1

There are two circles.

One circle has a radius of 4 cm, while the other circle has a radius of 10 cm.

How many times greater is the area of the second circle than the area of the first circle?

Video Solution

Step-by-Step Solution

The area of a circle is calculated using the following formula:

where r represents the radius.

Using the formula, we calculate the areas of the circles:

Circle 1:

π*4² =

π16

Circle 2:

π*10² =

π100

To calculate how much larger one circle is than the other (in other words - what is the ratio between them)

All we need to do is divide one area by the other.

100/16 =

6.25

Therefore the answer is 6 and a quarter!

Answer

614 6\frac{1}{4}

Exercise #2

The rectangle ABCD is shown below.

AB = X

The ratio between AB and BC is x2 \sqrt{\frac{x}{2}} .


The length of diagonal AC is labelled m.

XXXmmmAAABBBCCCDDD

Choose the correct answer.

Video Solution

Step-by-Step Solution

We know that:

ABBC=x2 \frac{AB}{BC}=\sqrt{\frac{x}{2}}

We also know that AB equals X.

First, we will substitute the given data into the formula accordingly:

xBC=x2 \frac{x}{BC}=\frac{\sqrt{x}}{\sqrt{2}}

x2=BCx x\sqrt{2}=BC\sqrt{x}

x2x=BC \frac{x\sqrt{2}}{\sqrt{x}}=BC

x×x×2x=BC \frac{\sqrt{x}\times\sqrt{x}\times\sqrt{2}}{\sqrt{x}}=BC

x×2=BC \sqrt{x}\times\sqrt{2}=BC

Now let's look at triangle ABC and use the Pythagorean theorem:

AB2+BC2=AC2 AB^2+BC^2=AC^2

We substitute in our known values:

x2+(x×2)2=m2 x^2+(\sqrt{x}\times\sqrt{2})^2=m^2

x2+x×2=m2 x^2+x\times2=m^2

Finally, we will add 1 to both sides:

x2+2x+1=m2+1 x^2+2x+1=m^2+1

(x+1)2=m2+1 (x+1)^2=m^2+1

Answer

m2+1=(x+1)2 m^2+1=(x+1)^2

Exercise #3

Given the rectangle ABCD

AB=X the ratio between AB and BC is equal tox2 \sqrt{\frac{x}{2}}

We mark the length of the diagonal A A with m m

Check the correct argument:

XXXmmmAAABBBCCCDDD

Video Solution

Step-by-Step Solution

Let's find side BC

Based on what we're given:

ABBC=xBC=x2 \frac{AB}{BC}=\frac{x}{BC}=\sqrt{\frac{x}{2}}

xBC=x2 \frac{x}{BC}=\frac{\sqrt{x}}{\sqrt{2}}

2x=xBC \sqrt{2}x=\sqrt{x}BC

Let's divide by square root x:

2×xx=BC \frac{\sqrt{2}\times x}{\sqrt{x}}=BC

2×x×xx=BC \frac{\sqrt{2}\times\sqrt{x}\times\sqrt{x}}{\sqrt{x}}=BC

Let's reduce the numerator and denominator by square root x:

2x=BC \sqrt{2}\sqrt{x}=BC

We'll use the Pythagorean theorem to calculate the area of triangle ABC:

AB2+BC2=AC2 AB^2+BC^2=AC^2

Let's substitute what we're given:

x2+(2x)2=m2 x^2+(\sqrt{2}\sqrt{x})^2=m^2

x2+2x=m2 x^2+2x=m^2

Answer

x2+2x=m2 x^2+2x=m^2

Exercise #4

There are 18 balls in a box, 23 \frac{2}{3} of which are white.

How many white balls are there in the box?

Video Solution

Answer

12

Exercise #5

In a box there are 28 balls, 14 \frac{1}{4} of which are orange.

How many orange balls are there in the box?

Video Solution

Answer

7

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