**Exploring Ratios :**

A ratio is a comparison of two quantities. It shows how many times as great one quantity is than another.

For example, the ratio of star-shaped beads to moon-shaped beads in a bracelet is 3 to 1. This means that for every 3 star beads, there is 1 moon bead.

The numbers in a ratio are called terms. Suppose that in a pet store there are 5 dogs for every 3 cats. The ratio of dogs to cats is 5 to 3.

The terms of the ratio are 5 and 3. The ratio can also be written as follows.

5 dogs to 3 cats or 5 to 3 or 5 : 3 or 5/3

A ratio can compare a part to a part, a part to the whole, or the whole to a part.

In this section, we are going to explore ratios with its properties.

1. Ratio exists only between quantities of the same kind.

2. Quantities to be compared (by division) must be in the same units.

3. The order of the terms in a ratio is important.

4. Both terms of a ratio can be multiplied or divided by the same (non–zero) number. Usually a ratio is expressed in lowest terms (or simplest form).

5. The order of the terms in a ratio is important.

6. To compare two ratios, convert them into equivalent like fractions.

7. If a quantity increases or decreases in the ratio a : b, then new quantity is

The fraction by which the original quantity is multiplied to get a new quantity is called the factor multiplying ratio.

8. One ratio is the inverse of another, if their product is 1. Thus a : b is the inverse of b : a and vice–versa.

9. A ratio a : b is said to be of greater inequality if a > b and of less inequality if a < b.

10. The ratio compounded of the two ratios a : b and c : d is ac : bd.

11. A ratio compounded of itself is called its duplicate ratio.

Thus a² : b² is the duplicate ratio of a : b. Similarly, the triplicate ratio of a : b is a³ : b³.

12. The sub–duplicate ratio of a : b is √a : √b and the sub triplicate ratio of a : b is ³√a : ³√b.

13 If the ratio of two similar quantities can be expressed as a ratio of two integers, the quantities are said to be commensurable; otherwise, they are said to be in-commensurable.

√3 : √2 cannot be expressed as the ratio of two integers and therefore, √3 and √2 are in-commensurable quantities.

14. Continued Ratio is the relation (or compassion) between the magnitudes of three or more quantities of the same kind. The continued ratio of three similar quantities a, b, c is written as a: b: c.

**Example 1 :**

The average age of three boys is 25 years and their ages are in the proportion 3:5:7. The age of the youngest boy is

**Solution :**

From the ratio 3 : 5 : 7, the ages of three boys are 3x, 5x and 7x.

Average age of three boys = 25

(3x+5x+7x)/3 = 25 ----------> 15x = 75 -----------> x = 5

Age of the first boy = 3x = 3(5) = 15

Age of the first boy = 5x = 5(5) = 25

Age of the first boy = 7x = 7(5) = 105

Hence the age of the youngest boy is 15 years.

**Example 2 :**

John weighs 56.7 kilograms. If he is going to reduce his weight in the ratio 7:6, find his new weight.

**Solution :**

Original weight of John = 56.7 kg (given)

He is going to reduce his weight in the ratio 7:6

His new weight = (6x56.7)/7 = 6x8.1 = 48.6 kg.

Hence his new weight = 48.6 kg.

**Example 3 :**

The ratio of the no. of boys to the no. of girls in a school of 720 students is 3:5. If 18 new girls are admitted in the school, find how many new boys may be admitted so that the ratio of the no. of boys to the no. of girls may change to 2:3.

**Solution :**

Sum of the terms in the given ratio = 3+5 = 8

So, no. of boys in the school = 720x(3/8)= 270

No. of girls in the school = 720x(5/8)= 450

Let "x" be the no. of new boys admitted in the school.

No. of new girls admitted = 18 (given)

After the above new admissions,

no. of boys in the school = 270+x

no. of girls in the school = 450+18 = 468

The ratio after the new admission is 2 : 3 (given)

So, (270+x) : 468 = 2 : 3

3(270+x) = 468x2 (using cross product rule in proportion)

810 + 3x = 936

3x = 126

x = 42

Hence the no. of new boys admitted in the school is 42.

**Example 4 :**

The monthly incomes of two persons are in the ratio 4:5 and their monthly expenditures are in the ratio 7:9. If each saves $50 per month, find the monthly income of the second person.

**Solution :**

From the given ratio of incomes ( 4 : 5 ),

Income of the 1st person = 4x

Income of the 2nd person = 5x

(Expenditure = Income - Savings)

Then, expenditure of the 1st person = 4x - 50

Expenditure of the 2nd person = 5x - 50

Expenditure ratio = 7 : 9 (given)

So, (4x - 50) : (5x - 50) = 7 : 9

9(4x - 50) = 7(5x - 50)

(using cross product rule in proportion)

36x - 450 = 35x - 350

x = 100

Then, income of the second person is

= 5x = 5(100) = 500.

Hence, income of the second person is $500.

**Example 5 :**

If the angles of a triangle are in the ratio 2:7:11, then find the angles.

**Solution :**

From the ratio 2 : 7 : 11,

the three angles are 2x, 7x, 11x

In any triangle, sum of the angles = 180

So, 2x + 7x + 11x = 180°

20x = 180 -------> x = 9

Then, the first angle = 2x = 2(9) = 18°

The second angle = 7x = 7(9) = 63°

The third angle = 11x = 11(9) 99°

Hence the angles of the triangle are (18°, 63°, 99°)

**Example 6 :**

In a business, if A can earn $ 7500 in 2.5 years, find the unit rate of his earning per month.

**Solution : **

Given : Earning in 2.5 years = $ 7500

1 year = 12 months

2.5 years = 2.5 x 12 = 30 months

Then, earning in 30 months = $ 7500

Therefore, earning in 1 month = 7500 / 30 = $ 250

Hence, the unit rate of his earning per month is $ 250

**Example 7 :**

If David can prepare 2 gallons of juice in 4 days, how many cups of juice can he prepare per day ?

**Solution : **

No of gallons of juice prepared in 4 days = 2 gallons

1 gallon = 16 cups

So, no. of cups of juice prepared in 4 days = 2 x 16 = 32 cups

Therefore, David can prepare 32 cups of juice in 4 days.

Then, no. of cups of juice prepared in 1 day = 32 / 4 = 8

Hence, David can prepare 8 cups of juice in 1 day.

**Example 8 :**

If John can cover 360 miles in 3 hours, find the number of miles covered by John in 1 minute.

**Solution : **

No of miles covered in 3 hours = 360

Then, no. of miles covered in 1 hour = 360 / 3 = 180

1 hour = 60 minutes

So, no. of miles covered in 60 minutes = 180

Then, no. of miles covered 1 minute = 180 / 60 = 3

Hence, John can cover 3 miles in 1 minute.

**Example 9 : **

Shanel walks 2/ 5 of a mile every 1/7 hour. Express her speed as a unit rate in miles per hour.

**Solution : **

Given : Shanel walks 2/ 5 of a mile every 1/7 hour

We know the formula for speed.

That is, Speed = Distance / time

Speed = (2/5) / (1/7)

Speed = (2/5) x (7/1)

Speed = 14 / 5

Speed = 2.8 miles per hour.

Hence, the speed of Shanel is 2.8 miles per hour

**Example 10 : **

Declan use 2 /35 of a gallon of gas for every 4 /5 of a mile that he drives. At this rate, how many miles can he drive on one gallon of gas?

**Solution : **

Given : In 2 /35 of a gallon of gas, 4 /5 of a mile is traveled

Then, in 1 gallon of gas = (4/5) x (35/2) miles traveled.

= 14 miles traveled.

Hence, Declan can drive 14 miles in 1 gallon of gas

After having gone through the stuff given above, we hope that the students would have understood ratios.

Apart from the stuff given in this section, if you need any other stuff in math, please use our google custom search here.

HTML Comment Box is loading comments...

You can also visit the following web pages on different stuff in math.

**WORD PROBLEMS**

**Word problems on simple equations **

**Word problems on linear equations **

**Word problems on quadratic equations**

**Area and perimeter word problems**

**Word problems on direct variation and inverse variation **

**Word problems on comparing rates**

**Converting customary units word problems **

**Converting metric units word problems**

**Word problems on simple interest**

**Word problems on compound interest**

**Word problems on types of angles **

**Complementary and supplementary angles word problems**

**Trigonometry word problems**

**Markup and markdown word problems **

**Word problems on mixed fractrions**

**One step equation word problems**

**Linear inequalities word problems**

**Ratio and proportion word problems**

**Word problems on sets and venn diagrams**

**Pythagorean theorem word problems**

**Percent of a number word problems**

**Word problems on constant speed**

**Word problems on average speed **

**Word problems on sum of the angles of a triangle is 180 degree**

**OTHER TOPICS **

**Time, speed and distance shortcuts**

**Ratio and proportion shortcuts**

**Domain and range of rational functions**

**Domain and range of rational functions with holes**

**Graphing rational functions with holes**

**Converting repeating decimals in to fractions**

**Decimal representation of rational numbers**

**Finding square root using long division**

**L.C.M method to solve time and work problems**

**Translating the word problems in to algebraic expressions**

**Remainder when 2 power 256 is divided by 17**

**Remainder when 17 power 23 is divided by 16**

**Sum of all three digit numbers divisible by 6**

**Sum of all three digit numbers divisible by 7**

**Sum of all three digit numbers divisible by 8**

**Sum of all three digit numbers formed using 1, 3, 4**

**Sum of all three four digit numbers formed with non zero digits**