**Exploring Ratios Worksheet :**

Worksheet given in this section will be much useful for the students who would like to practice problems on ratios.

**Problem 1 :**

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

**Problem 2 :**

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

**Problem 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.

**Problem 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.

**Problem 5 :**

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

**Problem 6 :**

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

**Problem 7 :**

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

**Problem 8 :**

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

**Problem 9 : **

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

**Problem 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?

**Problem 1 :**

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

**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.

**Problem 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.

**Problem 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.

**Problem 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.

**Problem 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°)

**Problem 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

**Problem 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.

**Problem 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.

**Problem **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

**Problem **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, how to solve problems using ratios.

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