Ratio Maths Problems with Solutions and Explanations for Grade 9
Detailed solutions and full explanations to ratio maths problems for grade 9 are presented.

There are 600 pupils in a school. The ratio of boys to girls in this school is 3:5. How many girls and how many boys are in this school?
Solution
In order to obtain a ratio of boys to girls equal to 3:5, the number of boys has to be written as 3 x and the number of girls as 5 x where x is a common factor to the number of girls and the number of boys. The total number of boys and girls is 600. Hence
3x + 5x = 600
Solve for x
8x = 600
x = 75
Number of boys
3x = 3 × 75 = 225
Number of girls
5x = 5 × 75 = 375

There are r red marbles, b blue marbles and w white marbles in a bag. Write the ratio of the number of blue marbles to the total number of marbles in terms of r, b and w.
Solution
The total number of marbles is
r + b + w
The total ratio of blue marbles to the total number of marbles is
r / (r + b + w)

The perimeter of a rectangle is equal to 280 meters. The ratio of its length to its width is 5:2. Find the area of the rectangle.
Solution
If the ratio of the length to the width is 5:2, then the measure L of the length and and the measure W of the with can be written as
L = 5x and W = 2x
We now use the perimeter to write
280 = 2(2L + 2W) = 2(5x + 2x) = 14x
Solve for x
280 = 14x
x = 280 / 14 = 20
The area A of the rectangle is given by
A = L × W = 5x × 2x = 10x^{2} = 10×20^{2} = 4000 square meters

The angles of a triangle are in the ratio 1:3:8. Find the measures of the three angles of this triangle.
Solution
If the ratio of the three angles is 1:3:8, then the measures of these angles can be written as x, 3x and 8x. Also the sum of the three interior angles of a triangle is equal to 180°. Hence
x + 3x + 8x = 180
Solve for x
12x = 180
x = 15
The measures of the three angles are
x = 15°
3x = 3 × 15 = 45°
8x = 8 × 15 = 120°

The measures of the two acute angles of a right triangle are in the ratio 2:7. What are the measures of the two angles?
Solution
If the ratio of the two angles is 2:7, then the measures of two angles can be written as 2x and 7x. Also the two acute angles of a triangle is equal to 90°. Hence
2x + 7x = 90
9x = 90
x = 10
Measures of the two acute angles are
2x = 2 × 10 = 20°
7x = 7 × 10 = 70°

A jar is filled with pennies and nickels in the ratio of 5 to 3 (pennies to nickels). There are 30 nickles in the jar, how many coins are there?
Solution
A ratio of pennies to nickels of 5 to 3 means that we can write the number of pennies and nickels in the form
number of pennies = 5x and number of nickels = 3x
But we know the number of nickels, 30. Hence
3x = 30
Solve for x
x = 10
The total number of coins is given by
5x + 3x = 8x = 8 × 10 = 80

A rectangular field has an area of 300 square meters and a perimeter of 80 meters. What is the ratio of the length to the width of this field?
Solution
Let L and W being the length and the width (with L > W) of the rectangular field. The area and the perimeter are given; hence
L × W = 300 (I)
2L + 2W = 80 (II) which is equivalent to L + W = 40 (III)
We need to find the ratio L / W. Equation (I) gives
W = 300 / L
Substitute W by 300 / L in equation (III)
L + 300 / L = 40
Multiply all terms in the above equation by L and simplify
L^{2} + 300 = 40L
Rewrite the equation in standard form, factor and solve
L^{2}  40 L + 300 = 0
(L  10)(L  30) = 0
Solutions: L = 10 and L = 30
We now calculate W
For L = 10 , W = 300 / L = 300 / 10 = 30 m
For L = 30 , W = 300 / L = 300 / 30 = 10
Since L > W, we select the soultion
L = 30 and W = 10
and the L / W is equal to
30 / 10 = 3 / 1 or 3:1

Express the ratio 3 2/3 : 7 1/3 in its simplest form.
Solution
We first convert the mixed numbers 3 2/3 and 7 1/3 into fractions
3 2/3 = 3+ 2 / 3 = 3 × 3 / 3 + 2 / 3 = 9 / 3 + 2 / 3 = 11 / 3
7 1/3 = 7 + 1 / 3 = 7 × 3 / 3 + 1 / 3 = 22 / 3
The ratio 3 2/3 : 7 1/3 can be expressed as
11 / 3 ÷ 22 / 3 = 11 / 3 × 3 / 22
Simplify
= 11 / 22 = 1 / 2
The ratio is 1 / 2 or 1:2

The length of the side of square A is twice the length of the side of square B. What is the ratio of the area of square A to the area of square B?
Solution
Let x be the length of the side of square A and y be the length of the side of square B with x = 2 y. Area of A and B are given by
A = x^{2} and B = y^{2}
But x = 2y. Hence
A = (2y)^{2} = 4 y^{2}
The ratio of A to B is
4 y^{2} / y^{2} = 4 / 1 or 4:1

The length of the side of square A is half the length of the side of square B. What is the ratio of the perimeter of square A to the perimeter of square B?
Solution
Let 2 x be the side of square B and x be the side of square A (half).
Perimeter of square A = 4 x , perimeter of square B = 4 (2 x) = 8 x
The ratio R of the perimeter of A to the perimeter of B is
R = 4 x / 8 x = 1 / 2

At the start of the week a bookshop had science and art books in the ratio 2:5. By the end of the week, 20% of each type of books were sold and 2240 books of both types were unsold. How many books of each type were there at the start of the week?
Solution
Let S and A be the number of science and art books respectively at the start of the week. Hence
S / A = 2 / 5
If 20% of each type of books were sold then 80% of each were unsold at the end of the week and their total is known: 2240. Hence
80% S + 80 % A = 2240 or 0.8 S + 0.8 A = 2240
We now need to solve the two equations in S and A obtained above to find the number of books of each type. Use the cross product on the equation S / A = 2 / 5 to obtain
5 S = 2 A
Solve the system of equations
0.8 S + 0.8 A = 2240 and 5 S = 2 A
to obtain
S = 800 and A = 2000

At the start of the month a shop had 20inches and 40inches television sets in the ratio 4:5. By the end of the month, 200 20inches and 500 40inches were sold and the ratio of 20inches to 40inches television sets became 1:1. How many television sets of each type were there at the start of the month?
Solution
Let x and y be the number of 20inches and 40inches television sets respectively at the start of the month. Hence
x / y = 4 / 5
By the end of the month 200 and 500 were sold from the 20inches and 40inches television sets respectively. Therefore x  200 and y  500 were unsold and their ratio is known and equal to 1:1. Hence
(x  200) / (y  500) = 1 / 1
Use the cross product on both equations obtained above
5x = 4 y and x  200 = y  500
Solve the system of equations to obtain
x = 1200 and y = 1500

The aspect ratio of a tv screen is the ratio of the measure of the horizontal length to the measure of the vertical length. Find the horizontal length and vertical height of a tv screen with an aspect ratio of 4:3 and a diagonal of 50 inches.
Solution
Let H be the horizontal length and V be the vertical height of the tv. Their ratio is given. Hence
H / V = 4 / 3 or use cross product to obtain: 3 H = 4 V
The relationship between the horizontal length, the vertical height and the diagonal is given by Pythagora's theorem as follows:
H^{ 2} + V^{ 2} = 50^{ 2}
We now solve the system of equations
3 H = 4 V and H^{ 2} + V^{ 2} = 50^{ 2}
to obtain
H = 40 and V = 30

More References and Links
Middle School Math (Grades 6, 7, 8, 9)  Free Questions and Problems With Answers
High School Math (Grades 10, 11 and 12)  Free Questions and Problems With Answers
Primary Math (Grades 4 and 5) with Free Questions and Problems With Answers
Home Page