Ratio Word Problems
with Solutions and Explanations  Grade 9
Detailed solutions and full explanations to math word problems  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


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Updated: 30 March 2009 (A Dendane)