Ratio Word Problems
with Solutions and Explanations - Grade 9

Detailed solutions and full explanations to math word problems - grade 9 are presented.

  1. 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

  2. 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)

  3. 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 = 10x2 = 10202 = 4000 square meters

  4. 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

  5. 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

  6. 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

  7. 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

    L2 + 300 = 40L

    Rewrite the equation in standard form, factor and solve

    L2 - 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

  8. 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

  9. 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 = x2 and B = y2

    But x = 2y. Hence

    A = (2y)2 = 4 y2

    The ratio of A to B is

    4 y2 / y2 = 4 / 1 or 4:1

  10. 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

  11. 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

  12. At the start of the month a shop had 20-inches and 40-inches television sets in the ratio 4:5. By the end of the month, 200 20-inches and 500 40-inches were sold and the ratio of 20-inches to 40-inches 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 20-inches and 40-inches 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 20-inches and 40-inches 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

  13. 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)