Ratio Maths Problems with Solutions and Explanations for Grade 9

Detailed solutions and full explanations to ratio maths problems for 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, let the number of boys be \(3x\) and the number of girls be \(5x\), where \(x\) is a common factor. The total number of pupils is 600. Hence \[ 3x + 5x = 600 \] Solve for \(x\): \[ 8x = 600 \] \[ x = 75 \] Number of boys: \[ 3x = 3 \times 75 = 225 \] Number of girls: \[ 5x = 5 \times 75 = 375 \]
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 ratio of blue marbles to the total number of marbles is \[ \frac{b}{r + b + w} \]
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 length to width is 5:2, let \[ L = 5x \quad \text{and} \quad W = 2x \] The perimeter is given by \[ 2(L + W) = 280 \implies 2(5x + 2x) = 280 \] \[ 2 \cdot 7x = 280 \] \[ 14x = 280 \] \[ x = \frac{280}{14} = 20 \] The area \(A\) of the rectangle is \[ A = L \times W = 5x \times 2x = 10x^2 = 10 \times 20^2 = 4000 \text{ square meters} \]

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 \(180^\circ\). Hence \[ x + 3x + 8x = 180 \] Solve for \(x\): \[ 12x = 180 \quad \Rightarrow \quad x = 15 \] The measures of the three angles are \[ x = 15^\circ, \quad 3x = 3 \times 15 = 45^\circ, \quad 8x = 8 \times 15 = 120^\circ \]
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 the two angles can be written as \(2x\) and \(7x\). Also, the sum of the two acute angles of a right triangle is \(90^\circ\). Hence \[ 2x + 7x = 90 \] Solve for \(x\): \[ 9x = 90 \quad \Rightarrow \quad x = 10 \] The measures of the two acute angles are \[ 2x = 2 \times 10 = 20^\circ, \quad 7x = 7 \times 10 = 70^\circ \]
6. 
   A jar is filled with pennies and nickels in the ratio of 5:3 (pennies to nickels). There are 30 nickels in the jar. How many coins are there?

   Solution

   A ratio of pennies to nickels of 5:3 means that we can write \[ \text{number of pennies} = 5x, \quad \text{number of nickels} = 3x \] But we know the number of nickels is 30. Hence \[ 3x = 30 \quad \Rightarrow \quad x = 10 \] The total number of coins is \[ 5x + 3x = 8x = 8 \times 10 = 80 \]
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 \(W\) be the length and width (\(L > W\)) of the rectangular field. The area and perimeter are given; hence \[ L \times W = 300 \quad \text{(I)} \] \[ 2L + 2W = 80 \quad \Rightarrow \quad L + W = 40 \quad \text{(II)} \] We need to find the ratio \(L/W\). From (I): \[ W = \frac{300}{L} \] Substitute \(W = 300/L\) into (II): \[ L + \frac{300}{L} = 40 \] Multiply all terms by \(L\) and simplify: \[ L^2 + 300 = 40L \] Rewrite in standard form and factor: \[ L^2 - 40L + 300 = 0 \] \[ (L - 10)(L - 30) = 0 \] Solutions: \[ L = 10 \quad \text{or} \quad L = 30 \] Calculate \(W\): \[ L = 10 \Rightarrow W = \frac{300}{10} = 30, \quad L = 30 \Rightarrow W = \frac{300}{30} = 10 \] Since \(L > W\), we select \[ L = 30, \quad W = 10 \] The ratio \(L/W\) is \[ \frac{L}{W} = \frac{30}{10} = 3:1 \]

8. 
   Express the ratio \(3 \frac{2}{3} : 7 \frac{1}{3}\) in its simplest form.

   Solution

   First, convert the mixed numbers to improper fractions: \[ 3 \frac{2}{3} = 3 + \frac{2}{3} = \frac{9}{3} + \frac{2}{3} = \frac{11}{3} \] \[ 7 \frac{1}{3} = 7 + \frac{1}{3} = \frac{21}{3} + \frac{1}{3} = \frac{22}{3} \] The ratio can be expressed as \[ \frac{11}{3} \div \frac{22}{3} = \frac{11}{3} \times \frac{3}{22} = \frac{11}{22} = \frac{1}{2} \] Hence, the ratio is \[ 1:2 \]
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 side of square A and \(y\) be the side of square B. Then \(x = 2y\). The areas are: \[ A = x^2, \quad B = y^2 \] Substitute \(x = 2y\): \[ A = (2y)^2 = 4y^2 \] The ratio of areas is: \[ \frac{A}{B} = \frac{4y^2}{y^2} = 4:1 \]
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 \(x\) be the side of square A and \(2x\) the side of square B. Then: \[ \text{Perimeter of A} = 4x, \quad \text{Perimeter of B} = 4(2x) = 8x \] The ratio of perimeters is: \[ R = \frac{4x}{8x} = \frac{1}{2} \]
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 book 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. Then: \[ \frac{S}{A} = \frac{2}{5} \quad \Rightarrow \quad 5S = 2A \] 80% of each type was unsold: \[ 0.8S + 0.8A = 2240 \] Solve the system: \[ 5S = 2A, \quad 0.8S + 0.8A = 2240 \] \[ S = 800, \quad A = 2000 \]
12. 
    At the start of the month, a shop had 20-inch and 40-inch televisions in the ratio 4:5. By the end of the month, 200 20-inch and 500 40-inch TVs were sold, and the ratio became 1:1. How many televisions of each type were there at the start?

    Solution

    Let \(x\) and \(y\) be the number of 20-inch and 40-inch TVs. Then: \[ \frac{x}{y} = \frac{4}{5} \quad \Rightarrow \quad 5x = 4y \] After sales: \[ \frac{x - 200}{y - 500} = 1 \quad \Rightarrow \quad x - 200 = y - 500 \] Solve the system: \[ 5x = 4y, \quad x - 200 = y - 500 \] \[ x = 1200, \quad y = 1500 \]
13. 
    The aspect ratio of a TV screen is the ratio of horizontal length to vertical height. Find the horizontal length and vertical height of a TV with aspect ratio 4:3 and diagonal 50 inches.

    Solution

    Let \(H\) and \(V\) be the horizontal length and vertical height. Then: \[ \frac{H}{V} = \frac{4}{3} \quad \Rightarrow \quad 3H = 4V \] The diagonal satisfies Pythagoras’ theorem: \[ H^2 + V^2 = 50^2 \] Solve the system: \[ 3H = 4V, \quad H^2 + V^2 = 2500 \] \[ H = 40, \quad V = 30 \]

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