Ratio Math Problems: Questions and Step-by-Step Solutions

Practice Questions

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?

   View Step-by-Step Solution

   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. Therefore, we can set up the equation:

   \[ \begin{aligned} 3x + 5x &= 600 \\ 8x &= 600 \\ x &= 75 \end{aligned} \]

   Now, substitute \( x \) back into our expressions:

   • 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 \( w \).

   View Step-by-Step Solution

   First, find the total number of marbles in the bag:

   \[ \text{Total} = r + b + w \]

   The ratio of blue marbles (\( b \)) to the total number of marbles is:

   \[ \dfrac{b}{r + b + w} \]

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.

   View Step-by-Step Solution

   Since the ratio of length to width is 5:2, let the length \( L = 5x \) and the width \( W = 2x \).

   The perimeter formula is \( 2(L + W) \). Substitute the known values:

   \[ \begin{aligned} 2(5x + 2x) &= 280 \\ 2(7x) &= 280 \\ 14x &= 280 \\ x &= \dfrac{280}{14} = 20 \end{aligned} \]

   Now, calculate the area (\( A = L \times W \)):

   \[ \begin{aligned} A &= (5x) \times (2x) \\ A &= 10x^2 \\ A &= 10(20)^2 = 10(400) = 4000 \end{aligned} \]

   Answer: The area is 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.

   View Step-by-Step Solution

   Because the ratio of the three angles is 1:3:8, we can write their measures as \( x \), \( 3x \), and \( 8x \).

   The sum of the three interior angles of any triangle is always \( 180^\circ \). Hence:

   \[ \begin{aligned} x + 3x + 8x &= 180 \\ 12x &= 180 \\ x &= 15 \end{aligned} \]

   Calculate the measure of each angle:

   • \( x = 15^\circ \)
   • \( 3x = 3 \times 15 = 45^\circ \)
   • \( 8x = 8 \times 15 = 120^\circ \)

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?

   View Step-by-Step Solution

   In a right triangle, the two acute angles are complementary, meaning their sum is \( 90^\circ \).

   Given the ratio 2:7, let the measures of the two angles be \( 2x \) and \( 7x \).

   \[ \begin{aligned} 2x + 7x &= 90 \\ 9x &= 90 \\ x &= 10 \end{aligned} \]

   The measures of the two acute angles are:

   • \( 2x = 2 \times 10 = 20^\circ \)
   • \( 7x = 7 \times 10 = 70^\circ \)

6. A jar is filled with pennies and nickels in the ratio of 5 to 3. There are 30 nickels in the jar, how many coins are there in total?

   View Step-by-Step Solution

   A ratio of pennies to nickels of 5:3 means we can write the number of pennies as \( 5x \) and the number of nickels as \( 3x \).

   We are given that the number of nickels is 30:

   \[ 3x = 30 \quad \Rightarrow \quad x = 10 \]

   The total number of coins in the jar is the sum of pennies and nickels:

   \[ \text{Total coins} = 5x + 3x = 8x = 8(10) = 80 \]

   Answer: There are 80 coins in total.

7. A rectangle 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?

   View Step-by-Step Solution

   Let \( L \) be the length and \( W \) be the width, with \( L > W \). Set up the area and perimeter equations:

   \[ L \times W = 300 \quad \text{(I)} \] \[ 2L + 2W = 80 \quad \Rightarrow \quad L + W = 40 \quad \text{(II)} \]

   From equation (II), isolate \( W \):

   \[ W = 40 - L \]

   Substitute this into equation (I):

   \[ \begin{aligned} L(40 - L) &= 300 \\ 40L - L^2 &= 300 \\ L^2 - 40L + 300 &= 0 \\ (L - 10)(L - 30) &= 0 \end{aligned} \]

   The solutions are \( L = 10 \) or \( L = 30 \). Since length is typically the longer side (\( L > W \)), we select \( L = 30 \) and \( W = 10 \).

   The ratio of length to width is:

   \[ \dfrac{L}{W} = \dfrac{30}{10} = \dfrac{3}{1} \]

   Answer: The ratio is 3:1.

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

   View Step-by-Step Solution

   First, convert the mixed numbers to improper fractions:

   \[ 3 \dfrac{2}{3} = \dfrac{(3 \times 3) + 2}{3} = \dfrac{11}{3} \] \[ 7 \dfrac{1}{3} = \dfrac{(7 \times 3) + 1}{3} = \dfrac{22}{3} \]

   Now, write the ratio as division and solve:

   \[ \dfrac{11}{3} \div \dfrac{22}{3} = \dfrac{11}{3} \times \dfrac{3}{22} = \dfrac{11}{22} = \dfrac{1}{2} \]

   Answer: The ratio in simplest form 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?

   View Step-by-Step Solution

   Let \( y \) be the side length of square B. Since the side of square A is twice as long, its side length is \( 2y \).

   Calculate the area of both squares:

   \[ \text{Area of A} = (2y)^2 = 4y^2 \] \[ \text{Area of B} = y^2 \]

   The ratio of the area of square A to the area of square B is:

   \[ \dfrac{\text{Area of A}}{\text{Area of B}} = \dfrac{4y^2}{y^2} = \dfrac{4}{1} \]

   Answer: The ratio is 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?

    View Step-by-Step Solution

    Let \( x \) be the side length of square A. Since this is half the length of square B, the side length of square B is \( 2x \).

    Calculate the perimeter of both squares (Perimeter = 4 × side):

    \[ \text{Perimeter of A} = 4x \] \[ \text{Perimeter of B} = 4(2x) = 8x \]

    The ratio of the perimeter of square A to the perimeter of square B is:

    \[ \dfrac{\text{Perimeter of A}}{\text{Perimeter of B}} = \dfrac{4x}{8x} = \dfrac{1}{2} \]

    Answer: The ratio is 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?

    View Step-by-Step Solution

    Let \( S \) and \( A \) be the initial number of science and art books respectively.

    Because the ratio is 2:5, we can write \( S = 2x \) and \( A = 5x \). This means the total initial inventory was \( 7x \).

    If 20% of the books were sold, that means 80% of the books were unsold. The total number of unsold books is 2240:

    \[ \begin{aligned} 0.80(7x) &= 2240 \\ 5.6x &= 2240 \\ x &= \dfrac{2240}{5.6} = 400 \end{aligned} \]

    Now, calculate the initial amounts:

    • Science books: \( S = 2(400) = 800 \)
    • Art books: \( A = 5(400) = 2000 \)

12. At the start of the month a shop had 20-inch and 40-inch television sets in the ratio 4:5. By the end of the month, 200 20-inch and 500 40-inch TVs were sold and the ratio of 20-inch to 40-inch television sets became 1:1. How many television sets of each type were there at the start of the month?

    View Step-by-Step Solution

    Let the initial number of 20-inch TVs be \( 4x \) and 40-inch TVs be \( 5x \), reflecting the 4:5 ratio.

    After selling the respective amounts, the new quantities are \( 4x - 200 \) and \( 5x - 500 \). Since the new ratio is 1:1, the quantities are equal:

    \[ \begin{aligned} 4x - 200 &= 5x - 500 \\ 500 - 200 &= 5x - 4x \\ 300 &= x \end{aligned} \]

    Substitute \( x = 300 \) back into the initial expressions:

    • 20-inch TVs: \( 4(300) = 1200 \)
    • 40-inch TVs: \( 5(300) = 1500 \)

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.

    View Step-by-Step Solution

    Let the horizontal length \( H = 4x \) and the vertical height \( V = 3x \).

    The screen forms a right triangle with the diagonal as the hypotenuse. We apply the Pythagorean theorem (\( a^2 + b^2 = c^2 \)):

    \[ \begin{aligned} H^2 + V^2 &= 50^2 \\ (4x)^2 + (3x)^2 &= 2500 \\ 16x^2 + 9x^2 &= 2500 \\ 25x^2 &= 2500 \\ x^2 &= 100 \\ x &= 10 \end{aligned} \]

    Calculate the actual dimensions:

    • Horizontal length: \( H = 4(10) = 40 \) inches
    • Vertical length: \( V = 3(10) = 30 \) inches

More References and Links

• Middle School Math (Grades 6, 7, 8, 9) - Free Questions and Problems
• High School Math (Grades 10, 11 and 12) - Free Questions and Problems
• Primary Math (Grades 4 and 5) - Free Questions and Problems