Detailed Solutions to Questions on Math Proportions

Detailed solutions the questions on proportions are presented with full explanations.

  1. Find \( x \) if \( \dfrac{x}{2} = \dfrac{4}{8} \).

    Solution

    Multiply both terms of the proportion by the denominators 2 and 8.

    \( \color{Red} {2\times8}\,\dfrac{x}{2}=\color{Red} {2\times 8}\,\dfrac{4}{8} \)

    Simplify.

    \( {\color{Red} {\cancel{2}\times8}}\,\dfrac{x}{\cancel{2}}={\color{Red} {2\times \cancel{8}}}\,\dfrac{4}{\cancel{8}} \)

    Simplify.

    \( 8x = 8 \)

    Divide both sides of the equation by the factor of x and simplify.

    \( \dfrac{8x}{8}= \dfrac{8}{8} \)

    \( x = 1 \)
  2. Find \( p \) if \( \dfrac{3}{p} = \dfrac{1}{5} \).

    Solution

    Multiply both terms of the proportion by the denominators p and 5.

    \( \color{Red} {5\times p}\,\dfrac{3}{p}=\color{Red} {5\times p}\,\dfrac{1}{5} \)

    Simplify.

    \( {\color{Red} {5\times \cancel{p}}}\,\dfrac{3}{\cancel{p}}={\color{Red} {\cancel{5}\times p}\,\dfrac{1}{\cancel{5}} } \)

    Simplify.

    \( 15 = p \)
  3. If \( \dfrac{31}{5} = \dfrac{w}{15}\), then what is the value of \( w \)?

    Solution

    Multiply both terms of the proportion by the denominators 5 and 15.

    \( \color{Red} {5\times 15}\,\dfrac{31}{5}=\color{Red} {5\times 15}\,\dfrac{w}{15} \)

    Simplify.

    \( {\color{Red} {\cancel{5}\times 5}}\,\dfrac{31}{\cancel{5}}={\color{Red} {5\times \cancel{15}}\,\dfrac{w}{\cancel{15}} } \)

    but do not multiply numbers.

    \( 5 \times 31 = 5 \times w \)

    Divide both sides by 5

    \( \dfrac{5 \times 31}{5} = \dfrac{5 \times w}{5} \)

    and simplify.

    \( 31 = w \)
  4. Find \( k \) if \( \dfrac{2k}{3} = \dfrac{20}{6} \).

    Solution

    Multiply both terms of the proportion by the denominators 3 and 6.

    \( \color{Red} {3\times 6}\,\dfrac{2k}{3}=\color{Red} {3\times 6}\,\dfrac{20}{6} \)

    Simplify.

    \( \color{Red} {\cancel{3}\times 6}\,\dfrac{2k}{\cancel{3}}=\color{Red} {3\times \cancel{6}}\,\dfrac{20}{\cancel{6}} \)

    Simplify.

    \( 12k = 60 \)

    Divide both sides by the factor of k (12).

    \( \dfrac{12k}{12} = \dfrac{60}{12} \)

    and simplify.

    \( k = 5 \)
  5. Solve the proportion \( \dfrac{3}{7} = \dfrac{y}{0} \) if possible.

    Solution

    No solution because the division by zero is not allowed in mathematics.
  6. Solve the proportion \( \dfrac{1}{4} = \dfrac{0}{x} \) if possible.

    Solution

    The left side 1/4 is not equal to zero. The right side is either zero or undefined for x = 0. Hence the above proprotion does not have a solution.
  7. If \( \dfrac{m}{4} = \dfrac{3}{12} \), then what is the value of m?.

    Solution

    Note here that the fraction on the right side 3 / 12 may be reduced to 1 / 4 by dividing both numerator and denominator by 3 as follows.

    \( \dfrac{3}{12} = \dfrac{3 \div 3}{12\div 3} = \dfrac{1}{4} \)

    Hence the given proportion may be written as.

    \( \dfrac{m}{4} = \dfrac{1}{4} \)

    The two fractions have the same denominator and therefore their numerators must be equal. Hence

    \( m = 1 \)
  8. Find \(t \) if \( \dfrac{6}{14} = \dfrac{2t}{14} \).

    Solution

    Note the two fractions making the proportion given above have equal denominator and therefore their numerator must be equal. Hence

    \( 6 = 2 t \)

    Divide by the coefficient of t (2)

    \( 6 / 2 = 2 t / 2 \)

    and simplify

    \( 3 = t \)


More High School Math (Grades 10, 11 and 12) - Free Questions and Problems With Answers

More Middle School Math (Grades 6, 7, 8, 9) - Free Questions and Problems With Answers

More Primary Math (Grades 4 and 5) with Free Questions and Problems With Answers

Author - e-mail

Home Page

Search