Algebra: Cancellation Of Indices Term Practice 1
This section contains 4 focused practice questions designed to strengthen students’ understanding of the cancellation technique. Students will practise identifying common numerical and algebraic factors and cancelling terms correctly within algebraic fractions to simplify expressions efficiently.
Related Lessons:
Practice 1 - Question 1
Simplify completely \(\frac{5c}{2}\div\frac{20c^2}{d}\)
Solution:
\[
\begin{align*}
\frac{5c}{2}\div\frac{20c^2}{d}&=\frac{5c}{2}\times\frac{d}{20c^2}\\
&=\frac{d}{8c}
\end{align*}
\]
Practice 1 - Question 2
Simplify \(\left(\frac{c^2}{2a^3}\right)\times\left(\frac{-a^2}{bc}\right)^2\div\left(\frac{ac}{b}\right)\) completely.
Solution:
\[
\begin{align*}
\left(\frac{c^2}{2a^3}\right)\times\left(\frac{-a^2}{bc}\right)^2\div\left(\frac{ac}{b}\right)&=\frac{c^2}{2a^3}\times\frac{-a^2}{bc}\times\frac{-a^2}{bc}\times\frac{b}{ac}\\
&=\frac{1}{2bc}\\
\end{align*}
\]
Practice 1 - Question 3
Simplify as a single fraction in its simplest form \(\frac{3a^2}{7bc}\div\frac{9a}{14b}\)
Solution:
\[
\begin{align*}
\frac{3a^2}{7bc}\div\frac{9a}{14b}&=\frac{3a^2}{7bc}\times\frac{14b}{9a}\\
&=\frac{2a}{3c}\\
\end{align*}
\]
Practice 1 - Question 4
Simplify the expression \(\frac{5a}{9b}\times\frac{ac^2}{2b}\div\frac{c^3}{8b^3}\)
Solution:
\[
\begin{align*}
\frac{5a}{9b}\times\frac{ac^2}{2b}\div\frac{c^3}{8b^3}&=\frac{5a}{9b}\times\frac{ac^2}{2b}\times\frac{8b^3}{c^3}\\
&=\frac{20a^2b}{9c}\\
\end{align*}
\]
Step By Step Full Solution
Some students may arrive at the correct answer but using incorrect methods. That is why it is essential to understand clearly how each solution is developed step by step. Below is the complete solution for the four practice questions above.