Chapter 4.1, Problem 43PPS

To determine

To evaluate: $4\frac{1}{5}·5$ using distributive property

Answer to Problem 43PPS

  $21$

Explanation of Solution

Given information:

  $4\frac{1}{5}·5$

Concept Used:

Distributive Property: The product sum or difference of two or more numbers inside a parentheses with another number outside the parentheses is equivalent to the sum or difference of product of number outside the parentheses with each number inside the parenthesis. i.e.-

  $a\left(b+c-d\right)=ab+ac-ad$

Calculation:

We know $4\frac{1}{5}$ can be written as $4+\frac{1}{5}$

Therefore, the given expression can be written as-

  $\left(4+\frac{1}{5}\right)5$

We can write the given expression in distributive form as-

  $\left(4+\frac{1}{5}\right)5=\left(4\right)5+\left(\frac{1}{5}\right)5$

This can further be simplified as-

  $\begin{array}{l}\left(4\right)5+\left(\frac{1}{5}\right)5=20+1\\ \Rightarrow \left(4\right)5+\left(\frac{1}{5}\right)5=21\end{array}$