Problem 45 - Prealgebra

Question

Simplify each expression as much as possible. $$100 \div\left(\frac{5}{7}\right)^{2}+200 \div\left(\frac{2}{3}\right)^{2}$$

Step-by-Step Solution

Answer

The simplified expression is 646.

Step 1: Simplify the First Division

First, we need to evaluate the division part of the expression. We have \(100 \div \left(\frac{5}{7}\right)^{2}\). The division by a fraction can be done by multiplying by its reciprocal. Therefore, we first square the fraction: \(\left(\frac{5}{7}\right)^2 = \frac{25}{49}\). Now, take the reciprocal: \(\frac{49}{25}\). The expression becomes \(100 \times \frac{49}{25}\). Multiply: \(100 \times \frac{49}{25} = \frac{4900}{25} = 196\).

Step 2: Simplify the Second Division

Next, simplify the second part of the expression: \(200 \div \left(\frac{2}{3}\right)^{2}\). Again, we take the square: \(\left(\frac{2}{3}\right)^2 = \frac{4}{9}\). Take the reciprocal: \(\frac{9}{4}\). Multiply: \(200 \times \frac{9}{4}\). This simplifies to \(\frac{1800}{4} = 450\).

Step 3: Add the Results from Both Simplifications

Now that we have simplified both parts of the original expression, we add the results together: \(196 + 450\). This results in \(646\).

Key Concepts

Division of Fractions

When dividing by a fraction, the process may appear daunting at first. Thankfully, there is a simple trick: multiply by the reciprocal instead. This method simplifies division of fractions to a multiplication problem. Let’s take a close look at how it works.
Consider an example where we need to divide 100 by the square of \(\frac{5}{7}\). Squaring the fraction gives \(\left(\frac{5}{7}\right)^2 = \frac{25}{49}\). Now, instead of directly dividing, multiply by the reciprocal of \(\frac{25}{49}\), which is \(\frac{49}{25}\).

• This turns \(100 \div \frac{25}{49}\) into \(100 \times \frac{49}{25}\).
• Then calculate: \(100 \times \frac{49}{25} = \frac{4900}{25} = 196\).

Therefore, division of fractions using reciprocals makes this task approachable and simple.

Reciprocal

A reciprocal is like a mirror image for fractions; it flips the fraction upside down. The reciprocal of a fraction \(\frac{a}{b}\) is \(\frac{b}{a}\).
Using reciprocals is crucial when dealing with fraction division, as dividing by a fraction is equivalent to multiplying by its reciprocal. It ensures the multiplication process is straightforward, helping us avoid the complexity of direct division.

For instance, in our exercise, the fraction \(\frac{5}{7}\) was squared to create \(\frac{25}{49}\), and its reciprocal is \(\frac{49}{25}\). This reciprocal enables us to transform a complex division into a simple multiplication, further facilitating the simplification of expressions.

Squaring Fractions

Squaring a fraction involves multiplying the fraction by itself. This operation is key to our simplification process, especially when dealing with expressions involving division of squared fractions.

• Take \(\frac{5}{7}\) as an example. To square it, calculate \(\left(\frac{5}{7}\right)^2 = \frac{5}{7} \times \frac{5}{7} = \frac{25}{49}\).
• This squared fraction is then used directly in subsequent steps. In division, the reciprocal becomes necessary, hence \(\frac{49}{25}\), transforming division into multiplication.

Remember, squaring fractions requires squaring both the numerator and the denominator, which lays the foundation for further simplification operations.

Arithmetic Operations

Arithmetic operations, including addition, multiplication, and division, form the backbone of algebraic simplifications. Mastery of these basic operations is essential in simplifying complex mathematical expressions.

For the given problem, once both divisions were simplified using reciprocals, the final arithmetic operation was addition:

• After obtaining the results from division, i.e., 196 and 450, adding them yielded \(196 + 450 = 646\).

This process illustrates how fundamental arithmetic operations are applied sequentially to simplify expressions into their simplest form.
By breaking down operations into smaller steps, we achieve clarity and ease in computation, making comprehensive simplification attainable.