What is the product of $\frac{27}{5}$, $(3)^{-3}$ and ($\frac{1}{5})^{-1}$?

Analysis

To solve this problem, we need to find the product of three distinct terms:

1. $\frac{27}{5}$
2. $(3)^{-3}$
3. $(\frac{1}{5})^{-1}$

Concept

We will apply the Laws of Indices:

• Negative Index Rule: $a^{-n} = \frac{1}{a^n}$
• Fractional Negative Index Rule: $(\frac{a}{b})^{-n} = (\frac{b}{a})^n$

Step-by-Step Solution

Step 1: Simplify each term.

• The first term is already simplified: $\frac{27}{5}$
• For the second term $(3)^{-3}$: Using the negative index rule, $(3)^{-3} = \frac{1}{3^3} = \frac{1}{27}$
• For the third term $(\frac{1}{5})^{-1}$: Using the fractional negative index rule, $(\frac{1}{5})^{-1} = (\frac{5}{1})^1 = 5$

Step 2: Multiply the simplified terms together. Product = $\frac{27}{5} \times \frac{1}{27} \times 5$

Step 3: Perform the calculation. Product = $\left( \frac{27}{5} \right) \times \left( \frac{1}{27} \right) \times 5$

Notice that the $27$ in the numerator and the $27$ in the denominator cancel each other out: Product = $\frac{1}{5} \times 5$

Finally, the $5$ in the numerator and the $5$ in the denominator cancel out: Product = $1$

Conclusion

The product of the three terms is 1, which corresponds to option C.

Exam Tip

When dealing with products of fractions and indices, always look for opportunities to cancel out terms before multiplying the numerators and denominators. It saves time and reduces the risk of arithmetic errors!