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BECE Mathematics 2013 Paper

Question : 7

Total: 40

Simplify

\frac{2^2 \times 3^2}{4^2 \times 3^3}

Solution:

Concept:

Express all terms with prime bases and apply the laws of indices to simplify.

Explanation:

Rewrite

4

as a power of

2

4 = 2^2

Therefore,

4^2 = (2^2)^2 = 2^{2 \times 2} = 2^4

Substitute this into the expression:

\frac{2^2 \times 3^2}{4^2 \times 3^3} = \frac{2^2 \times 3^2}{2^4 \times 3^3}

Apply the division rule of indices:

\frac{a^m}{a^n} = a^{m-n}

For base

2

2^{2-4} = 2^{-2}

For base

3

3^{2-3} = 3^{-1}

So the expression becomes

2^{-2} \times 3^{-1}

Use the rule

a^{-b} = \frac{1}{a^b}

2^{-2} \times 3^{-1} = \frac{1}{2^2} \times \frac{1}{3} = \frac{1}{4} \times \frac{1}{3}

Multiply the fractions:

\frac{1}{4} \times \frac{1}{3} = \frac{1}{12}

Answer:

\frac{1}{12}

(Option A).

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