WAEC 2019 Mathematics · Question 20

If 2 $^{a}$ = $\sqrt{64}$ and $\frac{b}{a}$ = 3, evaluate a $^2 + b^{2}$

Options

A.250
B.160
C.90
D.48

Answer: C. 90

Explanation written with AI and checked against the answer key

Step 1: Analyze the question We need to find a from the first equation, then b from the second, then compute a² + b². Step 2: Find a √64 = 8, and 8 = 2³. So 2^a = 2³, which gives a = 3. Step 3: Find b b/a = 3, so b = 3a = 3 × 3 = 9. Step 4: Evaluate a² + b² a² + b² = 3² + 9² = 9 + 81 = 90 Answer: 90, which is option C. Tip: When you see an equation like 2^a = a number, rewrite the number as a power of the same base (here 8 = 2³). Then you can equate the powers directly. Also, remember that √64 = 8 is a quick number to recognize in exams.

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If 2a^{a}a = 64\sqrt{64}64​ and ba\frac{b}{a}ab​ = 3, evaluate a2 +b2^2 + b^{2}2 +b2

Options

Answer: C. 90

If 2 $^{a}$ = $\sqrt{64}$ and $\frac{b}{a}$ = 3, evaluate a $^2 + b^{2}$