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WAEC SSCE - General Mathematics - 1999 (Objective)

Question 1 Report

A car travel at x km per hour for 1 hour and at y km per hour for 2 hours. Find its average speed

\(\frac{2x + 2y}{3}kmh^{-1}\)

\(\frac{x + y}{3}kmh^{-1}\)

\(\frac{x + 2y}{3}kmh^{-1}\)

\(\frac{2x + y}{3}kmh^{-1}\)

Answer Details

The average speed of a car is given by the total distance traveled divided by the total time taken. Let's assume that the car traveled a distance of D km in the given time. Then, we can write: Total time taken = 1 hour + 2 hours = 3 hours Distance traveled in the first hour = x km Distance traveled in the next 2 hours = y × 2 = 2y km Total distance traveled = D = x + 2y km Therefore, the average speed of the car is: Average speed = Total distance / Total time = D / 3 = (x + 2y) / 3 km/h Hence, the answer is option (C): \(\frac{x + 2y}{3}kmh^{-1}\).

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Question 2 Report

Given that \(81\times 2^{2n-2} = K, find \sqrt{K}\)

\(4.5\times 2^{n}\)

\(4.5\times 2^{2n}\)

\(9\times 2^{n-1}\)

\(9\times 2^{2n}\)

Answer Details

To find $\sqrt{K}$ given that $81 \times 2^{2n-2} = K$ , we need to simplify the

expression step-by-step.

First,

rewrite $81$ in terms of powers of $3$ : $81 = 3^{4}$

So,

the given equation becomes: $K = 3^4 \times 2^{2n-2}$

Next,

we need to find $\sqrt{K}$ : $\sqrt{K} = \sqrt{3^4 \times 2^{2n-2}}$

Using the property of square roots $\sqrt{a \times b} = \sqrt{a} \times \sqrt{b}$ : $\sqrt{K} = \sqrt{3^4} \times \sqrt{2^{2n-2}}$

Since $\sqrt{3^4} = 3^2 = 9$ and $\sqrt{2^{2n-2}} = 2^{(2n-2)/2} = 2^{n-1}$ : $\sqrt{K} = 9 \times 2^{n-1}$

Thus, the correct answer is: $\boxed{9 \times 2^{n-1}}$ =

9 \times 2^{n - 1}

Read lesson note on Number Bases (WAEC)

Number Bases

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