WAEC 2019 · Paper 2 · Q6

(a)
Fred bought a car for $5,600.00 and later sold it at $90\%$ of the cost price. He spent $1,310.00 out of the amount received and invested the rest at $6\%$ per annum simple interest. Calculate the interest earned in 3 years.
(b)
Solve the equations $2^x(4^{-7}) = 2$ and $3^{-x}(9^{2y}) = 3$ simultaneously.
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

(a)

He sold the car for $90\%$ of $5,600: $0.9 \times 5600 = 5040$ , so he received $5,040.
After spending $1,310, he invested $5040 - 1310 = 3730$ , that is $3,730.
Simple interest for 3 years at $6\%$ : $\frac{3730 \times 6 \times 3}{100} = 671.40$ .
The interest earned is $671.40.

(b)

Write everything as powers of the same base.
$4^{-7} = (2^2)^{-7} = 2^{-14}$ , so the first equation is $2^x \times 2^{-14} = 2^1$ .
Add the powers: $2^{x - 14} = 2^1$ , so $x - 14 = 1$ and $x = 15$ .
Similarly $9^{2y} = (3^2)^{2y} = 3^{4y}$ , so $3^{-x} \times 3^{4y} = 3^1$ gives $-x + 4y = 1$ .
With $x = 15$ : $4y = 16$ , so $y = 4$ .

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