(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.
(a) Selling price = \(\frac{90 \times 5600}{100}\) = $5,040.00. Thereafter, to obtain the amount invested = 5,040 - 1310 = $3,730.00
The interest amount invested is \(\frac{3730 \times 6 \times 3}{100}\) = $671.40
(b) 2\(^x\)(2\(^{-14}\)) = 2. This will be simplified and solved for x to get x = 15.
Similarly, 3\(^{-x}\)(3\(^{4y}\)) = 3 and this will yield -x + 4y = 1.
Substituting for x and solving for y yielded y = 4.
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