(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.
Explanation
(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.