(a) A radio which a dealer bought for N6,000.00 and marked to give a profit of 30% was reduced in a sales by 10%. Find : (i) the final sales price ; (ii) the percentage profit.
(b) Solve the equation : \(2^{(2x + 1)} - 9(2^{x}) + 4 = 0\).
(b) Solve the equation : \(2^{(2x + 1)} - 9(2^{x}) + 4 = 0\).
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Correct Answer: Option n
Explanation:
(a) Cost price of a radio = N6,000.00
\(\therefore\) 30% profit = \(\N6000 \times \frac{30}{100} = N1800\)
\(\therefore\) Marked price = N(6000 + 1800) = N7800
Final Selling price = \(\frac{100 - 10}{100} \times N7800 = N7020\)
Profit = N(7020 - 6000) = N1020.
\(\therefore\) % profit = \(\frac{1020}{6000} \times 100% = 17%\)
(b) \(2^{(2x + 1)} - 9(2^{x}) + 4 = 0\)
\((2^{x})^{2} \times 2 - 9(2^{x}) + 4 = 0\)
Let \(2^{x} = y\)
\(2y^{2} - 9y + 4 = 0\)
\(2y^{2} - 8y - y + 4 = 0\)
\(2y(y - 4) - 1(y - 4) = 0\)
\((2y - 1)(y - 4) = 0\)
\(2y = 1 \implies y = \frac{1}{2}; y = 4\)
\(y = \frac{1}{2} \implies 2^{x} = \frac{1}{2} = 2^{-1} \)
\(x = -1\)
\(y = 4 \implies 2^{x} = 4 = 2^{2}\)
\(x = 2\)
\(\therefore x = -1, 2\)
(a) Cost price of a radio = N6,000.00
\(\therefore\) 30% profit = \(\N6000 \times \frac{30}{100} = N1800\)
\(\therefore\) Marked price = N(6000 + 1800) = N7800
Final Selling price = \(\frac{100 - 10}{100} \times N7800 = N7020\)
Profit = N(7020 - 6000) = N1020.
\(\therefore\) % profit = \(\frac{1020}{6000} \times 100% = 17%\)
(b) \(2^{(2x + 1)} - 9(2^{x}) + 4 = 0\)
\((2^{x})^{2} \times 2 - 9(2^{x}) + 4 = 0\)
Let \(2^{x} = y\)
\(2y^{2} - 9y + 4 = 0\)
\(2y^{2} - 8y - y + 4 = 0\)
\(2y(y - 4) - 1(y - 4) = 0\)
\((2y - 1)(y - 4) = 0\)
\(2y = 1 \implies y = \frac{1}{2}; y = 4\)
\(y = \frac{1}{2} \implies 2^{x} = \frac{1}{2} = 2^{-1} \)
\(x = -1\)
\(y = 4 \implies 2^{x} = 4 = 2^{2}\)
\(x = 2\)
\(\therefore x = -1, 2\)