(a) Lamin bought a book for N300.00 and sold it to Bola at a profit of x%. Bola then sold the same book at a profit of x%. If James paid \(N(6x + \frac{3}{4})\) more for the book than Lamin paid, find the value of x.
(b) Find the range of values of x which satisfies the inequality \(3x - 2 < 10 + x < 2 + 5x\).
(b) Find the range of values of x which satisfies the inequality \(3x - 2 < 10 + x < 2 + 5x\).
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Correct Answer: Option n
Explanation:
(a) S.P = \(C.P + \frac{% profit}{100} \times C.P\)
= \(300 + (\frac{x}{100} \times 300)\)
S.P = N(300 + 3x)
Therefore, Bola bought it at N(300 + 3x).
James paid \(N(6x + \frac{3}{4})\) extra from what Lamin paid, therefore Bola's S.P = \(N(300 + 6x + \frac{3}{4})\)
= N(300.75 + 6x).
Profit for Bola = \(N(300.75 + 6x - (300 + 3x)) = N(0.75 + 3x)\)
\(\frac{x}{100} \times N(300 + 3x) = N(0.75 + 3x)\)
\(300x + 3x^{2} = 75 + 300x\)
\(\implies 3x^{2} = 75\)
\(x^{2} = 25 \therefore x = 5\)
(b) \(3x - 2 < 10 + x < 2 + 5x\)
\(3x - 2 < 10 + x \implies 3x - x < 10 + 2\)
\(2x < 12 \implies x < 6\)
\(10 + x < 2 + 5x\)
\(x - 5x < 2 - 10\)
\(-4x < -8 \implies x > 2\)
The range = \(2 < x < 6\)
(a) S.P = \(C.P + \frac{% profit}{100} \times C.P\)
= \(300 + (\frac{x}{100} \times 300)\)
S.P = N(300 + 3x)
Therefore, Bola bought it at N(300 + 3x).
James paid \(N(6x + \frac{3}{4})\) extra from what Lamin paid, therefore Bola's S.P = \(N(300 + 6x + \frac{3}{4})\)
= N(300.75 + 6x).
Profit for Bola = \(N(300.75 + 6x - (300 + 3x)) = N(0.75 + 3x)\)
\(\frac{x}{100} \times N(300 + 3x) = N(0.75 + 3x)\)
\(300x + 3x^{2} = 75 + 300x\)
\(\implies 3x^{2} = 75\)
\(x^{2} = 25 \therefore x = 5\)
(b) \(3x - 2 < 10 + x < 2 + 5x\)
\(3x - 2 < 10 + x \implies 3x - x < 10 + 2\)
\(2x < 12 \implies x < 6\)
\(10 + x < 2 + 5x\)
\(x - 5x < 2 - 10\)
\(-4x < -8 \implies x > 2\)
The range = \(2 < x < 6\)