(a) A manufacturer offers distributors a discount of \(20%\) on any article bought and a further discount of \(2\frac{1}{2}%\) for prompt payment.
(i) if the marked price of an article is N25,000, find the total amount saved by a distributor for paying promptly. (ii) if a distributor pays N11,700 promptly for an article marked Nx, find the value of x.
(b) Factorize \(6y^{2} - 149y - 102\), hence solve the equation \(6y^{2} - 149y - 102 = 0\).
(i) if the marked price of an article is N25,000, find the total amount saved by a distributor for paying promptly. (ii) if a distributor pays N11,700 promptly for an article marked Nx, find the value of x.
(b) Factorize \(6y^{2} - 149y - 102\), hence solve the equation \(6y^{2} - 149y - 102 = 0\).
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Correct Answer: Option n
Explanation:
(a) (i) First discount = 20% of N25,000
= \(\frac{20}{100} \times N25,000 = N5,000\)
Further discount for prompt payment = \(2\frac{1}{2}%\) of N25,000.
= \(\frac{5}{200} \times N25,000 = N625\)
\(\therefore \text{Total amount saved by distributor} = N(5,000 + 625)\)
= \(N5625\)
(ii) Total %age discount for prompt payment = \((20 + 2\frac{1}{2})% = 22\frac{1}{2}%\)
Therefore, the distributor paid \((100% - 22\frac{1}{2}%) \times Nx\)
= \(77\frac{1}{2}% \times Nx\)
\(\implies 77\frac{1}{2}% \times x = 11,700\)
\(x = \frac{11,700 \times 200}{155}\)
= \(N15,096.77\)
(b) \(6y^{2} - 149y - 102 \)
= \(6y^{2} - 153y + 4y - 102\)
= \(3y(2y - 51) + 2(2y - 51)\)
= \((3y + 2)(2y - 51)\)
\(\therefore 6y^{2} - 149y - 102 = 0 \implies (3y + 2)(2y - 51) = 0\)
\(3y + 2 = 0 \implies 3y = -2 \)
\(y = -\frac{2}{3}\)
\(2y - 51 = 0 \implies 2y = 51\)
\(y = 25\frac{1}{2}\)
\(\therefore y = -\frac{2}{3} ; y = 25\frac{1}{2}\)
(a) (i) First discount = 20% of N25,000
= \(\frac{20}{100} \times N25,000 = N5,000\)
Further discount for prompt payment = \(2\frac{1}{2}%\) of N25,000.
= \(\frac{5}{200} \times N25,000 = N625\)
\(\therefore \text{Total amount saved by distributor} = N(5,000 + 625)\)
= \(N5625\)
(ii) Total %age discount for prompt payment = \((20 + 2\frac{1}{2})% = 22\frac{1}{2}%\)
Therefore, the distributor paid \((100% - 22\frac{1}{2}%) \times Nx\)
= \(77\frac{1}{2}% \times Nx\)
\(\implies 77\frac{1}{2}% \times x = 11,700\)
\(x = \frac{11,700 \times 200}{155}\)
= \(N15,096.77\)
(b) \(6y^{2} - 149y - 102 \)
= \(6y^{2} - 153y + 4y - 102\)
= \(3y(2y - 51) + 2(2y - 51)\)
= \((3y + 2)(2y - 51)\)
\(\therefore 6y^{2} - 149y - 102 = 0 \implies (3y + 2)(2y - 51) = 0\)
\(3y + 2 = 0 \implies 3y = -2 \)
\(y = -\frac{2}{3}\)
\(2y - 51 = 0 \implies 2y = 51\)
\(y = 25\frac{1}{2}\)
\(\therefore y = -\frac{2}{3} ; y = 25\frac{1}{2}\)