(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 th...
Question 1 Report
(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 \(6y^{2} - 149y - 102 = 0\).
Answer Details
(a)(i) Marked price \(=\)N\(25{,}000\). Trade discount of \(20\%\): \[0.20\times 25000=\text{N}5000,\] so price becomes \(25000-5000=\)N\(20{,}000\). Prompt-payment discount of \(2\tfrac12\%\) on this: \[0.025\times 20000=\text{N}500.\] Total amount saved \(=5000+500=\) N5,500.
(a)(ii) After both discounts the distributor pays \(80\%\) then \(97.5\%\) of the mark: \[0.80\times 0.975\,x=0.78x.\] So \(0.78x=11700\Rightarrow x=\dfrac{11700}{0.78}=\) 15,000. The article is marked N15,000.
(b) Factorize \(6y^{2}-149y-102\). Product of ends \(=6\times(-102)=-612\); two numbers with product \(-612\) and sum \(-149\) are \(-153\) and \(+4\): \[6y^{2}-153y+4y-102=3y(2y-51)+2(2y-51)=(2y-51)(3y+2).\] Setting \((2y-51)(3y+2)=0\): \[y=\frac{51}{2}=25\tfrac12\quad\text{or}\quad y=-\frac{2}{3}.\]
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