(a) A manufacturer offers distributors a discount of (20%) on any article bought and a further discount of (2frac{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).
Explanation
(a) (i) First discount = 20% of N25,000
= (frac{20}{100} times N25,000 = N5,000)
Further discount for prompt payment = (2frac{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 + 2frac{1}{2})% = 22frac{1}{2}%)
Therefore, the distributor paid ((100% – 22frac{1}{2}%) times Nx)
= (77frac{1}{2}% times Nx)
(implies 77frac{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 = 25frac{1}{2})
(therefore y = -frac{2}{3} ; y = 25frac{1}{2})