(a) Ms. Maureen spent \(\frac{1}{4}\) of her monthly income at a shopping mall, \(\frac{1}{3}\) at...
(a) Ms. Maureen spent \(\frac{1}{4}\) of her monthly income at a shopping mall, \(\frac{1}{3}\) at an open market and \(\frac{2}{5}\) of the remaining amount at a Mechanic workshop. If she had N222,000.00 left, find:
(i) her monthly income.
(ii) the amount spent at the open market.
(b) The third term of an Arithmetic Progression (A. P.) is 4m - 2n. If the ninth term of the progression is 2m - 8n. find the common difference in terms of m and n.
Explanation
(a) Let the salary be x
Shopping mall = \(\frac{1}{4}\) x \(x = \frac{x}{4}\)
Open market = \(\frac{1}{3} \times x = \frac{x}{3}\)
The remainder x - [\(\frac{x}{4} + \frac{x}{3}\)]
= x - [\(\frac{3x + 4x}{12}\)]
= \(\frac{x}{1} - \frac{7x}{12}\)
= \(\frac{12x - 7x}{12}\)
= \(\frac{5x}{12}\)
mechanic = \(\frac{2}{5} \times \frac{5x}{12} = \frac{x}{6}\)
\(\frac{x}{6} + \frac{x}{3} + \frac{x}{4}\) + 225,00 = x
\(\frac{2x + 4x + 3x}{12}\) + 225,000
\(\frac{9x}{12}\) + 225,000 = x
225,000 = \(\frac{x}{1} = \frac{9x}{12}\)
225,000 = \(\frac{12x - 9x}{12}\)
225,000 = \(\frac{3x}{12}\)
\(\frac{225,000 \times 12}{3}\) = x
N900,000 = x
(i) Her monthly income = N900,000
(ii) Amount in open market
= \(\frac{N900,000}{3}\)
= N300,000
(b) T\(_3\) = 4m - 2n
T\(_9\) = 2m - 8n
a + 2d = 4m - 2n .....(i)
a + 8d = 2m - 8n ......(ii)
a + 8d = 2m - 8n
-(a + 2d = 4m - 2n
\(\overline{6d = 6m - 6n}\)
6d = 6(m - n)
d = m - n
Common difference = m - n