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12 men working 8 hours a day can complete the work in 4 days 12 men working 1 hour a day will complete the same work in (8 x 4) days [working less hours a day means they have to work for more days] 1 man working 1 hour a day will complete same work in (8 x 4 x 12) days [fewer number of men working means the remaining ones will have to work for more days ] So, it will take 1 man working 1 hour a day to complete a piece of work in (8 x 4 x 12) days = 384 days Now, 1 man working 16 hours a day will complete the piece of work in (384/16) days = 24 days [working more hours a day means less days to complete the work] ∴ 4 men working 16 hours a day will complete the piece of work in (24/4) days = 6 days [more men working means fewer days to complete the work] ALTERNATIVELY 12 x 8 x 4 = 4 x 16 x d ⇒ d = \(\frac{12 \times 8 \times 4}{4 \times 16}\) ⇒ d = \(\frac{384}{64}\) ⇒ d = 6 ∴ 4 men working 16 hours a day will complete the piece of work in 6 days

Amount (A) = Principal (P) + Interest (I) i.e. A = P + I 1 = \(\frac {PTR}{100}\) A = 3P; T = 10 years (Given) Since A = P + I \(\implies 3P = P + \frac {P\times 10 \times R}{100}\) \(\implies 3P = P + \frac {10PR}{100}\) \(\implies 3P = P + \frac {PR}{10}\) \(\implies 3P - P = \frac {PR}{10}\) \(\implies 2P = \frac {PR}{10}\) \(\implies \frac {2P}{1} = \frac {PR}{10}\) \(\implies\) PR = 20P \(\therefore\) R = 20% Since the rate stays the same A = 5P; R = 20%;T =?; A =P + I \(\implies 5P = P + \frac {p \times T \times 20}{100}\) \(\implies 5P - P = \frac {2PT}{10}\) \(\implies 4P = \frac {2PT}{10}\) \(\implies 2P = \frac {PT}{10}\) \(\implies\) PT = 20P \(\therefore\) T = 20 years

First S.P = ₦230.00 % profit = 15% % profit = \(\frac{S.P - C.P}{C.P}\) x 100% ⇒ 15% = \(\frac{230 - C.P}{C.P}\) x 100% ⇒ \(\frac{15}{100}= \frac{230 - C.P}{C.P}\) ⇒15C.P = 100 (230 - C.P) ⇒15C.P = 23000 - 100C.P ⇒15C.P + 100C.P = 23000 ⇒115C.P = 23000 ⇒ C.P = \(\frac{23000}{115}\) = ₦200.00 Second S.P = ₦180.00 Since C.P is greater than S.P, therefore it's a loss % loss = \(\frac{C.P - S.P}{C.P}\) x 100% ⇒ \(\frac{200 - 180}{200}\) x 100% ⇒ \(\frac{20}{200}\) x 100% ⇒ \(\frac{1}{10}\) x 100% = 10% ∴ It's a loss of 10%

Principal (P) = ₦15,700 Rate (R) = 8 Number of years (t) = 2 A = P \((1+\frac{R}{100})^t\) ⇒ A = 15700 \((1+\frac{8}{100})^2\) ⇒ A = 15700 (1 + 0.08)\(^2\) ⇒ A = 15700 (1.08)\(^2\) ⇒ A = 15700 x 1.1664 ⇒ A = ₦18,312.48 Total amount, A = ₦18,312.48 A = P + CI ⇒ CI = A - P ⇒ CI = 18,312.48 - 15,700 ∴ CI = ₦2,612.48

Total cost price = ₦9,200.00 + ₦1,500.00 = ₦10,700.00 Selling price = ₦13,400.00 Gain = ₦13,400.00 - ₦10,700.00 = ₦2,700.00 ∴ % gain = \(\frac{₦2,700.00}{₦10,700.00}\times\) 100% = 25.23%

Let number of children's ticket at ₦250.00 each = \(x\) ∴ Number of adult tickets at ₦520.00 each = 5\(x\) Then, Total amount of money received from children's tickets = 250\(x\) Total amount of money received from adult tickets = 520(5\(x\)) ⇒ 250\(x\) + 520(5\(x\)) = 171,000 ⇒ 250\(x\) + 2600\(x\) = 171,000 ⇒ 2850\(x\) = 171,000 ⇒ \(x = \frac{171,000}{2850} = 60\) ∴ 60 tickets were sold at ₦250.00 and 300 tickets were sold at ₦520.00

Let \(x\) = number of VIP tickets sold and \(y\) = number of regular tickets sold Total number of tickets sold = 200 ⇒ \(x\) + \(y\) = 200 ---- (i) If it costs ₦1,200 for a VIP ticket, then it costs ₦1200x for \(x\) number of VIP tickets sold and If it costs ₦700 for a regular ticket, then it costs ₦700\(y\) for \(y\) number of VIP tickets sold The total amount realised from the sale of tickets = ₦180,000 ⇒ 1200\(x\) + 700\(y\) = 180000 ----- (ii) From equation (i) \(x\) = 200 - \(y\) ----- (iii) Substitute (200 - \(y\)) for \(x\) in equation (ii) ⇒ 1200(200 - \(y\)) + 700\(y\) = 180000 ⇒ 240000 - 1200\(y\) + 700\(y\) = 180000 ⇒ 240000 - 500\(y\) = 180000 Collect like terms ⇒ 240000 - 180000 = 500\(y\) ⇒ 60000 = 500\(y\) ⇒ \(y = \frac{60000}{500} = 120\) Substitute 120 for \(y\) in equation (iii) ⇒ \(x = 200 - 120\) ⇒ \(x = 80\) ∴ The total number of VIP tickets sold is 80 and regular is 120

y = \(\frac {3x - 9c}{4x + 5d}\) \(\frac {y}{1} = \frac {3x - 9c}{4x + 5d}\) = y(4x + 5d) = 3x - 9c = 4xy + 5dy = 3x - 9c = 4xy - 3x = -9c - 5dy = (4y - 3)x = -9c - 5dy = x = \(\frac {-9c - 5dy}{4y - 3}\) x = \(\frac {-(9c + 5dy)}{4y - 3}\)

3(x – 1) \le 2 (x – 3) = 3x – 3 \le 2x – 6 Add 3 to both sides = 3x – 3+ 3 \le 2x – 6+ 3 = 3x \le 2x – 3 Subtract 2x from both the sides = 3x – 2x \le 2x – 3 – 2x ∴ x \le -3

\(x^2 - x - 4 \le 2\) Subtract two from both sides to rewrite it in the quadratic standard form: = \(x^2 - x - 4 - 2 \le 2 - 2\) = \(x^2 - x - 6 \le 0\) Now set it = 0 and factor and solve like normal. = \(x^2 - x\) - 6=0 = \((x - 3)(x + 2)\)=0 \(x\) + 2 = 0 or \(x\) - 3 = 0 \(x\) = -2 or \(x\) = 3 So the two zeros are -2 and 3, and will mark the boundaries of our answer interval. To find out if the interval is between -2 and 3, or on either side, we simply take a test point between -2 and 3 (for instance, \(x\) = 0) and evaluate the original inequality. = \(x2 - x - 4 \le 2\) = \((0)^2 - (0) - 4 \le 2\) = \(0 - 0 - 4 \le 2\) \(−4 \le 2\) Since the above is a true statement, we know that the solution interval is between -2 and 3, the same region where we picked our test point. Since the original inequality was less than or equal, we include the endpoints. ∴ \(-2 \le x \le 3.\)

First, we find the equation of the boundary line using the two intercepts.The slope is m = \(\frac{4 - 0}{0 - 5} = {4}{5}\) The y-intercept is 4 The slope-intercept form of the equation is therefore y = -\(\frac{4}{5} x + 4\) \(\implies y + \frac{4}{5} x = 4\) Multiply both sides by \(\frac{5}{4}\) \(\implies\frac{5}{4}(y +\frac{4}{5} x) = 4\times\frac{5}{4}\) \(\implies\frac{5}{4} y + x = 5\) Multiply both sides by 4 ⇒ \(5y + 4x = 20\) The inequality is therefore either \(5y + 4x \le 20\) or \(5y + 4x \ge 20.\) Using the test point (0, 0) -The origin ⇒ 5(0) + 4(0) \le 20 ⇒ 0 \le 20 (True) ∴ The inequality is \(5y + 4x \le 20\)

16\(x^4 - y^4\) = 2\(^4x^4 - y^4\) = \((2x)^4 - y^4\) = \(((2x)^2)^2 - (y^2)^2\) Using a\(^2 - b^2\) = (a - b)(a + b) identity = ((2x)\(^2 - y^2)((2x)^2 + y^2)\) Using the identity one more time = \((2x - y)(2x + y)((2x)^2 + y^2)\) ∴ \((2x - y)(2x + y)(4x^2 + y^2)\)

Let \(p(x) = -2x^3 + 6x^2 + 17x - 21\) Using the remainder theorem Let \(x + 1 = 0\) ∴ \(x = -1\) Since, \((x + 1)\) divides \(p(x)\), then, remainder will be p(-1) ⇒ p(-1) = -2(-1)\(^3 + 6(-1)^2\) + 17(-1) - 21 ∴ p(-1) = -30

Given the sequence 3, 8, 13, 18, ... which is an arithmetic sequence a = 3 d = T\(_2\) - T\(_1\) = 8 - 3 = 5 The general term of an A.P is: T\(_n\) = a + (n - 1)d ⇒ T\(_n\) = 3 + (n - 1)5 = T\(_n\) = 3 + 5n - 5 ∴ T\(_n\) = 5n - 2

T\(_3\) = 6 T\(_5\) = 12 S\(_{12}\) = ? T\(_n\) = a + (n - 1)d ⇒ T\(_3\) = a + 2d = 6 ----- (i) ⇒ T\(_5\) = a + 4d = 12 ----- (ii) Subtract equation (ii) from (i) ⇒ -2d = -6 ⇒ d\(\frac{-6}{-2}\) = 3 Substitute 3 for d in equation (i) ⇒ a + 2(3) = 6 ⇒ a + 6 = 6 ⇒ a = 6 - 6 = 0 S\(_n\) = \(\frac{n(2a + (n - 1)d)}{2}\) ⇒ S\(_{12}\) = \(\frac{12(2 \times 0 + (12 - 1)3)}{2}\) ⇒ S\(_{12}\) = 6(0 + 11 x 3) ⇒ S\(_{12}\) = 6(33) ∴ S\(_{12}\) = 198