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m:n = \(2\frac{1}{3} : 1\frac{1}{5}\) = m : n = \(\frac{7}{3} : \frac{6}{5}\) \(\frac{7}{3} : \frac{6}{5}\) = \(\frac{7}{3} \div \frac{6}{5}\) \(\frac{m}{n}\) = \(\frac{7}{3} \times \frac{5}{6}\) \(\frac{m}{n}\) = \(\frac{35}{18}\) = m = \(\frac{35n}{18}\) n : q = \(1\frac{1}{2} : 1\frac{1}{3}\) = \(\frac{3}{2} : \frac{4}{3}\) \(\frac{n}{q}\) = \(\frac{3}{2} \times\frac{3}{4}\) \(\frac{n}{q}\) = \(\frac{9}{8}\) = q = \(\frac{8n}{9}\) q : m = \(\frac{8n}{9}\) : \(\frac{35n}{18}\) \(\frac{q}{m}\) = \(\frac{8n}{9} \div \frac{35n}{18}\) \(\frac{q}{m}\) = \(\frac{8n}{9}\times\frac{18}{35n}\) =\(\frac{q}{m} = \frac{16}{35}\) = q : m = 16 : 35

a. The total time taken for the entire journey is the sum of time taken for each journey. But Average speed = \(\frac { total distance covered}{ time taken}\) then Time taken = \(\frac { total distance covered}{ time taken}\) for the first journey(t1), d = 112km and speed = 70km/hr \(t_1 = \frac{112}{70} = \frac{8}{5}\)hrs for the second journey(t2), distance = 60km and speed = 50km/hr \(t_2 = \frac{60}{50} = \frac{6}{5}\)hrs Therefore, Total time taken = \( t_1 + t_2 = \frac{8}{5} + \frac{6}{5}\) = 2.8hrs. b. if \(\frac{x}{y}\) = 2 , then x = 2y and if \(\frac{y}{z}\) = 3 then z = \(\frac{y}{3}\), substituting, the values of x and z into \(\frac{ x + y}{y + z }\) = \(\frac{2y + y}{y + \frac{y}{3}} = 3y\div\frac{4y}{3} = 3y\times\frac{3}{4y} = \frac{9}{4}\) Therefore, \(\frac{ x + y}{y + z } = \frac{9}{4}\)

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%

\(M ∝ n^2\sqrt{q}\) \(M = Kn^2\sqrt{q}\) K = \(\frac{M}{n^2\sqrt{q}}\) K = \(\frac{24}{2^2\sqrt4}\) k = \(\frac{24}{8} = 3\) Now, let's find M when n = 5 and q = 9 M = \(Kn^2\sqrt{q}\) M = \( 3\times5^2\sqrt9\) \(M = 3\times25\times3\) Therefore, M = 225.

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

\(1 + \sqrt [3]{x-3} = 4\) = \(\sqrt[3]{x - 3} = 4 - 1\) \(\sqrt[3]{x - 3} = 3 \) take the cube of both sides = x - 3 = 27 x = 27 + 3 ∴ x = 30

Let Mr Manu be x years old and Adu be y years old. But Mr Manu is four times as old as Adu then, x = 4y. 7 years ago, the sum of their ages was 76. ( x - 7) + ( y - 7) = 76 x + y - 14 = 76 x + y = 76 + 14 = 90 But x = 4y Therefore, 4y + y = 90 5y = 90 y = \(\frac{90}{5}\) y = 18 Therefore Adu is 18 years old.

Let the two numbers be x and y \(\frac{1}{3}( x + y) = 12\) then x + y = 36 ..........i 2( x - y) = 12 x - y = 6 ............ii add equations i and ii 2x = 42 x = 21, put x = 21 into equation i x + y = 36 21 + y = 36 y = 36 - 21 = 15 therefore the numbers are 21 and 15

2x - 3y = -11 --- (i) 3x + 2y = 3 --- (ii) Multiply equation (i) by 3 and equation (ii) by 2 6x - 9y = -33 --- (iii) 6x + 4y = 6 --- (iv) Subtract equation (iii) from (iv) 13y = 39 y = \(\frac{39}{13}\) = 3 substitute (3) for y in equation (ii) 3x + 2(3) = 3 3x + 6 = 3 3x = 3 - 6 3x = -3 x = \(\frac{-3}{3}\) = - 1 Now, \((y - x)^2 = (3 - (-1))^2\) = \((3 + 1)^2\) = \(4^2\) = 16

let x = number of tickets sold to children and y = number of tickets sold to adults. Given: x + y = 400 ...................................................i 3x + 5y = 1700 ..............................................ii from Eqn(i) x = (400 - y) .........................(iii) put (400 - y) into Eqn ii 3(400 - y ) + 5y = 1700 1200 - 3y + 5y = 1700 1200 + 2y = 1700 2y = 1700 - 1200 2y = 500 y = \(\frac{500}{2}\) y = 250 Therefore, the number of tickets sold to adults is 250 b. Total number of tickets Mr. Johnson sold = 250 Number of tickets he sold to adults = 175 Then, the number of tickets he sold to children = 250 - 175 = 75 The price of a ticket for an adult is D5.00 and for a child is D3.00 Total sales = 175(5) + 75(3) = 875 + 225 = D1,100.00

a. Let the numbers be p, r and q respectively Given: p + r + q = 81 --- (i) r = 2p --- (ii) q = r + 6 --- (iii) Substitute (2p) for r in equation (iii) q = 2p + 6 --- (iv) Substitute (2p) for r and (2p + 6) for q in equation (i) ⇒ p + (2p) + (2p + 6) = 81 ⇒ 5p + 6 = 81 ⇒ 5p = 81 - 6 ⇒ 5p = 75 p = \(\frac{75}{5}\) = 15. From equation (ii) ⇒ r = 2(15) = 30 From equation (iv) ⇒ q = 2(15) + 6 ⇒ q = 30 + 6 = 36 ∴ The numbers are 15, 30 and 36 b. \(m_1 = \frac{ y_2 - y_1}{ x_2 - x_1}\) Given points are P( 3, 5) and Q( -5, 7) = P(\( x_1 , y_1)\) , Q(\( x_2 , y_2)\) the slope of line PQ = \((\frac{ 7 - 5}{ -5 - 3}) = \frac{2}{-8} = -\frac{1}{4}\) Given that R( x, y) is the midpoint of the line formed by the points P( 3, 5) and Q( -5, 7) R(x, y) = \(\frac{3 + (-5)}{2}, \frac{5 + 7}{2} = \frac{ 3 - 5}{2} , \frac{ 5+7}{2}\) = -\(\frac{2}{2} , \frac{12}{2}\) = (-1, 6) but the line passes through the point R( -1, 6) and is perpendicular to LINE PQ \(m_2 = \frac{1}{m_1} = -\frac{1}{\frac{-1}{4}}\) = 4 Equation of the line can be obtained using \(y - y_1 = m_2( x - x_1)\) y - 6 = 4( x - (-1)) = y - 6 = 4( x + 1 ) y - 6 = 4x + 4 y = 4x + 4 + 6 Therefore, y = 4x + 10