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Let the present ages of father be x and son = y five years ago, father = x - 5 son = y - 5 x - 5 = 3(y - 5) x - 5 = 3y - 15 x - 3y = -15 + 5 = -10 ......(i) x + y = 110......(ii) eqn(ii) - eqn(i) 4y = 120 y = 30 sub for y = 30 in eqn(i) x -10 = 3(30) x -10 = 90 x = 80

y = 2x2 + 9x - 35 2x2 + 9x = 35 x2 + \(\frac{9}{2}\) = \(\frac{35}{2}\) x2 + \(\frac{9}{2}\) + \(\frac{81}{16}\) = \(\frac{35}{2}\) = \(\frac{81}{16}\) (x + \(\frac{9}{4}\))2 = \(\frac{361}{16}\) x = \(\frac{-9}{4}\) + \(\frac{\sqrt{361}}{16}\) x = \(\frac{-9}{4}\) + \(\frac{19}{4}\) = 2.5 or -7 -7 < x < \(\frac{5}{2}\)

Let the cost of living = y. The new cost of living = \(y + \frac{15y}{100} = 1.15y\) The food bill now = \((1 - \frac{90}{100})(1.15y)\) = \(1.035y\) The fractional increase in food bill = \((1.035 - 1) \times 100% = 3.5%\) = \(\frac{35}{1000} = \frac{7}{200}\)

Soln. a*b = ab + a + b, a ♦ b = a + b + 1 a*c = ac + a + c (a*b) ♦ (a*c) = (ab + a + b + ac + a + c + 1) = ab + ac + 2a + b + c + 1

Sum of interior angles of a polygon = (2n - 4) x 90 where n is the no. of sides sum of interior angles of the quad. = ({2 x 4} -4) x 90 (8 - 4) x 90 = 4 x 90o = 360o (p + 10)o + (p - 30)o + (2p - 45)o + (p + 15)o = 360o p + 10o + p - 30o + 2p - 45o + p + 15 = 360o 5p = 360o + 50o = 82o

\(\frac{a - b}{a + b}\) - \(\frac{a + b}{a - b}\) = \(\frac{(a - b)^2}{(a + b)}\) - \(\frac{(a + b)^2}{(a - b0}\) applying the principle of difference of two sqrt. Numerator = (a - b) + (a + b) (a - b) - (a + b) = (a = b + a = b)(a = b - a = b) 2a(-2b) = -4ab = \(\frac{-4ab}{(a + b)(a - b)}\) = \(\frac{-4ab}{a^2 - b^2}\)

Force to the east = 5 units and angle to the east = 0º force to the Northeast = 4 units and angle to northeast = 45º Resolve these forces into x and y components FORCES x - component y - component 5 5cos0 = 5 5sin 0 = 0 4 4cos45 = 4 x \(\frac{\sqrt2}{2}\) 4sin45 = \(\frac{\sqrt2}{2}\) Total force 5 + 2\(\sqrt{2}\) 0 + 2\(\sqrt{2}\) Resultant force = \(\sqrt{(F_x)^2 + (F_y)^2}\) = \(\sqrt{ (5 + 2\sqrt{2})^2 + (2\sqrt{2})^2}\) = \(\sqrt{(25 + 20\sqrt{2} + 8) + 8}\) = \(\sqrt{33 + 20\sqrt{2} + 8}\) = \(\sqrt{41 + 20\sqrt{2}}\)

Given the curve \(y = x^{2} - 6x + 5\) At minimum or maximum point, \(\frac{\mathrm d y}{\mathrm d x} = 0\) \(\frac{\mathrm d y}{\mathrm d x} = 2x - 6\) \(2x - 6 = 0 \implies x = 3\) Since 3 > 0, it is a minimum point. When x = 3, \(y = 3^{2} - 6(3) + 5 = -4\) Hence, the turning point has coordinates (3, -4).

\(3x - (\frac{1}{4})^{-\frac{1}{2}} > \frac{1}{4} - x\) = \(3x - 4^{\frac{1}{2}} > \frac{1}{4} - x\) = \(3x - 2 > \frac{1}{4} - x\) = \(3x + x > \frac{1}{4} + 2 \implies 4x > \frac{9}{4}\) \(x > \frac{9}{16}\)

The canal's width = 10cm = 100mm (given) The speed of water = 1000mm 10mm = 1cm 1000mm = 100cm The adjacent sea must give speed x width = 1000 x 100 = 100,000

Let the average speed for the first 4 km = x km/h. Hence, the last 5 km, speed = (x + 2) km/h Recall: \(Time = \frac{Distance}{Speed}\) Total time = 1 hour. \(\therefore \frac{4}{x} + \frac{5}{x + 2} = 1\) \(\frac{4(x + 2) + 5x}{x(x + 2)} = 1\) \(9x + 8 = x^{2} + 2x\) \(x^{2} + 2x - 9x - 8 = 0 \implies x^{2} - 7x - 8 = 0\) \(x^{2} - 8x + x - 8 = 0\) \(x(x - 8) + 1(x - 8) = 0\) \(x = -1; x = 8\) Since speed cannot be negative, x = 8km/h.

The pyramid has 8 edges in itself while the cuboid has 12 edges. When merging the two shapes together, the edge of the base of the pyramid becomes same as the edges of the top of the cuboid. Hence, the new structure will have (12 + 8) - 4 = 16 edges.

\(T_1\) = 153 \(T_6 = \frac{17}{27}\) The equation for the nth term is ar\(^{n-1}\) \(T_6: 153r^5\) = \(\frac{17}{27}\) divide through by 153 r\(^5\) = \(\frac{17}{27}\) x \(\frac{1}{153}\) take 5th root of both sides r = \(\frac{1}{3}\) The sum of n terms is \(S_n = \frac{a(1 - r^n)}{1 - r }\) So, \(S_4 = \frac{153(1 - (\frac{1}{3})^4)}{1 - \frac{1}{3}}\) = \(\frac{680}{3}\)

1st term a = 11, 4th term = 32 nth term = a + (n - 1)d 4th term = 11 = (4 - 1)d = 11 + 3d = 32 3d = 21 d = 7 sn = \(\frac{n}{2}\)(2a + (n - 1)d) sn = \(\frac{9}{2}\)(2 \(\times\) 11) + (9 - 1)7 \(\frac{9}{2}\)(22 + 56) = \(\frac{9}{2}\) x 78 = 351

\(\frac{6}{\sin 30}\) = \(\frac{x}{\sin 135}\) \(\frac{6}{\sin 30}\) = \(\frac{x}{\sin 45}\) x = \(\frac{6 \times \sin 45}{\sin 30}\) = \(6 \sqrt{2}\)cm