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y = k\(\sqrt{x}\) when y = 1, x = 4 k = \(\frac{y}{\sqrt{x}}\) = \(\frac{1}{\sqrt{4}}\) = \(\frac{1}{2}\) = 0.5

From y = \(\frac{1}{2} \sqrt{x}\) when y = 2.5 or \(\frac{5}{2}\), x = P \(\frac{5}{2} \times \frac{1}{2} \sqrt{P}\) \(\sqrt{P} = \frac{10}{2} = 5\) P = \(5^2\) = 25

radius = \(\frac{diameter}{2}\) = \(\frac{1.4}{2}\) = 0.7m circumference = 2\(\pi r = 2 \times \frac{22}{7} \times 0.7m = 4.4m\) distance covered = 4.4m x 100 = 440m

(3y - 450 + (y + 25) + t = 360o(angles at a point) 3y - 45 + y + 25 + y = 360 5y = 360 + 20 5y = 380 y = \(\frac{380}{5}\) = 76o

Area of carpet = 2.5 x 2.4 = 6 Let F represent the floor 6 = 5% of F 6 = \(\frac{5}{100}\) x F 5F = 6 x 100 F = 6 x \(\frac{100}{5}\) = 120m2

Sin 65o = \(\frac{x}{16}\) x = 16x sin 65 = 16 x 0.9063 x = 14.5m

|VU| = |UT| = |TS| |VS| = (4.8) x 3 = 14.4cm |PQ| = |QR| = 4.8 |PR| = (4.8) x 2 = 9.6cm Since quad. PRSV is a trapezium of height 5cm Area of quad. PRSV = \(\frac{1}{2}(a + b)h\) = \(\frac{1}{2}(14.4 + 9.6) \times 5\) = \(\frac{1}{2}(24) \times 5\) 12 x 5 = 60cm2

PQR = QRS = Y(alt. amgles) PQR + x = 180o(angles on a straight line) y + x = 180o

PRQ = 90 - 58 = 32o(angle in a semi-circle) Since PRS = 90o(radius angular to tangent) QRS = 90 + 32 = 122o

Draw a line from P to Q o Also, o x + y + 5= = 180(angles in a triangle) x + y = 180 - 50 x + y = 130o

Q1 = 50(alternate angle) Q2 = 180o - 110o (straight line angle) Q2 = 70 PQR = Q1 + Q2 = 50o + 70o = 120o

area of larger circle = 100\(\pi\) \(\pi r^2 = 100\pi\) R2 = 100 R = 10(Radius) Area of smaller circle = \(49 \pi\) \(\pi r^2 = 49\pi\) r2 = 49 r = 7(radius) Since R = x + r x = R - r x = 10 - 7 = 3

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Source: Myschool Past Questions Archive (Record #36210)