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Let the tens digits of the number be x and the unit digit be y 3x = 2y - 2 3x - 2y = -2.......(i) If the digits are interchanged, the tens digit becomes y, the unit digit becomes x. Hence 2(10x + y) = 10y + x + 20 (20x + 2y) - (10y + x) = 20 19x - 8y = 20.....(ii) Multiply eqn.(i) by 8 and eqn.(ii) by 2 24x - 16y = -16......(iii) 38x - 16y = 40........(iv) eqn(iv) - eqn(iii) 14x = 56 x = 4 Sub. for x = 4 in eqn(i) 3(4) - 2y = -2 14 = 2y y = 7 So the original number is 10(4) + 7 i.e. 10x + y = 47

Area of circle 1 with diameter 26cm: \(\pi r^{2} = \pi \times (\frac{26}{2})^{2} \) = \(169 \pi cm^{2}\) Area of circle 2 with diameter 20 cm: \(\pi R^{2} = \pi \times (\frac{20}{2})^{2}\) = \(100 \pi cm^{2}\) Area between the two circles = \((169 - 100) \pi cm^{2}\) = \(69 \pi cm^{2}\)

\(\sin x = \frac{a}{b}\) \(\sin^{2} x + \cos^{2} x = 1\) \(\sin^{2} x = \frac{a^{2}}{b^{2}}\) \(\cos^{2} x = 1 - \frac{a^{2}}{b^{2}} = \frac{b^{2} - a^{2}}{b^{2}}\) \(\therefore \cos x = \frac{\sqrt{b^{2} - a^{2}}}{b}\) \(\sin (90 - x) = \sin 90 \cos x - \cos 90 \sin x\) = \((1 \times \frac{\sqrt{b^{2} - a^{2}}}{b}) - (0 \times \frac{a}{b})\) = \(\frac{\sqrt{b^{2} - a^{2}}}{b}\)

In \(\Delta DAC, \stackrel\frown{DAC} = \theta\) \(\sin \theta = \frac{5}{13}\) \(\theta = 22.6^\circ\) \(< DOC = 22.6^\circ \times 2 = 45.2^\circ\) \(\sin 45.2 = \frac{5}{r} \implies r = \frac{5}{\sin 45.2}\) \(r = 7.046cm\) = \(7\frac{1}{24} cm\)

1 hr, = 60 mins, 60 mins = 360o 30 mins = \(\frac{360^o}{1}\) c \(\frac{30}{60}\) = 180o 90 mins = 360 + 180o = 540o

Let XW = a, a :7 \(\frac{a}{5 - a}\) = \(\frac{3}{7}\) 7a = 15 - 3a 10a = 15 a = \(\frac{15}{10}\) = \(\frac{3}{2}\) = 1.5cm

Distance travelled by both x and y = 150 km. Speed of y = 60km per hour, Time = \(\frac{dist.}{Speed}\) Time spent by y = \(\frac{150}{60}\) = 2\(\frac{3}{6}\)hrs = \(\frac{1}{2}\) hrs if x arrived at p 25 minutes earlier than y, then time spent by x = 2 hour 5 mins = 125 mins x average speed = \(\frac{150}{125}\) x \(\frac{60}{1}\) = \(\frac{9000}{125}\) = 72 km\hr

Area = L x B L - B = 6 L = 6 + B area = 135 B(6 + B) = B\(^2\) + 6B - 135 = 0 (B - 9)(B + 15) = 0 B = 9 or - 15 L = 6 + 9 = 15

\(\frac{\sin^{2} x}{1 + \cos x} + \frac{\sin^{2} x}{1 - \cos x}\) \(\frac{\sin^{2} x (1 - \cos x) + \sin^{2} x (1 + \cos x)}{1 - \cos^{2} x}\) = \(\frac{\sin^{2} x - \cos x \sin^{2} x + \sin^{2} x + \sin^{2} x \cos x}{\sin^{2} x}\) (Note: \(\sin^{2} x + \cos^{2} x = 1\)). = \(\frac{2 \sin^{2} x}{\sin^{2} x}\) = 2.

b = a + cp....(i) r = ab + \(\frac{1}{2}\)cp2.....(ii) expressing b2 in terms of a, c, r, we shall first eliminate p which should not appear in our answer from eqn, (i) b - a = cp = \(\frac{b - a}{c}\) sub. for p in eqn.(ii) r = ab + \(\frac{1}{2}\)c\(\frac{(b - a)^2}{\frac{ab + b^2 - 2ab + a^2}{2c}}\) 2cr = 2ab + b2 - 2ab + a2 b2 = 2cr - a2

Sugar : flour = 2.5 : 4.5 Total = 7 sugar used = \(\frac{2.5}{7}\) x 24.5 = \(\frac{61.25}{7}\) = 8.75

Pentagon is a polygon with 5 sides. Sum of interior angles of polygon (regular) = (2n - 4) x 90o = 5 = (10 - 4) x 90o = 540o each interior angle = \(\frac{540}{5}\) = 108o = 180o let x represent ext. angle x + 108o = 180v x = 180o - 108o = 72o

Given: \(\theta\) = 2\(\frac{1}{2}\) radians, r = 8cm but 360º = 2 \(\pi\) radian. Using: L = length of an arc = \(\frac{\theta}{2 \pi}\) x 2\(\pi\) r = \(\frac{2.5}{2\pi}\) x 2 x \(\pi\) x 8 = 2.5 x 8 = 20cm.

Area of trapezium XYZW = \(\frac{1}{2}\)(a + b)h h = 6 sin 30º = 3 area = \(\frac{1}{2}\)(10 + 8) x 3 \(\frac{1}{2}\)(18) x 4 = 27cm\(^2\)

180 - x + a + b + c = 180(sum of interior angle in triangle) a + b + c = x.