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Since area of square PQRS = 100cm2 each lenght = 10cm Also TUYS : XQVU = 1 : 16 lengths are in ratio 1 : 4, hence TU : UV = 1: 4 Let TU = x UV = 1: 4 hence TV = x + 4x = 5x = 10cm x = 2cm TU = 2cm UV = 8cm But TU = SY and UV = YR YR = 8cm

Number f students that took the test = \(\sum f\). Where f is the frequencies = 2 + 5 + 0 + 3 + 4 + 0 + 0 + 2 + 3 + 1 + 2 = 20

\(P = {..., -5, -4, -3, -2, -1}\) \(Q = {..., -5, -3, -1, 1, 3, 5, ...}\) \(R = {-2, -1, 0, 1, 2, 3, 4, 5, 6}\) (a) \(Q \cap R = {-1, 1, 3, 5}\) (b) \(R' = {..., -5, -4, -3, 7, 8, ...}\) (c) \(P' = {0, 1, 2, 3, ...}\) \(P' \cup R' = {-5, -4, -3, 0, 1, 2, 3, ...}\) (d) \(Q = {..., -5, -3, -1, 1, 3, 5,...}\) \(P \cap Q = {..., -7, -5, -3, -1}\) \((P \cap Q)' = {..., -8, -6, -4, -2, 0, 1, 2, 3, ...}\)

(a) \(\frac{3}{m + 2n} - \frac{2}{m - 3n}\) = \(\frac{3(m - 3n) - 2(m + 2n)}{(m + 2n)(m - 3n)}\) = \(\frac{3m - 9n - 2m - 4n}{(m + 2n)(m - 3n)}\) = \(\frac{m - 13n}{(m + 2n)(m - 3n)}\) (b) Let the numbers in the digit be x and z. Hence the number is \(10x + z\). When it is interchanged, we have \(10z + x\). \(10z + x = 10x + z + 9\) \(10z - z + x - 10x = 9\) \(9z - 9x = 9 \implies z - x = 1 ... (1)\) \(x + z = 11 ... (2)\) From (1), \(z = 1 + x\) \(\therefore x + 1 + x = 11\) \(2x + 1 = 11 \implies 2x = 11 - 1 = 10\) \(\therefore x = 5\) \(z = 1 + x = 1 + 5 = 6\) \(\therefore\) The original number = 56.

(a) Given : Circle ABC, centre O. To prove: < AOB = < ACB Construction : Join CO produced to P. Proof: With the lettering in the figure, OA = OB (radii) \(x_{1} = x_{2}\) (base angles of isosceles triangle) \(\therefore < AOP = x_{1} + x_{2}\) (exterior angle of triangle AOC) \(\therefore < AOP = 2x_{2} (x_{1} = x_{2})\) (base angles of an isosceles triangle) Similarly, \(< BOP = 2y_{2}\) \(\therefore < AOB = 2x_{2} + 2y_{2} \) = \(2(x_{2} + y_{2})\) \(\implies < AOB = 2 \times < ACB\) (proved) (b) From the figure, \(< ACB = \frac{130^\circ}{2} = 65^\circ\) (angle subtended at the centre of the circle) \(< ACO = 26^\circ\) (Base angles of isosceles triangle ACO) \(\therefore < BCO = 65^\circ - 26^\circ = 39^\circ\) \(< OBC = 39^\circ\) (base angles of isosceles triangle OBC) \(\therefore < COB = 180^\circ - (39^\circ + 39^\circ) = 102^\circ\)

(\(\frac{3}{4} + \frac{1}{3}) \times 4\frac{1}{3} \div 3\frac{1}{4}\) = \(\frac{13}{12} \times \frac{13}{3} \times\frac{4}{13}\) = \(\frac{13}{9}\) or 1\(\frac{4}{9}\))

2y\(^2\) + xy - 3x\(^2\) 2y\(^2\) + 3xy - 2xy - 3x\(^2\) y(2y + 3x) - x(2y + 3x) = (y - x)(2y + 3x)

Identity(e) : a \(\ast\) e = a m \(\ast\) e = m...(i) m \(\ast\) e = me + m + e Because m \(\ast\) e = m : m = me + m + e m - m = e(m + 1) e = \(\frac{0}{m + 1}\) e = 0

U = {1,2,3,4}; P = {2,3}; Q = {2,4}; P∩Q = {2} (P∩Q)' = {1,3,4}