Question Library

Total possible outcome = 6 x 6 = 36 Required outcome = 15 ∴ Pr(E) = \(\frac{15}{36} = \frac{5}{12}\)

Total numbers from 40 to 50 inclusive = 11 (40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50). Prime numbers between 40 and 50 (inclusive) are 41, 43, and 47. There are 3 prime numbers in this range. therefore, Pr(prime numbers) = \(\frac{3}{11}\)

To find the point of intersection, equate the two equations ⇒ 2x\(^2\) + 10 = 4x + 16 ⇒ 2x\(^2\) + 10 - 4x - 16 = 0 ⇒ 2x\(^2\) - 4x - 6 = 0 Factorize ⇒ 2(x + 1)(x - 3) = 0 ∴ x = -1 or 3 The curves will intersect at -1 and 3 Area = \(\int^b_a\) [upper function] - [lower function] dx = \(\int^3_1 4x + 16 - (2x^2 + 10) dx\) = \(\int^3_1 -2x^2 + 4x + 6dx\) = \((\frac {-2x^3}{3} + 2x^2 + 6x)^3_1\) = \((\frac {-2(3)^3}{3} + 2(3)^2 + 6(3)) - (\frac {-2(-1)^3}{3} + 2(-1)^2 + 6(-1))\) = 18 - \((\frac {10}{3})\) = 18 + \(\frac {10}{3}\) = \(\frac {64}{3}\)

From ∆PSR |PS| = |SR| (If two tangents are drawn from an external point of the circle, then they are of equal lengths) ∴ ∆PSR is isosceles ∠PSR + ∠SRP + ∠SPR = 180\(^o\) (sum of angles in a triangle) Since |PS| = |SR|; ∠SRP = ∠SPR ⇒ ∠PSR + ∠SRP + ∠SRP = 180\(^o\) ∠PSR + 2∠SRP = 180\(^o\) 36\(^o\) + 2∠SRP = 180\(^o\) 2∠SRP = 180\(^o\) - 36\(^o\) 2∠SRP = 144\(^o\) ∠SRP = \(\frac {144^o}{2} = 72^0\) ∠SRP = ∠PQR (angle formed by a tangent and chord is equal to the angle in the alternate segment) ∴ ∠PQR = 72\(^0\)

1st bag → 4 W, 3 B = 7 marbles 2nd bag → 5 R, 6 B = 11 marbles The only possible way of picking marbles of the same colour is to pick a blue marble from each bag therefore, Pr( same colour) = \(\frac{3}{7}\times \frac{6}{11} = \frac{18}{77}\)

\(8 + 2x - x^2 = 0\) \(8 + 4x - 2x - x^2 = 0\) 4(2 + x) - x(2 + x) = 0 (4 - x)(2 + x) = 0 4 - x = 0 or 2 + x = 0 x = 4 or x =- 2 So, p = 4 and q = -2 ∴ p + q = 2

a. Since K is the centre and KR and KS are radii, ∆SKR is isosceles. ∠SRK + ∠RSK + ∠SKR = 180^\circ (sum of angles in a triangle is 180o) ⇒ ∠SRK + ∠RSK + 80^\circ = 180^\circ ⇒ ∠SRK + ∠RSK = 180^\circ - 80^\circ = 100^\circ Since ∠SRK = ∠RSK (base angles of an isosceles triangle are equal) ∠SRK = ∠RSK = \(\frac{100^0}{2} = 50^0\) ∠RSP = 50^\circ + 41^\circ = 91^\circ ∴ ∠ RQP = 180^\circ - 91^\circ = 89^\circ (opp. angles of a cyclic. quad. are supplementary) b. Since KR is a bisector of angle ∠SRQ: ∠SRK = ∠KRQ = 50^\circ ∠SRQ = 50^\circ + 50^\circ = 100^\circ ∴ ∠SPQ = 180^\circ - 100^\circ = 80^\circ (opp. angles of a cyclic. quad. are supplementary)

Check: (2a-3b) (2a+3b) 4a\(^2\) + 6ab - 6ab - 9b\(^2\) 4a\(^2\) - 9b\(^2\)

5x + 3y=4 5x-3y= 2 Using elimination method 5x + 3y=4 → 5x + 3y=4 -[5x-3y= 2] → -5x +3y= -2 6y = 2 y → 1/3 and x = 3/5 solving (25x\(^2\) -9y\(^2\)) 25 * [3/5]\(^2\) -9 * [1/3]\(^2\) 25 * \(\frac{9}{25}\) - 9 \(\frac{1}{9}\) 9 - 1 = 8

\(\frac{2-18m^2}{1+3m}\) = \(\frac{2[1-9m^2]}{1+3m}\) \(\frac{2[1-3m][1+3m]}{1+3m}\) 2[1-3m]

Since the mean of the two numbers x and y is: \(\frac{x+y}{2}\) = 4 x+y =2 * 4 x+y = 8 then we get: \(\frac{x+2x+y+2y}{4}\) =(\(\frac{3x+3y}{4}\)) = 3(\(\frac{x+y}{4}\)) → 3(\(\frac{8}{4}\)) = 3 * 2 = 6

Where sin \theta = \(\frac{opp}{hyp}\) → \(\frac{-3}{5}\) opp = -3, hyp = 5 using pythagoras formula hyp\(^2\) = adj\(^2\) + opp\(^2\) adj\(^2\) = hyp\(^2\) - opp\(^2\) adj\(^2\) = 5\(^2\) - 3\(^2\) → 25 - 9 adj\(^2\) = 16 adj = 4 cos \theta = \(\frac{adj}{hyp}\) → \(\frac{4}{5}\) In third quadrant: cos \theta is negative → - \(\frac{4}{5}\)

Using line LZ with angles sum = 180º : ∠MNY = 180º - (g + h)º or 180º - gº - hº

Given that sec\(^2\)\theta + tan\(^2\)\theta = 3 Where sec\(^2\)\theta = 1 + tan\(^2\)\theta : 1 + tan\(^2\)\theta + tan\(^2\)\theta = 3 2tan\(^2\)\theta = 3 - 1 2tan\(^2\)\theta = 2 divide both sides by 2 tan\(^2\)\theta = 1 tan\theta = \sqrt1 tan\theta = 1 \theta = tan\(^{-1}\) (1) \theta = 45º

hyp = 5cm, adj = \(\frac{8cm}{2}\) 4cm, opp = xcm Pythagoras theorem: hyp\(^2\) = opp\(^2\) + adj\(^2\) 5\(^2\) = x\(^2\) + 4\(^2\) x\(^2\) = 25 - 16 x = \sqrt9 x = 3cm