PAPER - III ANSWER
1. Principal for the 1st half year = Rs.2560 Interest for the 1st half year = 2560*1/2*61/4 = Rs.80 100 Principal for the 2nd half year = 2560+80 = Rs.2640 Interest for the 2ndhalf year = 2640*1/2*61/4 = Rs.82.50 100 Principal for 3rd half year = 2722.50*1/2*25/4*1/100 = Rs.85.08 Amount for the 3rd half year =2722.50+85.08 =2807.58
2. Amount in 1st year = Rs.1632 The principal = Rs.1536 \Interest = Rs.96 Hence r = 100*I = 100*96 = 25% P*t 1536*1 4 Principal for 2nd year = Rs.1632 Interest for 2nd year = 1632*1*25/4*1/100 = Rs.102 \ Amount for 2nd year = 1632+102 = Rs.1734
3. Proof:- In DAMB and DDNE \ÐB =ÐE (given) \ÐM = ÐN = 90 (construction) \ DAMB @DDNE A.A.A AM = AB = BC DABC ~DDEF DN DE EF A DABC = ½ BC. AM (formula) A DDEF ½ EF. DN BC = BC EF EF if DABC = BC2 = AB2 = AC2 DDEF EF2 DE2 DF2 For DABC and DDEF DABC = AB2 = BC2 = AC2 DDEF DE2 EF2 DF2
4. (i) 2sin67° + tan40° + cos0° cos23° cot50° = 2sin67° + tan40° + 1 cos(90-67°) cot(90-40°) = 2sin67° + tan40° + 1 sin67° tan40° = 2+1+1 = 4 (ii) BD = DC = 40cm (Altitude bisects the base of isoD) From (DADC, AD2+DC2 = AC2(DADC = 90°) \ AD2+402 = 412 \ AD2+1600 = 1681 \ AD2 = 81 \ AD = 9 Area of DABC = ½*BC*AD = 360CM2
5. construction
6. (i) (a+b)2 - 4ab = (a-b)2 (formula) \ 62 - 4 * 63/4 = (a-b)2 \ 36-4*27/4 = (a-b)2 \ a-b = Ö 3 (ii) The quadratic equation with p and 5p as roots is (x-p) (x-5p) = 0 x2-6px+45 = 0 5p2 = 45 p2 =9 \p = ± 3
7. (i) 5<6,5<7,5<8,6<7,6<8,7<8 Hence the relation `is less than' in å* å is {(5,6),(5,7),(5,8),(6,7),(6,8),(7,8)} 6>5,7>6,8>7 Hence the relation `is 1 more than' can be written as {(6,5),(7,6),(8,7)} (ii) BC/AB = tanA BC = 4.8*tan32°33¢ BC = 4.8*0.6383 = 3.06384cm. AB/AC = cosA 4.8/AC = cos32°33° \ AC = 4.8/0.8429 = 5.69 ÐC = 90° - ÐA = 90° - 32°33° = 57°27°
8. (a) é x-2y ù = é -3ù ë3x+2y û = ë 7 û x-2y = -3 3x+2y = 7 Solving we get x = 1, y = 2 (b) x2 - 2x =é 1 2 ù é 1 2 ù -2 é 1 2 ù ë2 1 ù ë2 1 ù 2 1 ù = é 3 0ù ë0 3 û = 3 é 1 0 ù = 3I ë 0 1 û (c) Arrange the numbers in ascending order 2,4,4,8,16,21,21,21,43 Total no. of observation = 9 = n \ Median = (9+1)/2)th observation = 5th observation \ Median = 16 Since 21 occurs more often than others, it is the mode.
9. (a) y = x y = -x+2 x 0 1 2 x 0 1 2 y 0 1 2 y 2 1 0 From graph, the lines meet at (1,1) Hence solution set = {1,1} (b) y = 3/2m*x + 4/2m y = -8/3m*x + 10/3m Now slope of line m1 = 3/2m and m2 = -8/3m Since lines are perpendicular to each other m1*m2 = -1 Þ 3/2m * (-8/3m) = -1 \ m = ± 2
10. (a)
No. of workers f
Mid values x
x =åfx/åf = 8958/266 = 33.67 (b) Let ÐB = x ÐBAM = 90-x Þ ÐMAC = x In DABM and DAMC ÐB = ÐMAC = x ÐM = 90° = ÐM Þ DMBA ~ DMAC [ A-A-A] In DAMB AND DABC ÐB = ÐB (common) ÐAMB = ÐBAC = 90° (given) DMBA ~ DABC
11. (a) construction
12. (a) m(arc PNQ) = 360° - 130° = 230° m ÐPQS = measure of the angle inscribed in arc PNQ = ½ m(arc PMQ) = ½ * 130 = 65° m ÐPQR = ½ * 230 = 115° (b)
Let Q be the centre of the circle. Let seg QA ^ chord MN \ AM = ½ MN = 5 In right angled triangle MAQ QM2 = QA2 + MA2 = 82 + 52 = 64 + 25 = 89 \QM = Ö89 \ The radius of the circle is Ö89 and the diameter is 2Ö89.
13. (a) Here r = 5cm and h = 12cm The slant height (l) of a cone can be obtained by using the relation. l2 = r2 + h2 l2 = (5)2 + (12)2 = 25 + 144 = 169 l = 13 Curved surface area of cone = Prl = 3.14 * 5 * 13 = 204.1cm2 Area of base of the cone = Pr2 = 3.14 * (5)2 = 78.5cm2 Total surface area of the cone = Area of base + curved surface area = 78.5 + 204.1 = 282.6cm2 (b)
Let O be the centre and seg OA and seg OB the two radii. Such that ÐAOB = 135°. Let the tangent at A meet the tangent at B, in point P. The sum of the measures of the four angles of a quadrilateral is 360°. mÐPAO = 90° and mÐOBP = 90° \ mÐAPB + 90° + 90° + 135° = 360° \ mÐAPB = 360° - 315° = 45° The angle between those tangents = 45°