ANSWER
1. The difference between CI and SI for 2 years = Rs.210-Rs.200 = Rs.10 \interest on 1st year's interest = Rs.10 \Rate = 100*I = 100*10 = 10% PT 100*1 Principal = 100*I = 100*200 = Rs.1000 R*T 10*2
2. P = Rs.27,000 R = 4% N = 9 Amount = P(1+R/100 )N A = 27,000*(1+4/100)9 Log A = log 27000 + 9log1.04 = 4.4314 + 9(0.0170) = 4.5844 Taking antilogs A = 38410 C.I. = 38410-27000 = Rs.11410
3. In D APC and D ABC ÐACP = ÐABC ÐA = ÐA (common) Þ DACP ~ DABC [a-a-a] ÞAP = PC = AC if AP = 8 = 6 AC BC AB 6 10 AB \AP = 4.8cm and AB = 7.5cm DACP = CP2 = 82 = 0.64 DABC BC2 102
4. ( i) L.H.S. = sin(90-q)cos(90-q) tanq = cosqsinq*1 tanq = cosqsinq*cosq sinq = cosq2 = 1-sin2q = R.H.S. (ii) In DABC BC2 = AB2+AC2 } ÐA = 90° DAXY XY2 = AX2+AY2 \ BC2+XY2 = AB2+AC2+AX2+AY2 = (AB2+AY2)+(AC2+AX2) = BY2+CX2 \ BC2+XY2 = BY2+CX2
5. CONSTRUCTION
6. ( i) 5(a+3)2 = 23(a+3)+10 substituting a+3 = x 5x2 = 23x+10 5x2-23x-10 = 0 5x2-25x+2x-10 = 0 (x-5)(5x+2) = 0 \ x = 5 or -2 5 \ a = 2 or -3 2 5 (ii) a=4, b=-12,c=9 use x = -b±Öb2-4ac 2a \x = 3/2
7. ( i) Consider a line P. `P is parallel to P' is true. Hence the relation is reflexive. Consider the lines P and q. if P is parallel to q then q is parallel to P is true. Hence relation is symmetric. Consider the lines P, q and r. if P is parallel to q and q is parallel to r then P is parallel to r is true. Hence the relation is transitive. From the above 3 it is an equivalence relation. (ii) Let the distance of the cliff from the ship be xm. And height of the cliff be hm. As shown in figure above. BM = DC = 10, CM = DB = x CM = tan70° \x = 27.47 BM AM = tan42° \AM = 24.74 CM \AB = AM+MB = 24.74+10 = 34.74
.
8. (a) A2 = é 1 0 ù \A2 ¹ A ë 0 1 û A3 = é 1 0ù \ A3 = A ë 0 -1û (b) é2 -3ù é 2 -3 ù = é 4 -3a -6 - 3b ù ëa bû ë a b û ë 2a+ab -3a+b2û = é 1 0 ù ë 0 1 û \ 4 -3a = 1, -6-3b = 0, 2a+ab = 0, -3a+b2 = 1 \ a = 1, b = -2 Substituting again we get a = 1 and b = -2 (c) Mean = Sfx = 2*3+4*5+6*6+8*4+10*2 Sf 3+5+6+4+2 = 5.2
9. (a)
(b) 2y = 4x-3 Þ y = 2x- 3 2 y = 2x- 3 « y = mx+c 2 \ slope (m) = 2 y-intercept (c)= -3 2
10. (a) x = 4*6+8*10+12*12+16*8+20*4 6+10+12+8+4 x = 456 = 11.4 40 (b) (i) DDFE ~ DDBC (AA) DF2 = area DDFE DB2 area DDBC DF = 3 \ FB = 1 FB 1 DF 3 \ DF = 3 DB 4 \ area DDFE = DF2 = 9 area DDBC DB2 = 16 (ii) area DOEF = EF2 area DODA AD2 EF = EF = DF = 3 (opp. Sides of 11gm.) AD BC DB 4 \ area DOEF = 9 area DODA 16
11. (a) construction
12. (a )
Proof : Surely the points P,Q,S are non-collinear \ A circle passes through P, Q, and S If the fourth vertex R of the quadrilateral, does not lie on circle PQS, then it lies either in the interior or in exterior of that circle.Suppose R lies in the exterior of circle PQS Let T be the point where line QR intersects the circle. Join S to T. PQTS will be cyclic.
13. (a) Here V = 1 litre = 1000 cm3, r = 5 cm. Volume of right cylinder = Pr2h \ h = 1000 78.5 \ log h = log (1000/78.5) log h = 1.1051 \ h = 12.74cm.
(b) Supposing T to be the point of contact of the circles and C to be the centre of the smaller circle, points A, C and T will be collinear with AC+CT = AT \ AC+5 = 18 \ AC = 13 AC2 = AP2+PC2 \ AP = 12 = 2 PB 6 1 \ P divides seg. AB in ratio 2:1