Maths last 10 year papers

13. Let ������������ be the height then ������������ = 1500 3. In 15 seconds, the aeroplane mo es from ������ to ������. ������ and ������ are the points here angles of ele ations are 600 and 300 . Let ������������ = ������ and ������������ = b ������������ = ������ + ������ In triangle ������������������, ������ 600 = ������������/������������ 3 = 1500 3/������ ������ = 1500 In triangle ������������������, ������ 30 0 = ������������/������������ 1/ 3 = 1500 3/(������ + ������) ������ = 3000 14. Downloaded from www.padhle.in

15. Gi en that the jar contains red, bl e and orange balls. For s ppose ������ be the n mber of red balls and ������\" be the n mber of bl e balls N mber of orange balls is gi en as 10 Total n mber of balls ill be = ������ + ������ + 10 Gi en that Probabilit of selecting red ball is= 1/4 implies ������/(������+������+10) = 4������ = ������ + ������ + 10 implies 3������ ������ = 10 ..(1) Probabilit of selecting red ball is= 1/3 implies ������/(������+������+10) = ⅓ 3������ = ������ + ������ + 10 implies 2������ ������ = 10 ..(2) Sol ing (1) and (2) e get, M ltipl eq. (2) be 3 then add it ith eq ation (1) 5������ = 40 implies ������ = 8 If ������ = 8 then ������ = 6. Total n mber of balls in the jar ill be eq al to 6 + 8 + 10 = 24. Downloaded from www.padhle.in

16. Gi en radi s of the circle is 14������ and center angle is 600 . We ha e the form la for finding the area. 17. ill be = 2.1 , Gi en parameters are Diameter of the tent = 4.2 then radi s Height of the c lindrical part is (ℎ) = 4 Height of the conical part is (ℎ1) = 2.8m We ha e the form la for slant height = 52.8m2 18. ol me of hemisphere =2/3 r = ol me of liq id present in bo l =2/3 22/7 (18) =2/3 22/7 18 18 18 cm no , according to q estion 10% liq id is asted b transfer . so, rest ol me of liq id =90 of ol me of liq id present in bo l Downloaded from www.padhle.in

again , 90 of ol me of liq id present in bo l = 72 ol me of each bottle 9/10 2/3 22/7 18 18 18 = 72 22/7 3 3 H H = 5.4 cm 19. The greatest diameter a hemisphere can ha e 10 cm. Total s rface area of the solid = S rface area of the c be + CSA of the hemisphere Area of the base of the hemisphere. Cost of printing the block at the rate of Rs 5 per sq. cm The cost of 678.57sq cm =678.57 5=3392.85 Hence, the amo nt is Rs 3393. 20. Downloaded from www.padhle.in

21. . Let s ass me that ������ be the longer length and ������ be the shorter length Gi en that diagonal of a rectang lar field is 16 ������ ������ more than the shorter side Implies diagonal= 16 + b. Longer side is 14 ������ ������ more than the shorter side implies������ = 14 + ������. B sing P thagoras theorem (16 + ������)2 = (14 + ������)2 + ������2 256 + ������ 2 + 32������ = 196 + ������ 2 + 28������ + ������ 2 ������ 2 4������ 60 = 0 (������ + 6)(������ 10) = 0 ������ = 6 or ������ = 10 Breadth ill be eq al to 10 and other length is = 24 . 22. =60-10=50 AP=8,10,12.... =8, =10-8=2 , = +( -1) ₀ =8+(60-1)2 =8+118=126 , ₀=60/2[2 8 +(60-1)2] =30[16+118] =4020 , Downloaded from www.padhle.in

₅₀=50/2[2 8+(50-1)2] ₀- ₅₀=4020=-2850 =1170 =25[16+98] =2850 S= 23. Speed can t be negati e and hence initial speed of the train is 36 /h. 24. In PQO and PRO PO= common QO=RO= Radi s of circle ∠PQO=∠PRO=90 o (Radi s tangent) Downloaded from www.padhle.in

∴ PQO PRO (RHS) and hence PQ=PR Hence pro ed (b CPCT) 25. In the fig re, ������ is the midpoint of the minor arc ������������,������ is the center of the circle and ������������ is tangent to the circle thro gh point ������. We ha e to sho the tangent dra n at the midpoint of the ������ ������ ������������ of a circle is parallel to the chord joining the end points of the ������ ������ ������������. We ill sho ������������ is parallel to ������������. It is gi en that ������ is the midpoint point of the ������ ������ ������������. So, ������ ������ ������������ = ������ ������ ������������ Implies ������������ = ������������ This sho s that the triangle ������������������ is an isosceles triangle. Th s, the perpendic lar bisector of the side ������������ of triangle ������������������ passes thro gh ������ ������ ������. The perpendic lar bisector of a chord passes thro gh the center of the circle. So the perpendic lar bisector of ������������ passes thro gh the center ������ of the circle. Th s perpendic lar bisector of ������������ passes thro gh the points ������ and ������. Implies ������������ is perpendic lar to ������������ ������������ is the tangent to the circle thro gh the point ������ on the circle. Implies ������������ is perpendic lar to ������������ The chord ������������ and the tangent ������������ of the circle are perpendic lar to the same line ������������. ������������ is perpendic lar to ������������. 26. Downloaded from www.padhle.in

27. if e ass me the distance of the clo d from point A. Downloaded from www.padhle.in

e.g., DE = h and CF = h - 20 no , from ADE, tan30 = DE/AD 1/ 3 = h/AD AD = 3h ..............(1) from triangle ADF, tan60 = DF/AD 3 = (FC + CD)/ 3h 3 3h = (h + 20 + 20) 3h = h + 40 h = 20m. hence, Distance of clo d from A = 20m 28. Let s ass me that S be the sample space of dra ing a card from a ell-sh ffled deck. (S) = 521C i) In a deck e ha e 13 spade 4 ace card a spade can be chosen in 13 a s. An ace can be chosen in 4 a s. B t e ha e alread chosen one ace card hich is incl ded in spade. So req ired probabilit = 13+4 1/52 = 16/52 = 4/13. ii) There are onl t o black king in a deck. A black king card is dra n in 2 a s. So req ired probabilit = 2/52 = 1/26 . iii) There are 4 jack and 4 king cards in deck. So there are 44 cards hich are neither jacks nor kings So req ired probabilit = 44/52 = 2/13 . i ) There are 4 kings and 4 q een cards in a deck. So there are 4 + 4 = 8 card hich are either king or q een. So req ired probabilit = 8/52 = 1/26 . 29. Downloaded from www.padhle.in

Area of the triangle formed b ( 1, 1 ), ( 2, 2) and ( 3, 3) is gi en b 30. Gi en that ������������������������ is a sq are. So each side is eq al and angle bet een the adjacent sides is a right angle. Also the diagonals perpendic larl bisect each other. Area of the shape ������������������ = ������ ������������ ������ ������������ ������������������ ������ ������������ ������ ������������ ������ ������ ������������������ = ( 42/2 ) 2 ( /2 1) = 441(0.57) = 251.37������ 2 Area of the flo er bed ������������������ = Area of the flo er bed ������������������ = 251.37������ 2 31. . F, = = 10 = 4.2 V = 176.4 Downloaded from www.padhle.in

V = 4/3 = 4/3 (4.2) = 98.784 98.8 T = 176.4 - 98.8 = 77.6 =T NF , D = 1.4 R = 1.4/2 = 0.7 T 158.4 77.6 = C 77.6 = (0.7) = 77.6/0.49 = 158.36 Downloaded from www.padhle.in

CBSE Maths 2014 1) All q estions are comp lsor . 2) The q estion paper consists of thirt q estions di ided into 4 sections A, B, C and D. Section A comprises of ten q estions of 01 mark each, Section B comprises of fi e q estions of 02 marks each, Section C comprises ten q estions of 03 marks each and Section D comprises of fi e q estions of 06 marks each. 3) All q estions in Section A are to be ans ered in one ord, one sentence or as per the e act req irement of the q estion. 4) There is no o erall choice. Ho e er, internal choice has been pro ided in one q estion of 02 marks each, three q estions of 03 marks each and t o q estions of 06 marks each. Yo ha e to attempt onl one of the alternati es in all s ch q estions. 5) In q estion on constr ction, dra ing sho ld be near and e actl as per the gi en meas rements. 6) Use of calc lators is not permitted. Q estions - 1. If the height of a ertical pole is 3times the length of its shado on the gro nd, then the angle of ele ation of the S n at that time is ______. 2. A bag contains cards n mbered from 1 to 25. A card is dra n at random from the bag. The probabilit that the n mber on this card is di isible b both 2 and 3 is 3. T o different coins are tossed sim ltaneo sl . The probabilit of getting at least one head is ____. 4. T o concentric circles are of radii 5 cm and 3 cm. Length of the chord of the larger circle (in cm), hich to ches the smaller circle is _______. Downloaded from www.padhle.in

5. In fig re, a q adrilateral ABCD is dra n to circ mscribe a circle s ch that its sides AB, BC, CD and AD to ch the circle at P, Q, R and S respecti el . If AB = cm, BC = 7 cm, CR = 3 cm and AS = 5 cm, find . 6. The perimeter of a triangle ith ertices (0, 4), (0, 0) and (3, 0) is _______. 7. A rectang lar sheet of paper 40 cm 22 cm, is rolled to form a hollo c linder of height 40 cm. The radi s of the c linder (in cm) is _______. 8. The ne t term of the A.P. 7,28,63, ... is EC ION B 9. In fig re , XP and XQ are t o tangents to the circle ith centre O, dra n from an e ternal point X. ARB is another tangent, to ching the circle at R. Pro e that XA + AR = XB + BR. Downloaded from www.padhle.in

10. Pro e that the tangents dra n at the ends of an diameter of a circle are parallel. 12. In Fig re , OABC is a q adrant of a circle of radi s 7 cm. If OD = 4 cm, find the area of the shaded region. [Use = 22/7 ] 13. Sol e for : 3 2 22 23 = 0. 14. The s m of the first n terms of an AP is 5n n2. Find the nthterm of this A.P. EC ION C 15. Points P, Q, R and S di ide the line segment joining the points A (1, 2) and B (6, 7) in 5 eq al parts. Find the coordinates of the points P, Q and R. 16. In the fig re, from a rectang lar region ABCD ith AB = 20 cm, a right triangle AED ith AE = 9 cm and DE = 12 cm, is c t off. On the other end, taking BC as diameter, a semicircle is added on o tside the region. Find the area of the shaded region. [Use = 3.14] Downloaded from www.padhle.in

17. In the fig re, ABCD is a q adrant of a circle of radi s 28 cm and a semi circle BEC is dra n ith BC as diameter. Find the area of the shaded region. [Use = 22/7] 18. A 5 m ide cloth is sed to make a conical tent of base diameter 14 m and height 24 m. Find the cost of cloth sed at the rate of Rs 25 per metre. [Use = 22/7] 19. A girl empties a c lindrical b cket, f ll of sand, of base radi s 18 cm and height 32 cm, on the floor to form a conical heap of sand. If the height of this conical heap is 24 cm, then find its slant height correct p to one place of decimal. 20. The s m of the first 7 terms of an A.P. is 63 and the s m of its ne t 7 terms is 161. Find the 28th term of this A.P. 21. T o ships are approaching a light-ho se from opposite directions. The angles of depression of the t o ships from the top of the light-ho se are 30 and 45 . If the distance bet een the t o ships is 100 m, find the height of the light-ho se. [Use 3= 1.732] 22. If 2 is a root of the q adratic eq ation 3 2+ p 8 = 0 and the q adratic eq ation 4 2 2p + k = 0 has eq al roots, find the al e of k. 23. Constr ct a triangle PQR, in hich PQ = 6 cm, QR = 7 cm and PR = 8 cm. Then constr ct another triangle hose sides are 45 times the corresponding sides of PQR. Downloaded from www.padhle.in

24. Find the al e(s) of p for hich the points (p + 1, 2p 2), (p 1, p) and (p 3, 2p 6) are collinear. EC ION D 25. The mid-point P of the line segment joining the points A ( 10, 4) and B ( 2, 0) lies on the line segment joining the points C ( 9, 4) and D ( 4, ). Find the ratio in hich P di ides CD. Also find the al e of . 26. A q adrilateral is dra n to circ mscribe a circle. Pro e that the s ms of opposite sides are eq al. 27. The angle of ele ation of the top of a chimne from the foot of a to er is 60 and the angle of depression of the foot of the chimne from the top of the to er is 30 . If the height of the to er is 40 m, find the height of the chimne . According to poll tion control norms, the minim m height of a smoke emitting chimne sho ld be 100 m. State if the height of the abo e mentioned chimne meets the poll tion norms. What al e is disc ssed in this q estion? 28. A hemispherical depression is c t o t from one face of a c bical block of side 7 cm, s ch that the diameter of the hemisphere is eq al to the edge of the c be. Find the s rface area of the remaining solid. [Use = 227] . 29. If Sn Denotes the s m of the first n terms of an A.P., pro e that S30 = 3 (S20 S10). 30. A metallic b cket, open at the top, of height 24 cm is in the form of the fr st m of a cone, the radii of hose lo er and pper circ lar ends are 7 cm and 14 cm respecti el . Find Downloaded from www.padhle.in

(i) the ol me of ater hich can completel fill the b cket. (ii) the area of the metal sheet sed to make the b cket. [Use = 22/7] 31. The s m of the sq ares of t o consec ti e e en n mbers is 340. Find the n mbers. 32. Pro e that the tangent at an point of a circle is perpendic lar to the radi s thro gh the point of contact. 33. A dice is rolled t ice. Find the probabilit that (i) 5 ill not come p either time. (ii) 5 ill come p e actl one time. 34. Downloaded from www.padhle.in

CBSE Ma h 2014 Sol ions :- 1. A me OA a he ole and OB i i hado . Le he leng h of he hado be . Le be he angle of ele a ion of he o of he ole f om he g o nd. Gi en ha he heigh of he ole (h) = 3 leng h of i hado ⇒h = 3 In OAB an = AO OB = h = 3 = 3 ⇒ an = an 60 ⇒ = 60 ∴ The angle of ele a ion of he S n i 60 . 2. To al n mbe of ca d = 25 To al n mbe of o ible o come = 25 A me E a he e en ha he n mbe on he ca d d a n i di i ible b bo h 2 and 3. ∴ E = 6, 12, 18, 24 ∴ N mbe of o ible o come fa o able o E = 4 P(E) = N mbe of o ible o come fa o able o ETo al n mbe of o come P(E) = 425. Downloaded from www.padhle.in

3. When o coin a e o ed im l aneo l , he o ible o come a e (H, H), (H, T), (T, H), (T, T) . ∴ To al n mbe of o ible o come = 4 A me he e en of ge ing a lea one head a 'E'. N mbe of o ible o come fa o able o E i (H, H), (H, T), (T, H) . ∴ N mbe of fa o able o come = 3 P(E) = N mbe of fa o able o come o ETo al n mbe of o ible o come P(E) = 34. 4. In he fig re, O is he common cen re, of he gi en concen ric circles. AB is a chord of he bigger circle s ch ha i is a angen o he smaller circle P. Since, OP is he radi s of he smaller circle. ∴ OP ⊥ AB => ∠APO = 90 Also, radi s perpendic lar o a chord bisec s he chord ∴ OP bisec s AB => AP = No , in righ APO, Downloaded from www.padhle.in

⠀ OA = AP + OP => 5 = AP + 3 => AP = 5 - 3 => AP = 4 => AP = 4cm => AB = 4 => AB = 2 4 = 8cm Hence, he req ired leng h of he chord AB is 8cm. 5. Tangen s dra n from an e ernal poin o a circle are al a s eq al. The q adrila eral, ABCD is dra n o circ mscribe a circle s ch ha i s sides AB, BC, CD and AD o ch he circle a P, Q, R and S respec i el . Therefore A,B,C and D are e ernal poin s from hich angen s AS=AP=5, BP=BQ, CQ=CR=3 and DR=DS are dra n BC=7. => BQ + CQ =BC=7 => BQ= 7-3=4 ∴ BP=4 AB= , => AP +BP = => 5+4 = => =9 6. A(0,4) B(0,0) C(3,0) Therefore perime er of riangle ABC is = AB+BC+AC AB= (0-0)^2+(0-4)^2 = (0)+(16) = +4,-4 BC= (3-0)^2+(0-0)^2 = 9+0 =+3,-3 AC= (3-0)^2+(0-4)^2 = 9+16 = +5,-5 Meas remen of sides canno be rejec ed herefore -4,-3,-5 is rejec ed Therefore AB=4 cm BC=3cmAC=5cm Downloaded from www.padhle.in

Perime er= 12 cm 7. A ea of he ec ang la hee = l b = (40 22) cm2 = 880 cm2 A me he adi of he c linde a ' '. Gi en ha heigh (h) of he c linde = 40 cm A ea of he ec ang la hee i ame a he c ed face a ea of he c linde . ∴ 2 h = A ea of he ec ang la hee ⇒ 2 22/7 40 = 880 ⇒ = 7/2 = 3.5 ∴Radi of he c linde i 3.5 cm. 8. 7, 28, 63... = 7, (7 2 2) , (7 3 3) ... = (7 1 1) , (7 2 2), (7 3 3) ... = 7, 27, 37... Gi en ha he firs erm of he gi en AP is 7 and he common difference is (27 7) = 7 So, a = d = 7 The ne erm of he seq ence, i.e. he 4 h erm = a + 3d = 37+ 7 = 47= 7 4 4 = 112. Downloaded from www.padhle.in

9. GIVEN: XP and XQ are angen s of he circle from i h cen re O and R is a poin on he circle. TO PROVE : XA + AR = XB + BR. PROOF: [X is an e ernal poin ].......(1) XP = XQ [A is an e ernal poin ]......(2) AP = AR [B is an e ernal poin ]......(3) BQ = BR From eq 1 XP = XQ (XA + AP) = (XB + BQ) XA + AR = XB + BR [From eq 2 & 3 , AP = AR ,BQ = BR ] Hence pro ed. 10. Downloaded from www.padhle.in

Le AB be a diame er of he circle. T o angen s PQ and RS are dra n a poin s A and B respec i el . Radi s dra n o hese angen s ill be perpendic lar o he angen s. Th s, OA ⊥ RS and OB ⊥ PQ ∠OAR = 90 ∠OAS = 90 ∠OBP = 90 ∠OBQ = 90 I can be obser ed ha ∠OAR = ∠OBQ (Al erna e in erior angles) ∠OAS = ∠OBP (Al erna e in erior angles) Since al erna e in erior angles are eq al, lines PQ and RS ill be parallel. 11. The n mbe ha o dice co ld ho a e 1, 2, 3, 4, 5, 6. The o ible o come hen o dice a e olled a e: (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6) ∴ N mbe of o ible o come = 6 6 = 36. A me 'E' a he e en ha he m of n mbe a ea ing on he o dice i 10. The o ible o come fa o able o e en , E a e (4, 6), (5, 5), (6, 4) . ∴ N mbe of fa o able o come o e en E = 3. P(E) = N mbe of o ible o come fa o able o ETo al n mbe of o ible o come = 336 = 12. Downloaded from www.padhle.in

12. Gi en ha , adi of he ci cle = 7 cm A ea of he ad an OABC = 1/4 2 = 1/4 22/7 7 7 cm2= 77/2cm2 = 33.5cm2 A ea of ODC = 1/2 OD OC = (1/2 4 7) cm2 (Since OC i he adi of he ci cle) = 14 cm2 A ea of he haded egion = A ea of he ad an OABC A ea of ODC A ea of he haded egion = 33.5 14 = 24.5cm2 13. Gi en he ad a ic e a ion a , 3 2 22 23 = 0 ⇒ 3 2 32 + 2 23 = 0 ⇒ 3 ( 6) + 2( 6) = 0 ⇒ (3 +2) ( 6) = 0 ⇒ = 2/3 o = 6 14. The m of n e m of an A.P. i gi en b 5n -n2 B e kno ha Sn= n2(a + n) Downloaded from www.padhle.in

he e a = fi e m Tn = n h e m So, e ha e 5n n2= n2 ( a + n) ⇒ n(5 n) = n2( a + n) ⇒ 10 2n = a + n ...(1) No , e ha e m of he fi e m S1= a ⇒ a = (5(1) 12) = 4 S b i e he al e of a in e a ion (1), e ge 10 2n = 4 + n ⇒ n= 6 2n . The efo e, n h e m of he A.P. i 6 2n. 15. I is gi en ha P, Q, R and S di ide he line segmen joining he poin s A(1, 2) and B(6, 7) in 5 eq al par s. Sec ion form la, P di ides he line AB in 1:4. Q di ides he line AB in 2:3. R di ides he line AB in 3:2. Downloaded from www.padhle.in

Therefore he coordina es of he poin s P, Q and R are P(2,3), Q(3,4) and R(4,5). 16. AED i an igh angle iangle, AD2= AE2+ED2. ⇒ AD2 = (92+ 122) = (81 + 144) = 225 cm2 ⇒ AD = 15 cm A ea of he ec ang la egion ABCD = AB AD = (20 15) = 300 cm2 A ea of AED = 1/2 AE DE = (1/2 9 12) cm2 = 54 cm2 In a ec angle AD = BC = 15 cm Since, BC i he diame e of he ci cle, adi of he ci cle = 15 cm A ea of he emi-ci cle = 1/2 2 = (1/2 3.14 15/2 15/2) = 88.3125 cm2 A ea of he haded egion = A ea of he ec angle + A ea of he emi-ci cle A ea of he iangle = (300 + 88.3125 54) cm2 Downloaded from www.padhle.in

= 334.3125 cm2. 17. Gi en: Radi ( ) of he ci cle = AB = AC = 28 cm A ea of ad an ABPC = 1/4 = (1/4 22/7 28 28) cm =22 28= 616 cm A ea of ABC = 1/2 AC AB = (1/2 28 28) = 392 cm A ea of egmen BPC = A ea of ad an ABPC A ea of ABC = (616 392) cm = 224 cm .............(1) In a igh -angled BAC BC = BA + AC (B P hago a heo em) BC = (28 + 28 ) cm BC = 784 +784 cm BC = 16 98 = 16 49 2 BC = 4 7 2= 28 2 BC= 28 2 cm BC(Diame e ) = 28 2 Radi of emici cle= 28 2/2= 14 2 cm A ea of emici cle BEC= 1/2 Downloaded from www.padhle.in

= ( 1/2 22/7 14 2 14 2) cm = 22 2 14 2 = 22 14 2 = 44 14 = 616 cm A ea of he haded o ion = A ea of emici cle BEC A ea of egmen BPC = 616 224 cm = 392 cm Hence, he a ea of he haded egion = 392 cm 18. Fo Conical en , b=5m adi = diame e /2 =14/2=7m heigh h=24 m lan heigh l = ((h^2)+( ^2))= (576)+(49)= 625=25 C.S.A of en = l =(22/7) 7 25 =550 m^2 A ea of clo h e i ed = C.S.A of en l b=550 l 5=550 l=110m Co of 1 m clo h= 25 Co of 110 m clo h= (110 25)= 2750 Downloaded from www.padhle.in

19. 20. Gi en : S7 = 63 and m of i ne 7 e m i 161 B ing he fo m la ,S m of n h e m , Sn = n/2 [2a + (n 1) d] S7 = (7/2) [ 2a + (7 - 1)d ] 63 = 7/2 [2a + 6d] 63 2/7 = [2a + 6d] 9 2 = 2a + 6d 2a + 6d = 18 ...(1) And S m of i ne 7 e m = 161(Gi en) Downloaded from www.padhle.in

S m of fi 14 e m = S m of fi 7 e m + S m of ne 7 e m . S14 = 63 + 161 = 224 S14 = 224 B ing he fo m la ,S m of n h e m , Sn = n/2 [2a + (n 1) d] S14 = (14/2) [ 2a + (14 - 1)d ] 224 = 7 [ 2a + 13d ] 224/7 = 2a + 13d 2a + 13d = 32 ... (2) On b ac ing e (1) f om (2) 2a + 13d = 32 2a + 6d = 18 (-) (-) (-) ------------------- 7d = 14 --------------------- d = 14/7 d=2 On ing he al e of d = 2 in e (1), 2a + 6d = 18 2a + 6(2) = 18 Downloaded from www.padhle.in

2a + 12 = 18 2a = 18 - 12 2a = 6 a = 6/2 a=3 Fo 28 h e m : B ing he fo m la ,an = a + (n - 1)d a28 = a + ( 28 - 1) d a28 = 3 + (27) 2 a28 = 3 + 54 a28 = 57 Hence, 28 h e m of hi A.P i 57. 21. AB = 100 Le AC = Le BC = 100- Downloaded from www.padhle.in

Le CD = h In EAD, EA = DC ED = AC an 45 = EA/ED 1 = h/ =h In DFB FB = CD DF = CB an 30 = FB/DF 1/ 3 = h/100- 1/ 3 = h/100-h 100-h = 3h 100-h = 1.732h 2.732h = 100 h = 100/2.732 h = 36.603m h = 36.6m 22. Gi en : 3 + - 8 = 0 .(1) 4 - 2 + k = 0 (2) On ing he al e of gi en oo i.e = 2 in e 1 . 3 + -8=0 3(2) + (2) 8 = 0 3 4+2 -8=0 12 + 2 8 = 0 Downloaded from www.padhle.in

4+2 =0 2 =-4 = - 4/ 2 = -2 =-2 Hence he al e of i - 2. On ing he al e of = - 2 in e 2, 4 -2 +k=0 4 - 2(- 2) + k = 0 4 +4 +k=0 On com a ing he gi en e a ion i h a + b + c = 0 He e, a = 4, b = 4 , and c = k D(di c iminan ) = b 4ac Gi en : Q ad a ic e a ion ha e al oo i.e D = 0 b 4ac = 0 4 4(4)(k) = 0 16 16k = 0 16 = 16k k = 16/16 = 1 k=1 Downloaded from www.padhle.in

23. S e of con c ion: S e 1: D a a a PX. S e 2: PQ = 6 cm a he adi , d a an a c f om oin P in e ec ing PX a Q. S e 3: Q a he cen e and adi e al o 7 cm, d a an a c. S e 4: P a he cen e and adi e al o 8 cm, d a ano he a c, c ing he e io l d a n a c a R. S e 5: Join PR and QR. The efo e, PQR i he e i ed iangle. S e 6: D a an a PY making an ac e angle i h PQ. S e 7: No loca e 5 oin P1, P2, P3, P4and P5 on PY o ha PP1=P1P2= P2P3= P3P4= P4P5. S e 8: Join P5Q and d a a line h o gh P4, hich i a allel o P5Q, o in e ec PQ a Q'. S e 9: D a a line h o gh Q', hich i a allel o line QR, o in e ec PR a R'. Then e ge , PQ'R' i he e i ed iangle. Downloaded from www.padhle.in

24. Gi en , Poin = ( +1 , 2 -2 ), ( -1 , ) and ( -3 , 2 -6) Fo he gi en oin ( ₁ , ₁ ) , ( ₂ , ₂ ) and ( ₃ , ₃) o be collinea hen [ ₁ ( ₂ - ₃ ) + ₂( ₃ - ₁ )+ ₃( ₁- ₂ )] = 0 He e, ₁ = +1 ₁=2 -2 ₂= -1 ₂= ₃ = -3 ₃=2 -6 S b i ing he al e in he fo m la , ( +1)( -(2 -6))+( -1)(2 -6-(2 -2))+( -3)(2 -2-( ))=0 ( +1)( - 2 +6))+( -1)(2 -6- 2 +2))+( -3)(2 -2- )=0 ( + 1 ) ( - + 6 ) + ( - 1 ) ( -4 ) + ( - 3 ) ( - 2 ) = 0 - - +6 +6-4 +4+ -3 -2 +6=0 - 4 + 16 = 0 4 = 16 Di iding bo h he ide b 4 4 / 4 = 16 / 4 =4 Hence, Fo he oin o be collinea , = 4 Downloaded from www.padhle.in

25. 26. Downloaded from www.padhle.in

Gi en:- Le ABCD be he ad ila e al ci c m c ibing he ci cle i h cen e O. The ad ila e al o che he ci cle a oin P,Q,R and S. To o e:- AB+CD+AD+BC P oof:- A e kno ha , leng h of angen d a n f om he e e nal oin a e e al. The efo e, AP=AS.....(1) BP=BQ.....(2) CR=CQ.....(3) DR=DS.....(4) Adding e a ion (1),(2),(3) and (4), e ge AP+BP+CR+DR=AS=BQ+CQ+DS (AP+BP)+(CR+DR)=(AS+DS)+(BQ+CQ) ⇒AB+CD=AD+BC Hence P o ed. 27. Downloaded from www.padhle.in

Le PQ be he chimne Le AB be he o er here AB = 40 m Also, gi en ha and Le s consider he riangle ABP Th s, e ha e, No , e shall consider he riangle APQ Hence, e ha e, Th s, he h​ eight of the chimney is 120 m Since, he minim m heigh of a smoke emi ing chimne sho ld be 100 m and he heigh of he chimne is 120 m. Th s, he heigh of he chimne mee s he poll ion norms. 28. Downloaded from www.padhle.in

S rface area of he remaining solid = S rface area of he c be - S rface area of he circle on one of he faces of he c be + S rface area of he hemisphere. 29. In an A.P , he m of n e m i Sn= n2[2a+ (n 1) d]. he e, a = Fi e m d = Common diffe ence n = N mbe of e m in an A.P. We ha e o o e ha , S30= 3(S20 S10). Take R.H.S., 3(S20 S10) = 3 [202 2a + (20 1)d 102 2a + (10 1)d ] = 3[10 (2a + 19d) 5(2a + 9d)] Downloaded from www.padhle.in

= 3[20a + 190d 10a 45d] = 3[10a + 145d] = 15[2a + 29d] No ake L.H.S., e ge S30= 302[2a + (30 1)d] S30= 15[2a + 29d] ∴ L.H.S. =R.H.S. Hence, o ed. 30. (i). ol me of f m of cone =⅓ h (R +R + ) = ⅓ 22/7 24 ( 14 + 14 7 + 7 ) = 22 8/7 ( 196 + 98 + 49 ) = 22 8/7 343 = 22 8 49 = 8624 cm (ii). lan heigh of f m l = ((R- ) +h ) = ( ( 14 - 7 ) + 24 ) = ( 7 + 24 ) = ( 49 + 576 ) = 625 = 25 cm a ea of me allic hee = CSA of f m + a ea of ba e = l (R+ )+ = 22/7 25 ( 14 + 7 ) + 22/7 7 = 550/7 21 + 22 7 = 550 3 + 154 = 1650 + 154 = 1804 cm Downloaded from www.padhle.in

31. Le he con ec i e e en n mbe be & + 2. A/ S a e of he e n mbe i 340. ( ) + ( + 2) = 340 + + 2 + 2 2 = 340 2 + 4 = 340 4 2 ( + 2) = 336 ( + 2) = 336/2 + 2 = 168 + 2 168 = 0 No , B ing he middle e m li ing me hod. + 2 168 + 14 12 168 ( + 14) 12 ( + 14) ( + 14) ( 12) ( + 14) = 0 o , ( 12) = 0 = 14 o = 12 Rejec he nega i e al e of and ake = 12 Hence, Re i ed n mbe a e = 12 and ( + 2) = 12+2 = 14 Ve ifica ion - Downloaded from www.padhle.in

12 + 14 = 340 144 + 196 = 340 340 = 340 32. Tangen i a line hich o che a ci cle e ac l a one oin . All o he line in e ec he ci cle a o oin . Take a ecan PQOR. Tangen be PT. Le he adi be R. Le PQ = . o e oin of P ci cle: . = PQ * PR = ( + 2R) = +2R = ( + R) - R = PO - R So o e oin of P ci cle de end onl on di ance PO and Radi Po e oin of P calc la ed along PT, ( ha i Q and R me ge a T): = PT * PT Hence, e ha e : PT = PO - R alid. So he iangle i igh In he iangle PTO, hi mean P hago a heo em i angled. Hence PT ⊥ OT. Downloaded from www.padhle.in

33. In a die ha 6 face n mbe ed 1,2,3,4,5 and 6. N mbe of o ible o come on olling he dice ice a e (1, 1), (1, 2), (1,3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6) To al n mbe of o come = 36. (i) Fa o able o come fo he e en ha 5 ill no ho ei he ime a e (1, 1), (1, 2), (1, 3), (1, 4), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 6) To al n mbe of fa o able o come = 25 ∴ P obabili of no ge ing 5 ei he ime = N mbe of fa o able o come To al N mbe of o come = 2536 (ii) Fa o able o come fo he e en ha 5 ill ho e ac l one ime a e (1, 5), (2, 5), (3,5), (4, 5), (5, 1), (5, 2), (5, 3), (5, 4), (5, 6), (6, 5) To al n mbe of fa o able o come = 10 ∴ P obabili of ge ing 5 ch ha i ill come e ac l one ime = N mbe of fa o able o come To al n mbe of o come = 1036 Downloaded from www.padhle.in

34. 3( 3x 1 ) 2( 2x+3 ) 5 2x+3 3x 1 Le ( 3x 1 ) = m 2x+3 I.e. 3m-2/m = 5 3m2 - 2 = 5m 3m2 - 5m - 2 = 0 3m2 - 6m + m - 2 = 0 3m(m-2) + (m-2) = 0 (3m+1)(m-2) = 0 m = 2 o m = -⅓ ( 3x 1 ) = 2 OR ( 3x 1 ) = -⅓ 2x+3 2x+3 3 -1/2 +3 = 2 9 -3 = -2 -3 = -7 =0 An e : = -7, 0 Downloaded from www.padhle.in

CBSE Maths 2013 1) All questions are compulsor . 2) The question paper consists of thirt questions divided into 4 sections A, B, C and D. Section A comprises of ten questions of 01 mark each, Section B comprises of five questions of 02 marks each, Section C comprises ten questions of 03 marks each and Section D comprises of five questions of 06 marks each. 3) All questions in Section A are to be answered in one word, one sentence or as per the e act requirement of the question. 4) There is no overall choice. However, internal choice has been provided in one question of 02 marks each, three questions of 03 marks each and two questions of 06 marks each. You have to attempt onl one of the alternatives in all such questions. 5) In question on construction, drawing should be near and e actl as per the given measurements. 6) Use of calculators is not permitted. Downloaded from www.padhle.in

Questions - EC ION A 1. The common difference of the A.P. 1/p, 1 p/p, 1 2p/p..................is: 2. In Fig. 1, PA and PB are two tangents drawn from an e ternal point P to a circle with centre C and radius 4 cm. If PA ⊥ PB, then the length of each tangent is: 3. In Figure,, a circle with centre O is inscribed in a quadrilateral ABCD such that, it touches the sides BC, AB, AD and CD at points P, Q, R and S respectivel , If AB = 29 cm, AD = 23 cm, ∠B = 90 and DS = 5 cm, then the radius of the circle (in cm.) is: 4. The angle of depression of a car, standing on the ground, from the top of a 75 m high tower, is 30 . The distance of the car from the base of the tower (in m.) is: 5. The probabilit of getting an even number, when a die is thrown once, is : Downloaded from www.padhle.in

6. A bo contains 90 discs, numbered from 1 to 90. If one disc is drawn at random from the bo , the probabilit that it bears a prime-number less than 23, is : 7. In Fig. 3, Find the area of triangle ABC (in sq. units) is: 8. If the difference between the circumference and the radius of a circle is 37 cm, then using = 22/7, the circumference (in cm) of the circle is: (A) 154 (B) 44 (C) 14 (D) 7 EC ION B 9. Solve the following quadratic equation for : 4 3 2+5 2 3= 0 Downloaded from www.padhle.in

10. How man three digit natural numbers are divisible b 7? 11. In Fig. 4, a circle inscribed in triangle ABC touches its sides AB, BC and AC at points D, E and F respectivel . If AB = 12 cm, BC = 8 cm and AC = 10 cm, then find the lengths of AD, BE and CF. 12. Prove that the parallelogram circumscribing a circle is a rhombus. 13. A card is drawn at random from a well shuffled pack of 52 pla ing cards. Find the probabilit that the drawn card is neither a king nor a queen. 14. Two circular pieces of equal radii and ma imum area, touching each other are cut out from a rectangular card board of dimensions 14 cm 7 cm. Find the area of the remaining card board. [ Use = 22/7] EC ION C 15. For what value of k, are the roots of the quadratic equation k ( 2) + 6 = 0 equal? 16. Find the number of terms of the A.P. 18, 151/2, 13, ...., - 491/2 and find the sum of all its terms. Downloaded from www.padhle.in

17. Construct a triangle with sides 5 cm, 4 cm and 6 cm. Then construct another triangle whose sides are 2/3 times the corresponding sides of first triangle. 18. The hori ontal distance between two poles is 15 m. The angle of depression of the top of first pole as seen from the top of second pole is 30 . If the height of the second pole is 24 m, find the height of the first pole. [ 3= 1.732 ] 19. Prove that the points (7, 10), ( 2, 5) and (3, 4) are the vertices of an isosceles right triangle. 20. Find the ratio in which the -a is divides the line segment joining the points ( 4, 6) and (10, 12). Also find the coordinates of the point of division. 21. In Fig.5, AB and CD are two diameters of a circle with centre O, which are perpendicular to each other. OB is the diameter of the smaller circle. If OA = 7 cm, find the area of the shaded region.[ use = 22/7] 22. A vessel is in the form of a hemispherical bowl surmounted b a hollow c linder of the same diameter. The diameter of the hemispherical bowl is 14 cm and the total height of the vessel is 13 cm. Find the total surface area of the vessel. [ = 22/7] 23. A wooden to was made b scooping out a hemisphere of same radius from each end of a solid c linder. If the height of the c linder is 10 cm, and its base is of radius 3.5 cm, find the volume of wood in the to . [ = 22/7] Downloaded from www.padhle.in

24. In a circle of radius 21 cm, an arc subtends an angle of 60 at the centre. Find: (i) the length of the arc (ii) area of the sector formed b the arc. [ Use = 22/7] 25. Solve the following for : 26. Sum of the areas of two squares is 400 cm2. If the difference of their perimeters is 16 cm, find the sides of the two squares. 27. If the sum of first 7 terms of an A.P. is 49 and that of first 17 terms is 289, find the sum of its first n terms. 28. Prove that the tangent at an point of a circle is perpendicular to the radius through the point of contact. 29. In fig, l and m are two parallel tangents to a circle with centre O, touching the circle at A and B respectivel . Another tangent at C intersects the line l at D and m at E. Prove that ∠DOE = 90 Downloaded from www.padhle.in