Maths last 10 year papers

3. In figure, O is the centre of a circle, PQ is a chord and PT is the tangent at P. If ∠POQ = 70 , then ∠TPQ is equal to ________. 4. In Figure,, AB and AC are tangents to the circle ith centre O such that ∠BAC = 40 . Then ∠BOC is equal to ________. 5. The perimeter (in cm) of a square circumscribing a circle of radius a cm, is _________. 6. The radius (in cm) of the largest right circular cone that can be cut out from a cube of edge 4.2 cm is _________. 7. A to er stands verticall on the ground. From a point on the ground hich is 25 m a a from the foot of the to er, the angle of elevation of the top of the to er is found to be 45 . Then the height (in meters) of the to er is _________ m. 8. If P(a2, 4) is the mid-point of the line-segment joining the points A ( 6, 5) and B( 2, 3), then the value of a is Downloaded from www.padhle.in

9. If A and B are the points ( 6, 7) and ( 1, 5) respectivel , then the distance. 2AB is equal to __________. 10. A card is dra n from a ell-shuffled deck of 52 pla ing cards. The probabilit that the card ill not be an ace is ________. EC I N B 11. Find the value of m so that the quadratic equation m ( 7) + 49 = 0 has t o equal roots. 12. Find ho man t o-digit numbers are divisible b 6. 13. In Figure 3, a circle touches all the four sides of a quadrilateral ABCD hose sides are AB = 6 cm, BC = 9 cm and CD = 8 cm. Find the length of the side AD. 14. Dra a line segment AB of length 7 cm. Using ruler and compasses, find a point P on AB such that AP/AB= 3/5. Downloaded from www.padhle.in

15. Find the perimeter of the shaded region in Figure 4, if ABCD is a square of side 14 cm and APB and CPD are semicircles. [Use = 22/7 ] 16. T o cubes each of volume 27 cm3 are joined end to end to form a solid. Find the surface area of the resulting cuboid. (EITHER Q.15 OR Q.16) 17. Find the value of for hich the distance bet een the points A(3, 1) and B(11, ) is 10 units. 18. A ticket is dra n at random from a bag containing tickets numbered from 1 to 40. Find the probabilit that the selected ticket has a number hich is a multiple of 5. EC I N C 19. Find the roots of the follo ing quadratic equation: 2 35 +10=0 20. Find the A.P. hose fourth term is 9 and the sum of its si th term and thirteenth term is 40. 21. In Figure 5, a triangle PQR is dra n to circumscribe a circle of radius 6 cm such that the segments QT and TR into hich QR is divided b the point of contact T, are of lengths 12 cm and 9 cm respectivel . If the area of PQR = 189 cm 2 , then find the lengths of sides PQ and PR. Downloaded from www.padhle.in

22. Dra a pair of tangents to a circle of radius 3 cm, hich are inclined to each other at an angle of 60 . 23. A chord of a circle of radius 14 cm subtends an angle of 120 at the centre. Find the area of the corresponding minor segment of the circle. Use = 22/7 and 3= 1.73. 24. An open metal bucket is in the shape of a frustum of a cone of height 21 cm ith radii of its lo er and upper ends as 10 cm and 20 cm respectivel . Find the cost of milk hich can completel fill the bucket at Rs. 30 per litre. [Use = 22/7] 25. Point P( , 4) lies on the line segment joining the points A( 5, 8) and B(4, 10). Find the ratio in hich point P divides the line segment AB. Also find the value of . 26. Find the area of the quadrilateral ABCD, hose vertices are A( 3, 1), B ( 2, 4), C(4, 1) and D (3, 4). 27. From the top of a vertical to er, the angles of depression of t o cars, in the same straight line ith the base of the to er, at an instant are found to be 45 and 60 . If the cars are 100 m apart and are on the same side of the to er, find the height of the to er. [Use 3= 1.73] Downloaded from www.padhle.in

28. T o dice are rolled once. Find the probabilit of getting such numbers on the t o dice, hose product is 12. EC I N D 29. Prove that the tangent at an point of a circle is perpendicular to the radius through the point of contact. 30. The first and the last terms of an A.P. are 8 and 350 respectivel . If its common difference is 9, ho man terms are there and hat is their sum? 31. A train travels 180 km at a uniform speed. If the speed had been 9 km/hour more, it ould have taken 1 hour less for the same journe . Find the speed of the train. 32. In the given figure, three circles each of radius 3.5 cm are dra n in such a a that each of them touches the other t o. Find the area enclosed bet een these three circles (shaded region). [Use = 22/7 ] Downloaded from www.padhle.in

33. Water is flo ing at the rate of 15 km/hour through a pipe of diameter 14 cm into a cuboidal pond hich is 50 m long and 44 m ide. In hat time ill the level of ater in the pond rise b 21 cm? 34. The angle of elevation of the top of a vertical to er from a point on the ground is 60 . From another point 10 m verticall above the first, its angle of elevation is 30 . Find the height of the to er. Downloaded from www.padhle.in

CBSE Ma h 2011 Sol ions: 1. Gi en ad a ic e a ion: 2 3 m(m+3)=0. I can be i en a 2 (m+3) m m(m+3)=0 2 (m+3) +m m(m+3)=0 - (m + 3) + m - (m + 3) = 0 ( + m) - (m + 3) =0 The efo e, = - m and = (m + 3) a e he e oe of he e a ion. 2. Le a be he fi e m of he A.P. and d be he common diffe ence. n h e m of an A.P. = an=a+(n 1)d a20 a15= [a + (20 1)d] [a + (15 1)d] = 19d 14d = 5d =5 3 = 15. Downloaded from www.padhle.in

3. Con ide he iangle in he ci cle. i.e., OPQ: OP = OQ (a bo h a e adii of he ci cle) ⇒ ∠OQP = ∠OPQ (E al ide of a iangle ha e e al angle o o i e o hem) ∠POQ + ∠OPQ + ∠OQP = 180 (ba ed on he angle m o e of iangle ) ⇒ 70 + 2∠OPQ = 180 ⇒ 2∠OPQ = 180 70 = 110 ⇒ ∠OPQ = 55 The angen i e endic la o he adi h o gh he oin of con ac . ∴ ∠OPT = 90 ⇒ ∠OPQ + ∠TPQ = 90 ⇒ 55 + ∠TPQ = 90 ⇒ ∠TPQ = 90 55 = 35 . 4. The angen o a ci cle i e endic la o he adi h o gh he oin of con ac . ∴ ∠ABO = ∠ACO = 90 In ad ila e al ABOC, acco ding o he angle m o e : ∠ABO + ∠BOC + ∠ACO + ∠BAC = 360 ⇒ 90 + ∠BOC + 90 + 40 = 360 ⇒ ∠BOC + 220 = 360 ⇒ ∠BOC = 360 220 = 140 . Downloaded from www.padhle.in

5. Le he adi of he ci cle be a cm The diame e of he ci cle become o ime he adi 2 a cm = 2a cm. The efo e, he ide of he ci c m c ibing a e = Diame e of he ci cle = 2a cm. ∴ Pe ime e of he ci c m c ibing a e = 4 2a cm = 8a cm. 6. In a c be, he la ge igh ci c la cone ha a diame e ha i e al o he ide of a c be. Simila l , i heigh i e al o he ide of he c be. So, if he adi of he cone i cm, ∴ 2 = 4.2 cm. ⇒ = 4.22 cm = 2.1 cm. 7. In he fig e abo e AB i he o e and C i he oin on he g o nd 25 m a a f om he foo of he o e in ch a a ha ∠ACB = 45 . The efo e, in ABC, an 45 = AB AC ⇒1 = AB 25 ⇒AB = 25 m The efo e, he heigh of he o e AB i 25 m. Downloaded from www.padhle.in

8. P(a2, 4) i he mid- oin of he line egmen joining oin A ( 6, 5) and B ( 2, 3). ∴ Coo dina e of he oin P can be fo nd b ing he mid oin fo m la. P = ( 6+( 2)2, 5+32)= ( 82, 82) = (-4, 4) Gi en ha P(a2,4) i he mid- oin , he efo e (-4, 4) = (a2,4) ∴ a2 = - 4 ⇒a = 8 So, he al e of a i 8. 9. Gi en ha he coo dina e of he oin A and B a e ( 6, 7) and ( 1, 5) e ec i el . AB = ( 6 ( 1))2+(7 ( 5))2 = ( 6+1)2+(7+5)2 = ( 5)2+(12)2 = 25+144 = 169 = 13 ∴ 2AB = 2 13 = 26. Hence, he di ance 2AB i 26 ni . Downloaded from www.padhle.in

10. N mbe of ace ca d in a deck of 52 la ing ca d i 4. ∴ N mbe of non-ace ca d in he ack = 52 4 = 48 A me E a he e en of d a ing a non-ace ca d. To al n mbe of o ible o come = 52 N mbe of fa o able o come = 48 P(E) = N mbe of fa o able o come To al n mbe of o come = 4852= 1213. 11. ad a ic e a ion m ( -7)+49=0 ha o e ha e o find he al e of m o ha he e al oo . Com a e abo e e a ion i h he gene al ad a ic e a ion , e ge a=m, b=-7m, c=49 A e kno if he ad a ic e a ion ha e al oo hen he di c iminan ill be e al o 0 i.e ⇒ 4, 0 ⇒ im lie Downloaded from www.padhle.in

12. The o digi n mbe ha a e di i ible b 6 a e 12, 18, 24, , 96 He e, Fi e m = a = 12 Common diffe ence d = 18 12 = 6, La e m = an = 96 Since, n h e m i gi en b : an = a + (n - 1)d 96 = 12 + (n 1)6 96 12 = (n 1)6 84 = 6n 6 84 + 6 = 6n 90 = 6n n = 15 The e a e 15 o-digi n mbe di i ible b 6. 13. He e, ABCD i a ad ila e al. Le a me ha he ad ila e al ABCD o che he ci cle a oin P, Q, R and S. Th , AB, BC, CD and AD a e angen o he ci cle P o e of angen a e ha \"Tangen d a n f om an e e nal poin o a ci cle a e eq al in leng h\". Th , e ge AP = AS Le AP = AS = Al o, BP = BQ, CQ = CR and DR = DS Downloaded from www.padhle.in

Since AP = Th , BP = AB - AP = No , BQ = BP = Al o, CQ = CB - BQ = No , CR = CQ = Al o, DR = CD - CR = No , DS = DR = Finall , AD = AS + DS AD = + 5 - AD = 5 cm Th , leng h of ide AD i 5cm 14. Con c ion gi en belo : Downloaded from www.padhle.in

15. ABCD i a a e of ide 14 cm. ∴ ∴ AB = BC = CD = AD = 14 cm The e a e o emici cle in he a e of adi = 14/2 = 7 cm Pe ime e of he haded egion = BC + AD + Pe ime e of emi-ci cle APB + Pe ime e of emi-ci cle CPD Pe ime e of emici cle APB = 1/2 2 adi = 1/2 2 7 = 22/7 7 = 22 cm. Pe ime e of emici cle CPD = 1/2 2 adi = 1/2 2 7 = 22/7 7 = 22 cm. ∴ Hence, he e ime e of he haded egion = 22 + 22 + 14 + 14 = 72 cm. Downloaded from www.padhle.in

16. 17. Coo dina e of he gi en oin a e A (3, 1) and B (11, ). Di ance be een he oin A and B = AB = 10 ni ∴(11 3)2+[ ( 1)]2 = 10 ⇒(64)+( +1)2=10 ⇒ 64 + ( +1)2=100 ⇒( +1)2 = 100 64 ⇒( +1)2= 36 ⇒( + 1) = 6 ⇒( + 1) = 6 o ( + 1) = - 6 ⇒ =6 1=5o 6 1= 7 Hence, he al e of = 5 o 7. Downloaded from www.padhle.in

18. A me he e en of d a ing a m l i le of 5 a E. Since he e a e 40 icke , he o al n mbe of o come = 40 The o come of e en E a e 5, 10, 15, 20, 25, 30, 35 and 40. So, he n mbe of fa o able o come of e en E = 8 P obabili ha he elec ed icke ha a n mbe ha i a m l i le of 5: P(E) = N mbe of fa o able o come To al n mbe of o come = 840 = 15 Hence, he obabili of d a ing a icke hich i a m l i le of 5 = 15. 19. Com a e he gi en ad a ic e a ion i h he anda d fo m of ad a ic e a ion i.e., a 2 + b + c = 0. a = 1, b = 35, c = 10 D =b2 4ac = ( 35)2 4 1 10 = 45 40 =5 ∴ = ( b D)/2a = 35 5/2 1 = 35+ 5/2 OR 35- 5/2 Downloaded from www.padhle.in

20. A me he fi e m of he A.P. a 'a' and he common diffe ence a 'd'. 4 h e m of he A.P. = 4= 9. The m of he i h and hi een h e m i 40 ⇒ 6+ 13= 40. If 4= 9 ⇒a + (4 1)d = 9 [ n = a + (n 1)d] ⇒a + 3d = 9 ---------- (1) 6+ 13= 40 ⇒ a + (6 - 1)d + a + (13 - 1)d = 40 ⇒ a + 5d + a + 12d = 40 ⇒2a + 17d = 40 ---------- (2) Sol ing he linea e a ion (1) and (2) F om (1): a = 9 3d S b i ing he al e of a in (2): 2 (9 3d) + 17d = 40 ⇒ 18 + 11d = 40 ⇒ 11d = 22 ⇒d=2 ∴a=9 3 2=3 Hence, he gi en A.P. = a, a + d, a + 2d , he e a = 3 and d = 2. So, he A.P. i 3, 5, 7, 9 Downloaded from www.padhle.in

21. Since he iangle i ci c m c ibing he ci cle, he ci cle i he inci cle of he iangle. All he ide of he iangle a e angen o he ci cle. Gi en RT= 9cm and QT = 12cm Since RT and RU a e angen o a ci cle f om a oin R, he a e e al Th RT = RU = 9cm imila l , QT = QS = 12cm Le PS = PU = cm adi of ci cle, = 6cm ide of iangle a e: PQ = (12+ ) cm QR = 12+9 = 21cm PR = (9+ ) cm Gi en ha a ea = 189cm Downloaded from www.padhle.in

h 21+ = 31.5 ⇒ = 31.5 - 21 ⇒ = 10.5 Side of iangle a e: PQ = (12+ ) cm = 12+10.5 = 22.5cm QR = 12+9 = 21cm PR = (9+ ) cm = 9+10.5 = 19.5cm 22. Downloaded from www.padhle.in

23. The adi of he ci cle i 14 cm. The b end angle i 120 deg ee a he cen e. The a ea of mino ec o i A ea of iangle i The fo m la o find he a ea of mino egmen i Downloaded from www.padhle.in

The efo e he a ea of mino egmen of he ci cle i 120.46 cm . 24. Heigh of f m of cone, h = 21 cm Radi of lo e end, = 10 cm Radi of e end, R = 20 cm Vol me of f m of cone = 1/3 h(R2 + 2 + R ) Vol me of milk e i ed o fill he b cke = 1/3 22/7 21 [(20)2+(10)2+20 10]cm3 = 22 (400 + 100 + 200) = 22 700 = 15400 cm3 = 15400/1000 li e = 15.4 li e Co of milk = R 30 e li e Hence, he co of milk fo filling he b cke i R (30 15.4) i.e. R 462. 25. Downloaded from www.padhle.in

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AB i he o e The angle of de e ion of o ca in he ame aigh line i h he ba e of he o e a an in an a e fo nd o be 45 deg ee and 60 deg ee i.e. ∠ACB = 60 and ∠ADB = 45 The ca a e 100 m a a i.e. CD = 100 m Le CB be So, DB = DC+CB = 100+ In ABC ---1 In ABD ------2 Wi h 1 and 2 So, Downloaded from www.padhle.in

Hence he heigh of o e i 236.602 m 28. Li o he am le ace of he gi en e e imen S = (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) The efo e, n (S) = 36 A me he e en of ge ing he n mbe ho e od c i 12 a E. E = (2, 6), (3, 4), (4, 3), (6, 2) The efo e, n (E) = 4 Hence, he e i ed obabili = n(E)n(S)= 436= 19. 29. Gi en : A ci cle C (0, ) and a angen l a oin A. To o e : OA ⊥ l Con c ion : Take a oin B, o he han A, on he angen l. Join OB. S o e OB mee he ci cle in C. P oof: We kno ha , among all line egmen joining he oin O o a oin on l, he e endic la i ho e o l. OA = OC (Radi of he ame ci cle) Downloaded from www.padhle.in

No , OB = OC + BC. ∴ OB > OC ⇒ OB > OA ⇒ OA < OB B i an a bi a oin on he angen l. Th , OA i ho e han an o he line egmen joining O o an oin on l. He e, OA ⊥ l 30. n h e m of AP = a+(n-1)d he e a i he fi e m , d i he common diffe ence. 350= 8+(n-1)9 342=(n-1)9 n-1 = 38 n = 39 N mbe of e m = 39. S m = (39/2)(8+350) = (39/2)(358) = 39*179 = 6981 Downloaded from www.padhle.in

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32. Con c ion:- D a a line connec ing he cen e of each ai of adjacen ci cle c ea ing a e ila e al iangle of ide leng h 7 cm. No , A ea of iangle = The e a e fo a ea in ide of hi e ila e al iangle. [Th ee 60 ci cle ec o .] A ea of ec o = A ea of h ee ec o = Th , a ea of Shaded o ion = a ea of e ila e al iangle - a ea of h ee ec o Downloaded from www.padhle.in

33. adi =7cm=7/100 h=21cm=21/100m ol me of i e = ol me of ond 2h=l b h 22/7 7/100 7/100 h=50 44 21/100 22/100 7/100 h=50 44 21/100 h=50 44 21 100 100 22 7 100 h=30000m h=30km heigh of i e = di ance a elled b a e ime=di ance/ eed ime=30/15 ime=2 ho 34. Downloaded from www.padhle.in

Le he heigh of he o e be h cm. No , In And, In Since, AB = CD, SO, e a ion (2) become , E a ing e a ion (1) and (3), e ge , Hence, he heigh of he o e ill be 15 cm. Downloaded from www.padhle.in