CHALLENGING QUESTIONS FOR VARIOUS MATH COMPETITIONS

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1 CHALLENGING QUESTIONS FOR VARIOUS MATH COMPETITIONS. Show that there are ifiitely may positive primes.. Solve the followig equatio : Prove that the equatios (p + ) + py + 5 = 0 ad p (p + ) y 8 = 0 caot have a ifiite umber of solutios for ay real values of p. 4. The speed of a trai is reduced to 5 of its origial speed because of some defect i its egie, after it 6 covers a distace of 40 km with a uiform speed. As a result, the trai reaches its destiatio late by 4 miutes. If this defect i the egie would occur after coverig a distace of 0 km more, the the trai would have reached its destiatio 8 miutes late. Fid the speed of the trai ad the distace of the jourey. 5. Solve the followig system of equatio i ad y : (a) a + by = (b) (a b) + (a + b)y = a ab b ab b ay a b (a + b) ( + y) = a + b 6. Fid the values of a ad b so that a + b is divisible by If ad are the zeroes of the quadratic polyomial f() = a + b + c, the evaluate : (a) (b) (c) 8. Solve the followig equatio for : (a) ; (c) (d) (d) (b) 5 5 (e) 4 4 MATHEMATICS X CHALLENGING QUESTIONS ; 0 4 (e) 4 40 (f) ; Fid the least ad the greatest values of for all real value of. 0. If the sum of the roots of the equatio p is zero, show that the product of the roots is a b ( ). a b

2 . Show that the roots of the equatio : ( a)( b) ( b)( c) ( c)( a) 0 are always real ad these caot be equal uless a = b = c.. A swimmig pool is filled with three pipes with uiform flow. The first two pipes operatig simultaeously, fill the pool i the same time durig which the pool is filled by the third pipe aloe. The secod pipe fills the pool five hours faster tha the first pipe ad four hours slower tha the third pipe. Fid the time required by each pipe to fill the pool seperately.. The product of two umbers is 6. Whe their sum is added to the sum of their squares, we get 46. Fid the umbers. 4. The vertices of a triagle ABC are A(5, 6), B(a, ) ad C(8, b). Fid the values of a ad b so that the side AB is parallel to y-ais ad BC is parallel to the -ais. Also fid ABC. 5. The co-ordiates of the opposite vertices of a square are (0, ) ad (4, ). Fid the co-ordiates of the remaiig vertices. 6. A equilateral triagle has oe verte at (, 4) ad aother at (, ). Fid the coordiates of the third verte. 7. If the vertices of a triagle have itegral coordiates, prove that the triagle caot be equilateral. 8. If a a b b is the arithmetic mea betwee a ad b, fid the value of. 9. If p, q, r are i A.P., prove that,, are also i A.P. q r r p p q 0. If the m th term of a A.P. is ad the th term is m, show that the sum of m terms is ( ). m. The sum of the first p, q, r terms of a A.P. are a, b, c respectively. Show that a b c ( q r) ( r p) ( p q) 0. p q r. Each of the A.P. s, 4, 6, 8,... ad, 6, 9,,... is cotiued to 00 terms. How may terms of these two A.P. s are idetical?. The iterior agles of a polygo are i A.P. The smallest agle is 0 ad the commo differece is 5. Fid the umber of sides of the polygo. 4. The sum of terms of two A.P. s are i the ratio ( ) : ( + 5). Fid the ratio of their 0 th terms. 5. Two cars start together i the same directio from the same place. The first goes with uiform speed of 0 km/h. The secod goes at a speed of 8 km/h i the first hour ad icreases the speed by km i each succeedig hour. After how may hours will the secod car overtake the first car if both cars go ostop? 6. I the give figure, the side AC of ABC is produced to G so that CG = AC. If D is the midpoit of BC, GD is produced to meet AB i E ad DF is draw parallel to AB, prove that DE = GE. B E A D 446 CHALLENGING QUESTIONS MATHEMATICS X F C G

3 7. Prove that three times the sum of the squares of the sides of a triagle is equal to four times the sum of the squares of the medias of the triagle. 8. I trapezium ABCD, AB DC ad AB = DC. EF draw parallel to AB cuts AD i F ad BC i E such that BE. Diagoal DB itersects EF at G. Prove that AB 7. EC 4 FE 0 9. If is a acute agle ad ta + cot =, fid the value of ta 0 + cot Fid a acute agle, whe cos si. cos si. If cosec si = m ad sec cos =, prove that (m ) / + (m ) / =.. A equilateral triagle is iscribed i a circle of radius 4 cm. Fid its side.. If a si + b cos = c, the show that a cos b si a b c 4. Prove that : si 5 si 0... si 85 si ( y) 5. If ad y are two uequal real positive quatities, show that the equatios (i) si ad 4y (ii) cos are both impossible. 6. If ( + si ) ( + si y) ( + si z) = ( si ) ( si y) ( si z), prove that each is equal to ± cos cos y cos z. 7. Two statios due south of a leaig tower which leas towards the orth are at distaces a ad b from its foot. If be the elvatios of the top of the tower from these statios, prove that its icliatio to b cot a cot the horizotal is give by cot. b a 8. From a widow, h metre high above the groud, of a house i a street, the agles of elevatio ad depressio of the top ad the foot of aother house o the opposite sides of the street are ad respectively. Show that the height of the opposite house is h ( + ta cot ). 9. At the foot of a moutai the elevatio of its summit is 45, after ascedig 000 m towards the moutai up a slope of 0 icliatio, the elevatio is foud to be 60. Fid the height of the moutai. 40. A ladder rests agaist a wall at a agle to the horizotal. Its foot is pulled away from the wall through a distace a so that it slides a distace b dow the wall makig a agle with the horizotal. Show that a cos cos. b si si 4. If the diameter of the cross-sectio of a wire is decreased by 5%, how much percet will the legth be icreased so that the volume remais the same? 4. The three vertices of a rhombus lie o a circle with cetre O. If O is the fourth verte of the rhombus ad if the area of the rhombus is 50 cm, fid the radius of the circle. MATHEMATICS X CHALLENGING QUESTIONS 447

4 4. I the give figure, a semi-circle is draw with lie-segmet PR as a diameter. Q is the mid-poit of the lie-segmet PR. Two semi-circles with lie-segmets PQ ad QR as diameters are draw. A circle is draw which touches the three semi-circles. If PR = 4 cm, fid the area of the shaded regio. P B Q C R 44. A copper wire 4 mm i diameter is evely woud about a cylider whose legth is 4 cm ad diameter 0 cm so as to cover the whole surface. Fid the legth ad weight of the wire assumig the specific gravity to be 8.88 gm/cm. 45. A coical vessel of radius 6 cm ad height 8 cm is completely filled with water. A sphere is lowered ito the water ad its size is such that whe it touches the sides, it is just immersed as show i the give figure. What fractio of water overflows? O 46. The height of a right circular coe is trisected by two plaes draw parallel to the base. Show that the volumes of the three portios startig from the top are i the ratio : 7 : A solid metallic cylider with base radius 6 cm ad height 4 cm is give. O oe side of the cylider, a hemisphere shape of solid ad o the other side, a coical shape of solid are removed. If the base radius of the hemisphere is equal to the radius of the cylider ad the base radius of the coe is cm ad its height is 5 cm, fid the volume ad the total surface area of the remaiig solid. 4 cm 5 cm 6 cm cm 6 cm 8 cm 448 CHALLENGING QUESTIONS MATHEMATICS X

5 48. The followig table shows the distributio of the life-time of 50 radio tubes. Life-Time (i hrs.) No. of tubes Assum that the class frequecies are uiformly distributed withi the correspodig classes, fid the percetage of tubes that have life time : (a) greater tha 760 hours. (b) less tha 50 hours. 49. If 4 ad y 6, fid the probability that y Usig the digits,, 5 ad 6 oly oce, all possible 4-digit umbers are formed. A umber is selected at radom. What is the probability that the selected umber is eactly divisible by 5? ANSWERS. = 4. speed of trai = 60 km/hr, distace = 60 km a b 5. (a), y a b a b 7. (a) b ac a (b) abc b a ab (b) a b, y 6. a =, b = 7 a b (c) b ac ac MATHEMATICS X CHALLENGING QUESTIONS 449 (d) 8. (a) =, (b) 5 or (c) = 9 (d) 5 abc b c 9. least value = 5 ad greatest value = 4. 5 hr, 0 hr, 6 hr. 7 7 ad 5. (, 0) ad (, 4) or 7 ad 9 4. a = 5, b = ; ABC = 90 (e) ( b ac) a c 4 a,, (e) = (f), or , or, 8. = 0. = 66. = : 4 5. t = 9 hours = cm m % early 4. r = 0 cm 4. 0 cm cm, 46.4 gm 45. : cm, (67 4) cm 48. (a) 4.57% (b).4%

6 HINTS/SOLUTIONS TO CHALLENGING QUESTIONS. Let if possible, there are fiite umber of positive primes p, p, p,..., p such that p < p < p <... < p. Let p = + p p p...p. Clearly, p p p...p is divisible by each of p, p, p,..., p. p = + p p p...p is ot divisible by ay oe of p, p, p,..., p. p is a prime or it has prime divisors other tha p, p, p,..., p which meas there eists a positive prime differet from p, p, p,..., p, which cotradicts that there are fiite umber of positive primes. Hece, the umber of positive primes is ifiite.. Give equatio ca be rewritte as : Takig LCM ad simplifyig gives ( ) ( ) For ifiite umber of solutios, ( p ) p 5 p ( p ) 8 = As. p p from the first two ratios, we get 6( p ) p 6 6, which is ot possible. Hece, p is ot real ad so the two equatios caot have ifiite umber of p p solutios. y 4. Let origial speed of the trai is km/h ad distace be y km. The, the scheduled time, t hr...() 40 y 40 By give first coditio, we have the total time take hr Also, this distace would have bee covered by the trai i t hr t hr y 40 y hece, we have : t [ usig ()] CHALLENGING QUESTIONS MATHEMATICS X

7 Simplifyig, we get y + 40 = 0...() Agai, i secod case, 70 y 70 8 Total time hr t hr t hr Simplifyig, we get y = 0...() Solvig () ad (), we get = 60, y = 60. Hece, speed of trai = 60 km/hr ad distace = 60 km. 5. (a) a by 0 ab b ay 0 a b By cross multiplicatio, we have y ab ab a a b b b a ( ) a b ( ) a b a b y a ab a b b a b a b a b a ( a b ) ( ) ad y b a b a b a b a b a b a b ad y a b a b (b) Give system of equatios may be writte as (a b) + (a + b) y (a ab b ) = 0 (a + b) + (a + b) y (a + b ) = 0 Now use cross-multiplicatio to get = a + b ad ab y a b y hr 0 6. Here, remaider = 0. 4 O dividig, we get +) a + b ( Quotiet = ad remaider = (a ) + (b 7). Remaider = 0 (a ) + (b 7) = a + b + a 0 ad b 7 = ( a ) + b a = ad b = ( a ) + ( b 7) MATHEMATICS X CHALLENGING QUESTIONS 45

8 7. here, b c, a a (a) ( ) b c b ac a a a (b) ( ) ( ). Now put values of + ad. (c) (d) (e) ( ) ( ) ( ). ( ) ( ). Now put values of + ad. Now put values of + ad. 4 4 ( ) [( ) ] ( ). Now put values of + ad. 8. (a) Give equatio (b) ( ) 4 0 =, ( ) ( ) = 5 or = 5 (c) Squarig both sides, 5 8 ( 5)( 8) 7 ( 5)( 8). Agai squarig, we get ( 5)( 8) or 9. but = 4 do ot satisfy the give equatio. = 9. (d) From the give equatio, we have CHALLENGING QUESTIONS MATHEMATICS X

9 Let y y y 0 y 0 or y. 0 or = 0 or = 0 = ± or =, As. (e) From the give equatio, we have ( ) = 0 40 = 0 40 = 0 Let y y y 40 0 y + 5y 48 y 40 = 0 (y + 5) (y 4) = 0 5 y or y 4 5 or 4 = = ( as > 0 for all real ) ( f ) Let y Give equatio reduces to 6 y 40 y + 9 = 0 (4y ) (4y 9) = 0 9 y or y or 4 4 or =, or As. 8 4 MATHEMATICS X CHALLENGING QUESTIONS 45

10 Let y Cross multiplyig ad rearragig the terms, we get ( y) (4 y) (9 y) 0..() here, discrimiat, D = (4 y) 4 ( y) ( 9 y) = 60 8y 8y 0. 8( y y 0) 8( y 5)( y 4) Now, sice is real, hece two roots of equatio () are real. Hece, D 0. 8 (y + 5) (y 4) 0 (y + 5) (y 4) 0 either y ad y () or y 5 0 ad y () from eq. (), y 5 ad y 4 5 y 4 from eq. (), y 5 ad y 4, which is ot possible. 5 y 4. least value = 5 ad greatest value = 4. p a b ( b) ( a ) p( a )( b) simplifyig we get p ap bp abp a b ( ) 0 Let are roots of this equatio. ap bp ( a b) p the, p p Sice 0 ( a b) p 0 p a b...() ( ) Now, abp a b a ab b ab ( a b). a b [ usig ()] p p ( ) a b 454 CHALLENGING QUESTIONS MATHEMATICS X

11 . Simplifyig the give equatio, we get ( a b c) ( ab bc ca) 0 discrimiat, d [ ( a b c)] 4 ( ab bc ca) 4( a b c ab bc ca) (a b c ab bc ca) [( a b ab) ( b c bc) ( c a ca)] [( a b) ( b c) ( c a) ] 0, beig the sum of three perfect squares. Hece the roots are real. Now, [( a b) ( b c) ( c a) ] will vaish oly whe a = b = c. So, the roots will be equal, whe a = b = c.. Let V be the volume of the pool ad the umber of hours required by the secod pipe aloe to fill the pool. Parts of the pool filled by the first, secod ad third pipes i oe hour are respectively V V V, ad. 5 4 Let the time take by first ad secod pipes to fill the pool simultaeously be t hours, which is same time take by third pipe. V V V. t. t Simplifyig, we get = 0,. = 0 ( ca t be egative) Required timig for first, secod ad third pipe are 5 hours, 0 hours ad 6 hours respectively.. Let the umbers be ad 6. The, Let y. The y + y 7 = 0 y = 7 or y = 6 MATHEMATICS X CHALLENGING QUESTIONS 455

12 ad, or 7. Required umbers are ad or 7 ad 9. As. 4. ABC is clearly a right-agled triagle with ABC = 90. The abscissae of A ad B will be equal, sice AB is parallel to the y-ais ad the ordiates of B ad C will be equal, sice BC is parallel to the -ais. 5. Here, AB = ( 0) + (y ) = + y y + BC = ( 4) + (y ) = + y 8 6y + 5 AC = (4 0) + ( ) = 0. Sice, ABCD is a square, we have AB = BC AB = BC + y y + = + y 8 6y + 5 y = 6...() Now, B = 90 AB + BC = AC + y y y 8 6y + 5 = 0 + y 4 4y + = 0 + (6 ) 4 4 (6 ) + = 0 [ usig eq. ()] 4 + = 0 = or = whe =, from eq. (), y = 0 whe =, from eq. (), y = 4 co-ordiates of the other vertices are (, 0) ad (, 4). As. 6. Here, BC = AB = AC BC = AB = AC Now, BC = ( ) + ( 4) = 6 AB = ( ) + (y 4) AC = ( + ) + (y ) D C(4, ) A(0, ) B (, y) A(, y) B(, 4) C(, ) BC = AB ( ) + (y 4) = 6 + y 6 8y = 0...() BC = AB ( + ) + (y ) = 6 + y + 4 6y = 0...() Subtractig () from (), 0 + y = 0 y = y = 0 y = 6 5 Usig value of y i eq. (), we get (6 5 ) 6 8(6 5 ) CHALLENGING QUESTIONS MATHEMATICS X

13 7 5 y 6 5 The third verte has the coordiates as , or, As. 7. Let A (, y ), B(, y ) ad C(, y ) be the vertices of a triagle ABC where i, y i, ; i =,, are itegers. The, area of ABC is give by (y y ) + (y y ) + (y y ) = A ratioal umber If possible, let ABC be a equilateral. The its area is give by (side) (A positive iteger) 4 4 = A irratioal umber. This is a cotradictio to the fact that the area is a ratioal umber. Hece, the triagle caot be equilateral. 8. Accordig to questio, a b a b a b ( a b ) ( a b)( a b ) Simplifyig, we get ( a b )( a b) 0 a b 0 as a b a a a a b b b b 9. Sice p, q, r are i A.P., we have 0 = 0 As. q p r...() Now,, ad will be i A.P. q r r p p q if i.e. if r p q r p q p q r r p ( q r)( p q) i.e. if (q + r) (p + q) = (r + p) (p + q + r) Simplifyig, we get i.e. if p + r = q, which is true by eq. (). MATHEMATICS X CHALLENGING QUESTIONS 457

14 0. Let a be the first term ad d be the commo differece of the give A.P. The, a m a ( m ) d...() ad a a ( ) d m m...() Subtractig eq. () from eq. (), we get m ( m ) d ( m ) d d m m m Puttig d i eq. (), we get a m m m Now, Sm [ a ( m ) d] Usig a d i S m, we get S m m ( m ). Let A be the first term ad D be the commo differece of the give A.P. The, p a a [ A ( p ) D] [ A ( p ) D]...() p q b b [ A ( q ) D] [ A ( q ) D]...() q r c c [ A ( r ) D] [ A ( r ) D]...() r multiplyig eq. (), () ad () by (q r), (r p) ad (p q) respectively ad addig, we get the desired result.. For the first A.P., 4, 6,... we have a 00 = + (00 ) = 400. For the secod A.P., 6, 9,... we have a 00 = + (00 ) = 600. Observe that the idetical terms i the two A.P. s will be upto 400 oly, sice 400 < 600. Clearly, the idetical terms of two A.P. s will be the multiples of = 6. Hece, terms will be 6,, 8,... Let th term of this A.P. be less tha or equal to 400. i.e. a ( ) Sice must be a positive iteger, = CHALLENGING QUESTIONS MATHEMATICS X

15 . We kow sum of all the iterior agles of a polygo of sides is ( ) 80. Clearly, here a = 0, d = 5. S [ a ( ) d] [0 ( ) 5 ] [5 5 ] accordig to give questio, [5 5 ] ( ) 80 Simplifyig, we get or 9. whe = 6, the 6th agle = 0 + (6 ) 5 = 95 > 80, which is ot possible. = Let a ad a be the first terms ad d ad d be the commo differeces of the two A.P. s. Let S ad S be the sums of the two A.P. s respectively. The, S [ a ( ) d] ad S [ a ( ) d ] S a ( ) d S a ( ) d 5 Now, ratio of the 0 th terms of the two A.P. s a (0 ) d a 9d a 58 d a (0 ) d a 9d a 58d from () ad (), we observe Puttig = 59 i eq. (), we get...[give]...() a 58d 59 9 = 9 : 4 As. a 58d Suppose the secod car overtakes the first car after t hours. The, the two cars travel the same distace i t hours. Distace travelled by the first car i t hours = 0 t km. Distace travelled by the secod car i t hours = sum of t terms of a A.P. with a = 8, d. t t 8 ( t ) ( t ) 4 t accordig to questio, 0 t ( t ) 4 t 9 t = 0 = t = 0 or t = 9 t = 9 hours ( t 0) MATHEMATICS X CHALLENGING QUESTIONS ()

16 6. Clearly AF = FC = AC also, GFD ~ GAE GF GD FD GA GE AE...() GD GF...() GE GA Now, GD = GE DE GF = CG + CF = AC + AC = AC ad, GA = CG + CA = AC + AC = AC Hece, from eq. (), we have GE DE AC GE AC DE DE GE GE DE GE (AA similarity) 7. Accordig to Apolloius theorem, if AD is a media of ABC, the AB + AC = BC + AD....() By symmetry, we get AB + BC = AC +.BE...() ad, AC + BC = AB +.CF...() addig (), () ad (), we get (AB + BC + AC ) = (AB + BC + AC ) + (AD + BE + CF ) (AB + BC + AC ) = (AD + BE + CF ) (AB + BC + AC ) = 4 (AD + BE + CF ) 460 CHALLENGING QUESTIONS MATHEMATICS X

17 8. Clearly, DFG ~ DAB (AA similarity) DF FG...() DA AB I trapezium ABCD, EF AB DC AF BE DF EC AF AF DF 4 DF 4 AF+DF 7 AD 7 DF 4...() DF 4 DF 4 AD 7 from () ad (), FG 4 FG 4 AB...() AB 7 7 Agai, BEG ~ BCD BE EG BC CD EG 7 CD ( AA similarity) BE (give) EC 4 D A B F E G BE EC 4 EC 4 BC 7 EC 4 BE BE BE EG = 7 CD = 7 AB [ CD = AB (give)] EG = 6 AB...(4) 7 addig () ad (4), we get 4 6 FG EG AB AB AB 7 EF AB 7 FE 0 9. We have, ta + cot = ta ta ta ta ta ta + = 0 (ta ) = 0 ta = 0 ta = = ta 45 = 45 ta 0 + cot 0 = ta cot 0 45 = () 0 + () 0 = + =. As. C MATHEMATICS X CHALLENGING QUESTIONS 46

18 0. cos si. cos si cos si cos cos si cos ta ta ( dividig umerator ad deomiator by cos ) ta = (o comparig two sides) ta = ta 60 = 60 As.. We have, cosec si = m ad sec cos = si m ad cos si cos si cos m ad si cos cos m si ad si cos Now, substitute the values of m ad i LHS to get the desired result.. Here, OA = OB = OC = 4 cm. Draw OD BC ad OE AC. The OD bisects BC ad OE bisects AC. Hece, BD = DC = CE = EA ( BC = CA) A Clearly, ODC OEC (RHS cogruece rule) OCD = OCE (cpct) But DCE = 60 OCD = OCE = 0 i.e. OC is the bisector of BCA. Similarly, OB ad OA are bisectors of ABC ad BAC respectively. BD BD Now, I OBD, cos 0. OB 4 BD BC = BD = 4 cm.. Cosider, (a si + b cos ) + (a cos b si ) = a si + b cos + ab si cos + a cos + b si ab si cos = a (si + cos ) + b (si + cos ) = a + b c + (a cos b si ) = a + b ( asi bcos c) 46 CHALLENGING QUESTIONS MATHEMATICS X B O E 0 D 0 (a cos b si ) = a + b c a cos b si = a b c C

19 4. LHS = si 5 + si si 40 + si (90 40 ) si (90 5 ) + si 5 si 0... si 40 cos cos 5 (si 5 cos 5 ) (si 0 cos 0 )... (si 40 cos 40 ) terms 5. (i) we have, ( y) 4 y ( y) 0 if ad y are uequal. ( ) 4 y y ( y) 4y Sice, 0 si whe (ii) We have, i.e. si which is impossible, Now, ( ) cos, which is impossible, Sice, 0 cos whe We have ( + si ) ( + si y) ( + si z) = ( si ) ( si y) ( si z) multiplyig both sides by ( si ) ( si y) ( si z) ( si ) ( si y) ( si z) = [( si ) ( si y) ( si z)] cos. cos y cos z = [( si ) ( si y) ( si z)] ( si ) ( si y) ( si z) = ± cos cos y cos z. ( si ) ( si y) ( si z) = ( + si ) ( + si y) ( + si z) = ± cos cos y cos z. 7. Let AB be the leaig tower ad let C ad D be two give statios. h I AEB, we have ta h cot...() B h D C a A E b MATHEMATICS X CHALLENGING QUESTIONS 46

20 I CEB, we have ta h h cot a a...() I DEB, we have ta h h cot b b a from () ad (), we get h cot h cot a h cot cot b from () ad (), we get hcot hcot b h cot cot a b from (4) ad (5), we get cot cot cot cot b cot a cot rearrage ad simplify to get cot b a 8. Let B be the widow h m above the groud. D, the top ad C, the foot of a house. I DBE, we have ta y cot...() y h I BAC, we have ta y h cot...() y from () ad (), we get cot hcot h ta cot...() Height of the house = DC = DE + EC = ( + h) m = (h ta cot + h) = h ( + ta ta ). 9. Let C be the foot ad A be the summit of the moutai ABC such that BCA = 45 BAC = 45 AB = BC = h Draw DE BC ad DF AB. CE I DEC, cos 0 CE 000 CD m 500 m...() Also, DE si 0 DE m BF CD Now, AF = h BF = (h 500) m I AFD, DF AF h 500 cot 60 DF AF BC = CE + BE = CE + DF = 500 h h C B h m A m...()...(4)...(5) y E D 60 A D F B m C E h m 464 CHALLENGING QUESTIONS MATHEMATICS X

21 Now, I ABC, AB ta 45 BC 000 h 000 h BC ( ) h 000 h h m = 65 m As. 40. Let AB be the ladder leaig agaist vertical wall AC at a agle with the horizotal. Whe its foot B is pulled away from the wall at B, the its top A slides dow the wall to A so that BB = a ad AA = b, ABC = ad ABC =. A Let AB = AB = l. B a b y I ABC, we have si l...() cos l...() y I ABC, we have si l...() a cos l...(4) from () ad (), we have y b y b si si...(5) l l l a a from () ad (4), we have cos cos...(6) l l l cos cos a l a cos cos dividig (6) by (5), we get si si l b b si si 4. Let d be the origial diameter ad l be the origial legth of the wire. The, the origial volume of the wire d l. 4 Let the legth be icreased by %. The, the ew legth l 00 l l B MATHEMATICS X CHALLENGING QUESTIONS 465 l l b A y C

22 The, the ew volume 9 00 d l 0 00 accordig to give questio, 900 Simplifyig, we get required percet is 0.8% early d d l l Clearly, OA = AB = OB = OC = BC = r, the rhombus is divided ito two equilateral triagles OAB ad OBC, each of side r cm, by the diagoal OB. Now, area of OAB cm 4 r Area of rhombus OABC = 4 r cm = r r r = 0 cm 4. here, radius of the semi-circle PEQ = The radius of the semi-circle QFR 4 cm 6 cm. 4 r cm. Also, ABQ ACQ (SSS cogruece rule) AQB = AQC (cpct) also, AQB + AQC = 80 AQB = AQC = 90 Now, I ABQ, AB = AQ + BQ (r + 6) = ( r) + 6 ( AB = AE + EB = r + 6, AQ = DQ AD = r) r = 4 cm radius of the circle with cetre at A is 4 cm. E D A r cm r cm r cm P B Q C R Now, Area of the circle with cetre A = (4) cm = 6 cm Area of the semi-circle with cetre B = area of the semi-circle with cetre at C (6) cm 8 cm 466 CHALLENGING QUESTIONS MATHEMATICS X F C r r O r r B r A

23 Area of the semi-circle with cetre at Q () cm 7 cm Hece, the required area of the shaded regio = The area of the semi-circle with cetre at Q The area of the semi-circle with cetre at B The area of the semi-circle with cetre at C The area of the circle with cetre at A = (7 8 6 ) cm = 0 cm As. 44. Clearly, oe roud of wire covers 4 mm = (0.4 cm) i thickess of the surface of the cylider. o. of rouds to cover 4 cm Legth of wire i completig oe roud = r = 0 cm = 0 cm. Legth of wire i coverig the whole surface = 0 60 cm = 00 cm Now, radius of copper wire = mm = 0. cm Volume of wire = (0.) 00 cm = 48 cm So, weight of wire gm = 46.4 gm As I VOA, ta si r r I PVO, si r cm VO 5 8 r V = Volume of sphere 4 () cm 6 cm V = Volume of the water = Volume of the coe (6) 8 cm 96 cm Fractio of water that flows out V :V 6 : 96 : 8 As. 46. here, VL=LO h. clearly, VOA ~ VOA VO OA r h r = r VO OA r h also, VOA ~ VLC VO OA r h r = r VL LC r h r V = Volume of coe VCD r h h r h 7 A P r O V 6 cm O r B Q 8 cm A A C r r r L V O O D B B h h h MATHEMATICS X CHALLENGING QUESTIONS 467

24 V = Volume of the frustum ABDC ( r r r r ) h 4r r r 7 h V = Volume of the frustum ABBA = ( r r rr ) h required ratio = V : V : V = : 7 : 9 As. 47. Volume of cylider = (6) 4 cm = 504 cm r h 4r r 9 r h r h 9 7 Volume of coe which is removed () 5 cm 5 cm Volume of hemi-sphere which is removed (6) cm 44 cm Required volume of the remaiig solid = ( ) cm = 45 cm. Agai, CSA of the cylider = (6) 4 cm = 68 cm CSA of the coe 5 cm 4 cm CSA of the hemisphere = (6) cm = 7 cm Area of the metallic portio betwee the two circles at the top of the cylider = [ (6) () ] cm = 7cm Total area of the remaiig metallic portio is ( ) cm 48. (i) No. of tubes with the life-time greater tha 760 hours = (67 + 4) π cm 6 4 ( ) , sice o. of tubes ca t be fractioal required percetage 5 00 = (ii) o. of tubes with life less tha 50 hours required percetage CHALLENGING QUESTIONS MATHEMATICS X

25 49. Clearly 4, y 6 represets a rectagle with the coordiates (, ), (4, ), (, 6), (4, 6). Let s eamie the graph of the total regio i which the successful regio is shaded. The regios we are comparig are areas. Now, P(usuccessful outcome) Area of usuccessful regio = Area of total regio Area of FAE = Area of ABCD 4 6 P (successful outcome) D(, 6) C(4, 6) F(,4) A(, ) E(, ) 4 B(4, ) 50. Clearly uit digit ca be ay of the four digits give. Te s digit ca be ay of the remaiig three digits (as oe digit is used up at uits place ad repetitio is ot allowed). Similarly, hudreds ad thousads digit ca be ay of the remaiig two digits ad oe digit respectively. Total possible umbers formed = 4 = 4. A umber is divisible by 5 if uit digit is 5. Now, out of these 4 umbers, si umbers starts with digit at thousads place, out of which two umbers have 5 at their uit place. Clearly, total umber havig 5 at their uit place = = 6 (si umbers starts with digit 5 also). 6 Required probability 4 As. 4 y + =5 MATHEMATICS X CHALLENGING QUESTIONS 469