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Class 10 CBSE Board Exam 2015 : Mathematics

19 pages, 31 questions, 10 questions with responses, 28 total responses,    0    0
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SET-1 H$moS> Z . Series RLH 30/1 Code No. amob Z . narjmWu H$moS >H$mo C ma-nwp VH$m Ho$ _wI-n >na Ad ` {bIo & Roll No. Candidates must write the Code on the title page of the answer-book. H $n`m Om M H$a b| {H$ Bg Z-n _o _w{ V n > 11 h & Z-n _| Xm{hZo hmW H$s Amoa {XE JE H$moS >Z ~a H$mo N>m C ma -nwp VH$m Ho$ _wI-n > na {bI| & H $n`m Om M H$a b| {H$ Bg Z-n _| >31 Z h & H $n`m Z H$m C ma {bIZm ew $ H$aZo go nhbo, Z H$m H $_m H$ Ad ` {bI| & Bg Z-n H$mo n T>Zo Ho$ {bE 15 {_ZQ >H$m g_` {X`m J`m h & Z-n H$m {dVaU nydm _| 10.15 ~Oo {H$`m OmEJm & 10.15 ~Oo go 10.30 ~Oo VH$ N>m Ho$db Z-n H$mo n T>|Jo Am a Bg Ad{Y Ho$ Xm amZ do C ma-nwp VH$m na H$moB C ma Zht {bI|Jo & Please check that this question paper contains 11 printed pages. Code number given on the right hand side of the question paper should be written on the title page of the answer-book by the candidate. Please check that this question paper contains 31 questions. Please write down the Serial Number of the question before attempting it. 15 minute time has been allotted to read this question paper. The question paper will be distributed at 10.15 a.m. From 10.15 a.m. to 10.30 a.m., the students will read the question paper only and will not write any answer on the answer-book during this period. g H${bV narjm II SUMMATIVE ASSESSMENT II J{UV MATHEMATICS {ZYm [aV g_` : 3 K Q>o A{YH$V_ A H$ : 90 Time allowed : 3 hours 30/1 Maximum Marks : 90 1 P.T.O. gm_m ` {ZX}e : (i) g^r Z A{Zdm` h & (ii) Bg Z-n _| 31 Z h Omo Mma I S>m| A, ~, g Am a X _| {d^m{OV h & (iii) I S> A _| EH$-EH$ A H$ dmbo 4 Z h & I S> ~ _| 6 Z h {OZ_| go `oH$ 2 A H$ H$m h & I S> g _| 10 Z VrZ-VrZ A H$m| Ho$ h & I S> X _| 11 Z h {OZ_| go `oH$ 4 A H$ H$m h & (iv) H $bHw$boQ>a H$m `moJ d{O V h & General Instructions : (i) All questions are compulsory. (ii) The question paper consists of 31 questions divided into four sections A, B, C and D. (iii) Section A contains 4 questions of 1 mark each. Section B contains 6 questions of 2 marks each, Section C contains 10 questions of 3 marks each and Section D contains 11 questions of 4 marks each. (iv) Use of calculators is not permitted. I S> A SECTION A Z g `m 1 go 4 VH$ `oH$ Z 1 A H$ H$m h & Question numbers 1 to 4 carry 1 mark each. 1. `{X { KmV g_rH$aU H$s{OE & px2 2 5 px + 15 = 0 If the quadratic equation px2 p H$m _mZ kmV 2 5 px + 15 = 0 has two equal roots, then find the value of p. 30/1 Ho$ Xmo g_mZ _yb hm|, Vmo 2 2. AmH ${V 1 _|, EH$ _rZma AB H$s D $MmB 20 _rQ>a h Am a BgH$s ^y{_ na naN>mB b ~mB 20 3 _rQ>a h & gy` H$m C Vm e kmV H$s{OE & BC H$s AmH ${V 1 In Figure 1, a tower AB is 20 m high and BC, its shadow on the ground, is 20 3 m long. Find the Sun s altitude. Figure 1 3. Xmo {^ nmgm| H$mo EH $gmW CN>mbm J`m & XmoZm| nmgm| Ho$ D$nar Vbm| na AmB g `mAm| H$m JwUZ\$b 6 AmZo H$s m{`H$Vm kmV H$s{OE & Two different dice are tossed together. Find the probability that the product of the two numbers on the top of the dice is 6. 4. AmH ${V 2 _|, O H|$ dmbo d m H$s PQ EH$ QPT = 60 h , Vmo PRQ kmV H$s{OE & Ordm h VWm PT EH$ ne aoIm h & `{X AmH ${V 2 30/1 3 P.T.O. In Figure 2, PQ is a chord of a circle with centre O and PT is a tangent. If QPT = 60 , find PRQ. Figure 2 I S> ~ SECTION B Z g `m 5 go 10 VH$ `oH$ Z 2 A H H$m h & Question numbers 5 to 10 carry 2 marks each. 5. AmH ${V 3 _|, Xmo ne aoImE RQ VWm RP d m Ho$ ~m {~ X R go ItMr JB h & d m H$m Ho$ O h & `{X PRQ = 120 h , Vmo {g H$s{OE {H$ OR = PR + RQ. AmH ${V 3 In Figure 3, two tangents RQ and RP are drawn from an external point R to the circle with centre O. If PRQ = 120 , then prove that OR = PR + RQ. Figure 3 30/1 4 6. AmH ${V 4 _|, 3 go_r { `m dmbo EH$ d m Ho$ n[aJV EH$ { ^wO ABC Bg H$ma ItMm J`m h {H$ aoImI S> BD VWm DC H$s b ~mB`m H $_e 6 go_r VWm 9 go_r h & `{X ABC H$m jo \$b 54 dJ go_r h , Vmo ^wOmAm| AB VWm AC H$s b ~mB`m kmV H$s{OE & AmH ${V 4 In Figure 4, a triangle ABC is drawn to circumscribe a circle of radius 3 cm, such that the segments BD and DC are respectively of lengths 6 cm and 9 cm. If the area of ABC is 54 cm2, then find the lengths of sides AB and AC. Figure 4 7. {Z Z { KmV g_rH$aU H$mo 2 4x + 4bx (a 2 x Ho$ {bE hb H$s{OE : 2 b )=0 Solve the following quadratic equation for x : 4x2 + 4bx 8. (a2 b2) = 0 EH$ g_m Va lo T>r Ho$ W_ n nXm| Ho$ `moJ\$b H$mo Sn mam Xem `m OmVm h & Bg lo T>r _| `{X S5 + S7 = 167 VWm S10 = 235 h , Vmo g_m Va lo T>r kmV H$s{OE & In an AP, if S5 + S7 = 167 and S10 = 235, then find the AP, where Sn denotes the sum of its first n terms. 9. {~ X A(4, 7), B(p, 3) VWm C(7, 3) EH$ g_H$moU { ^wO Ho$ erf h , {Og_| g_H$moU h & p H$m _mZ kmV H$s{OE & B na The points A(4, 7), B(p, 3) and C(7, 3) are the vertices of a right triangle, right-angled at B. Find the value of p. 30/1 5 P.T.O. 10. `{X {~ X H$s{OE & A(x, y), B( 5, 7) VWm C( 4, 5) maoIr` hm|, Vmo x VWm y _| g ~ Y kmV Find the relation between x and y if the points A(x, y), B( 5, 7) and C( 4, 5) are collinear. I S> g SECTION C Z g `m 11 go 20 VH$ `oH$ Z 3 A H$ H$m h & Question numbers 11 to 20 carry 3 marks each. 11. EH$ g_m Va lo T>r H$m 14dm nX CgHo$ 8d| nX H$m X JZ w m h & `{X CgH$m N>R>m nX Vmo CgHo$ W_ 20 nXm| H$m `moJ\$b kmV H$s{OE & The 14th term of an AP is twice its 8th term. If its 6th term is find the sum of its first 20 terms. 12. x Ho$ {bE hb H$s{OE 3 x2 8 h , 8, then : 2 2x 2 3 =0 Solve for x : 3 x2 13. 2 2x 2 3 =0 YamVb Ho$ EH$ {~ X A go EH$ hdmB Ohm O H$m C `Z H$moU 60 h & 15 goH$ S H$s C S>mZ Ho$ n MmV , C `Z H$moU 30 H$m hmo OmVm h & `{X hdmB Ohm O EH$ {Zp MV D $MmB 1500 3 _rQ>a na C S> ahm hmo, Vmo hdmB Ohm O H$s J{V {H$bmo_rQ>a/K Q>m _| kmV H$s{OE & The angle of elevation of an aeroplane from a point A on the ground is 60 . After a flight of 15 seconds, the angle of elevation changes to 30 . If the aeroplane is flying at a constant height of 1500 3 m, find the speed of the plane in km/hr. 14. `{X ( 2, 2) VWm (2, {ZX}em H$ kmV H$s{OE O~{H$ 4) P H $_e {~ X aoImI S> AB A VWm B na h VWm Ho$ {ZX}em H$ h , Vmo {~ X AP = P Ho$ 3 AB. 7 If the coordinates of points A and B are ( 2, 2) and (2, 4) respectively, 3 find the coordinates of P such that AP = AB, where P lies on the line 7 segment AB. 30/1 6 15. EH$ Oma _| Ho$db bmb, Zrbr VWm Zma Jr a J H$s J|X| h & `m N>`m EH$ bmb a J H$s J|X Ho$ {ZH$mbZo H$s m{`H$Vm {ZH$mbZo H$s m{`H$Vm 1 3 1 4 h & Bgr H$ma Cgr Oma go `m N>`m EH$ Zrbr J|X Ho$ h & `{X Zma Jr a J H$s Hw$b J|X| 10 h , Vmo ~VmBE {H$ Oma _| Hw$b {H$VZr J|X| h & The probability of selecting a red ball at random from a jar that contains 1 only red, blue and orange balls is . The probability of selecting a blue 4 1 ball at random from the same jar is . If the jar contains 10 orange balls, 3 find the total number of balls in the jar. 16. 14 go_r { `m dmbo d m Ho$ Cg bKw d mI S> H$m jo \$b kmV H$s{OE, {OgH$m H|$ r` [ = 22 br{OE ] 7 Find the area of the minor segment of a circle of radius 14 cm, when its central angle is 60 . Also find the area of the corresponding major 22 segment. [Use = ] 7 H$moU 60 h & g JV XrK d mI S> H$m jo \$b ^r kmV H$s{OE & 17. AMmZH$ ~m T> AmZo na, Hw$N> H$ `mUH$mar g WmAm| Zo {_b H$a gaH$ma H$mo Cgr g_` 100 Q>|Q> bJdmZo Ho$ {bE H$hm VWm Bg na AmZo dmbo IM H$m 50% XoZo H$s noeH$e H$s & `{X `oH$ Q>|Q> H$m {ZMbm ^mJ ~obZmH$ma h {OgH$m `mg 4.2 _r. h VWm D $MmB 4 _r. h VWm D$nar ^mJ Cgr `mg H$m e Hw$ h {OgH$s D $MmB 2.8 _r. h , Am a Bg na bJZo dmbo H $Zdg H$s bmJV < 100 {V dJ _r. h , Vmo kmV H$s{OE {H$ BZ g WmAm| H$mo {H$VZr am{e XoZr hmoJr >& BZ g WmAm| mam {H$Z _y `m| H$m Xe Z {H$`m J`m ? 22 br{OE ] [ = 7 Due to sudden floods, some welfare associations jointly requested the government to get 100 tents fixed immediately and offered to contribute 50% of the cost. If the lower part of each tent is of the form of a cylinder of diameter 4.2 m and height 4 m with the conical upper part of same diameter but of height 2.8 m, and the canvas to be used costs < 100 30/1 per sq. m, find the amount, the associations will have to pay. What values 22 ] are shown by these associations ? [Use = 7 7 P.T.O. 18. EH$ A Jmobr` ~V Z H$m Am V[aH$ `mg 36 go_r h & `h Vab nXmW go ^am h & Bg Vab H$mo 72 ~obZmH$ma ~moVbm| _| S>mbm J`m h & `{X EH$ ~obZmH$ma ~moVb H$m `mg 6 go_r hmo, Vmo `oH$ ~moVb H$s D $MmB kmV H$s{OE, O~{H$ Bg {H $`m _| 10% Vab {Ja OmVm h & A hemispherical bowl of internal diameter 36 cm contains liquid. This liquid is filled into 72 cylindrical bottles of diameter 6 cm. Find the height of the each bottle, if 10% liquid is wasted in this transfer. 19. go_r ^wOm dmbo EH$ KZmH$ma bm H$ Ho$ D$na EH$ AY Jmobm aIm h Am h & AY Jmobo H$m A{YH$V_ `mg `m hmo gH$Vm h ? Bg H$ma ~Zo R>mog Ho$ g nyU n >r` jo H$mo n|Q> H$admZo H$m < 5 {V 100 dJ go_r H$s Xa go `` kmV H$s{OE & [ = 3.14 br{OE ] 10 A cubical block of side 10 cm is surmounted by a hemisphere. What is the largest diameter that the hemisphere can have ? Find the cost of painting the total surface area of the solid so formed, at the rate of < 5 per 100 sq. cm. [ Use = 3.14 ] 20. go_r `mg VWm 3 go_r D $Mo 504 e Hw$Am| H$mo {nKbmH$a EH$ Ymp dH$ Jmobm ~Zm`m J`m & Jmobo H$m `mg kmV H$s{OE & AV BgH$m n >r` jo \$b kmV H$s{OE & 3 .5 [ = 22 7 br{OE ] 504 cones, each of diameter 3.5 cm and height 3 cm, are melted and recast into a metallic sphere. Find the diameter of the sphere and hence 22 find its surface area. [Use = ] 7 I S> X SECTION D Z g `m 21 go 31 VH$ `oH$ Z 4 A H$ H$m h & Question numbers 21 to 31 carry 4 marks each. 21. EH$ Am`VmH$ma IoV H$m {dH$U BgH$s N>moQ>r ^wOm go 16 _rQ>a A{YH$ h & `{X BgH$s ~ S>r ^wOm N>moQ>r ^wOm go 14 _rQ>a A{YH$ h , Vmo IoV H$s ^wOmAm| H$s b ~mB`m kmV H$s{OE & The diagonal of a rectangular field is 16 metres more than the shorter side. If the longer side is 14 metres more than the shorter side, then find the lengths of the sides of the field. 30/1 8 22. g_m Va lo T>r 8, 10, 12, ... H$m 60dm nX kmV H$s{OE, `{X Cg_| Hw$b Bg lo T>r Ho$ A {V_ 10 nXm| H$m `moJ\$b kmV H$s{OE & 60 nX h & AV Find the 60th term of the AP 8, 10, 12, ..., if it has a total of 60 terms and hence find the sum of its last 10 terms. 23. EH$ aobJm S>r nhbo 54 {H$bmo_rQ>a H$s X ar {H$gr Am gV Mmb go MbVr h VWm CgHo$ ~mX H$s 63 {H$bmo_rQ>a H$s X ar nhbo go 6 {H$bmo_rQ>a {V K Q>m A{YH$ H$s Am gV Mmb go MbVr h & `{X Hw$b X ar 3 K Q>o _| nyar hmoVr h , Vmo aobJm S>r H$s nhbr Mmb `m h ? A train travels at a certain average speed for a distance of 54 km and then travels a distance of 63 km at an average speed of 6 km/h more than the first speed. If it takes 3 hours to complete the total journey, what is its first speed ? 24. {g H$s{OE {H$ d m Ho$ ~m {~ X go d m na ItMr JB ne aoImE b ~mB _| g_mZ hmoVr h & Prove that the lengths of the tangents drawn from an external point to a circle are equal. 25. {g H$s{OE {H$ d m H$s {H$gr Mmn Ho$ _ `-{~ X na ItMr JB ne aoIm, Mmn Ho$ A ` {~ X Am| H$mo {_bmZo dmbr Ordm Ho$ g_m Va hmoVr h & Prove that the tangent drawn at the mid-point of an arc of a circle is parallel to the chord joining the end points of the arc. 26. EH$ { ^wO ABC H$s aMZm H$s{OE {Og_| AB = 6 go_r, A = 30 VWm B = 60 . EH$ A ` { ^wO AB C H$s aMZm H$s{OE Omo {H$ { ^wO ABC Ho$ g_ $n hmo VWm {OgH$m AmYma AB = 8 go_r hmo & Construct a ABC in which AB = 6 cm, A = 30 and B = 60 . Construct another AB C similar to ABC with base AB = 8 cm. 27. EH$ Prb _| nmZr Ho$ Vb go 20 _rQ>a D $Mo {~ X A go, EH$ ~mXb H$m C `Z H$moU 30 h & Prb _| ~mXb Ho$ {V{~ ~ H$m A go AdZ_Z H$moU 60 h & A go ~mXb H$s X ar kmV H$s{OE & At a point A, 20 metres above the level of water in a lake, the angle of elevation of a cloud is 30 . The angle of depression of the reflection of the cloud in the lake, at A is 60 . Find the distance of the cloud from A. 30/1 9 P.T.O. 28. A N>r Vah go \|$Q>r JB EH$ Vme H$s J >r go EH$ n mm `m N>`m {ZH$mbm J`m & m{`H$Vm kmV H$s{OE {H$ {ZH$mbm J`m n mm (i) h Hw$_ H$m n mm h `m EH$ B $m h & (ii) EH$ H$mbo a J H$m ~mXemh h & (iii) Z Vmo Jwbm_ h VWm Z hr ~mXemh h & (iv) `m Vmo ~mXemh h `m ~oJ_ h & A card is drawn at random from a well-shuffled deck of playing cards. Find the probability that the card drawn is (i) (ii) a black king. (iii) neither a jack nor a king. (iv) 29. a card of spade or an ace. either a king or a queen. Ho$ _mZ kmV H$s{OE {OZgo jo \$b 24 dJ BH$mB hmo &$ k (1, 1), ( 4, 2k) VWm ( k, 5) erfm] dmbo { ^wO H$m Find the values of k so that the area of the triangle with vertices (1, ( 4, 2k) and ( k, 30. 5) is 24 sq. units. AmH ${V 5 _|, PQRS EH$ dJm H$ma bm Z h {OgH$s ^wOm PQ = 42 _rQ>a h & Xmo d mmH$ma \y$bm| H$s `m[a`m ^wOm PS VWm QR na h {OZH$m Ho$ Bg dJ Ho$ {dH$Um] H$m {V N>oXZ {~ X O h & XmoZm| \y$bm| H$s `m[a`m| (N>m`m {H$V ^mJ) H$m Hw$b jo \$b kmV H$s{OE & AmH ${V 5 30/1 1), 10 In Figure 5, PQRS is a square lawn with side PQ = 42 metres. Two circular flower beds are there on the sides PS and QR with centre at O, the intersection of its diagonals. Find the total area of the two flower beds (shaded parts). Figure 5 31. EH$ R>mog YmVw Ho$ ~obZ Ho$ XmoZmo| {H$Zmam| go Cgr `mg Ho$ A Jmobo Ho$ $n _| YmVw {ZH$mbr JB & ~obZ H$s D $MmB 10 go_r VWm BgHo$ AmYma H$s { `m 4.2 go_r h & eof ~obZ H$mo {nKbmH$a 1.4 go_r _moQ>r ~obZmH$ma Vma ~ZmB JB & Vma H$s b ~mB kmV H$s{OE & [ = 22 7 br{OE ] From each end of a solid metal cylinder, metal was scooped out in hemispherical form of same diameter. The height of the cylinder is 10 cm and its base is of radius 4.2 cm. The rest of the cylinder is melted and converted into a cylindrical wire of 1.4 cm thickness. Find the length of 22 ] the wire. [Use = 7 30/1 11 P.T.O. QUESTION PAPER CODE 30/1 EXPECTED ANSWERS/VALUE POINTS SECTION - A Q.No. Marks 1. p=3 1m 2. 30o 1m 3. 1 9 1m 4. 120o 1m SECTION - B 5. POR = 90 60 = 30 o m PRO 1 sin30 o OR 2 PR OR 2 = PR + QR 6. m Let AF = AE = x AB = 6 + x, AC = 9 + x, BC = 15 1 15 6 x 9 x 3 54 2 x = 3 AB = 9 cm, AC = 12 cm and BC = 15 cm 2 m 1m m 7. 4x + 4bx + b a = 0 (2 x + b) (a) = 0 2 2 2 2 2 m (2x + b + a) (2x + b a) = 0 x 8. m a b a b , x 2 2 S5 S7 167 + m 5 2a 4d 7 2a 6d 167 2 2 24a + 62d = 334 or 12a + 31d = 167 .............................(i) S10 235 5 2a 9d 235 or 2a 9d 47 ..............(ii) Solving (i) and (ii) to get a = 1, d = 5. Hence AP is 1, 6, 11, ......... 9. 2 2 2 Here, AB + BC = AC 2 2 2 + m 2 p = 7 or 4 1m since p 7 p 4 10. m m (4) + (p 4) + (7 p) = (3) + ( 4) 2 m m Using ar ( ABC) = 0 m x (7 5) 5 (5 y) 4 (y 7) = 0 1m 2x 25 + 5y 4y + 28 = 0 2x + y + 3 = 0 m SECTION - C 11. a14 = 2 a 8 a + 13d = 2 (a + 7d) a6 = 8 a + 5d = 8 a= d 1m m solving to get a = 2, d = 2 m S20 = 10 (2a + 19d) = 10 (4 38) = 340 3 1m 12. 3 x2 2 2 x 2 3 0 3 x2 3 2 x 2 x 2 3 0 x 6 x 6, x 3x 2 0 1+1 m 2 3 + m Let AL = x 13. BL tan 600 x 1500 3 x Fig. 3 x 1500 m. m 1m 1 CM tan 300 AL LM 3 1500 + LM = 1500 (3) = 4500 LM = 3000 m. Speed = 14. AP 3 AB AP : PB 3 : 4 7 A 2, 2 P (x, y) 3: 4 B 2, 4 3000 = 200 m./s. = 720 Km/hr. . 15 x 6 8 2 7 7 12 8 20 7 7 20 2 P , 7 7 P Red 1m m m 1 1 , P blue 4 3 P orange 1 m 1m y 15. 1m 1 1 5 4 3 12 1 m 5 Total no. of balls 10 12 Total no. of balls m 10 12 24 5 4 1m 16. 0 r = 14 cm. = 60 Area of minor segment r2 360 1 2 r sin 2 m 22 60 1 3 14 14 14 14 7 360 2 2 m 308 49 3 cm 2 or 17.89 cm 2 or 17.9 cm 2 Approx. 3 1m Area of Major segment 308 r2 49 3 3 m 1540 49 3 cm 2 or 598.10 cm 2 3 m or 598 cm 2 Approx. 17. Slant height ( ) 2.8 2 2.1 2 Area of canvas 2 for one tent 3.5 cm. 22 22 2.1 4 2.1 3.5 7 7 = 6.6 (8 + 3.5) = 6.6 11.5 m2 m m Area for 100 tents = 66 115 m2 Cost of 100 tents = Rs. 66 115 100 50% Cost = 33 11500 = Rs. 379500 Values : Helping the flood victims 18. Volume of liquid in the bowl = 2 3 18 cm 3 3 Volume, after wastage = Volume of liquid in 72 bottles = 2 h 3 18 3 2 90 3 cm3 18 3 100 3 2 h 72 cm 3 9 10 5.4 cm. 2 3 72 m m 1m m m m +1m 5 19. Laergest possible diameter = 10 cm. of hemisphere 1m radius = 5 cm. Total surface area = 6 (10) + 3.14 (5) 2 Cost of painting 20. 1m 678.5 5 Rs. 3392.50 100 100 = Volume of metal in 504 cones 504 2 33.9250 1m 33.93 1 22 35 35 3 cm. 3 7 20 20 4 22 1 22 35 35 r 3 504 3 3 7 3 7 20 20 1m m r = 10.5 cm. diameter = 21 cm. Surface area 4 21. m 1m 22 21 21 21 1386 cm 2 7 7 2 2 Let the length of shorter side be x m. length of diagonal = (x + 16) m m and, lenght of longer side = (x + 14) m m x + (x + 14) = (x + 16) 1m 2 2 2 x = 10 m. 1m length of sides are 10m and 24m. + m 2 x 4x 6 = 0 6 22. t60 = 8 + 59 (2) = 126 1m sum of last 10 terms = t 51 t 52 .......... t 60 1m t51 = 8 + 50 (2) = 108 m Sum of last 10 terms = 5 [108 + 126] 1m = 1170 23. m Let the original average speed of (first) train be x km./h. 54 63 3 x x 6 1 m 54x + 324 + 63x = 3x (x + 6) x 33x 108 = 0 2 1m Solving to get x = 36 1m First speed of train = 36 km/h. 24. m For correct Given, To Prove, const. and figure x 4=2 m For correct proof 25. 2m B is mid point of arc (ABC) 1 = 2 OAF 1m m OCF SAS. AFO CFO 900 AFO DBO 90 0 But these are corresponding angles AC DE 7 Correct Fig. m m m m m 26. Constructing ABC 1 m Constructing AB C 2 m 27. correct figure 1m 1 h tan 30 0 x 3 h. x 3 40 h tan 600 x 3h 3 x 40 b h 20 m. 3 x 20 3 m AC 28. 40 h 3 20 2 20 13 3 4 52 13 (i) P(spade or an ace) (ii) P(a black king) (iii) P(neither a jack nor a king) 52 8 44 11 52 52 13 (iv) P(either a king or a queen) 4 4 8 2 52 52 13 2 1 52 26 3 2 m m m m 40 m. 1m 1m 1m 8 1m 1m 29. 1 1 2k 5 4 5 1 k 1 2k 24 2 30. 2m 2k 2 3k 27 0 Solving to get k = 3, k 1m 9 2 1m Radius of circle with centre O is OR let OR = x x 2 x 2 42 2 x 21 2 m. 1m Area of one flower bed = Area of segment of circle with 0 centre angle 90 22 90 1 21 2 21 2 21 2 21 2 7 360 2 1m = 693 441 = 252 m2 31. + + m Area of two flower beds = 2 252 = 504 m2 Total Volume of cylinder m 22 42 42 10 cm 3 7 10 10 = 554.40 cm. Volume of metal scooped out 4 42 42 3 7 10 = 310.46 cm3 m m 3 m m Volume of rest of cylinder = 554.40 310.46 = 243.94 cm3 m If is the length of were, then 22 7 7 24394 7 10 10 100 1m = 158.4 cm. m 9

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