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Paper 4 Solution

Class 10 paper4 and solutions

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Paper 4 Solution

Class 10 paper4 and solutions

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satyaisfree
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56 a @ @ a oy o o ® SECTION Answer according to instruction. (Question no. 1 ta 16) (1 mark each) 116] State whether the following statements are True or False: For any positive integers a and b. H.CF (a, 6) * LCM (a, 6) a ¥ b, (ch.3) a and B are two zeroes of quadratic polynomial > coefficient of x Pls) ~ ae + be +e. + B=. (cha) Product of cot and A is cot A. (cha) For any angle 6 sin 0 (cha) (4) False + is possible (1) Truc (2) False (3) False Fill in the blanks so as to make each of the following statement true : Given two positive integers a and b there exist unique integers q and r such that w ~ dy +r, which upton ove font the following ? (ch) (o 4g © arated (D) none of th If B ~ 4ac > 0 in quadratic equation a? + bx +. = 0 them (ch) (A) we do not get real roots (B) we get real roots (©) we get equal roots (D) none of them o ao) ay (12) a3) (14) «asy (16) Ans. \Vraj STANDARD 10 : QUESTION PAPERS | MATHEMATICS (BASIC) | [ETEUTTEZUTTEEE |FULL sowurion| For the following options which is known as section formula if coordinates af the P (x, ») divides the fine segment joining the points A (x), y,) and B (x, »3) in the ratio. mim, ? (cn7) mt tm, time mat my FRE TA ® (setae mma) mom "mM o (Be, ta) mm? m tm my, + omy | If P(E) = 0.05 then probability of P(E) is (eni25) (@) 065 @) 095 (©) 056 (0) 1.95 = Fillin the blanks so as to make cach of the following statement true : tangents can be drawn on a point of circle. (one and any one, two, one) (cho) There are exactly taugents to a ciicle through a point lying outside the circle. (one, one and any one, two) (ch.10) Write formula to find length of are af sector of eile if r radius and sustain angle 0. 8 8 8 (sep 7 GB 2m, GBS x amr) ghatay radius of circle if circumference is 22 cme 117 (er) (cvi33) ‘The abscissa of the point of intersection of the less than type and of the more than type cumulative frequency curves of a grouped data gives its (median, mean, mode) (ch.1a) ‘The probability of rise the sun into the west is“ (zer0, one, two) (C15) (11) one and any one (12) two, ay 2 13) 369 * 28 oa 5 (15) median (16) zero 40 (21) Find values of 2 cut? 45° + cos? 60" sin®30°.(ch.8) Ans. 2 cot? 45° + cos? 60° — sin? 30° sor FG) 1) 1 20 (i) 4 2 OR @1) If P,Q, R are angles of triangle then prove tan (EEE) = cot 3. (cna) Atk In APQR, P+ R + Q= 180° P+R=180-Q FER - 108 iviging by 2 on both sides) P+R 180 - Q tan (PSE) ~ tan (MZ 9) 7 (dividing by tan on both sides) P+R)_, (180 Q ve tan (FSF) =n (FP - 9) om (8) =m 9} P+R) a tan (PS*) = ot : (complimentary angle uf tan is cot.) (22) An observer 1.5 m tall 1s 28.5 m away from a chimney. The angle of elevation of the top of the G2 Himney from her cyes is 45°, What is the height of the chimney? (ch.) ‘Ans... Here, AB is the chimney, CD is the observer and ZADE s=athe angle of elevation (see Fig). In this ease, ADE is a “-tnangle, right-angled at E and we are required to find @ the height of the chimney. 45° We have, AB - AE + BE = AE +15 @3) Ans. 23) ‘Vraj_STANDARD 10 : QUESTION PAPERS DE ~ CD ~ 28.5 m. To determine AE, we choose a trigonometric ratio, which involves both AE and DE. Let us choose the tangent of the angle of elevation, and Now, AE = 285 So, the height of the chimney AB = (28.5.1 BS) m = 30m, - If radii of two concentric circle are 21, ame’ and 29 cm. If chord of larger circle touches smi fcle (ch.20) them find length of chord. OP As radius of small curele = 71 em OB is radius of large circle ~ 29 cm AB touches ©(0, 21) in point P. . OPLAD (OB is hypotenuse In AOPB ZOPB = 90 op! - op! + pp? (O92 =(12 + PB? (Pythagoras 841 - 441 + PB? - 840-441 = PB? S400 = PB? + (20? = BP PB= 20 AB = 2 x PB=2* 20= 40cm oR Prove tha, the perpendicular drawn from thea <<. of contact to tangent is passing through cetiter, of the circle. 10) bam Given : AB is a tangent for ©(0, 7) at C. To prove : The perpendicular from point of contact to tangent is passing through centre O. Constructions : Join OC and O'C. ” © B Proof : Take any point O' such that OC LAB => ZOCB = 90° ‘raj STANDARD 10 : QUESTION PAPERS. compare x-coordinate of eg. (i) Gm, — 3m, m, +m, = m, ~ m, = 6m, ~ 3m, — Lm, ~ 6m, myim,=2:7 compare y-coordinate of eq, (2) + 10(7) 247 _ 716 +70 9 = =He6 LHS = RHS. (21), Evaluate the following + ‘Sos? 60° + 4sec? 30° ~ tan? 45° (chs) sin? 30° + cos? 30° 2 60° 4 Asec? 30° — tan? 45° ang Sez 6 + dee” 20" — on? as sin? 30° + cos? 30° s(t) (2) 08 2 3 (: _ ste | 5, 16 -4aty 2 vy OR 21) Iftana = cotB, prove that A + B= 90° Ans. tan A = cot B is given > cot (9 — A) = cot B['" tan 8 = cot (90° ~ 6)] = 90°-A=B > = ALB A+ B= 90 (Ch8) (22) Acircus artist Is climbing a 20 m long rope. which is tightly stretched and tied from the top of a vertical pole to the ground. Find the height of the pole, if the angle made by the rope with the ground level is 30° (see Figure). (ch) In given figure AB is a height of pole length of tope AC = 20 m which is tightly stretched and tiét trom the top of pole to the ground at C. In right angled triangle ABC, Paattimde) AB AB H (hypotenusey AC ~ 20 sin 30° ar) 1 = 2° sin 3085 20 2 height of pole is 10 meter. OR In right angled triangle opposite side to 30° angle is half of hypotenuse. 1 AB = > AC AB= 10m 1 = 5 %20 AB = 10 meter The length of a tangent from a point A at distance 5.cm from the centre of the circle is 4 cm, Fii@'the radius of the circle, (ch.10) ‘We know that the tangent at any point of a citele is perpendicular to the radius at point of contact, ZOTA = 90° _ = (ne ~~ Tt 4cm A 3) In Fight angled ‘langle OTA. OA OT? + aT? 8 =or+? or=s-# OT?= (5-4) (5 + 4) OP=1x9=9-37 or =3 Yuyvuy radius of circle is 3 cm. OR ANNUAL EXAMINATION : MATHEMATICS (BASIC) (23) Two concentric circles are of radii 3 em and $ cm. From an external point P, Gangents PA and PB are drawn to circle with centre O. If PA = 12 cm, then find the tangent PB which touches the smaller eirele. | (ch.10) Ans. In AQAP ZA = 90° : OA LAP) According to Pythagoras Theorem op? = 0a? + OF (12% = (6) + OP* 144 — 25 ~ oP? e, 19 = OP? Jn AOBP ZB = 90° OB L PR) According to pythagoras theorem 9 OP? = OB? + PB? 119 = G+ PBP 119 = 9 + PB? 119 9 ~ PB 0 Fit a] (24). Mayank made a bird-bath for his garden in the shape ‘of a cylinder with a hemispherical depression at one ‘ond (see Figure). The height of the cylinder is 1.45 m © and its radius is 30 cm. Find the total surface area (ch.13) we of the ‘Ans.* Let h be height of the cylinder, andr the common radius ‘of the cylinder and hemisphere. 59 ‘Then, the total surface area of the bird-bath = CSA of cylinder + CSA of hemisphere = 2h | 2x? = tar (th +n) = 2% 2B «30 (145 + 30) em? ~ 33000 en? =33 me OK (24) A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to 1.¢m and the helght of the cone Is equal to Its radius. Find the volume of the solid in terms of x. (Ch.13) Ans. -—_—+ radius of cone height of cone A = 1 om volume of solid box = radius of hemisphere = 1 cm volume of cone + volume of hemisphere 1 2a! 2 = Sars Fae =F naps Zaye = (m+2m) _ 3x . 5 y= Rom -2,2 “3 3” (25) Yor any claslal distribution data in ajols wus Jo20 ope D2 nw 68 and h=20 folk Ped median of the data. {(h.14) ‘Ans. Formula to find median of data Median 125 + (34 ~ 22) 129 + 12 3 Median = 137 (26) A box contains 5 red marbles, 8 white marbles and 4 green marbles. One marble is taken out of the box at random. What is the probability that the marble taken out will be (i) red 2 (ii) white ? (iii) not green? (ch.s5) Ans, Total marbles in a box = 5 red + 8 white + 4 green [Question no. 27 to 34) (3 marks oach) = 17 marbles. @) Let even (A) > marble is red Pay= & ‘oil) Let even (B) : marble is white Pe)- 5 (ii) Let event (C) marble is not green means marhle is red or white, 1 - PO) = 37 SECTION-C ‘Answer the following as required with calculations : ay (Sted 18 white ~ 13 marbles) (2i)»-Appart of monthly hostel charges is fixed and the remaining depends on the number of days one has taken food in the mess. When a student A takes food for 20 days she has to pay € 1000 as hostel charges whereas a student B, who takes food for 26 days, pays € 1180 as hostel charges. Find the fixed charges and the cost of food per day. (ch.3) Let fixed monthly charge ot hostel = © y and cost of one day's food = © x. For student A 20 + y= 1000 @ For student B 26x + y = 1180 . (ii) stbtrect eq. (i) from and eq, (i), 26x - 20x = 1180 ~ 1000 = 6r= 1805 x~ 30 Put x = 30 in eq. (i) 20 30 + y= 1000 => y = 1000 - 20 « 30 = 1000 ~ 600 ~ 400 So, fixed monthly charge y = % 400 and cost of ‘one day food x ~ ¥ 30 OR @n (28) ‘raj STANDARD 10 : QUESTION PAPERS On comparing the ratios = and * , find out a ts Be and Zt whether the lines representing the following pales of Linear equations intersect at a point, are parallel or coincident : 4y +8 ~ 0, Ix + by 9-0 9x + 3y +12 =0, he + 6y +240 (il) 6x — 39 +10 = 0, 2 y+ 9=0 {ch3) xziven pair of Tinear equation of two variables are Sx 4y +8=0 0) and x + 6y-9=0 ii) above equations are compare to general form So, line (i) and (ii) interseets in one point given pair of linear equations are 9x +3y + 12=0 ® and 18x + 6 +240 Ox +3y + 12-0 ti) hove equations compare to general form. we get 9, by = 3, ¢) = 12 and ay ~ 9, by ~ 3, 6— 12. BL = A S50 both tines are coitident a he (iii) Given pair of linear equation. or~3y + 10=0 of) w-yt9=0 Co) Pair of linear equation compare to general form we get above compare a = 6, 6) ~ -3, ¢, = 10 and 4, = 2, b=“, = 4-45 Ya a ae So both fines are parallel ‘The diagonal of a rectangular field is 60. metres more than the shorter side. If the longer side is 30 ‘metres more than the shorter side, find the sides of the field. (ch) Let shorter side QR = x meter of rectangular field PORS, diagonal PR = (c+ 60) ete, eo Be (e+ 20) meter ANNUAL EXAMINATION ; MATHEMATICS (BASIC) longer side PQ rectangle is 30 meter more than shorter side. PQ = (r+ 30) meter In right angled triangle POR according to pythayoras theorem. PR? = PQ? + QR? (+ 607 = (a + 307 + x? 27 + 120x + 3600 ~ x7 + 60x + 900 + 37 be a@r a+ Qab + 0 (2x2 — x2) + (60x ~ 120%) | (900 ~ 3600) = 0 x2 — 60x - 2700-0 x7 — O0r + 30x ~ 2700 = 0 x (90) + 30 (e- 90) = 0 (r= 90) & + 30) = x-90=00R r+ 30-0 x= 900Rx=~30 Here, negative value of x is not possible because ide’s value is not negative. x- 90 © Breadth of rectangular feld ~ 90 meter and length of ([Factorisation] rectangular field = 90 + 30 = 120 m. OR (28) Represent the following situations in the form of ‘quadratic equations = (@ The area of a rectangular plot is 528 m?, The Tength of the plot (in metres) is ane mare than twice its breadth. We need to find the length and breadth of the plot. Gi) “The product af twa consecutive pasitive integers is 306. We need to find the integers. (ii) Rohan's mother is 26 years older than him. ‘The product of their ages (in years) before 3 years is 360, We would like to find Rohan’s Present age. (cha) Ans. (i) Let breadth of plot xm length is one more than twice its breadth length = 2x + 1) m Area of rectangular plot = 528 m? engin * breadin = 328 2 Qet lx x= 528 Wo Wee 61 (ii) Let consecutive possitive integers x and x + 1 product of two consecutive possitive integers = 306 x (e+ 1) = 306 P+ x= 306 [+x 306 =0 i) Let present age of Rohan = x years, Present age of Rohan’s mother = (x+26) years before 3 years age of Rul (4 ~ 3) yeayigs before 3 years his mother’s age = x + 26-2 = (x + 23) yee before 3 years product of their ages = 360 (&~ 3) & +23) = 360 x + 23x ~ 3x - 69 = 360 2 (29) The first term of an AP is 5, the lst term is 45 and the sum of AP is 400. Find the number of terms (chs) and the common difference. Let for given AP first term is (a) and common is (d). First term a= 5 Last term = 45 Sum of » term S,, = 400 + SF @+O 400 = £ (5 +45) 400 x 7-= 50m _ 400 x2 > nS oo =8%2 S =16 Now, (= 43 > at(n-Nd=45 f. > SH(I6-Id = I5d=45-5=40 =-4_8 2 8 15 = 3 the number of terms of AP is 16. and common difference is *. \Vaaj_ STANDARD 10 : QUESTION PAPERS (30) If the sum of the first m terms ot an AP is 4n ~ 7, What is the first term (that is S) ? What is sum of first two terms ? What is the second term ? Similarly, find the 3", the 10 and the n"™ terms. (Ch.5) Sum of first » numbers S,=4n WF (i) is given put n= 1 in eq. (1) Sp=4x- So, first term = put n= 2 in eg. () 8) =4%2-2 -4-4 So, sum of first two terms Second terms = S, — 4am eq, (I) x3-3= 12-953 53-8, put m Ss (third term put n ~ 9 in eq. (I) S, =4% 9-92-36 81-45 then, put = 10 in eq, (1) Sig = 4X 10 WF = 40 — 100 = — 60 10" term = Sy ~ S, 60 - 45) ~= 601 45-15 Put, m1 in place of m in e9. (1) 4(n—1)- (ny 4n- 4-9 + 2n=1 w+ 6n-5 nt term of AP a, = S, ~ al tem = 8,3, _ 4 = 4n—P -Cr? s 6n~5) dn ant 6n+5 5-1 ni term of AP is (5 — 21) G1) IEA and B are (-2, -2) and (2, ~4), respectively, find the co-ordinates of P such as AP = 3 AB and P lcs on the line segment AB. (en.7) Ans. Acconting to problem AP = 3 AR ABT = AP 3 6D AP+ PBR 7 > ‘AP z P is on line segment AB -. AB = AP + PB] So, from section formula my + my) mem) (ma em, Covntinate of P= [ me BB 45D 3491 4x63)] ron = PS 3H = (G84) - (2 T4 TF Coon of poi tis (2,2) TT OR Find the values of y for which the distance hetween the points P(2, ~ 3) and Q(10, y) is 10 units. (Ch.7) Let Points P(2, ~3) and Q (10, y). distance between two points PQ = fun~7)? s[y-CayP = Vero = ort torso Wie but PQ = 10 is given +6ye 7-10 +6473 = 100 > yPtey-77=0 = yay t 9-27 (squaring both sides) (splitting the middle term) = 0 DG+9=0 > y-ts0 Synd o = yt9-0 => possible values of y ~ 3 or ~9. ANNUAL EXAMINATION : MATHEMATICS (BASIC) 63 (32) The following table gives the literacy rate (in percentage) of 35 cities. Kind the mean literacy rate. (Ch.14) [Literacy vate Gn 90) | sss | ss—os [ 0875 3 - may OP classmark (x) Sei | os | “30 ae 55-65 10 ou 10% =e 65-75 u ux0+0 15-85 8 80 BxI= 3 90 as Here a = 70 class Iength 4 — 10 ea e 47 10 vv The given data is represented as following table. =2 70 + 10 |e - tere - ~70+(4) F~ ~ 69.4285 = 69.43 ‘mean of literacy rate is 69.43 % lity of getting Gay “Tike cards ~ the ten, jack, queen, king and ace | (34) A dieis thrown once. Find the probal of diamonds, are well-shuffled with thelr face ‘ownwards. One card is then picked up at random. (What ie the probability that the card is te queen? (Wan 04 number eas) ii) If the queen is drawn and put aside, what is or + soupy tht te second cerd picked up | ANS St numbers on dice ae 1, 23,4, Sand 6, is (a) an ace? (b) a queen? (ch) (Suppose event (E,) = Prime number on dice © prime number (Wi) @ number lying between 2 and 6 Ans. Total cards = 5 ala Total outcomes = 5 PEE,)= & = 7+ Prime numbers are 2,3 and) ii) Suppose event (E3) = numbers between 2 and 6. (Pogson car) = ¥ (+ Here queen is 1) ° 1 (ii) Let queen is drawn and put aside 4 cards are left = PEY= =a which are ten. iack. king and ace. (ce 3, 4 and 5 are between numbers of 2and 6) 1 2) Peace = iii) Suppose event (E,) = numbers are oddon. dice A $ (89) Proveeay “s queen is put aside) PE) = 5 = 1, and § odd nurhbers) ‘Veaj STANDARD 10 : QUESTION PAPERS *® Answer the following as required with calculations: (Question mo. 35 to 39) (4 marks each) [20 (35) An acroplane leaves an airport and flies due north at a speed of 1000 km per hour, At the same time, another ‘aeroplane leaves the same airport and flies due west at a speed of 1200 km per hour. How far apart will be the tee plus ater 1 haar? fens Ans, Let O is an airport. Aplane covers distance to the north side of airport in 1 hour with speed 1000 km/hour OA = 1000 « 3 (Ce distance = speed * time) A plane covers distance to the west side of airport in 14 hour with 1200 kar OB ~ 1200 $ ~ 1800 fn (C= distance = speed x time) Now, North and West sides ate perpendicular to each other In AAOB use pythagoras theorem N 4 4 Ww E a 7 fos Lo. 8 5 ‘West oO AB? = OB? + OA‘ = AB! - (1800)? + (1500 _ =2 AB = 3240000 + 2250000 = 5490000 o AB = 30046 km [Taking square root} 1 So, After 17 hour distance between two plane 1s 300 J6T km. oR (B5)_ Write down the Pythagoras theorem and prove it. (ch) ‘Ans Theorem : In aright angle angle, the suare of the hypotenuse is eal to the sum ofthe squares ofthe ott sides. He 7 Fl From eg. (i) ANNUAL EXAMINATION : MATHEMATICS (BASIC) 67 (37) A metallic right cireular cone 20 cm high and whose vertical angle is 60° is cut into two parts at the middle of Its height by a plane parallel to its base. If the frustum so obtained be drawn into a wire of diameter “cm Bad the length of he wt. (Ch) ‘Ans. Let radi of frustun of cone of top and bottom part age andr, respectively. Height of cone ~ 20 om is given eight frstum of cone = 10 em To find radius of fruetum of cone in AAOE soem oem oem ||, ‘ =O Li OF sot 60° = a0 To [+ 0 = 60° is given] 1 2 ot hat an= em [ree and In AAO’C oc 5 ears £0 160" = Sor = 3h -: 0 = 60° is given} a 0 cosy be Bons Bom Volume of frustum of cone = Fh opt rptrr) Fro (atte. me 3 TSS 7000 % 3 9 Now lets length of wire ish. 1 diameter of wire = 75 cm is given radius () = Fy om Volume of wire = wh = x[s5] xh ‘According to question volume of wire frtlstum of cone volume of th _ 7000" > EH 9 7000 32 x32 = a= om 9 py ~ 100032 «32 1 = =e Lem= 9 ™ hh ~ 1964.44 m length of wire is 7964.44 m OR (37) Metallic spheres of radii 6 cm, 8 cm and 10%en, respectively, are melted (o form a single solid sphere. Find the radius of the resulting sphere. _{Ch.13) Ans, Let radii of dhree meowlic spebere is rpry and ry respectively and Radius of bigger sphere is R. r= 6 om, y= 8 and r ~ 10 em Volume of first sphere (V;) = 4 r? 4 zOP Volume of second sphere (V;)~ tm m3 = 42 Volume of third sphere (V3) to question volume of three metals spheres = volume of single solid sphere. J = VIVA 2048 3 Sto m= ime= a9 => R= 12m «radius of resulting sphere is 12 cm. ¢ © Method : 2 s Volume of bigger sphere = Volume of three éniiiler sphere 4 434 4 Sar? = taf tar, + tn} =P, +P, +P, (Divide by $9) = (6) + (8) + 10° = 216 + 512 + 1000 R3 = 1728 R= (12P R= 120m (88)z"The following table gives the distribution of the life time of 400 neon lamps [-Litetime | Number of lamps | “vm. [1500-2000 | 14 rtard | 60 70 2s 12 tat be} ut ae | —aet ‘Total observations = 730) 150 +a +b ~ 230 at b~ 80 Al) Median (M) = 46 Median class = 40 ~ 50 1340 S09 g=4aita h=10 0 M=46 [Eg] 16 = 40+ [3 +9) «10 3 46-0 = (524) x10 6x65 OBA Bea a-7B-39 a= 34 34 put in equation a + b = 80 3446 = x0 b- 80-34 (o= 46)

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