© SAMPLE © QUESTION 5 PAPER General Instructions: 1. The Question Paper contains three sections A, B and C. MATHEMATICS (Standard Level) (Solved) Time: 90 Min. Max. Marks : 40 2. Section A consists of 20 questions of 1 mark each, Any 16 questions ae tobe attempted 3. Section B consists of 20 questions of 1 mark each. Any 16 questions are to be attempted, 4. Section C consists of 10 questions based on two Case Studies. Attempt any 8 questions 5. There is no negative marking, SECTIO! Directions (QNos. 1 to 20): Section ‘A’ consists of 20 { 8. The pair of equations y =3and y = has: questions of 1 mark each. Any 16 questions are to be | attempted. In case more than desirable number of questions | are attempted, only first 16 will be considered for evaluation, Q1. Ifpandqare co-prime numbers, then p? andg?are: | a.co-prime not co-prime ; ceven d.odd i 2, The value of k for which the system of equations | x-+3y —3=Oand 4x + ky +7=0has no solution, | a0 b7 03 a2 3, Aladder 25 m tong just reaches the top of a building 24 mhigh from the ground. What is the distance of the foot of ladder from the base of the building? (cBsE2020) aSm b6m 7m dBm 4, The sides of a triangle are 30, 70 and 80 units. I an! alttude is droped upon the side of length 80 units, the | larger segment cutoff on ths side is: ' a.62units b.63units cS4units d.65units | Q5. Two different coins are tossed simultaneously. The | Drobablty of getting at Least oe tals, 1 at al 7 az 4 Mg °% a Q6.1f the corresponding medians of two similar ‘tiangles are in the ratio 5 :7, then the ratio oftheir corresponding sides i (c8se 2015) 25:49 bS:7 7:5 6.49:25 07, The value of @in'S sin? @- cos? 0 =2is: en a 8. The decimal expansion of the rational number “qagp Willterminate after: (NCERTEKEMPLAR) | b.two decimal places} d.four decimal places | 8.one decimal place three decimal places one solution b.two solutions infinitely many solutions dino solution 10, The distance between the points (0, 5)and(-5, (NCERTEXEMPLAR) as b5vZ cS 10 ant, Viis: a.an integer b.arational number C.anirrational number d. None of these 12, The HCF and LCM of two numbers are 33 and 264 respectively. When the first number is completely divided by 2 the quotient is 33. The other number is: a6 | BBO ct 6 13. In ABC right-angled at point 8, if tan A = # then the value of sin A-cosC + cosA-sinC is: (NCERT EXERCISE) ao | cal a; 14, In AOPQ right-angled at P, OP 0Q~ PQ=1cm, then the value of sinQis: (NCERT EXERCISE) em, and b. 1 5 ae $35 5 Q 15, Atoothed wheel of diameter 50 cm is attached to a smaller wheel of diameter 30 cm. How many revolutions will the smaller wheel make when the larger one makes 15 revolutions? a23 b.24 625 4.26 VUalIicu wi26 | Seenadhoon of MATHEMATICS (Standard Level) cx | Term = = 35cm, CA=25em | 18C ~ ADEF.IFAB= 4em, BC = 35¢m, Oe at pF = 75en, the perimeter of ADEF is: O19. The pair of equations Sx ay =a bx 5y = 2 hasmave: (cbse 2017) | eRe 20cm 300m 35cm d.d7em | coolio, ty, 17. In the given figure, if DE {I BC, A i b.two solutions, AE = Bem, EC = 2emand BC = 6 { infinitely many solutions cm,then DEis: (CBSE 2014) 0, e 4.no solution a2zem —b.32cm | 020, Aletter is chosen at cadcm 48cm 8 c! ; random from the eters ; Word ‘ASSASSINATION; then the probatiy, 018. If V3 sind = cos6, then the value i ability, cos!0+ 2080, ' the letter chosen Isa vowelis in the form of 4 Seas +2 (case2015) | then xis equal to: tt acoso b.3cose ' as 6.6 3sind d.¥3sine . ! 7 4B SECTION-B Directions (QNos. 21 to 40): Section ‘8" consists of 20 Q30, In the given figure, ABC is a s questions of 1. mark each. Any 16 questions are to be | triangle and GHED is a A ‘attempted. In case more than desirable number of questions rectangle. BC=12 cm, i ‘re attempted, only first 16 will be considered for evaluation. | HE=6 cm, FC=8F and 17k 21, Ifthe sum of LCM and HCF of two numbers is 1260! altitude AF is 24 em. The i] and their LCM is 900 more than their HCF, then the | _area of rectangle is: feb product of two numbers is: ta.5eem? t 2.203400 6.194400 198400 205400 | © b.suem? Boe = ' © 60cm 022. Ifax + by =a? — b? and bx +ay =0, then value of oe xtyis: ao+b ba-b, cO 1 Q23. If tan?0+cot?0=2, 0 is an acute angle, then tan? 6+ cot? Gis equal to: eee. ce Q24, If47x + 34y = 18and31x + 47y = 60, then value of rey i al bo c23 G47 ' Q25. If A=2n+13, B=n+7, where n is a natural | ‘number then HCF of A and Bis: ' ey | 26. A card is drawn from a well shuffled pack of 52 | playing cards. The probability that the card is a red { 3 7 oe Q27. Two dice are thrown together. The probability that ‘sum of the two fumbers ‘will be a multiple of 4, 1 1 1 1 Se Q.26. If x = (sec A — tan) (sec B - tanB) (secC — tanC) (sec A + tanA) (sec B+ tanB) (secC + tanC), then ao 29. The coordinates of the fourth vertex of the rectangle formed by the point (0,0), 2,0) and (0,3) are: (30) ba cl dil b(02) (23) 4.(3.2) 31. The coordinates of vertices A, Band of tian ABC are (0,2), (4, 1) and (0, 4) respectivaly. Fin the length of the median through &. al b2 63 a4 (092, If sec + tand = my then the value of sec'®-tn's ~2sec6 tandis: am u : m et d. None ofthese m 033, The Least number which when divided by 18,243) and 42 will leave in each case the same remain 1, would be: 2.2520 2519 ©2521 6. None of the above Q34. If D, E, F are the mid-points of sides BC, C4 ae respectively of AABC, then the ratio of the are igles DEF and ABC is: 4 baz 4:5 al: 62:3 . ae 35. A straight line is drawn joining the points (5 oe (5, 6) Ifthe line is extended, the ordinate Point on the line, whose abscissa is~ 15! a} bo «1 a2 veamicu wil Ca36. ABCDEF is any hexagon with different vertices | A.B,C,D,E and F as the centres of circles with | same radius rare drawn. The area of the shed | The area of the shaded | Q37. In the given figure, ABC is an A equilateral triangle inscribed in a circle of radius 4 cm with centre 0, then the area of the shaded region c Directions (Q.Nos. 41 to 50): Case study based questions: Section ‘C’ consists of 10 questions of 1 mark each, Any 8 questions are to be attempted. In case more than desirable ‘number of questions are attempted, only first 8 will be considered for evaluation. Case Study 1 Naveeka everyday goes to swimming. One day, Naveeka noticed the water coming out of the pipes to fill the pool. She then told her brother that the shape of the path of the water falling is like that of a parabola and also that a parabola can be represented by a quadratic polynomial which has atmost two zeroes. Based on the given information, answer the following questions. 41, The number of zeroes J (x)= (+2)? + 6 can have is: ao bt c2 a3 Q42.1f the product of the zeroes of the quadratic polynomial f(x) = ax? — 6x — 6is 4, then the value ; of als: that _ polynomial ri | 29 ‘Sample Question Pape! 5, /3)cm? a oe ~3V3) b Stan -303)em? Been B)em? 4.2 (0-33) em? 038. if the zeroes of the quadratic polynomial x? (+1) x+ bare 4 and -3, then a - bis: ar b.10 “7 at 0.39, A circular wire of radius 42 em is cut and bent into the form of a rectangle whose sides are in the ratio SECTION-C of 6 :5. The smaller side of the rectangle is: a.30cm b.60em 70cm adem 040, If2**Y =2%-” = V6, then the value of x and y are: 3 3 B baz ao 2 2 415 2 2 3 0.43, The flow of the water in the pool Is represented by x? 2x —8 then its zeroes are: 32.4 b.4.-2 o-2-2 4-4-4 Q 44, If cand B be the zeroes of the polynomial x? — 4, then the value of = + — ap a0 ot al a 45. A quadratic polynomial whose one zero is -3 and product of zeroes is 0, is: a3x7+3 b.x?-3x x? +3x 3x? -3 Case Study 2 A garden bed is shown below. It is a rectangle with semi-circular ends. It has a concrete circular fountain at the centre as shown, Grass is grown in the garden bed at all places except the central fountain region. wu owiu| Samnachoan of MATHEMATICS (Standard Level) class x | Term-1 30 | Som 48, Area of the garden b , — : asiem? b294 fe ee “a i 149, The outer perimeter ofthe lawns; ’™ at o p )faam aasm bm CBOm ame 5 Suppose we havea pavement all arcund ng bed of which 1 m as shown below, ‘Sit, The pavement is: ea of, Based on the above information, answer the following questions: 46, The area of the concrete circular fountain at the centre is: : aTim? b.308m? 154m? 4.616m (47, The area of one semi-circular end of the garden bed is: : 2 a. (21+ 152) m? b.(42 + 152)m2 308m? b.154m? 616m? 77m (42 -15n)m 4. (84 + 305) m? OMR Answer Sheet H c Instructions * Only a black or blue ballpoint pen has to be * Don't put a tick mark or a cross mark in the used for filling the bubbles. bubble. ¢ Fill the bubbles completely and appropriately. « Multiple markings are invalid. * Half filled-or overiilled bubbles will not be © Do not write anything on the OMR sheet and evaluated, make sure that it is not damaged. oe Se WAOOOOD a6 ©OOO a3 OOOO au. OOOO 2 OOOO a7. OOOOG az @OOO@ 247. ©OO@ 2 OOOO a8 OOHO@ a3. OOO 48. @OOO 4OOOO av. @OO@ 234. ©OO® qa. OOOO BOOOO@ ew. @OO@ 23. O©OHO® as. ©OO@ 2% OOOO au O©OO@ 36 ©@OO@ 27.© OOO a2 OOO 37. ©OO@ 2%OOOO@ e2. ©OO@ 38. OOO@ 2 OOOO eo. OOO@ 3. ©OOO@ UOOOD es. OOO 240. ©G©O@ 22 OOOO 2% @©OO® aa. ©0O0O@ Q2@OOD az. OOO@ 242. OD O@ 23O2OOO a2 @OO@ 243. ©OO@ 24 ®OOOD a ©OO® aaa ®OO@ AOOOD 93. O©O™O@ W485. ©OO®@ hing below this line C_) Marks scored ith dat@ VURAISE Wii ooSection-A ‘ 1. (a) Since p and q are co-prime. So, theirs i 7 squares Le, | p? and q? will also have no common factor ie, they | are also co-prime. 2. (d) The given equations are x +3y-3-0 and 4xthy +720, Here, 9) =1.b,=3.¢, = 3.09 = 4b, The system of equations has no solution. if Hb G13 a Bh teeta 3, (¢)Let ABbe the ladder and CA be the building in which Cis the base of the building. Also, and TRICK Inaright triangle, the square of the hypotenuse is equal tothe sum of the squares ofthe other two sides. ‘Now. in right ACB, AB? = BC? + AC? (By Pythagoras theorem) = BC? =AB?- AC? f (25)? ~(24)? Y ly # lz = (25-24) (25+24)=49 3 Bl m B. ic So. the distance of the foot of the ladder from the base of the building is 7 m. 4 (a) A B (80x) units Now 10 -(B0-x)? Also bh? =70?-x? = 30?-(80-x)?=70?-x? 2 160x=10400 = x=65units. 5. (€) Total possible outcomes are (HH, HT, TH!TT} ie. 4 innumber. Favourable outcomes of getting at least one tall are {HT.TH,T7} ie, 3 in number. + Required probability =3 / 4 & (b) Let ABC and POR are two similar triangles with Medians AD and PS respectively, Then, TRICK caught of the medians of two similar triangles is 010 the ratio oftheir corresponding sides. (Given) (ABC ~ APQR) xunits ' A | Z\ s Hence, the ratio of corresponding sides is 5:7. 7.(b) —5sin?@-cos*o=2 = sin? ~(I-sin?o) =2 = 6sin?9=3 = sind = =sinas? = 0=45° 14587 _ 14587 _ 14587 | (2)? 8. (0) We have. 355 “ahest “Tas! “Oy 14587 xB _ 116696 101000 ~ 10000 Hence, given rational number will terminate after four decimal places. 9. (@) Both the lines represented by the equations y =3 and y =8 are parallel to x-axis. So, parallel to each other. The given pair of equations has no solution. 10. (b) Distance between A(O. 5) and 8(-5.0}. AB= y+ (0-5 50 =5V2 Th (We have, Vid = Jax3 = 243 Here, 2is rational and 3 is irrational. And, the product of a rational and an irrational is irrational. 23 ie, V2 is an irrational number. 12, (¢) First number =2 x33 =66 +. Other number = HCFXLEM _ 33264 _13 stnumber~ 66 116696 13. (b) + tanA= 5 =tan30° = A=30° TRICK Sum of all interior angles in a triangle = 180* * A+B+C= 180° = 30° +90° + C= 180° = C= 180° - 120° «sin A-cosC + cosA-sin A =sin30°-cos60° + cos30"-sin6o° [> 11,88 °48 | Sarnadhaan of MATHEMATICS (Standard Level) class x | Term-1 ' 8 De = i 48 = DE aS = 10 “i748, 14 (¢ Given, 2P =90°and OP= 7em Also, 0Q-PQ=1cm_ (I) => Number of revolutions = 25 16, (b) Given, ABC ~ AEF TIP The ratio of the perimeters of two similar triangles is equal to the ratio of their corresponding sides. a 2 and 0; =36, and on | Common /i\Error ———" | | Most students are not able to prove MADE ~ ayy, 2 H 3c0s?0 + 2c050 bd Of 18 e050 +2 (ro? -b? =(0 +b) (0-b)) | T!1pP—_—_—____ = trom ea. (1) | [ Fottow step by step simplification to avoid ery, re | Sx(ano) + 2xBsno From eqs. tana 2 we get i am Oe 50 = 09=25 ' i and ae cs m 1 x3sin?0 +2V/3sind _ V3sing oF ' ~ WHasind +2 (Being 5 $0, sing=9P -7 ! ) 09°35 ' sino 1. (c) Circumference of smaller wheel =30x cm 49, (¢ Given. 5x -15y = Band3x - Sy = 2 Circumference of bigger wheel =50x cm | "Now. 15 50x =number of revolutions x 30% 1 Here, 0) =5.b,=-15.¢,=8 which shows that the given equations have ifaw: many solutions. 4em/- \asem 75 em 20. (b) There are 13 letters in the word ‘ASS ‘out of which one letter can be chosen in 13 8 ce F Hence, total number of outcomes =13 oy There are 5 vowels in the word ASSISSNAT So, there are 6 ways of selecting a vowel Hence, required probability = © Perimeter of AABC_ AC Perimeter of ADEF ~ DE TRICK B Perimeter of any triangle is equal to the sum of the But given that erro! length of al its three sides, ele 3 Misra 2x+1=B = 2x=22 Berimeter of ADEF ~ OF Section-B ze) = eg 8 21, (b) Let the two numbers be oand b a : Given, LOM (a,b) + HCF (o,b) = 1260 ( = Perimeter of ADEF "3 and LCM (0,6) = 900 + HCF (a8) = Perimeter of ADEF =3x10=30 ont : On solving (1) and (2), we have LCM (a, b) =1080 and HCF (a, b)=180 We know that, product of two numbers = HCF xt = axb=180x1080 = 194400 Hence, perimeter of ADEF is 30 em. 17. (d) In BADE and aAgc, TIP Students should know about AAA criteria for similarity | 22. () lven equations are ofttriangles. x + by =a? —b? { and bx +ay=0 # ZOAE = ZBAC (Common angle) an On adding b ZADE = ZABC (Corresponding angles) ex 9) 4 ita), aoe ADE ~ AABC (By AA similarity) eb cnce 2 (048) x+(0+b)yaa?—w? = (0+6)(x+y)=(o+b)(o-0) Hence, e6ae=b AE OE 8 _DE AC BC B+ VUGaIIicu wieCommon (\\ Error Students do error in simplifying these type of equations. 23, (a) tan? + cot?@ =2 = tan?@ +cot?@ =2 tano.coto = (tane - cote)? = tand—cot@=0 = tand =cote = 02 45° 2s tan?@ +cot?@ = tan? 45° 4 cot? 45° = (1)? + (1)? =2 24.(@)Given, 47x43 =18 “) and Bix + 47y =60 (2) ‘Adding eqs. (1) and (2). we get 78x + 78y = 78 = x+y =1 (divide both sides by 78) Common A Error ‘Students do error in simplifying these type of equations. They solve these equation by elimination method. 25. (b) Taking different values of n, we find that A and B are coprime. re HCF =1 26, () Total number of possible outcomes =52 Number of red cards or clubs = 26 +13=39 So, number of favourable cua es =39 2, Required probability -2 a (3.1) (2.2). (1.3), (6.2). (5.3). (4.4). (3.5). 27. (d) Here. (2.6).(6.6)} Total number of outcomes =36 Number of favourable outcomes =9 Hence, P (sum of two numbers wil be multiple of 4) 3 (35° 4 28, (d) x =(sec A ~ tan A) (secB - tan8) (sec — tanC) =(6ec A + tanA) (secB + tanB)(secC + tanc) = (secA+ tan A) (secB + tanB) (secC + tanC) (sec A — tan A) (secB— tanB) (sec - tan) = {(secA + tan) (secB + tan8) (sec A + tan ))® = (sec? A—tan? A) (sec?B - tan?) (sec? - tan? C) =x? = sl = x=dl 28, (c) Let A(0, 0). B(2, 0), C(a, b) and D(O. 3) be the vertices of the rectangle ABCD. Mid-point of AC = Mid-point of BD (Ce2.0e2) (240,043) 2 2 2 3 b_3 = and = 5 = a=2andb=3 Hence, the coordinates of fourth vertex are (2.3) 30, (b) In AAFC, AF || HE A = AAFC ~ AEHC = AF LEC HE HC 9 E = 46 6 AC | el foe ne solutions | 49 = FH =6-3/2=9/2cm Similarly, GF =9/2 GH 2°2 ‘Area of rectangle GHED = GH xHE =9x6 = 54cm? 31, (d) Since BE Is the median. So, E's the mid-point of AC. TIP Concept of median and mid-point formula should be understood adequately Coordinates of E are A(0.-2) +0 244 C2. 4)-00 +. Length of metian through B=0e = (0-47 + 0-1 =4 Common aN Error ‘Some students ore not aware about that the medion divides the opposite side of triangle in equal parts. 32. (b)- TRICK seca - tan?e=1 1 san 0,4) sec + tan@ =m sec -tand m Now, sec @—tan*@-2sece-tand ec? — tan? 9) (sec? + tan? 8) -2 secd-tané =1-(sec26 + tan?@) -2sec8-tano = (seca - tano)? I (ea? +b? -20b =(0-b))) 33, (¢) We have, 18 = 2 x32, 24 =2? x3 30=2 x3x5,42=2 3x7 UCM =2? x3? x5 x7=2520 So, required least number is 2520 + 1=2521 34, (a) We know that the line A segment joining the ‘mid-points of two sides of the triangle is half the length of the third side 1 DE =< Al e a ° 1 1 {fin two triangles, corresponding sides are in the same ratio, then their corresponding angles are equal and hence the triangles are similar (SSS). vvai wu owiu50 | Samachaon of MATHEMATICS (Standard Level) class x | Term-1 ADEF ~ SABC (By SSS similarity criterion) | ‘Thus 2 ? : ot aDEF) _ (OF ek i eaaae) (AC. 274 35. (b) 1 AGB4) BS, 6) chy) TRICK The coordinates of point which divides the line segment joining the points (xy.3) ond (x,y) internally in the ratio m, zm, are By te Py msm "my +m Let Une is extended to C(-Ly)such that AB:EC =K:1 . (Coxks3 | . Ixk+3..5 and ye 36. (b) Internal angles of areguar Heagon =120" 120" ‘Area of one sector ==> Be cnx ada . Area of 6 sectors =6 x 277 =2ar? 37. (b) Wehave, R= 4m Hence, AB=BC =CA=R/3 = 43 [ 2 8 ince. R =2 I <2, Since, R =$handh = 3 0,Hence, (ZAOC =2ZABC =2x60°=120° Now, Required area =} (Area ofthe circle ~ Area of ABC) J fea? . 7 x (seer = Required area = lequired area = 5 fi 8 8? = (Gn -12V5)= Stan ~3\3)cm?, ‘3B. (b) It is given that 4 and -3 are the zeroes of the polynomial x? +(a +1) x +b. fort =-0-1 ‘Sum of zeroes =~ =» ' And product of zeroes =b (4)(-3)=6 be=-12 ie a-b=-2-(-12)=-24 12219 (0) Length of wie = 242222 44), Let side are 6x and 5x 3 6x +5x)=264 2x5 =60cm Va ae 40. (a) Given, ‘Adding both equations, we get 2x= Section-C 4A, (a) We have, f(x)=(x +2) +6 2444 hxs Genie 1as no factorisation. No real value of xis possible ie. no So, option (a) is correct. 42. (a) We have, F(x) = ax? - 6x-6 Constant ter Coefficient of = Now, Product of zeroes = So, option (a) is correct. 43, (b) We have, f(x) = x? ~2x ~B = x? ~4x = x(x —4)+2(x-4) (4-4) For the zeroes of f(x), x-4=00rx+2=0 o x -2 TR !¢K ——______—_—— B=2x4=8x1 + We will take 2 and 4 as a factors of 8. So, middle term is -2 ~442 So, option (b) is correct, axe 44, (ebuctel) x1 Coefficientof jow, sum of zeroes = — "ore entaf x = a+p= Constant term and product of zeroes = 7 Facet of = vuafitiec’' wi Caap” op So, option (a) is correct 45. (c) Let aandB be the zeroes of the required polynomial a=-3anda-p=0 -3-B=0 = B=0 Now. required polynomial = x? =x2-(-3+0)x40 x7 43x So, option (¢) is correct. 46, (Q Diameter of the concrete circular fountain at the centre = Width of the rectangle =14 m. :. Radius of the concrete circular fountain = 7m So, Area of the concrete circular fountain = xr? 2x 7x7 =154m? +8) x +08 So. option (c) is correct. 47. (d) Diameter of semi-circular region of the garden bed =l4m adius of the semi-circular end = 7m 1 ag? daca’ Z So, Area of one semi-circular end 1,22 ex x7 2 2 = T1m? So, option (d) correct. ‘48. (b) Area of the garden bed where grass is grown = Area of two semi-circular ends + Area of rectangular bed ~ Area of central circular region. Length of rectangular bed = 35 -2 xRadius of circular Semivend =35 -2 x7 =34-14 =21m and breadth = Mi Solutions | 54 +: Required area =2 xx nr? = xb —ar? alxb=21x%% 234m? So, option (b) is correct. 49, (a) Outer perimeter of the lawn =2 (Length of rectangle) +2 x Circumference of semi-circular ends =2x(t+1x: =2x(l+Jx2x1) -2x(a1-22x7) =2x43=86m So, option (a) is correct. 50. (b) We have FG =1+ 14 +1=16m —__ 5m ———> ‘Area of the pavement = Area of rectangle EFGH + 2 x (Area of semi-circle with diameter 16 m) - (Area of rectangle ABCD+ 2x (Area of semicircle with diameter 14 m)) 2 2 wer xF6+2x44(3) -[prcenge(i] 2 = 21x16 + x x (8)? ~(21x14 +n x(7)?) 2336 + 64m -294 - 490 =(42 + 15x) m2 9. option (b) ct. So, option (b) is correc Sa ied with Ca