Page 1 Sample Paper 6 Solutions Sample Paper 5 Solutions Class- X Exam - 2023-24 Mathematics - Standard ‘Time Allowed: 3 Hours General Instructions : 1, This Question Paper has 5 Sections A-E, 2, Section A has 20 MCQs carrying 1 mark each, 3. Section B has 5 questions carrying 02 marks each, Section C has 6 questions carrying 03 marks each, Section D hus 4 questions carrying 05 marks each, Scotion E has 3 case based integrated ‘and 2 marks each respectively. Maximum Marks : 80 of aswessmont (04 marks exch) with sub-parts of the values of 1, 1 7. All Questions are compulsory. However, an intemal choice in 2 Qs of 5 marks, 2 Qs of 3 marks and 2 Questions of 2 marks has been provided, An internal choice has been provided in the 2 marks questions of Section E. 8, Draw neat figures wherever required, Take x = wherever req Section - A Section A consists of 20 questions of 1 mark each. 1. If the distance between the points A(4, p) and B(1,0) ix 5 units thon the value(s) of p is(are) (a) 4 only (b) =4 only () +4 (ao Ans Given, points are A(4, p) and (1,0). d = Janay + FW) 5 =V(1-4)*+ (0p)? 25 =945" pi = 25-9 =16 potd ‘Thus (c) is correct option, 2. For which value(s) of p, will the lines represented by the following pair of linear equations be parallel Sr-y-5 =0 6e-2y-p =0 (a) all real values except 10 (b) 10 © 5p (aie Ans We have, 3r=y=5 =0 and Here, and Since given lines are parallel Hehge 7b * In figuro, on a citcle of radius 7 cm, tangent PTis drawn from a point P such that PT = 24 em. If 0 is tho contre of the circle, then the length of PR is (a) 30 em (b) 28 cm, (©) 2am (a) 25 em, Ans ‘Tangent at any point of a circle is perpendicular to the radius at the point of contact.Page 2 Thus or. Pr Now in right-angled triangle PTO OP = OT + PT = (7)? + (24)? = 494576 = 625 Thus OP = 25 em Since OR = OT because of radii of circle, PR = OP+ OR =25+7 =32 cm Tins (c) is correct option Each root of = b+ ronulting equation in 2222-4 1 = 0, the (@) (yb © (ab Ans For 2#— br+ ¢=0 we have +6 =b and of =e Now a-2+B-2satf—4=0-4 (28-2) = 08-20 +8)+4 =e-2b+4 For 2-22-4150 we have Qab-dad and lae-244 sen 2x6+4 ac-8 e=14+8=9 ‘Thus (a) is correct option, ‘The centroid of the triangle whose vertices are (3, ~ 7) (= 8,6) and (5, 10) is (a) (0,9) (b) 0.3) (©) (1.3) (a) 6.5) Ans ve (BEGINS Ta g-10) (9.9) ‘Thus (b) is correct option. 3) is docreasod by 2. The Sample Paper 5 Solutions If the sum of the zeroes of the quadratic polynomial ke? +22- 3k is equal to their produet, then equals 1 (a) 3 3 ©} ha Wohave pla) =ke +224 38 Comparing it byad’+ bt e, we get a =k, b and ¢ = 3k costae, ots ated Protuetafzerm, of = mS ms According to qeotin, we bare a+B =a8 2 2 ~a8 eka} ‘Thus (d) is correet option. cd Tho n torm of the AP «, 8a, 5a, is (a) na (b) @n=1)a (©) Qntia (a) 2a Ans Given AP is a, 34, 5a First term iv a and d = 3a—a = 2a ni term a =at(n-Id =0+(n=1)20 = 0+ 2na—2a 2na— (Q2n-1)a ‘Thus (b) is correet option. In an AP, if a= 3.5, d= 0 and n= 101, then a, will (a) 0 (0) 35 (c) 103.5 (a) 104.5, Foran AP, a =a¢(n~1)d 3.5+(101-1)x0 =35 ‘Thus (b) is correct option,Page 3 Sam « Paper 6 Solutions If one zero of the quadratic polynomial 2+ 32+ ix cose" ~ (15° ~ 6] - see(15* ~ 8) ~ tans" +6) + ot" ~ (5° +4) 2, then the value of Fis ‘ isa U8 @i0 ihe = sec(15* - 6) — sec(15* — #) — tan(55* + 0) (c) -7 (d) -2 +tan(55° +0) Ans =0 We have pa) = 2432+ k 12, If the angle of depression of an object from a 75 m 162 iv a zero of p(2), thon we have high tower is 30°, then the distance ofthe object from ws tho tower is p(2) =0 (0) 573 m (b) 5078 m (2)°+3Q2) +k =0 (©) V3 m (a) 150 m 4464k =0 Aus +k =05k=-10 ‘Thus (b) is correct option. Wehave — tan30' 10. In the adjoining figure, OABC ix a square of wide Zem. OAC is quadrant of a circle with O as centre. ‘The area of the shaded region is OB =T5V/3m = A ° c Be = Tower B Olobject) Thus (¢) ia correet option, A B (a) 10.5 cm? (b) 38.5em? (©) 49em? (a) 1.5 en? Ans 13. The 2 digit number which becomes $th of itself when digits are reversed. The difference in the digits of Ragnar eneam(3- Le) eat the number being 1, then the two digits mumber ix _ (ax Bx) (a) 45 (b) 54 a(pot, 2 * (a) None of these =(7-4x? 2 (A) None of th - (0 ete eee IC the two digs are 2 and y, then the mu ‘Thus (a) is correct option ete. LL, The value of the expression Now — Sort y) = Oy +2 cosec(75° + 6) — sec(15" — 8) - tan(55* +6) + cot(35* — 6) is Solving, we got 4Az-+ 55y (a) =1 (b) 0 z£ =} wt @y y oe sa Solving thom, wo get 2= 5 and y= 4. nber is 54 nm Thus (b) is correct option. conoc(75* +) ~sec(15* ~ 0) ~ tan(55* + 8) + cot(35" —Page 4 ua 15. 16. In the given figure, DEI BC. The value of EC is QD SY B Cc () Lem (b) Sem (c) 2em (@) 1m An: Since, DE WBC AD _ AE DB BC ‘Thus (c) is correct option, If the radius of the sphere is increased by 100%, the volume of the corresponding sphere is increased by (a) 200% (b) 500% (c) 700% (a) 800% Ans: Let 1 be the original radius of sphere. If we increased radius by 100 %. it will be 2r vader Me = $a (an)? a4axse = faxer ‘Thus now volume is 8 times of original volume, Hence when the radins is increased by 100%, the corresponding volume becomes 800% and thus increase is 700%. Thus (c) is correct option The median and mode respectively of a frequency dlistribution are 26 and 29, Then its mean is (a) 275 (b) 245 (©) 284 (a) 258 Aus W. 18. Sample Paper 5 Solutions Wo have 3My— 2M 3 x 26-2 2M =78-29=49 49 M a2 = 45 Thus (b) is correct option. An event is very unlikely to happen. Its probability is clonest to (a) 0.0001 (b) 0.001 (6) 001 (aor Ans ‘The probability of an event which is very unlikely to happen is closest to zero and from the given options 0.0001 is closest to zero, ‘Thus (a) is correct option. (24 IF 2 =0 has (a) four real roots (c) no real roots (b) two real roots (4) one real root Ans Webave (#4 1-2 =0 #+14+27-27 =0 #+241 50 (}+241=0 Let # = y then we have wt+ytl=0 Comparing with ay’ + by +e = 0 wo get a= 1, and ¢=1 D =#—4ac =(1F-4@)0) nee, D< 0, 9+ y+ 1 =0 has no real roots. io #4241 20 oF (P+ 1F—2 = 0 has no real roots Thus (¢) is correct option. Discriminant, DOWNLOAD FREE INDIA'S BEST STUOY MOBILE APP '= 30 Sample Paper wih Solutons '= Chapterwise Question Bank of All Subject '= NCERT and Previous 16 Years Solved Papers "= Case Study Questions with full Solutions '= Word File of Material for Teachers So get Mobile app Link Whatsapp at 89056 20060Page 5 DIRECTION : Inthe question number 19 and 30, a statement of Ansertion (A) i followed by a statement of Reavon (R) Choose the correction option, 19, Assertion : The value of y is 6, for which the distance between the points P(2,— 3) and Q(10,y) is 10. Reason : Distance between two given points ‘AC, 4) and B(x, y.) is given, AB =V(n- 2) +(n- WF (a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A). (b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A), (c) Amertion (A) ix true but reason (R) is false, (@) Assertion (A) is false but reason (R) in true, An PQ =10 Poel Pg =100 = (10-2f + (y+37 = 100 (+3) = 100-64 = 36 y+3 =+6 ya-36 y=3,-9 Assertion (A) is false but reason (R) is true. Tins (s) is correct option. 20, Assertion : z13- iv a term wating decimal fraction. 3. m5 Reason : If q=2°5" where m,n are nownegative integers, then £ is a terminating decimal fraction. (a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (a). (b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of ausertion (A). (c) Assertion (A) is true but reason (R) is false, (a) Assertion (A) is false but reason (R) is true, Aus We have 8125 = st = 5x2 Sinco the factors of the denominator 3125 SES is of the form 2" x 5%, ship is a terminating ‘decimal Sample Paper 5 Solutions Both sswertion (A) and reaso (R) is the correet explanati ‘Thus (a) is correct option, Section - B Section B consists of 5 questions of 2 marks each. 21, In AABC,AD 1 BO, such that AD'= BD x CD. Prove that 4 ABC is right angled at A. Ans [As per given condition we have drawn the figure below. A B D C Wo have AD! = BD x CD AD _ BD CD ~ AD Since 2D = 90", by SAS we have AADC -ABDA and 4BAD = ZACD; Since corresponding angles of similar triangles are equal ZDAC = ZDBA £BAD+ £ACD+ 4 DAC+ 2 DBA = 180° 22 BAD+22DAC = 180° ZBAD+ ZDAC = 90" 2A =90° Thus AABC is right angled at A, DOWNLOAR FREE INDIA'S BEST STUOY MOBILE APP '= 30 Sample Paper with Solutons ‘Chaplerwise Question Bank ofAll Subject NCERT and Previous 15 Years Solved Pepers = Case Study Questions with full Scltlons ‘= Word File of Materiel for Teachers 0 get Mobile app Link Whatsapp at 89056 29969Page 6 22. In figure, a circle touches all the four sides of a quadrilateral ABCD. If AB=6 em, BC=9 em and CD=8 em, then find the length of AD. c Ans [As per given information we have redrawn the figure below. Tangents drawn from an external point to a circle are equal in length. ‘Thus AP = AS and let it be Similarly, BP = BQ, CQ = CR and RD= DS Now BP = 4B-AP = BP = BQ =6- CQ = BC- BQ =9- (6-2) =342 Now, CQ =CR =3+2 RD = CD-CR =8-(3+2) Now, RD =DS = AD = AS+SD =2+5 find the value of A. Ans We have tan2A = cot(A~18") cot (90° = 24) = cot (A — 18°) Tow 90°24 = A-18* Sample Paper 6 Solutions 3A = +18" = 108" A =36" Find the mean the following distribution cas [35 [or [ro [ou [irs Frequency [5 [10 [10 |7__ [8 Ans Glass [Frequency [Mid-Vatue | £2, i) (a) 35 |s 4 20 sz__|w 6 oo v9 {to 8 80 ou |i 10 70 ui3_|s 2 96 Di=00 Din = 926 a Dla _ 326 Mean M a AIE = OP = 8.15 Find the mode of the following date : lass: [0 ]20- [40- [60 [80- | 100. |120- 20 |40 |60 {80 }100 [120 | 140 Frequency |6_[a_[1o [12 [6 Ans (Class 60-80 has the maxin del class, um frequency 12, therefore 26-6, $= 6 and h = 20 Moe, ae = 14 (AGE) 12-10 O43 I= 10Page 7 2%, Explain == why = (7X 13x 11)+11 and (XEX5XAXIxX2x1)4+3 are composite numbers, Ans (7x 13 x1) 411 = x (7 13-41) =11x (+1) aux 92 and (TXOX5XAX 3x2 1)43 =3(7X 6X 5x 4x 2x 141) =3 x (1681) =3 x 41 x 41 Since given numbers have more than two prime factors, both number are composite. or Explain whether 3x 12 x 101 +4 is « prime number ‘or a composite number, Ans greater than 1 that cannot be formed by 1 two smaller natural numbers, A natural number greater than 1 that is not prime is called a composite number. For example, 5 is prime becanse the only ways of writing it ax a product, 1x 5 or 5x1, involve 5 itself, However, 6 is composite because i is the product of two numbers (2 x 3) that are both smaller than 6. Every composite mmber can be written as the product of two or more (not necessarily listinet) primes. 3X 12x 10144 =4(3 x 3x 10141) = 2x 2x (10x 7x 13) 2x 2K 2K5XTX ID = 4(909 +1) = 4(010) = a composite number (DOWNLOAD FREE INDUS BEST STUDY MOBILE APP '= 30 Sample Paper with Solutions = Chapterwise Gestion Benk of ll Subject "= NCERT and Previous 15 Years Solved Papers = Case Study Questions with full Schitons = Word File of Material for Teachers ‘So get Mobile pp Link Whatsapp at 89086 20060 Sample Paper 5 Solutions Section - C Section C consists of 6 questions of 3 marks each. 26, Tho sum of four consecutive mumber in AP is 32 and the ratio of the product of the first and last term to tho product of two middle terms is 7 : 15, Find the numbers Ans Let the four conswentive terms of AP be (a—3d), (a~d), (a+ d) and (a+ 30) As per question statement we have a-3d+a-d+a+d+a+3d =32 4a =32> 0 =8 z and -$ we 5 960 — 135d" = 448 — 7 TE = 135d? = 448 - 960 1280 =-512 @ ated et? Hence, the number are 2, 6, 10 and 14 or 14, 10, 6 and 2. we that ; ot0-+eoseeO—1 _ 1+ cot 2B, Prove that : Sty E ese FT aint Ans cot + cosecd=1 2S = cot — cosec@ +1 atta! Z-aa1 sin 0(cos0 + 1 ~ sind) Sin O(cos= 1+ sind) = sincos0 + sind — sin’ “sin8(cos0-+ sind ~ 1) in Ocos + sin — (1 — cos*é) O(cos6 + sind= 1) ssin®(cos6+ 1) ~ (1 ~ cos*0) = sin (cos + sin0= 1) (1 + cos) (sind — 1 + cos6) sind (cos + sind — 1)Page 8 ular park. u. Find the area of the whose cireumference is 88 1m be the width of road. Circumference of « circular park, 2nr = 88m. Inner radins of park, + = 88 2 88x7 On “Ix =2x7=Mm Outer radius of park including road width Ra=rtw =M+T=21m ‘Area of the road, (R=) =x(R+ (R=) = Bear + 1421-14) 2 95 x7 = 770m? Hence, the arca of the rond is 70m’, In Figure, PQ and AB are two ares of concentric circles of radii 7 em and 3.5 cm respectively, with centre O. If 2POQ= 30", then find the area of shaded Ans redraw the given figure as below. Sample Paper 5 Solutions Q Po P Area of shaded region 8 2272 (5,5)9.30% alt= Agde = Fir'- 05) = Br + 3.5)(7- 358) xy Bei 1 Bxio5x35 xy = 9.625 em? 29, Compute the mode for the following frequency distribution: Size of items [o- [4 [8 [12 ]16- Jao. Jas. (in em) 8_|12 [16 |20 | 24 |28 Frequency [5 9 fiz [ia fro Jo Ans = 124246 = 14.46 30, Find the ratio in which P(4,m) divides the segment Joining the points A(2,3) and B(6,~3). Hence find Ans Let P(2,y) be the point which divide AB in k:1 ratio, (ym) k 1 Gam) A(2,3) P(4,m) (6-3) Now my, + 143, ‘me msPage 9 4 N41) =49H1) + dk+d = 6k42 6k 4k = 4-2 2k =2ak point P divides the line segment AB in 1:1 wnt may v= me me m =1X(-9)4109) T+ =2343 29 aae8 Thus m=0, In the given figure 4 ABC is an equilateral triangle of side 3 units, Find the co-ordinates of the other two vertices Ans The co-ordinates of B will be (2+3,0) or .(5,0) Let coordinates of C be (zy). Since triangle ix ‘equilateral, wo have AG = BC (e= 27 (= OF = (2-5) +(y-0F Peindety 7425-1004 OSE Gr =21 =i ty and (2-2F +(V- OF (2-2) +7 = ft =9 or, =9-$ Sample Paper 5 Satins Hesee © (5, 248). 31. Given that ¥‘5 is irrational, prove that 2V'5 ~ 3 is an jonal munber. Ans Assume that 2¥5 = 3 is a rational umber. Therefore, ‘we ean write it in the form of F where p and g are co-prime integers and q # 0. low B-3—ek N 2v5-3 : whore q #0 and p and q are co-prime integers, Rewriting the above expression as, =2. 2Vv5 b+3 +3 % Hero 243 iy rational because p and q are co egors, thus’ should be a rational number. But V5 iy irrational. This contradicts the given fact that V5 is irrational. Hence 25-3 is an irrational umber. 4 Section - D Section D consists of 4 questions of 5 marks each. 82, For what value of &, which the following pair of linear equatious have infinitely many solutions 2r-+ By = 7 and (k-+ I)r-+ (k= 1)y= 4k+1 Ans Wo have Qe By 7 and (k+ 1)r4(2k—1)y = 4k+1 Hoe a2 be 3 a “EFT b= @k=1) ant i eat @ ~ Sake) ~ Gee) For infinite many solutions oh oe a ha For 4 =4 we have @ eo == ri 2(4k-+1) = 7(+1) Bk+2 =7h47Page 10 kes Hence, the valne of & is 5, for which the given equation have infinitely many solutions. or The cost of 2 kg of apples and 1kg of grapes on a day was found to be Rs. 160. After a month, the cost of kg of apples and 2kg of grapes is Rs, 300. Represent the situations algebraically and geometrically. Ans Let the cost of 1 kg of apples be Rs.: and cost of 1 kg of grapes be Rs. y. The given conditions can be represented given by the following equations Bet y = 160 (1) Ary = 300 (2) From equation (1) y = 160-2 z 50 5 y oo 70 From equation (2) y = 150-22 5 50 40 y 50 70 Plotting these points on graph, we get two parallel Tine as shown below. Sseesezssae Ba. Sample Paper 6 Solutions 3B. Prove that opposite side of a quadrilateral circumseribing a circle subtend supplementary angles fat the centre of the circle, Ans A circle centre O is inscribed in a quadrilateral ABCD ‘as shown in figure given below. Since OE and OF are radius of circle, OE =0F ‘Tangent drawn at any point of a circle is perpendicular to the radius through the point contact. Thus ZOEA = 2OFA= 90° Now in 4 ABO and 4 AFO, OE = OF 4OEA = ZOFA=90° OA = 0A (Common side) ‘Thus AAEO = AAFO (SAS congruency) 27 =28 Similarly, 4122 “a=c4 25=26 Since angle around a point is 360°, L14 224-234 £44 254 264-274 21 2L142L842244225 = 360" 214 £84 £44 25 = 180" (214 28)4(244 25) = 180" 4AOB+ ZCOD = 180" Hence Proved. 34. The angles of depression of the top and hottom of an 8 m tall building from top of a multi-storeyed building are 30° and 45°, respectively. Find the height of multi- ing and distance between two buildingsPage 11 As per given in question we have drawn figure below. B A © Here AE= CD =8m BE= AB- AE =(h-8) and AC= DE = Also, ZFBD = ZBDE= 30° ZFBC = 2BCA = 45° In right angled 4 CAB we have (1) In right angled 4 DB + = BE tango* = 25 1. Wyre 2 = V3(i-8) 2) From (1) and (2), we get ha VBh-8v8 8/3 =V3h-h nw BYE y Bt1 Va-1* V341 = 4Y3(V3 +1) =(12-44V3) m a = (1244/3) Distance = (1244/3) m Sinco, 2 =48+¥3) m Hence the height of multi storey building is 4(3-+ 3) Sample Paper 6 Solutions Two poles of equal heights are standing opposite to each other on either side of « road, which is 80m wide, From a point between them on the road, angles of elevation of their top are 30° and 60°. Find the hoight of the poles and distance of point from poles. Ans Let the distance between pole AB and man E be 2, As per given in question we have drawn figure below. 4 c A h b NA d B—1—-E D Here distance between pole CD and In right angle triangle A ABE, tango" = A 4 Al) Wr (a) In angle triangle A CDE, tango" = h V8 = WF h es0v3—2V3 -Q) Comparing (1) and (2) we have w = 8073-23 2 s80X3-2x3 de = 240 240 s=teoon Substituting this value of x in (1) we have h = Dn 203 Hence, height of the pole is 34.64 mPage 12 35. A sold is in the form of a cylinder with hemispherical nd. The total height of the solid is 20 em and the ‘diameter of the cylinder is 7 em, Find the total volume of the solid, (Use w= #) Ans [As per given information in question we have drawn the figure given below. 3.5em 35 em re b+ -_—— tea + 20m ——4 Height of the eylinder, h = (20-7) cm Radius of cireular part, Bem r=tem =F Volume of solid, = Volume of eylinder +2 x Volume of hemisphere V anrth+2x (3x) Section - E Case study based questions are compulsory. 36. Maximum Profit : An automobile manufacturer ean produce up to 300 cars per day. Tho profit made from the sale of these vehicles ean be modelled by the function P(2) =~ +350r~ 6600 where (2) is the profit in thousand Rupees and is the number ‘of automobiles made and sold, Answer the following questions based on this model: (3) When no ears are produce what is a profit/los? (Gi) What is the break even point ? (Zero profit point is called break even) ? (Gli) What is the profit/loss if 175 cars are produced Sample Paper 5 Solutions What is the profit if 400 cars are produced ? Ans (i) When no cars are produced, P(0) =0+0~6600 6600 ‘Thus when no cars are produced, there is loss of Rx 6600 Thousand, (i) At break-even point P(2) = 0, thus 0 =~ + 3502—6600 2+ 3502+ 6600 = 0 2? — 3302-2024 6600 = 0 2(2~ 330) ~ 20(r4330) =0 (2= 330) (220) = 0 x = 20, 330 ‘Thus break even oceur if 20 of $30 ears are produced, (iii) When 175 cars are produce, P(1T5) == (175)? + 350 (175) ~ 6600 24025 Thore is profit of Rs 24025 thousand if 175 car are produced. or (iv) When 400 cars are produce, P(400) == (400)? + 350 (400) ~ 6600 =~ 26600 There is loss of Rs 26600 thousand if 400 cars are produced. DOWNLOAD FRE INDLA'S BEST STUBY MOB APP 90 Sample Paper wih Solitons '= Chaplerwise Question Bank of All Subject ‘= NCERT and Previous 15 Years Solved Pap 12 ' Case Study Questions with full Solutions = Word File of Materiel for Teachers 80 get Nobile app Link Whatsapp at 20056 20969Page 13, 37. Rohan ix vory intelligent in maths, He always try to relate the concept of maths in daily life. One day he it walking aivay from the base of « lamp post at a speod Of mf Lamp is 45. above the ground ~ B (i) If after 2 second, length of shadow is 1 meter, ‘what is the height of Roban (ii) What is the minimum time after which his shad- ‘ow will become larger than his original height? What is the distance of Rehan from pole at this point ? (iii) What will be the length of his shadow after 4 seconds? Ans: (i) As per question statement we make the diagram at following. 45 ml a Dl E At t= 2 seconds, BD =2x1=2m DE =1m Since, ABE ~ 4 CDE AB _ BE _ BD+ DE CD ~ DE ‘DE 2+ (ii) At point where shadow is equal to her heigh CD =DE =15m AB _ BE _ BD+DE cb = DE =~ DE Sample Paper 6 Solutions 45 = BDL 45 = BD+15 BD =45-15 =3m Time to reach at BD, t=} =3 wee As calculated in part (ii) we have BD = 3 m (iii) After 4 sec, BD =1x 4 =4 AB _ BE _ BD DE cb > DE ~~ BE 45 44 DE Su ApDE 3DE =4+DE » DE =2m jeal survey questions are questions asked to the opinions and attitudes of potential voters ical survey questions help you identify supporters and understand what the public needs. Using such questions, a pobtical candidate or ax orgunization can formulate policies to gain support from these people. A survey of 100 voters was taken to gather inforn oon critical issues and the demographic information collected is shown in the table, Ono out of the 100 voters is to he drawn at random to be interviewed on the India Today News on prime time. Women [Men | Totals Republican [17 20 37 Democrat [22 7 39 Indopendent [8 15 Green Party [6 3 5 Totals 53 a 100Page 14 (i) What is the probability the person is a woman or a Republican ? What is the probability the person is a Democrat (ii) What is the probability the person i a Independ- at men ? (iii) What is the probability the person is a Independ- nt men or green party men ? Ans : Total posible outcomes in all case is 100 because there are total 100 person and out of which we have to select one for interview. n(8) = 100 (j) Person is a woman (W) or Republican (R), Let B. be the event that the selected person is woman or a Republican, Favourable outcome, (i) Probal Person is a democrat, Let E; be the event that the selected person is, acest Favourable outcome, (i) = 39 Probability p(n) = = y= 0.30 i) Person is a Independent n Let B be the event that th Independent selected pernon is Favourable outcome, n() =7 mB) __7 3) ~ 100 (iii) Person is a Independent men or green party men, Let E, be the event that the selected person is Independent men or green party men, Probability, P(E) Favourable outeon Probability, Sample Paper 5 Solutions NODIA APP All NCERT MATERIAL SOLVED a ee CW Uae AL) See id eel