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MATHEMATICS STANDARD 2024 == SECTION -A 20 Marks (This section consists of 20 questions of 1 mark each) 1..A bag contains 3 red balls, 5 white balls and 7 balck balls. The probability that a ball drawn from the bag at random will be neither red nor black is: @ 5 (OMe © & @ & 1 2. The probablity of getting a bad egg in a lot of 400 eggs is 0.045. The number of good eggs in the lot is: @ 18 © 382 (b) 180 @ 220 1 3. Perimeter of a sector of a circle whose central angle is 90° and radius 7 cm is: (a) 35cm (b) 11. cm «) 22cm (d) 25 em 1 4. A pair of irrational numbers whose product is. a rational number is: @ (16,74) (&) (5,72) © 3,727) () (/36,V2) 1 5. Maximum number of common tangents that can be drawn to two circles intersecting at two distinct points is: @ 4 (b) 3 © 2 @i a 6. From a point on the ground, which is 30 m away from the foot of a vertical tower, the angle of elevation of the top of the tower is found to be 60°. The height (in metres) of the tower is: @ 10/3 (b) 30/3 © 60 (d) 30 1 7. Given HCF (2520, 6600) = 40, LCM (2520, 6600) = 252 x k, then the value of k is: (@) 1650 (b) 1600 © 165 @ 1625 1 8. The mean of five numbers is 15. If we include one more number, the mean of six numbers becomes 17. The included number is: (@) 27 (b) 37 © 17 (@) 25 1 9. The pair of linear equations x + 2y + 5 = 0 ¥ and - 3x = 6y - 1 has: (@)_ unique solution (b) exactly two solutions (© infinitely many solutions (d) no solution 1 10. In the given figure, O is the centre of the circle. MN is the chord and the tangent ML at point M makes an angle of 70° with MN. The measure of ZMON is: (a) 120° (b) 140° © 70° @ 90° 1 11. In AABC, DE || BC (as shown in the figure). If AD = 2 cm, BD = 3 cm, BC = 7.5 cm, then the length of DE (in cm) is: A D E B c (@) 25 (b) 3 © 5 @6é 1 12. Two dice are thrown together. The probablity that they show different numbers is: @ 2% ® % © 4% @ 2 1 13. If sina = 2 cos = AB. then tan a. tan B. is: @ v3 ® % © 1 @o fi 14. What should be added from the polynomial - 5x + 4, so that 3 is the zero of the resulting polynomial? @ 1 (2 x © 4 @5 aa 15. The smallest irrational number by which ¥20 should be multipled so as to get a rational number, is: @ 20 (b) ¥2 © 5 @ v5 1 16. If the diagonals of a quadrilateral divide each other proportionally, then it is a: (a) parallelogram (b) rectangle (©) square (d) trapezium 1 17. The common difference of the A.P. 1 1-4 1-8x a Dx De ett SE (@) -2x (b) -2 © 2 (d) 2x 1 18. In the given figure, if PT is a tangent to a circle with centre O and ZTPO = 35°, then the measure of 2x is ae (@) 110° (b) 115° © 120° (d) 125° 1 Directions: Question number 19 and 20 are Asssertion and Reason based questions carrying SECTION - B 1 mark each. Two statements are given, one labelled as Assertion (A) and the other is labelled as Reason (R). Select the correct answer to these questions from the codes (A), (B), (C) and (D) as given below: (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A). (8) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A). (©) Assertion (A) is true, but Reason (R) is false. (D) Assertion (A) is false, but Reason (R) is true. 19. Assertion (A): The point which divides the line segment joining the points A (1, 2) and B (-1, 1) internally in the ratio 1: 2is (34) 373): Reason (R): The coordinates of the point which divides the line segment joining the points A(xz, yx) and B(x, yz) in the ratio m;: m2 are (meetin mae mat) mtm ' mtm al 20. Assertion (A): In a cricket match, a batsman hits a boundary 9 times out of 45 balls he plays. The probalitiliy that in a given ball, he does not hit the boundary is 4. Reason (R): P(E) + P(not E) = 1. 1 10 Marks (This section consists of § questions of 2 marks each) 24. If 2x + y = 13 and 4x - y = 17, find the value of (x - y). OR Sum of two numbers is 105 and their difference is 4S. Find the numbers. 2 : Stan 60° 22. Evaluate: [3,?60" + cos*60°)tan30" 4 23. In the given figure, £4-£8, prove that AEAB ~ AECD. D 24. One card is drawn at random from a well shuffled deck of 52 cards. Find the probablity that the card drawn (A) is queen of hearts; (B) is not a jack. 2 25. Find a relation between x and y such that the point P(x, y) is equidistant from the points A(7, 1) and B(3, 5). OR Points A(-1, y) and B(5, 7) lie on a circle with centre O(2, -3y) such that AB is a diameter of the circle. Find the value of y. Also, find a _the radius of the circle. 2 37. Age 15-19 | 20-24 25-29 (in years) Number of 62 132 96 participants From the above answer the following questions: (A) What is the lower limit of the modal class of the above data? at (8) Find the median class of the above data. OR Find the number of participants of age less than 50 years who undergo vocational training. Ps (©) Give the empirical relationship between mean, median and mode. 1 Teaching Mathematics through activities is a powerful approach that enhances students’ understanding and engagement. Keeping this in mind, Ms. Mukta planned a prime number game for class 5 students. She announces the number 2)in.her class and asked the first student to multiply it by a prime number and then pass it to second student. Second student also multiplied it by a prime number and passed it to third student. In this way by multiplying to.a prime number, the last student got 173250. Now, Mukta asked some questions as given below to the students: (A) What is the least prime number used by students? 1 (®) How many students are in the class? OR What is the highest prime number used by students? 2 30-34 35-39 40-44 | 45-49 | 50-54 13 11 10 4 (C) Which prime number has been used maximum times? 1 38. A stable owner has four horses. He usually tie these horses with 7 m long rope to pegs at each corner of a square shaped grass field of 20 m length, to graze in his farm. But tying with rope sometimes results in injuries to his horses, so he decided to build fence around the area so that each horse can graze. A 7m 20 m| B c Based on the above, answer the following questions: (A) Find the area of the square shaped grass field. 1 (8) Find the area of the total field in which these horses can graze. OR If the length of the rope of each horse is increased from 7 m to 10 m, find the area grazed by one horse. (Use x = 3.14) 2: *(C)What is area of the field that is left ungrazed, if the length of the rope of each horse is 7 m? 1 * The unit mentioned at the end of question is incorrect in Board Paper and hence the entire question is solved in metre (m). SOLUTIONS SECTION -A ib 1. @ 3 Explanation: No. of red balls = 3 No. of white balls = 5 No. of black balls = 7 Total balls = 15 Probability that ball drawn is neither red nor black = & = + 1S ceed @ Concept Applied ‘“ The probability of drawing neither red nor black balls will be obtained by removing red and black balls from total balls. 2. (c) 382 Explanation: Probability of getting bad eggs in the lot = 0.045 Let the no. of bad éggs = x ~. Probability of bad eggs _ No.of bad eggs Total eggs = 0.045 = ~_ 400 > x= 400 x 0.045 = x=18 No. of bad eggs = 18 No. of good eggs = 400 - 18 = 382 3. @25cm Explanation: A Ex § f/ ae E 2nr0 Perimeter of sector = 2r + 360° =2x7+ ay oe ve 180° =14411 = 25cm A Caution \ Students may take only length of arc which can lead the wrong answer 11 cm which is option (b) a distractor. 4. © V3.N27 Explanation: Here 3 and ¥27 both are irrational numbers. The product of 3x27 = /3*27 = vei +. 9 is a rational number. A Caution Question should be read carefully as it is saying product of two irrational numbers. In option (a) and (@) numbers are not irrationals while in option (b), the product of numbers is irrational. 5. @2 Explanation: <== cme R Ss Here, circle with centre O and O' are intersecting at two distinct points A and B. So, in this situation PQ, RS are the tangents which can be drawn. 6. () 30V3 Explanation: Let BC be the tower and A be the observation point. ‘C L) Cc A<30m—> B AB = 30m ZCAB = 60° x Let, BC =hm ; A Caution © Students generally make mistakes in applying comparison of ratios. They write wrong ratios like AD _ OE ‘BD UBC which will give them wrong result. Option (©) which is a distractor. 12. @ 2 Explanation: Total outcomes, when two dice are thrown = {(1, (1, 2)(2, 3), 4)(2, 5). 6) (2, 1)(2, 2)(2, 3)(2, 4)(2, 5)(2, 6) (3, YG, 2B, 3)(3, HG, 5)G, 6) (4, 1)(4, 2)(4, 3)(4, 4)(4, 5)(4, 6) 6. YG. YG. 36, 4G, 5), 6) 196, 26, 36, 4), 56. 6} Favourable outcomes={(1, 2)(1, 3)-(2, 1), (2, 3). (3, (3, 2)(3, 4)..(4, D... (6, 5) No. of favourable outcomes = 30 Total outcomes = 36 30 So, P® = =~ 0, Cc. . ey 6 13. (1 Explanation: sina = —— 2. => sin a = sin 60° => a = 60° And cos B = ae = cos B = cos 30° = B =30° tana .tan B = tan 60° tan 30° 1 Pees ¥3 =1 14, (b)2 Explanation: Let, fix) =x? - 5x44 Let p should be added to Ax) then 3 becomes zero of polynomial. So, 3) +p =0 = 37-5«3+4+p=0 > 9+4-15+p =0 = -2+p=0 => p=2 So, 2 should be added. Concept Applied ‘ if any number is a zero of polynomial then value of the polynomial becomes zero at that number. 15.) /5 Explanation: @ 20x20 =20= Zqz0 But 20 is not the smallest among all options. (&) 20x /2= 40 = 275 is irrational © ¥20x5= 10Y5;is irrational @ V20xV5 = Vi00 10_P se =5.q#0 Th eqe Hence, option (d) is correct. 16. (d) trapezium Explanation: t AO BO Given: =a oc OD 41 =22 [vertically opposite angles] AAOB ~ ADOC So, 23 =24 ie, alternate 4s are equal AB || CD Hence, ABCD is a trapezium. 17. @)-2 Explanation: 18. (a) 125° Explanation: Ps ZOTP = 90° [Line from centre is 1 to tangent at point of contact] x= ZTPO + ZOTP [Exterior Angle Property] x = 35° + 90° = 125° @ Concept Applied + Exterior angle theorem states that when side of a triangle is extended, the resultant exterior angle formed is equal to the sum of the two opposite interior angles of the triangle. 19. (d) Assertion (A) is false, but Reason (R) is true. MY, + MAY, m,+m, 1x1+2x2 142 144 8 = 3 15 So point of intersection is (53) but. in -15 assertion, it is (33): So, assertion is wrong and reason is true. 20. (b) Both Assertion (A) and Reason (R) are true but Reason (R) is the correct explanation of Assertion (A). Explanation: Total balls (outcomes) = 45 No. of times boundaries are hit = 9 eS se 9 Explanation: Assertion says that point (2 | PAE = hating the bode) ae. divides the line joining the points A(1, 2) and 1 BC-1, 1) in 1:2. = e oS Aart Xp HM X, oe 1 -. By section formula, x = m,+m, P(E = not hitting the boundary) = 1- = Pee el eT 4 5 3 BS) 5 SECTION - B 21. 2x+y=13 10) ‘on solving eq. (i) and (ii) ; 4x-y=17 ¢ wii) 2x = 150 On adding ea. (i) ie) = x=75 : z 2 Put the value of x in eq. (i) Put the value of x in eq. () 2 Jot NOS 2eSey 213 => y=105-75 = 10+y =13 = y= 30 y=3 The numbers are 75 and 30. oo Re Ula 5 5 tan60° Sx ae * (Gin?60°+cos?60°)tan30° 4, 1 OR < V3 Let the numbers be x and y; x > y ee _ xeyed0e o* V3 x8 The difference of two numbers is 45. =5x3 & x-y=45 --(ii) =15 ; OR => a=1 s R z S29 = 22 [241 + (20-1) x2] pee ee oe Ea =10(2« 19 x 2] = 10[2 +38 Here PR:QR=2:1 Rees) = 40x10 So, mim, =2:1 = 400 eo; TI Now, m,+m, ) Concept Applied Bu eeec on formuld) Sum of n terms of an AP, S,= [2a + (n-1)d] 23 + 1x2 Be qe See! Let the AP be a,a+d,a + 2d, = 832 To 2 a 73 peg Od 4 my, +My, = ae st Deane ass a+29d 3 , +m, = 2x24 1x5 = 3a+27d=a+29d Se = 2a-2d=0 445 3 = 2a = 2d meek = =3 a=d (i) y=3 Now, Sg = 42 4 So, R= (#3) $i20-+(6-1)q]= 42 27. $7 =49 = Zia-+(7—14\ =49 = ene bd 7 = 2a+6d =14 = a+3d=7 S17 = 289 = Zpa+i6d) = 289 = 2a 46d S eee2 17 > 2(a + Bd) = 34 => a+8d=17 From eq (i) and (i) -5d=-10 2 d=2 Put the value of d in eq. (), a+3x2=7 => a+6=7 — a=7~-6 => 3[2a + Sd] = 42 => 2a+ Sd=14 => a+ Sa=14 [From eq. (i)] =) Ja= 14 => ae : 7 0) : eo So First term = 2 Common difference = 2 28. x-y+1=0 wi) x+y=5 (ii) From eq. (i) xsy-1 mC) at y=1, x=0 y=2, x=1 Usa we 1 2: 3 10) 1 2 From eq. (ii) xs xty=5 = x=5-y at Y-=0-x => On putting the values on graph we get the solution x = 2, y = 3 as the lines intersect at (2, 3). y x=+ V15 Zeroes willbe a= Vi5,B=- Vi5 Verification: Given polynomial x? -15 On comparing above polynomial with ax? + bx +c we have a=1,b=0,c=-15 Sum of zeros = a +B =vis-Vi5 -¢- Product of zeros = af vis (-v15) = alee Tur ig Hence, verified. A Caution ‘“ When students find the value of x they should write — x = 4V 15. Sometime they write x = 15 which is incorrect. 30. Join OA and OB. In AOAP and AOBP, OA = OB [Radii of same circle] PA = PB [Tangents from external point] OP = OP [Common] So AOAP=AOBP _[By SSS rule] A=2 [By CPCT] Now, In AATP and ABTP, PA =PB [Tangents from external point] [By CPCT] ZATP = ZBTP [By CPCT] _ ..(ii) AT = BT Since, ATB is a straight line. ZATP + ZBTP’= 180° = ZATP+ ZATP = 180° [From (ii) > 2ZATP = 180° = ZATP = 90° m(D) From (i) and (ii) we can say that OP is 1 bisector of AB. Z\ caution © Students may apply similarity concepts which is wrong and leads to deduction of marks. 31. LHS.=(cosec 6- sin 6) (sec 8 — cos 6) (tan8 + cot 8) = (cosec @ - sin 8) (sec 6 — cos 6) (tan 6 + cot 6) _ {1-sin?@ |{ 1-cos?@ |/ sin’@+cos?0 | sine cos@
[email protected] © cos*@ sin’ 1 = eae sin®@ cos@
[email protected] 0 1 = sin 8.cos 6B————— sin 6.cos 8 [-- sin?e + cos?6 = 4] =1=RHS. 3 ‘Hence, proved. Surface area of capsule = curved surface area of two hemispheres + curved surface area of cylinder =2x2 m+2nrh = 2nr [2r +h] S 2x22 x214+10) nae 7 = 176mm? Volume of capsule = volume of 2 hemisphere + volume of cylinder = Ane earth 3 [-- Volume of 2 hemisphere = volume of a sphere] =a (Sr) 3 2259x2[4.2410 7 3 = x 4x| 2410 7 3 2438 7 3 88x38 3344 pire 2” El 159.24 mm3 34. Let the speed of aircraft be x km/hr. Time taken to cover 2800 km at speed of xkmjhr = 2892p, New speed is (x — 100) km/hr So time taken to cover 2800 km at the speed 2800 of (x — 100) km/hr = x- 100 hr ATQ, 2800 _ 2800 _ 1 x-100 xX 2 x-x+100 4 Sal ( 1) 2 5 i007 = 84 ye -100x 2*2800 > 560000 = x? - 100x = x?—100x - 560000 = 0 = — x?-800x + 700x = 560000 = 0 > x(x - 800) + 700(x - 800) = 0 = (x = 800) (x + 700) = 0 x = 800, - 700 (Neglect) x =800 Speed = 800 km/hr 2800 Time = 800 = 3 hr 30 min OR Let the numerator be x. Denominator = 2x + 1 Fraction = 2x+1 2xt+1 16 ATQ, = 26 2 2x41 vs x err Let, Beas Z 2x41 Then, the equation will become pe yo 21 2 = yest _ 58 y ai = 21y? +21 = 58y > — -21y? - 58y + 21 =0 > 21y? - 49y - 9y + 21=0 = 7y(3y - 7) - 33y-7)=0 = (y-7(7y-3)=0 yee 37 z is 7 3 ~. Required fraction will be 3 and 7: & Concept Applied Students should assume y = —* which makes calculation easy. 2x+1 35. AD LBC and AD? = BD x DC ie, AD x AD = BD x DC AD _ BD Dc AD and ZADB = ZCDA [Each 90°] = AADB ~ ACDA Z1=22 [By Corresponding angles of similar triangles] 13224 0) In AADC, 23+ ZADC + 21 = 180° 243+90° + 41 = 180° AE =180"S 905-73 41=90°- 43 ZBAC = 21+ 24 =90°- 43+ 43 [From eq. (i)] ie, ZBAC = 90° Hence, proved. SECTION - E 36. First convert the given table in exclusive form. Subtract 0.5 from lower limit and add 0.5 to upper limit, so the new table will be: Age (in years) No. of Participants 145 -19.5 62 19.5 - 24.5 132 245 - 29.5 96 29.5 - 34.5 37 345 - 39.5 13 39.5 - 44.5 11 445-495 10 495-545 4 (A) Modal class is the class with highest frequency, so, it is 19.5 - 24.5. Hence, lower limit-will be ‘19.5’. ®) Age No. of (Coin) Participants C.F. @ 145-195 62 62 19,5- 24.5 132 194 245-295 96 290 295-345 7 327 345-395 13 340 395-445 11 351 445-495 10 361 495-545 4 365 ‘Ngee 1656. 2 Medium class will be 19.5 - 24.5 OR Approximately 361 participants are there in Class Interval 44.5-49.5 showing 361 cumulative frequency. Hence 361 participants are less than 50 year of age who undergo vocational training. (© Empirical relationship between mean, median and mode, Mode = 3 Median - 2 Mean 37. (A) 173250 86625 28875 9625 1925 385 77 11 an [Fle ml gs eS ak So least prime no. used by students = 3 (B) As the last student got 173250=2«3x3 x3x5xS5x5x7x1l there are 7 factors other than 2, which is announced by teacher. So, number of student = 7 OR Highest prime number used by student = 11 (C) Prime number 5 is used maximum times ie. 3 times. 38. (A) Area of square shaped field = 20x 20 = 400 sq.m (B) Area of 4 quadrant = area of a circle of radius 7 m = nr? pa. = AF xT xT = 154m? OR New radius = 10m So, area grazed by one horse = 4 (Area of circle with radius 10 m) Fn x (10)? _ 3:14%10%10 = 4 = 78.5 m? (© Area of ungrazed portion = Area of square field ~ Area of circle with radius 7m = 20x20- 2277 = 400-154 = 246 m?
