Page 1 :
CLICK HERE, , Download INDIA’s Best Study App having PDFs of, All Sample Paper and Question Bank, , Page 22, , Sample Paper 23 Solutions, , CBSE Maths Basic X, , Sample Paper 23 Solutions, Class – X Exam 2021-22 (TERM – II), Mathematics Standard (041), Time Allowed: 120 minutes, , Maximum Marks: 40, , General Instructions:, 1. The question paper consists of 14 questions divided into 3 sections A, B, C., 2. All questions are compulsory., 3. Section A comprises of 6 questions of 2 marks each. Internal choice has been provided in two questions., 4. Section B comprises of 4 questions of 3 marks each. Internal choice has been provided in one question., 5. Section C comprises of 4 questions of 4 marks each. An internal choice has been provided in one question. It, contains two case study based questions., , SECTION A, 1., , be d and number of terms be n ., Let an = 100, Here a = 25, d = 28 − 25 = 31 − 28 = 3, , If ½ is a root of the equation x2 + kx − 54 = 0 , then, what is the value of k ?,
,
Sol :, We have, , 100 = 25 + ^n − 1h # 3, , 100 − 25 = 75 = ^n − 1h # 3, , x2 + kx − 5 = 0, 4, , 25 = n − 1, , Since, ½ is a root of the given quadratic equation, it, must satisfy it., Thus, , n = 26, Since 26 is an whole number, thus 100 is a term of, given AP., , 2, b 1 l + kb 1 l − 5 = 0, 2, 2, 4, , 1+k −5 =0, 4 2 4, , 3., , 1 + 2k − 5 = 0, 4,

o, Find the nature of roots of the quadratic equation, 3x2 + 4 3 x + 4 .,

Sol :, , Now, , 7th term from the original end,, a7 = a + 6d, , 3x2 + 4 3 x + 4 = 0, , Here,, , a = 3 , b = 4 3 and c = 4, , Now, , D = b2 − 4ac = ^4 3 h − 4 ^3 h^4 h, , a7 = 184 + 6 ^− 3h, , , = 184 − 18 = 166 ., , 2, , Hence, the equation has real and equal roots., Find, 100 is a term of the AP 25, 28, 31, ...... or not.,
,
Sol :, Let the first term of an AP be a , common difference, , d = 7 − 10 =− 3, a = 184, n = 7, , Hence, 166 is the 7th term from the end., , , = 48 − 48 = 0, , 2., , Find the 7th term from the end of AP 7, 10, 13, .... 184.,

Sol :, Let us write AP in reverse order i.e., 184, ..... 13, 10, 7, Let the first term of an AP be a and common, difference be d ., , 2k - 4 = 0 & k = 2, , We have, , an = a + ^n − 1h d,, , Now, , 4., , Draw a line segment of length 5 cm and divide it in, the ratio 3 | 7 .,

Sol :, Step of Construction :, 1. Draw a line segment AB of length 5 cm., , CLICK HERE To Purchase Hard Books of CBSE Online Sample Papers., 20 Sample Paper in Each Subject and Rs 500/- For 4 Subjects

Page 2 :
CLICK HERE, For 30 Sample Papers With Solution, You have to download NODIA SolutionApp, , CBSE Maths Basic X, 2., 3., 4., 5., , Sample Paper 23 Solutions, , Draw any ray AX making on acute angle with, AB ., Mark ten points A1, A2, A3, ...A10 on AX such that, AA1 = A1 A2 = ... = A9 A10 ., Join BA10 ., At point A3 draw a line PA3 parallel to BA10 ., Hence AP | PB = 3 | 7, , Page 23, , Class, Frequency c.f. xi, Interval (fi), , Ui = x − a, h, , fi ui, , 55-70, , 6, , 18 62.50= 0, a, , 0, , 70-85, , 6, , 24 77.5, , 1, , 6, , 85-100, , 6, , 30 92.5, , 2, , 12, , / fi = 30, Mean,, , x = a+, , / fi ui =− 1, , / fi ui h, / fi #, , , = 62.5 + − 1 # 15, 30, , = 62.5 − 1 = 62.5 − 0.5 = 62, 2, , 5., ,

o, Observations of some data are x5 , x , x3 , 23x , x4 , 25x and, 3x, 4 where x > 0. If the median of the data is 4, then, what is the value of x ?, , A sphere of maximum volume is cut out from a solid, hemisphere of radius 6 cm. Find the volume of the, cut out sphere.,

Sol :, ,
,
Sol :, Given observations are x5 , x , x3 , 23x , x4 , 25x and 34x, where x > 0. On arranging the above observations in, ascending order, we get, x , x , x , 2x , 2x , 3x , x, 3, 5 4 3 5, 4, , Here diameter of sphere is equal to the radius of, hemisphere which is 6 cm., Diameter of sphere = Radius of hemisphere, , = 6 cm, Radius of sphere = 3 cm, , Here, total number of observations are 7, which is, odd., , V = 4 πr 3, 3, , = 4 # 22 # 33 cm3., 3, 7, , th, Median = b n + 1 l observation, 2, , Volume,, , th, , = b 7 + 1 l observation, 2, th, , = 4 observation = 2x, 5, 2, x, Median =, =4, 5, , , = 113.14 cm3., 6., , Find the mean of the following distribution :, Class, , 1025, , 2540, , 4055, , 5570, , 7085, , 85100, , Frequency, , 2, , 3, , 7, , 6, , 6, , 6, , x = 4 # 5 = 10, 2, , Section B, ,
,
Sol :, Let a = 62.5 be assumed mean., , 7., , Class, Frequency c.f. xi, Interval (fi), , Ui = x − a, h, , fi ui, , 10-25, , 2, , 2, , 17.5, , -3, , -6, , 25-40, , 3, , 5, , 32.5, , -2, , -6, , 40-55, , 7, , 12 47.5, , -1, , -7, , The angles of depression of the top and bottom of a, 50 m high building from the top of a tower are 45º, and 60º respectively. Find the height of the tower, and the horizontal distance between the tower and, the building. (Use 3 = 1.73),
,
Sol :, As per given in question we have drawn figure below., Here AC is tower and DC is building., , CLICK HERE To Purchase Hard Books of CBSE Online Sample Papers., 20 Sample Paper in Each Subject and Rs 500/- For 4 Subjects

Page 3 :
CLICK HERE, , Download INDIA’s Best Study App having PDFs of, All Sample Paper and Question Bank, , Page 24, , Sample Paper 23 Solutions, , 9., We have, , tan 45º = h − 50, x, x = h − 50, , and, , ...(1), , tan 60º = h, x, , CBSE Maths Basic X, , A right circular cone of radius 3 cm, has a curved, surface area of 47.1 cm2. Find the volume of the cone., (Use π = 3.14 ),
,
Sol :, We have,, , r = 3, πrl = 47.1, , Thus, , l =, , 3 =h, x, , h =, , x = h, 3, , ...(2), , 52 − 32 = 4 cm, , Volume of cone,, 1 πr2 h = 1 3.14 3 3 4, # # #, 3#, 3, , From (1) and (2) we have, h - 50 = h, 3, , , = 37.68 cm3, , 3 h - 50 3 = h, , 10. The following table shows the weights (in gms) of a, sample of 100 apples, taken from a large consignment, :, , 3 h - h = 50 3, , h ^ 3 - 1h = 50 3, , 50 ^3 +, h = 50 3 =, 2, 3 −1, , 3h, , , = 25 (3 + 3 ), , = 75 + 25 3 = 118.25 m, Thus h = 118.25 m., 8., , 47.1 = 5, 3 # 3.14, , Construct a pair tangents PQ and PR to a circle of, radius 4 cm from a point P outside the circle 8 cm, away from the centre. Measure PQ and PR .,

Sol :, Steps of Construction :, 1. Draw a line segment OP of length 8 cm., 2. Draw a circle with centre O and radius 4 cm., 3. Taking OP as diameter draw another circle which, intersects the first circle at Q and R ., 4. Join P to Q and P to R . On measuring, we get, PQ = PR = 5 cm, , Weight, (in, gms), , 5060, , 6070, , 7080, , 80- 9090 100, , 100110, , 110120, , 120130, , No. of, Apples, , 8, , 10, , 12, , 16, , 14, , 12, , 10, , 18, , Find the median weight of apples.,
,
Sol :, C.I., , 5060, , 6070, , 7080, , 8090, , 90100, , 100- 110- 120110 120 130, , f, , 8, , 10, , 12, , 16, , 18, , 14, , 12, , 10, , c.f., , 8, , 18, , 30, , 46, , 64, , 78, , 90, , 100, , N = 100 ; N = 50, 2, Cumulative frequency just greater than N2 is 64 and, the corresponding class is 90-100. Thus median class, is 90-100., We have, , Now, l = 90 , f = 18 , F = 46 and h = 10, Median,, , Md = l + d, , CLICK HERE To Purchase Hard Books of CBSE Online Sample Papers., 20 Sample Paper in Each Subject and Rs 500/- For 4 Subjects, , N, 2, , −F, h, f n

Page 4 :
CLICK HERE, For 30 Sample Papers With Solution, You have to download NODIA SolutionApp, , CBSE Maths Basic X, , Sample Paper 23 Solutions, , Page 25, AD = BC = 3000 3 m, , , = 90 + b 50 − 46 l # 10, 18, , Let the speed of the aeroplane be x ., AB = DC # 30 # x = 30x m ...(1), , , = 90 + 40 = 92.2, 18, , In right TAED, we have, , , = 92.2 gm., , tan 60º = AD, DE, , Thus median weight is 92.2., , 3 = 3000 3, DE, ,

o, Find the mean of the following data :, Class, , Less, than, 20, , Frequency 15, , Less, than, 40, , Less, than, 60, , Less, than, 80, , Less, than, 100, , 37, , 74, , 99, , 120, , DE = 3000 m, , ...(2), , In right TBEC,, tan 30º = BC, EC, 1 = 3000 3, DE + CD, 3, ,
,
Sol :, We prepare following table to find mean., , DE + CD = 3000 # 3, , C.I., , fi, , xi, , xi fi, , 3000 + 30x = 9000, , 0-20, , 15, , 10, , 150, , 30x = 6000, , 20-40, , 22, , 30, , 660, , 40-60, , 37, , 50, , 1850, , Hence, speed of plane is 200 m/s, , 60-80, , 25, , 70, , 1750, , 80-100, , 21, , 90, , 1890, , , = 200 # 18 = 720 km/hr, 5, Hence, the speed of the aeroplane is 720 km/hr., , / fi = 120, Mean, , M =, , / fi xi, / fi, , / xi fi = 6, 300, = 6300 = 52.5, 120, , Section C, 11. The angle of elevation of an aeroplane from a point, on the ground is 60º. After a flight of 30 seconds the, angle of elevation becomes 30º. If the aeroplane is, flying at a constant height of 3000 3 m, find the, speed of the aeroplane.,
,
Sol :, , x = 200 m/s, , 12. a, b and c are the sides of a right triangle, where c, is the hypotenuse. A circle, of radius r , touches the, sides of the triangle. Prove that r = a + b − c ., 2,
,
Sol :, As per question we draw figure shown below., , As per given in question we have drawn figure below., Here, , Let the circle touches CB at M , CA at N and AB, at P ., Now OM = CB and ON = AC because radius is, always perpendicular to tangent, +AED = 60º, +BED = 30º, , OM and ON are radius of circle, thus, OM = ON, , CLICK HERE To Purchase Hard Books of CBSE Online Sample Papers., 20 Sample Paper in Each Subject and Rs 500/- For 4 Subjects

Page 5 :
CLICK HERE, , Download INDIA’s Best Study App having PDFs of, All Sample Paper and Question Bank, , Page 26, , Sample Paper 23 Solutions, , CM and CN are tangent from C , thus, CM = CN, Therefore OMCN is a square. Let, Let OM = r = CM = CN = ON, Since length of tangents from an external point to a, circle are equal,, AN = AP, CN = CM and BM = BP, Now taking, , CBSE Maths Basic X, , 13. Sales Goals : At the time that I was newly hired, 100, sales per month was what I required. Each following, month the last plus 20 more, as I work for the goal of, top sales award. When 2500 sales are thusly made, I, get a holiday package., (i) How many sales were made by this person in the, seventh month?, (ii) What were the total sales after the 12th month?, , AN = AP, , AC - CN = AB − BP, b - r = c − BM, , b - r = c − ^a − r h, b-r = c−a+r, , 2r = a + b − c, r =a+b−c, 2, , Hence Proved., ,

o, In given figure, PA and PB are tangents from a, point P to the circle with centre O . At the point, M , other tangent to the circle is drawn cutting PA, and PB at K and N . Prove that the perimeter of, TPNK = 2PB ., ,
,
Sol :, Sales form a arithmetic sequence with a 1 = 100 ,, d = 20 and a n = 2500, (i) Sales in the seventh month,, a n = a 1 + (n − 1) d, a 7 = 100 + (7 − 1) (20), (, = 100 + 6 20) = $220, (ii) Total sales after the 12th month,, a 12 = 100 + (12 − 1) (20), , = 100 + 11 (20) = 320, ,
,
Sol :, Since length of tangents from an external point to a, circle are equal,, PA = PB, KM = KA, MN = BN, Now, , n (a 1 + a n), 2, 12 (100 + 320), 12 (420), =, =, = 2520 Yes, 2, 2, , Sn =, , KN = KM + MN, , , = KA + BN, Now perimeter of TPNK, p = PN + KN + PK, , = PN + BN + KA + PK, , = PB + PA, , = 2PB, , (PA = PB), , S12, , 14. Salary Cut : Our personal and professional lives have, taken quite a turn in the light of the coronavirus, pandemic. Worklife has been revamped across the, globe, in that working from home has become the, norm for those employees and organizations that, continue to carry on with their operations and, functioning in the lockdown. Increments, incentives,, businesses have all been impacted., The Covid-19 pandemic has stalled economic, activity at an unprecedented scale globally, raising, the spectre of job losses and salary cuts. Most of, companies decided to bring down the salary by upto, 50 %., , CLICK HERE To Purchase Hard Books of CBSE Online Sample Papers., 20 Sample Paper in Each Subject and Rs 500/- For 4 Subjects

Page 6 :
CLICK HERE, For 30 Sample Papers With Solution, You have to download NODIA SolutionApp, , CBSE Maths Basic X, , Sample Paper 23 Solutions, , The following table shows the salaries (in percent), received by 80 employee during lockdown., Salary Received, (in %), , 40- 50- 60- 70- 80- 9050 60 70 80 90 100, , Number of employee 4, , 10, , 14, , 20, , 24, , 8, , Based on the above information, answer the following, questions., (i) What is the mean salary (in %) received ?, (i) We prepare following cumulative frequency table., (ii) What is the lower limit of mode class of salary (in, %) received? Which is the median class of salary, (in %) received?,
,
Sol :, C.I., , fi, , c.f., , xi, , fi xi, , 40-50, , 4, , 4, , 45, , 180, , 50-60, , 10, , 14, , 55, , 550, , 60-70, , 14, , 28, , 65, , 910, , 70-80, , 20, , 48, , 75, , 1500, , 80-90, , 24, , 72, , 85, , 2040, , 90-100, , 8, , 80, , 95, , 760, , / f i = 80, , / fi xi = 5940, , / fi xi, / fi, , = 5940 = 74.25, 80, (ii) Class 80-90 has the maximum frequency 24,, therefore this is model class. Lower limit is 80., Cumulative frequency just greater than, 80, N, 2 = 2 = 40 is 48 and the corresponding class is, M =, , 70-80. Thus median class is 70-80.,
,
**********, , CLICK HERE To Purchase Hard Books of CBSE Online Sample Papers., 20 Sample Paper in Each Subject and Rs 500/- For 4 Subjects, , Page 27