Arihant Maths (standard) class 10 term 2

CBSE Term II Mathematics X (Standard) 143 Given that, and speed of water flowing through the pipe = (15 × 1000) = 15000 mh−1 external length of an open box (l) = 36 cm external breadth of an open box (b) = 25 cm Now, volume of water flow in 1 h = π R 2H and external height of an open box (h ) = 16.5 cm ∴ External volume of an open box = lbh = ⎝⎜⎛ 22 × 7 × 7 × 15000⎞⎟⎠ 7 100 100 = 36 × 25 × 16. 5 = 14850 cm 3 = 231 m 3 Since, the thickness of the iron (x) = 1. 5 cm So, internal length of an open box (l1) = l − 2x Since, 231 m3 of water falls in the pond in 1 h. = 36 − 2 × 1. 5 So, 1 m3 water falls in the pond in 1 h. = 36 − 3 = 33 cm 231 Therefore, internal breadth of an open box (b2) = b − 2x = 25 − 2 × 1. 5 Also, 462 m3 of water falls in the pond in ⎜⎛⎝ 1 × 462⎠⎞⎟ h = 2 h = 25 − 3 = 22 cm 231 and internal height of an open box (h 2) = (h − x) = 16. 5 − 1. 5 = 15 cm Hence, the required time is 2 h. So, internal volume of an open box 38. (i) Length of the cuboid (L) = 30 cm = (l − 2x)⋅(b − 2x)⋅ (h − x) Breadth of the cuboid (B) =20 cm Height of the cuboid (H) = 10 cm = 33 × 22 × 15 Then, = 10890 cm 3 Area covered by the cuboid Therefore, required iron to construct an open box = L × B + 2H(L + B) = External volume of an open box = 30 × 20 + 2 × 10(30 + 20) = 600 + 1000 − Internal volume of an open box = 1600 cm2 = 14850 − 10890 = 3960 cm 3 (ii) We know that, Hence, required iron to construct an open box is 3960 cm3. Given that, 1 cm3 of iron weights = 7. 5 g = 7. 5 kg C.S.A of the cylinder = 2πRH1 C.S.A of the cylinder = 2 × 22 × 7 × 60 1000 7 = 0.0075 kg ∴ 3960 cm3 of iron weights = 3960 × 0.0075 = 29.7 kg = 2 × 22 × 60 = 2640 cm2 37. Given, length of the pond = 50 m and width of the pond In cone, height is given and radius of the cone is equal to radius of the cylinder, = 44 m Q L2 = H 2 + R2 2 21 Depth required of water = 21 cm = 100 m ⇒ L2 = 7 × 7 + 24 × 24 ∴ Volume of water in the pond = l × b × h ⇒ L2 = 49 + 576 = 625 = ⎜⎛⎝ 50 × 44 × 12010⎠⎞⎟ ∴ L = 25 cm C.S.A. of cone = πrl = 22 ×7 × 25 = 550 2 = 462 m 3 7 cm Also, given radius of the pipe = 7 cm = 7 m ⇒ C.S. A of the cylinder 2640 = 24 : 5 100 = C.S. A. of the cone 550 14 cm (iii) (b) Given structure is based on the concept of surface area and volume. Pipe 39. (i) Given, r = 7 cm, h = 10.5 cm = 21 cm 44 m 22 Volume of cylindrical cup = πr2h = 22 × 7 × 7 × 21 7 22 2 Tank 21 m = 11 × 7 × 21 100 4 = 404.25cm 3 50 m

144 CBSE Term II Mathematics X (Standard) (ii) Given, r = 7 cm (ii) Given edge of cube = 7 cm 2 ∴Volume of cube = 7 × 7 × 7 = 343 cm 3 ∴ Volume of hemispherical cup = 2 × 22 × 7 × 7 × 7 (iii) Given, radius (r) = 7 cm 3 7 222 ∴Curved surface area of hemisphere = 2πr2 = 2 × 22 × 7 × 7 = 89.83 cm 3 7 (iii) Area of canvas provided = 551m 2 = 44 × 7 = 308 cm 2 Area of remained after westage = 551 − 1 = 550 m 2 (iv) Given, radius (r) = 7 cm and height (h ) = 24 cm So, area of conical tent = πrl Slant height (l) = ? Here, r = 7 m ∴ l2 = r2 + h 2 = (7)2 + (24)2 ∴ πrl = 550 ⇒ 22 × 7 × l = 550 ⇒ l = 25 m = 49 + 576 = 625 ⇒ l = 625 = 25 cm 7 Now, h = l2 − r2 = 625 − 49 (v) TSA of cone = πrl + 2πr2 = 22 × 7 × 25 + 2 × 22 × 7 × 7 = 576 = 24 m 77 40. (i) Given, edge of cube = 6 cm = 550 + 308 = 858 cm 2 ∴Diagonal of cube = 3 × edge of cube = 3 × 6 = 6 3 cm

Chapter Test Multiple Choice Questions (iii) The curved surface area of the conical cavity 1. How many cubes of side 2 cm can be made from a so formed is (b) 2200 cm2 (a) 2250 cm2 (d) 2400 cm2 solid cube of side 10 cm? (c) 1800 cm2 (a) 100 (b) 125 (c) 175 (d) 200 (iv) External curved surface area of the cylinder 2. 2 cubes, each of volume 125 cm3, are joined end to is (a) 2876 cm2 end. Find the surface area of the resulting cuboid. (c) 4224 cm2 (b) 1250 cm2 (d) 3824 cm2 (a) 100 cm2 (b) 200 cm 2 (c) 225 cm 2 (d) 250 cm 2 3. The radius of a sphere (in cm) whose volume is (v) Volume of conical cavity is 12π cm3, is (a) 6232 cm3 (b) 7248 cm3 (a) 3 (b) 3 3 (c) 32/ 3 (d) 31/ 3 (c) 5380 cm3 (d) 9856 cm3 4. A solid spherical ball fits exactly inside the cubical box Short Answer Type Questions of side 2a. The volume of the ball is (a) 16 πa3 (b) 1 πa3 7. A hemispherical depression is cut out from one face of 3 6 a cuboidal block of side 7 cm such that the diameter of (c) 32 πa3 (d) 4 πa3 the hemisphere is equal to the edge of the cube. Find 3 3 the surface area of the remaining solid. 5. A cone and a cylinder have the same radii but the 8. The capacity of a cylindrical glass tumbler is 125.6 cm3. height of the cone is 3 times that of the cylinder, then If the radius of the glass tumbler is 2 cm, then find its height. (Use π = 3.14) the ratio of their volumes. (a) 1 : 2 (b) 2 : 1 (c) 1 : 1 (d) None of these 9. In given figure, a solid toy is in the form of a Case Based MCQs hemisphere surmounted by a right circular cone. The 6. One day Aakash was going home from market saw a height of the cone is 2 cm and the diameter of the carpenter working on wood. He found that he is base is 4 cm. Determine the volume of the toy. 22 ⎤ carving out a cone of same height and same diameter 7 ⎦⎥ from a cylinder. The height of the cylinder is 48 cm and ⎡⎢⎣take π = base radius is 14 cm. While watching this, some questions came into Aakash’s mind. Help Aakash to 2 cm find the answer of the following questions. 14 cm 2 cm 48 cm Long Answer Type Questions (i) After carving out cone from the cylinder, 10. From a solid cylinder whose height is 2.4 cm and (a) Volume of the cylindrical wood will decrease (b) Height of the cylindrical wood will increase diameter 1.4 cm, a conical cavity of the same height (c) Volume of cylindrical wood will increase and same diameter is hollowed out. Find the total (d) Radius of the cylindrical wood will decrease surface area of the remaining solid to the nearest cm2. 11. A well of diameter 3 m is dug 14 m deep. The earth taken out of it has been spread evenly all around it in (ii) Find the slant height of the conical cavity so the shape of a circular ring of width 4 m to form a formed. platform. Find the height of the platform. (a) 28 cm (b) 60 cm ⎣⎡⎢take π = 22 ⎤ 7 ⎦⎥ (c) 40 cm (d) 50 cm Answers 1. (b) 2. (d) 3. (c) 4. (d) 5. (c) 6. (i) (a) (ii) (d) (iii) (b) (iv) (c) (v) (d) For Detailed Solutions Scan the code 7. 332.465 cm2 8. 10 cm 9. 25.12 cm3 10. 18 cm2 11. 1.125 m

146 CBSE Term II Mathematics X (Standard) CHAPTER 07 Statistics In this Chapter... l Mean of Grouped Data l Mode of Grouped Data l Median of Grouped Data Arithmetic Mean or Mean or Average In general, for the ith class interval, we have frequency f i corresponding to the class mark xi . The sum of the values in The arithmetic mean of a set of observations is obtained by the last column gives us Σf i xi , so the mean x of the given data dividing the sum of the values of all observations by the total is given by number of observations. x = Σf i xi . Thus, the mean of n observations x1 , x2 , x3 , K , xn , is Σf i defined as 2. Assumed Mean Method n The cases, in which numerical values of xi and f i are large and ∑ xi computation of product of xi and f i becomes tedious and time consuming, assumed mean method is used. In this method, Mean (x) = x1 + x2 + x3 + … + xn = i = 1 first of all, one among xi ’s is chosen as the assumed mean nn denoted by ‘a’. After that, the difference di between a and where, the Greek letter ‘Σ’ (sigma) means ‘Summation.’ each of the xi ’ s, i.e. d i = xi − a is calculated. Then, arithmetic mean is given by Let x1 , x2 , . . . , xn be n observations with respective frequencies f1 , f 2 , . . . , f n . This means observation x1 occurs f1 times, x2 occurs f 2 times and so on. n i∑ fixi = Σ fi xi =1 Σ fi x = a + Σf i d i Mean (x) n Σf i ∑∴ = fi i =1 where, di = xi − a Mode Method of Calculating Mean of Grouped Data The observation, which occurs most frequently among the given observations, i.e. the value of the observation having 1. Direct Method maximum frequency is called mode. e.g. Mode of the numbers 2, 3, 4, 4, 6, 6, 6, 6, 7 and 9 is 6 because it is repeated In this method, we find the class marks of each class interval. These class marks would serve as the maximum number of times, i.e. 4 times. representative of whole class and are represented by xi .

CBSE Term II Mathematics X (Standard) 147 Modal Class Case II If n is even, then In a grouped frequency distribution, it is not possible to Median = Mean of value of ⎜⎝⎛ n ⎞⎟⎠ th and ⎛⎜⎝ n + 1⎠⎞⎟th determine the mode by looking at the frequencies. So, here 2 2 we first locate a class with the maximum frequency. This class is called modal class. e.g. observations Class interval 0-10 10-20 20-30 30-40 40-50 50-60 = 1 × Value of ⎣⎡⎢⎛⎜⎝ n2 ⎞⎟⎠ th + ⎜⎛⎝ n + 1⎠⎞⎟ th⎤⎦⎥ 2 2 Number of students 2 9 14 20 22 8 observations Here, the highest frequency is of the class 40-50, which is 22. e.g. If six girls of different heights are made to stand Hence, the modal class is 40-50. in a row, in descending order of their heights, then the mean height of third and fourth girl from either Mode of Grouped Data end is the median height. In grouped data, mode is a value that lies in the modal class Since, n = 6 is even. and it is given by the formula, So, n = 6 = 3rd observation ⎧ f1 − f0 ⎫ 22 ⎨ 1− f0 − ⎬ Mode = l + ⎩ 2f f2 ⎭ × h and ⎝⎜⎛ n +1⎞⎠⎟ 6 +1 6+2 4th observation 2 2 2 = = = where, l = lower limit of the modal class ∴ Median = Mean of 3rd and 4th observations h = size of the class intervals (assuming all class sizes to be equal) Cumulative Frequency f1 = frequency of the modal class The frequency of an observation in a data refers to how many f 0 = frequency of the class preceding the modal class times that observation occur in the data. Cumulative f 2 = frequency of the class succeeding the modal class frequency of a class is defined as the sum of all frequencies upto the given class. Median Less than type and more than type. Formation of these two Median is defined as the middle-most or the central distributions can be understood with the help of following observation, when the observations are arranged either in example. ascending or descending order of their magnitudes. e.g. Consider a grouped frequency distribution of marks Median divides the arranged series into two equal parts, i.e. obtained out of 100, by 58 students, in a certain examination, 50% of the observations lie below the median and the as follows: remaining are above the median. Let n be the total number of observations and suppose that Marks Number of students they are arranged in ascending or descending order. 0-10 5 10-20 7 Median of the data depends on the number of 20-30 4 observations (n). 30-40 2 40-50 3 Case I If n is odd, then 50-60 6 60-70 7 Median = Value of ⎛ n 2+1⎠⎟⎞th observation 70-80 9 ⎝⎜ 80-90 8 90-100 7 e.g. If five girls of different heights are made to stand in a row, in descending order of their heights, then the height of the third girl from either end is median height. Since, n = 5 is odd. Cumulative frequency distribution of the less than type Here, the number of students who have scored marks less ∴ Median = ⎛ n + 1⎞ th observation = 5 +1 than 10 are 5. The number of students who have scored ⎝⎜ 2 ⎠⎟ 2 marks less than 20 includes the number of students who have scored marks from 0-10 as well as the number of students = 6 = 3rd observation who have scored marks from 10-20. 2

148 CBSE Term II Mathematics X (Standard) Thus, the total number of students with marks less than 20 is Let n be the total number of observations (sum of 5 + 7 , i.e. 12. So, the cumulative frequency of the class 10-20 is 12. frequencies), then median of the data depends on the number Similarly, on computing the cumulative frequencies of the of observations (n). other classes, which is shown in the table. If n is odd, then Median = Value of ⎜⎝⎛ n + 1⎞⎟⎠ th observation. 2 Marks obtained Number of students If n is even, then (cumulative frequency) Median = Mean of ⎜⎛⎝ 2n⎠⎟⎞ th and ⎝⎛⎜ n + 1⎠⎟⎞ th observations Less than 10 5 2 Less than 20 5 + 7 = 12 1 ⎣⎡⎢⎝⎛⎜ n ⎞⎟⎠ ⎜⎛⎝ n + 1⎠⎞⎟ th⎥⎦⎤ 2 2 2 Less than 30 12 + 4 = 16 = × Value of th + observations Less than 40 16 + 2 = 18 Here, for the value of observation, first look at the cumulative frequency just greater than (and nearest to) the position of Less than 50 18 + 3 = 21 required observations. Then, determine the corresponding value of the observation. Less than 60 21 + 6 = 27 Less than 70 27 + 7 = 34 Less than 80 34 + 9 = 43 Median for Grouped Data Less than 90 43 + 8 = 51 Less than 100 51 + 7 = 58 In a grouped data, we may not find the middle observation by looking at the cumulative frequencies, since the middle Cumulative frequency distribution of the more than type observation will be some value in a class interval, so it is necessary to find the value inside a class that divides the For this type of distribution, we make the table for the whole distribution into two halves. number of students with scores, more than or equal to 0, more than or equal to 10, more than or equal to 20 and so on. For this, we find the cumulative frequencies of all the classes From the example, we observed that all 58 students have and then determine n, where n = number of observations. scored marks more than or equal to 0. 2 There are 5 students scoring marks in the interval 0-10, it Now, locate the class whose cumulative frequency is greater shows that there are 58 − 5 = 53 students getting more than or than (i.e. nearest to) n and this class is called median class. equal to 10 marks. In the same manner, the number of students scoring 20 marks or above = 53 − 7 = 46 students, 2 and so on. After finding the median class, use the following formula for calculating the median. Marks obtained Number of students ⎧ N − cf ⎫ (cumulative frequency) ⎪ 2 f ⎪ More than or equal to 0 Median = l + ⎨ ⎬ × h More than or equal to 10 58 ⎪ ⎪ More than or equal to 20 58 − 5 = 53 ⎩⎭ More than or equal to 30 53 − 7 = 46 More than or equal to 40 46 − 4 = 42 where, l = lower limit of median class More than or equal to 50 42 − 2 = 40 N = sum of frequencies More than or equal to 60 40 − 3 = 37 cf = cumulative frequency of the class preceding More than or equal to 70 37 − 6 = 31 the median class More than or equal to 80 31 − 7 = 24 More than or equal to 90 24 − 9 = 15 f = frequency of the median class 15 − 8 = 7 h = class width (assuming class sizes to be equal) Relationship among Mean, Median and Mode Median for Discrete Series There is an empirical relationship among the three measures of central tendency, which is given by A series having observations x1 , x2 , x3 , K , xn with respective frequencies f1 , f 2 , f 3 K , f n is known as discrete series. Mode = 3(Median) − 2(Mean) or Mean = 3 (Median) − Mode Method to Find the Median of the Discrete Series Firstly, we arrange the data in the ascending or descending 2 order of xi , then we find the cumulative frequencies of all the or Median = Mode + 2 (Mean) observations. 3

CBSE Term II Mathematics X (Standard) 149 Solved Examples Example 1. Find the mean of the following data. Sol. The table for given data is x 10 30 50 70 89 Class Frequency (fi) Mid-value (xi ) fi xi f 7 8 10 15 10 2-4 6 3 18 Sol. Table for the given data is 4-6 8 5 40 xi fi fi xi 6-8 15 7 105 10 7 70 30 8 240 8-10 p 9 9p 50 10 500 70 15 1050 10-12 8 11 88 89 10 890 Total Σ fi = 50 Σ fi xi = 2750 12-14 4 13 52 Σfi = p + 41 Σ fi xi = 9p + 303 Here, Σfi = 50 and Σfixi = 2750 Given, mean = 7.5 ∴ Mean (x) = Σfi xi ∴ Σfi xi = 7. 5 ⇒ 9p + 303 = 7. 5 Σfi Σfi p + 41 ⇒ 9p + 303 = 7.5p + 307. 5 = 2750 = 55 ⇒ 9p − 7.5p = 307.5 − 303 50 ⇒ 1. 5p = 4. 5 Hence, mean of the given data is 55. ⇒ 4. 5 Hence, value of p is 3. p= =3 1.5 Example 2. Calculate the mean of the scores of 20 Example 4. The weights of tea in 70 packets are shown students in a mathematics test in the following table Marks 10-20 20-30 30-40 40-50 50-60 Weight (in gm) Number of packets Number of students 24761 200-201 13 201-202 27 Sol. We first, find the class marks xi of each class and then 202-203 18 proceed as follows 203-204 10 204-205 1 Marks Class marks (xi ) Frequency (fi ) fi xi 205-206 1 10-20 15 2 30 Find the mean weight of packets. 20-30 25 4 100 30-40 35 7 245 Sol. First, we find the class marks of the given data as follows. 40-50 45 6 270 50-60 55 1 55 Weight Number of Class Deviation fi di Σ fi = 20 Σ fixi = 700 (in gm) Packets (fi ) marks (xi ) (di = xi − a) Therefore, mean (x) = Σfi xi = 700 = 35 200-201 13 200.5 − 3 − 39 Σfi 20 201-202 27 201.5 − 2 − 54 202-203 18 202.5 − 1 − 18 Hence, the mean of scores of 20 students in mathematics test 203-204 10 a = 203.5 00 is 35. 204-205 1 204.5 11 205-206 1 205.5 22 Example 3. Find the value of p, if the mean of the N = ∑ fi = 70 ∑ fi di = − 108 following distribution is 7.5. Classes 2-4 4-6 6-8 8-10 10-12 12-14 Frequency (fi) 6 8 15 p 8 4

150 CBSE Term II Mathematics X (Standard) Here, assume mean (a) = 203. 5 Sol. Here, the number of students who have scored marks less ∴ Mean (x) = a + ∑ fi di than 10 are 10. The number of students who have scored marks less than 20 includes the number of students who ∑ fi have scored marks from 0-10 as well as the number of = 203.5 − 108 students who have scored marks from 10-20. 70 Thus, the total number of students with marks less than 20 is 10 + 8, i.e. 18. So, the cumulative frequency of the class = 203.5 − 1.54 10-20 is 18. = 201.96 Hence, the required mean weight is 201.96 gm. Similarly, on computing the cumulative frequencies of the other classes, i.e. the number of students with marks less Example 5. The following distribution gives cumulative than 30, less than 40, … less than 100, we get the distribution which is called the cumulative frequency frequencies of ‘more than type’: distribution of the less than type. Marks obtained 5 10 15 20 (More than or equal to) Number of students Number of students 30 23 8 2 Marks obtained (cumulative frequency) (cumulative frequency) Less than 10 10 Change the above data into a continuous grouped Less than 20 10 + 8 = 18 Less than 30 18 + 7 = 25 frequency distribution. [CBSE 2015] Less than 40 25 + 4 = 29 Less than 50 29 + 6 = 35 Sol. Given, distribution is the more than type distribution. Less than 60 35 + 8 = 43 Less than 70 43 + 5 = 48 Here, we observe that, all 30 students have obtained marks Less than 80 48 + 9 = 57 more than or equal to 5. Further, since 23 students have Less than 90 57 + 5 = 62 obtained score more than or equal to 10. So, 30 − 23 = 7 Less than 100 62 + 8 = 80 students lie in the class 5-10. Similarly, we can find the other classes and their corresponding frequencies. Now, we construct the continuous grouped frequency distribution as Class (Marks obtained) Number of students 5-10 30 – 23 = 7 Here, 10, 20, 30,…, 100 are the upper limits of the respective class intervals. 10-15 23 – 8 = 15 Example 7. In a class of 72 students, marks obtained by 15-20 8–2=6 the students in a class test (out of 10) are given More than or equal to 20 2 below: Example 6. Consider a grouped frequency distribution Marks obtained 1 23467 9 10 (Out of 10) 3 5 12 18 23 8 21 of marks obtained out of 100, by 70 students in a certain examination, as follows: Number of students Marks Number of students 0-10 10 Find the mode of the data. 10-20 8 Sol. The mode of the given data is 6 as it has the maximum frequency, i.e. 23 among all the observations. 20-30 7 Example 8. The weight of coffee in 70 packets are 30-40 4 shown in the following table 40-50 6 Weight (in gm) Number of packets 50-60 8 200-201 12 60-70 5 201-202 26 70-80 9 202-203 20 80-90 5 203-204 9 90-100 8 204-205 2 Form the cumulative frequency distribution of less 205-206 1 than type. Determine the modal weight.

CBSE Term II Mathematics X (Standard) 151 Sol. In the given data, the highest frequency is 26, which lies in Sol. In a given data, the highest frequency is 41, which lies in the the interval 201-202 interval 10000-15000. Here, l = 201, f1 = 26, f0 = 12 , f2 = 20 and h = 1 Here, l = 10000, f1 = 41, f0 = 26, ∴ Mode = l + ⎛ f1 − f0 ⎞ × h f2 = 16 and h = 5000 ⎜⎝ − f0 − f2⎠⎟ 2 f1 ⎛ f1 − f0 ⎞ ⎜⎝ − f0 − f2⎠⎟ ⎛ 26 − 12 ⎞ ∴ Mode = l + × h ⎝⎜ 2 × 26 − 12 − 20⎠⎟ = 201 + ×1 2 f1 ⎛ 14 ⎞ 14 = 10000 + ⎛ × 41 − 26 ⎞ × 5000 ⎝⎜ 52 − 32⎟⎠ 20 ⎜⎝ 2 41 − 26 − 16⎟⎠ = 201 + = 201 + = 201 + 0.7 = 201.7 gm = 10000 + ⎛ 15 ⎞ × 5000 Hence, the modal weight is 201.7 gm. ⎝⎜ 82 − 42⎟⎠ Example 9. Find the mode of the following distribution = 10000 + ⎛⎝⎜ 1450⎟⎞⎠ × 5000 Marks 0-10 10-20 20-30 30-40 40-50 50-60 = 10000 + 15 × 125 Number of 4 6 7 12 5 6 = 10000 + 1875 students = ` 11875 Hence, the modal income is ` 11875. Sol. Given, distribution table is Example 11. Find the median of the following data. Marks Number of students 0-10 4 Marks obtained 20 29 28 42 19 35 51 10-20 6 Number of students 3457923 20-30 30-40 7 (f0 ) Sol. Let us arrange the data in ascending order of xi and make a 40-50 12 (f1) cumulative frequency table. 50-60 5 (f2) Marks Number of Cumulative 6 obtained (x i) students (fi) frequency (cf ) The highest frequency in the given distribution is 12, whose 19 9 9 corresponding class is 30 - 40. 20 3 9 + 3 = 12 Thus, 30-40 is the required modal class. 28 5 12 + 5 = 17 29 4 17 + 4 = 21 Here, l = 30, f1 =12 , f0 =7, f2 = 5 and h = 10 35 2 21 + 2 = 23 ∴ Mode = l + f1 − f0 × h 42 7 23 + 7 = 30 51 3 30 + 3 = 33 2f1 − f0 − f2 = 30 + 12 − 7 × 10 2 × 12 −7 − 5 = 30 + 50 = 30 + 50 = 30 + 4.17 = 34.17 Here, n = 33 (odd) 24 −12 12 Hence, mode of the given distribution is 34.17. ∴ Median = Value of ⎛⎜⎝ n + 1⎠⎞⎟ th observation 2 Example 10. The monthly income of 100 families are = Value of ⎜⎝⎛ 33 + 1⎞⎟⎠ th observation given as below 2 Income (in `) Number of families = Value of 17th observation 0-5000 8 Corresponding value of 17th observation of cumulative frequency in xi is 28. Hence, median is 28. 5000-10000 26 Example 12. 200 surnames were randomly picked up 10000-15000 41 from a local telephone directory and the frequency 15000-20000 16 distribution of the number of letters in English alphabets in the surnames was obtained as follows: 20000-25000 3 25000-30000 3 30000-35000 2 Number of letters 0-5 5-10 10-15 15-20 20-25 Number of surnames 20 60 80 32 8 35000-40000 1 Calculate the modal income. Find the median of the above data.

152 CBSE Term II Mathematics X (Standard) Sol. The cumulative frequency table of given data is = 50 + (45 − 40 − p) × 10 20 Number of Number of Cumulative letters surnames (fi ) frequency (cf) ⇒ 50 = 50 + ⎝⎜⎛ 5 − p⎟⎠⎞ ⇒ 0 = 5 − p [Median = 50] 2 2 0-5 20 20 5-10 60 ∴ p=5 [given] 10-15 80 (= f) 20 + 60 = 80 (cf) Also, 78 + p + q = 90 15-20 32 80 + 80 = 160 20-25 8 160 + 32 = 192 ⇒ 78 + 5 + q = 90 Total N = 200 192 + 8 = 200 ⇒ q = 90 − 83 ∴ q=7 Since, the cumulative frequency just greater than 100 is 160 Example 14. The median of the following data is 525. Find the values of x and y, if total frequency is 100. and the corresponding class interval is 10-15. Class 0- 100- 200- 300- 400- 500- 600- 700- 800- 900- 100 200 300 400 500 600 700 800 900 1000 ∴ N = 200; ∴ N = 200 = 100 22 Frequency 2 5 x 12 17 20 y 9 7 4 Here, l = 10, cf = 80, h = 5 and f = 80 ⎧N − cf ⎫ Sol. Given, frequency table is f ⎪ Now, median = l + ⎪ 2 ⎬ × h = 10 + ⎧⎨⎩1008−0 80 ⎭⎬⎫ × 5 Class Frequency (f1) Cumulative Frequency (cf) ⎨ ⎪ ⎪ 2 ⎩⎭ 7 0-100 2 7+x 19 + x = 10 + ⎜⎝⎛ 2800⎠⎞⎟ × 5 = 10 + 1.25 = 11.25 100-200 5 36 + x (cf) 56 + x 200-300 x 56 + x + y 65 + x + y Example 13. The median of the following data is 50. 300-400 12 72 + x + y 76 + x + y Find the values of p and q, if the sum of all the 400-500 17 frequencies is 90. Marks Frequency 500-600 20 (f) 20-30 p 600-700 y 30-40 15 700-800 9 40-50 25 50-60 20 800-900 7 60-70 q 70-80 8 900-1000 4 80-90 10 Given, total frequency is 100. ∴ 2 + 5 + x + 12 + 17 + 20 + y + 9 + 7 + 4 = 100 ⇒ 76 + x + y = 100 Sol. ⇒ x + y = 24 …(i) Marks Frequency (fi ) Cumulative frequency (cf) It is given that the median is 525. 20-30 p p Clearly, 525 lies in the class 500-600. So, 500-600 is the 30-40 15 median class. 40-50 25 15 + p 50-60 20 (= f) 40 + p = cf Here, l = 500, h = 100, f = 20 and cf = 36 + x 60-70 q 70-80 8 60 + p ∴ N = 100 80-90 10 60 + p + q 68 + p + q N − cf 78 + p + q Q Median = l + 2 f ×h Given, N = 90 ⇒ 525 = 500 + 50 − 36 − x × 100 20 ∴ N = 90 = 45 ⇒ 525 = 500 + (14 − x) × 5 22 ⇒ 525 = 500 + 70 − 5x which lies in the interval 50-60. ⇒ 5x = 570 − 525 Here, l = 50, f = 20, cf = 40 + p and h = 10 ⇒ 5x = 45 ⇒ x = 45 = 9 5 ⎛⎝⎜ N − cf⎞⎟⎠ Put x = 9 in Eq. (i), we get 2 Q Median = l + × h 9 + y = 24 ⇒ y = 24 − 9 = 15 f Hence, x = 9 and y = 15

CBSE Term II Mathematics X (Standard) 153 Chapter Practice PART 1 8. If the arithmetic mean of the following distribution Objective Questions is 47, then the value of p is Class interval 0-20 20-40 40-60 60-80 80-100 Frequency 8 15 20 p 5 G Multiple Choice Questions (a) 10 (b) 11 (c) 13 (d) 12 1. Which of the following is a measure of central 9. The times (in seconds) taken by 150 atheletes to run tendency? a 110 m hurdle race are tabulated below (a) Frequency (b) Cumulative frequency (c) Mean (d) Class-limit Class 13.8- 14- 14.2- 14.4- 14.6- 14.8- 14 14.2 14.4 14.6 14.8 15 2. While computing mean of grouped data, we assume that the frequencies are Frequency 2 4 5 71 48 20 (a) evenly distributed over all the class (b) centred at the class marks of the class The number of atheletes who completed the race in (c) centred at the upper limits of the class less than 14.6 s is (d) centred at the lower limits of the class (a) 11 (b) 71 (c) 82 (d) 130 3. While computing the mean of grouped data, we 10. For the following distribution assume that the frequencies are Marks Number of students (a) evenly distributed over all the class (b) centred at the class marks of the class Below 10 3 (c) centred at the upper limits of the class (d) centred at the lower limits of the class Below 20 12 4. If the difference of mode and median of a data is 24, Below 30 27 then the difference of median and mean is Below 40 57 (a) 12 (b) 24 (c) 8 (d) 36 Below 50 75 n Below 60 80 5. If x is the mean of x’s, then the value of ∑ x i is i =1 The modal class is (a) x (d) x (b) 2 x (c) n x (a) 10-20 (b) 20-30 (c) 30-40 (d) 50-60 2n 11. Consider the following distribution 6. If xi ’s are the mid-points of the class intervals of grouped data, f i ’s are the corresponding Marks obtained Number of students frequencies and x is the mean, then Σ (f i xi − x )is More than or equal to 0 63 equal to More than or equal to 10 58 (a) 0 (b) −1 (c) 1 (d) 2 More than or equal to 20 55 7. In the formula x = a + Σf i di , for finding the mean More than or equal to 30 51 Σf i More than or equal to 40 48 of grouped data di ’s are deviation from a of More than or equal to 50 42 (a) lower limits of the class (b) upper limits of the class The frequency of the class 30-40 is (c) mid-points of the class (a) 3 (b) 4 (c) 48 (d) 51 (d) frequencies of the class marks

154 CBSE Term II Mathematics X (Standard) 12. For the following distribution 18. The mean, mode and median of grouped data will always be Marks Number of Marks Number of students students (a) same Below 10 Below 40 (b) different Below 20 3 Below 50 57 (c) depends on the type of data Below 30 Below 60 (d) None of the above 12 75 19. The mean and median of a distribution are 14 and 28 80 The modal class is 15 respectively. The value of mode is (a) 0-20 (b) 20-30 (c) 30-40 (d) 50-60 (a) 16 (b) 17 13. A student noted the number of cars passing through (c) 13 (d) 18 a spot on a road for 100 periods each of 3 min and summarised in the table given below. G Case Based Study Number of cars Frequency 20. Analysis of Water Consumption in a Society 0-10 7 10-20 14 An inspector in an enforcement squad of department 20-30 13 of water resources visit to a society of 100 families 30-40 12 and record their monthly consumption of water on 40-50 20 the basis of family members and wastage of water, 50-60 11 which is summarise in the following table. 60-70 15 70-80 8 Monthly Consumption 0- 10- 20- 30- 40- 50- Total (in kWh) 10 20 30 40 50 60 Number of Families 10 x 25 30 y 10 100 Then, the mode of the data is (a) 34.7 (b) 44.7 (c) 54.7 (d) 64.7 14. Mode of the following grouped frequency distribution is Class 3-6 6-9 9-12 12-15 15-18 18-21 21-24 Frequency 2 5 10 23 21 12 3 (a) 13.6 (b) 15.6 (c) 14.6 (d) 16.6 15. If the number of runs scored by 11 players of a cricket team of India are 5, 19, 42, 11, 50, 30, 21, 0, 52, 36, 27, then median is (a) 30 (b) 32 (c) 36 (d) 27 16. Consider the following frequency distribution Class 0-5 6-11 12-17 18-23 24-29 Frequency 13 10 15 8 11 Based on the above information, answer the following questions. The upper limit of the median class is (i) The value of x + y is (a) 17 (b) 17.5 (c) 18 (d) 18.5 (a) 50 (b) 42 (c) 25 (d) 200 17. Consider the following frequency distribution Class 65- 85- 105- 125- 145- 165- 185- (ii) If the median of the above data is 32, then x is 85 105 125 145 165 185 205 equal to Frequency 4 5 13 20 14 7 4 (a) 10 (b) 8 (c) 9 (d) None of these The difference of the upper limit of the median (iii) What will be the upper limit of the modal class? class and the lower limit of the modal class is (a) 40 (b) 60 (c) 65 (d) 70 (a) 0 (b) 19 (c) 20 (d) 38

CBSE Term II Mathematics X (Standard) 155 (iv) If A be the assumed mean, then A is always (iv) If a machine work for 10 h in a day, then (a) > (Actual mean) (b) < (Actual Mean) approximate time required to complete the work (c) = (Actual Mean) (d) Can’t say for a machine is (a) 3 days (b) 4 days (c) 5 days (d) 6 days (v) The class mark of the modal class is (v) The measure of central tendency is (a) 25 (b) 35 (a) Mean (b) Median (c) 30 (d) 45 (c) Mode (d) All of these 21. As the demand for the products grew a 22. Direct income in India was drastically impacted manufacturing company decided to purchase more due to the COVID-19 lockdown. Most of the machines. For which they want to know the mean companies decided to bring down the salaries of the time required to complete the work for a worker. employees upto 50%. The following table shows the frequency The following table shows the salaries (in percent) distribution of the time required for each machine received by 50 employees during lockdown. to complete a work. Salaries received (in %) 50-60 60-70 70-80 80-90 Time (in hours) 15-19 20-24 25-29 30-34 35-39 Number of employees 18 12 16 4 Number of machines 20 35 32 28 25 Based on the above information, answer the following questions. Based on the above information, answer the (i) Total number of persons whose salary is reduced following questions. by more than 20% is (a) 40 (b) 46 (c) 30 (d) 22 (i) The class mark of the modal class 30-34 is (ii) Total number of persons whose salary is reduced by atmost 40% is (a) 17 (b) 22 (a) 32 (b) 40 (c) 27 (d) 32 (c) 46 (d) 18 (ii) If xi ’s denotes the class mark and f i ’s denotes the corresponding frequencies for the given data, then (iii) The modal class is the value of Σxi f i equals to (a) 50-60 (b) 60-70 (a) 3600 (c) 70-80 (d) 80-90 (b) 3205 (iv) The median class of the given data is (c) 3670 (a) 50-60 (b) 60-70 (d) 3795 (c) 70-80 (d) 80-90 (iii) The mean time required to complete the work for a (v) The empirical relationship among mean, median worker is and mode is (a) 27.10 h (a) 3 Median = Mode + 2 Mean (b) 3 Median = Mode −2 Mean (b) 23 h (c) Median = 3 Mode −2 Mean (d) Median = 3 Mode + 2 Mean (c) 24 h (d) None of the above

156 CBSE Term II Mathematics X (Standard) PART 2 Find the mean mileage. Subjective Questions The manufacturer claimed that the mileage of the G Short Answer Type Questions model was 16 kmL −1 . Do you agree with this claim? 9. An NGO working for welfare of cancer patients, maintained its records as follows: 1. Find the class marks of the class 15-35 and Age of patients (in years) 0-20 20-40 40-60 60-80 class 45-60. 2. What is the arithmetic mean of first n natural Number of patients 35 315 120 50 numbers? Find mode. [CBSE 2016] 3. Calculate the mean of the following data. 10. Find the mode of the following distribution. Class 4-7 8-11 12-15 16-19 Class 10-15 15-20 20-25 25-30 30-35 35-40 Frequency 5 4 9 10 Frequency 45 30 75 20 35 15 4. Find the mean of the distribution. 11. Find the mode of the following distribution. Class 1-3 3-5 5-7 7-10 Class 25-30 30-35 35-40 40-45 45-50 50-55 17 Frequency 20 36 53 40 28 14 Frequency 9 22 27 5. The following table gives the number of pages 12. Compute the mode for the following frequency written by Sarika for completing her own book for distribution. 30 days. Number of pages 16-18 19-21 22-24 25-27 28-30 Size of items 0-4 4-8 8-12 12-16 16-20 20-24 24-28 written per day (in cm) Frequency 5 7 9 17 12 10 6 Number of days 13 4 9 13 13. Find the mode of the following frequency distribution. Find the mean number of pages written per day. 6. The mean of the following data is 14. Find the value Class 15-20 20-25 25-30 30-35 35-40 40-45 Frequency 3 8 9 10 3 2 of k. 14. The set of data given below shows the ages of x 5 10 15 20 25 participants in a certain summer camp. Draw a cumulative frequency table for the data. f 7k845 7. The mean of the following frequency distribution is Age (in years) 10 11 12 13 14 15 18. The frequency f in the class interval 19-21 is missing. Determine f . Frequency 3 18 13 12 7 27 Class interval 11-13 13-15 15-17 17-19 19-21 21-23 23-25 15. Given below is a cumulative frequency distribution showing the marks secured by 50 students of a Frequency 3 6 9 13 f 5 4 class 8. The mileage (kmL −1 ) of 50 cars of the same model Marks Below 20 Below 40 Below 60 Below 80 Below was tested by a manufacturer and details are 100 tabulated as given below. Number of 17 22 29 37 50 Mileage (kmL−1) 10-12 12-14 14-16 16-18 students Number of cars 7 12 18 13 Form the frequency distribution table for the data.

CBSE Term II Mathematics X (Standard) 157 16. The following table shows the cumulative 19. From the following distribution, find the median frequency distribution of marks of 800 students in an examination. Class Frequency 500-600 36 Marks Number of students 600-700 32 Below 10 10 700-800 32 Below 20 50 800-900 20 Below 30 130 900-100 30 Below 40 270 Below 50 440 20. Size of agricultural holdings in a survey of 200 Below 60 570 families is given in the following table Below 70 670 Below 80 740 Size of agricultural Number of Below 90 780 holdings (in hectare) families Below 100 800 10 0-5 15 Construct a frequency distribution table for the 5-10 30 data above. 10-15 80 15-20 40 17. The following distribution of weights (in kg) of 20-25 20 40 persons. 25-30 5 30-35 Weight (in kg) Number of persons Compute median and modal class of the holdings. 40-45 4 45-50 4 21. If median = 137 units and mean = 137.05 units, then 50-55 13 find the mode. 55-60 5 60-65 6 G Long Answer Type Questions 65-70 5 70-75 2 22. The weights (in kg) of 50 wrestlers are recorded in 75-80 1 the following table. Construct a cumulative frequency distribution (of Weight (in kg) Number of the less than type) table for the data above. wrestlers 100-110 18. Form the frequency distribution table from the 110-120 4 following data 120-130 14 130-140 21 140-150 8 3 Marks (Out of 90) Number of candidates Find the mean weight of the wrestlers. More than or equal to 80 4 More than or equal to 70 6 23. If mode of the following series is 54, then find the More than or equal to 60 11 value of f. More than or equal to 50 17 More than or equal to 40 23 Class 0-15 15-30 30-45 45-60 60-75 75-90 More than or equal to 30 27 interval More than or equal to 20 30 More than or equal to 10 32 Frequency 3 5 f 16 12 7 More than or equal to 0 34 Find the modal class in which the given mode lies and find the value of f by using the formula, Mode =l + ⎧ f1 −f 0 ⎫ × h . ⎨⎩2 1− f0 ⎬ f − f 2 ⎭

158 CBSE Term II Mathematics X (Standard) 24. Find the mode of the following distribution. 29. Weekly income of 600 families is tabulated below Classes 0-20 20-40 40-60 60-80 80-100 Weekly income (in `) Number of families Frequency 10 8 12 16 4 0-1000 250 1000-2000 190 25. The following are the ages of 300 patients getting 2000-3000 100 medical treatment in a hospital on a particular day 3000-4000 40 Age (in years) 10-20 20-30 30-40 40-50 50-60 60-70 4000-5000 15 Number of 60 42 55 70 53 20 5000-6000 5 patients Total 600 Form Compute the median income. (i) less than type cumulative frequency distribution. 30. A survey regarding the heights (in cm) of 51 boys of Class X of a school was conducted and the following (ii) more than type cumulative frequency distribution. data was obtained: 26. Find the unknown entries a, b, c, d, e and f in the Heights (in cm) Number of boys following distribution of heights of students in a class Less than 140 4 Height (in cm) Frequency Cumulative Less than 145 11 frequency Less than 150 29 a 150-155 12 25 Less than 155 40 155-160 b c 160-165 10 43 Less than 160 46 165-170 d 48 170-175 e f Less than 165 51 175-180 2 Total 50 Find the median height. 31. Find the missing frequencies in the following frequency distribution table, if N = 100 and median is 32. CBSE 2019 Marks obtained 0-10 10-20 20-30 30-40 40-50 50-60 Total 27. The maximum bowling speeds (in km/h) of Number of 10 ? 25 30 ? 10 100 33 players at a cricket coaching centre are given as students follows Number of 32. The table below shows the salaries of 280 persons. players Speed (in km/h) 11 Salary (in ` thousand) Number of persons 5-10 49 85-100 9 10-15 133 100-115 15-20 63 115-130 8 20-25 15 130-145 25-30 6 5 30-35 7 35-40 4 Calculate the median bowling speed. 40-45 2 45-50 1 28. Obtain the median for the following frequency distribution. x 123456789 y 8 10 11 16 20 25 15 9 6 Calculate (i) median of the data, (ii) mode of the data.

CBSE Term II Mathematics X (Standard) 159 G Case Based Questions (i) Estimate the mean time taken by a student to finish the race. 33. The men’s 200 m race event at the 2020 Time Number of Students Tokyo Olympic took place 3 and 4 August. (in seconds) A stopwatch was used to find the time that it took a 8 group of Athletes to run 200 m. 0-20 10 20-40 13 Time (in seconds) 0-20 20-40 40-60 60-80 80-100 40-60 6 60-80 3 Number of Students 8 10 13 6 3 80-100 (ii) What is the sum of lower limits of median class and modal class. (iii) How many students finished the race within 1 min? SOLUTIONS Objective Questions 8. (d) Let us construct the following table for finding the arithmetic mean 1. (c) Mean is the measure of central tendency. 2. (b) In computing the mean of grouped data, the frequencies Class Frequency (fi ) Class mark f i xi are centred at the class marks of the class. interval (xi ) 80 3. (b) While computing mean of grouped data, we assume that 0-20 8 10 the frequencies are centred at the class marks of the classes. 4. (a) We have, Mode − Median = 24 20-40 15 30 450 We know that, Mode = 3 Median − 2 Mean ∴ Mode − Median = 2 Median − 2 Mean 40-60 20 50 1000 ⇒ 24 = 2(Median − Mean) ⇒ Median − Mean = 12 60-80 p 70 70p 80-100 5 90 450 5. (c) We know that, Total Σfi = 48 + p Σfi xi = 1980 + 70p n ∑ xi Now, x = Σfi xi Σfi Q x = i =1 n = 1980 + 70p 48 + p n ⇒ ∑ xi = nx ⇒ 47 = 1980 + 70p i =1 48 + p 6. (a)Q x = Σ fi xi n ⇒ 2256 + 47p = 1980 + 70p ∴ Σ (fi xi − x) = Σ fi xi − Σx = nx − nx [Q Σx = nx ] ⇒ 276 = 23p =0 ⇒ p = 12 9. (c) The number of atheletes who completed the race in less 7. (c) We know that, di = xi − a than 14.6 = 2 + 4 + 5 + 71 = 82 i.e. di ’s are the deviation from a of mid-points of the classes.

160 CBSE Term II Mathematics X (Standard) 10. (c) Q Mode = l + f1 − f0 × h 2f1 − f0 − f2 Marks Number of students Cumulative frequency Below 10 23 − 10 3=3 3 = 12 + 46 − 10 − 21 × 3 10-20 20-30 12 − 3 = 9 12 = 12 + 13 × 3 = 12 + 13 = 12 + 2.6 = 14.6 30-40 15 5 40-50 27 − 12 = 15 27 50-60 57 − 27 = 30 57 15. (d) Arranging the terms in ascending order, 0, 5, 11, 19, 21, 27, 30, 36, 42, 50, 52 75 − 57 = 18 75 Here n = 11 (odd) 80 − 75 = 5 80 Q Median = ⎜⎝⎛ n + 1⎞⎠⎟ th 2 Here, we see that the highest frequency is 30, which lies in the interval 30-40. Median value = ⎜⎝⎛ 11 + 1⎠⎞⎟ th = 6th value = 27 2 11. (a) Given, the distribution table Marks obtained Number of students 16. (b) Given, classes are not continuous, so we make 0-10 63 − 58 = 5 continuous by subtracting 0.5 from lower limit and adding 10-20 58 − 55 = 3 0.5 to upper limit of each class. 20-30 55 − 51 = 4 30-40 51 − 48 = 3 Class Frequency Cumulative frequency 40-50 48 − 42 = 6 −0.5-5.5 13 13 50 42 5.5-11.5 10 23 11.5-17.5 15 38 Frequency of the modal class 30-40 is 3 from the above table. 17.5-23.5 8 46 23.5-29.5 11 57 12. (c) Let us first construct the following frequency distribution table. Here, N = 57 = 28. 5, which lies in the interval 11.5-17.5. 22 Marks Number of students Hence, the upper limit is 17.5. 0-10 3 17. (c) 10-20 9 Class Frequency Cumulative frequency 65-85 4 4 20-30 16 85-105 5 9 105-125 13 22 30-40 29 125-145 20 42 145-165 14 56 40-50 18 165-185 7 63 185-205 4 67 50-60 5 Since, the maximum frequency is 29 and the class corresponding to this frequency is 30-40. So, the modal class is 30-40. 13. (b) Here, modal class is 40-50. Since, it has maximum Here, N = 67 = 33. 5 which lies in the interval 125 -145. frequency which is 20. 22 ∴ l = 40, f1 = 20, f0 = 12, f2 = 11 and h = 10 Hence, upper limit of median class is 145. Here, we see that the highest frequency is 20 which lies in Q Mode = l + ⎛ f1 − f0 ⎞ × h 125-145. Hence, the lower limit of modal class is 125. ⎜⎝ − f0 − f2⎠⎟ ∴ Required difference = Upper limit of median class 2 f1 − Lower limit of modal class = 40 + ⎝⎜⎛ 402−01−21−211⎞⎟⎠ × 10 = 145 − 125 = 20 18. (c) No, the value of these three measures can be the same, it = 40 + 80 17 depends on the type of data. = 40 + 4.7 = 44.7 19. (b) Given, mean = 14 and median = 15 By using empirical relationship, 14. (c) We observe that the class 12-15 has maximum frequency. Mode = 3 Median −2 Mean = 3 × 15 − 2 × 14 = 45 − 28 = 17 Therefore, this is the modal class. We have, l = 12, h = 3 , f1 = 23 , f0 = 10 and f2 = 21

CBSE Term II Mathematics X (Standard) 161 Case Based Study (iv) (a) Approximate time = 27.10 10 20. (i) (c) Given x and y are the frequencies of class intervals = 2.710 10-20 and 40-50, respectively. Then, …(1) ~− 3 days 10 + x + 25 + 30 + y + 10 = 100 (v) (d) Measure of central tendency are mean, median and ⇒ x + y = 25 mode. (ii) (c) Median is 32, which lies in 30-40. So, the median class is 30-40. 22. (i) (b)∴Required number of employees = 18 + 12 + 16 = 46 ∴ l = 30, h = 10, f = 30, N = 100 (ii) (a)∴Required number of employees = 12 + 16 + 4 and cf = 10 + x + 25 = x + 35 = 32 ⎛N − cf ⎞ (iii) (a) The maximum frequency is 18 and the corresponding ⎜ f ⎟ class is 50-60. Now, median = l + ⎜ 2 ⎟ × h Hence, modal class is 50-60. ⎜ ⎟ ⎝⎠ (iv) (b) Consider the table ⇒ 30 + ⎡{50 − (x + 35)} × 10⎥⎦⎤ = 32 ⎣⎢ 30 Salaries received Number of Cumulative ⇒ 30 + (15 − x) = 32 (in %) employees (fi ) frequency (cf) 3 50-60 18 18 ⇒ (15 − x) = 6 ⇒ x = 9 60-70 12 18 + 12 = 30 70-80 16 30 + 16 = 46 Put x = 9 in Eq. (1), we get 80-90 4 46 + 4 = 50 Total Σfi = 50 y = 16 Hence, x = 9 and y = 16 (iii) (a) Since, the maximum frequency is 30, so the modal class is 30-40. Hence, upper limit of the modal class is 40. N 50 2 2 (iv) (d) The value of assumed mean can be less, more or equal Here, = = 25 than the actual mean. The cumulative frequency more than 25 lies in 60-70. (v) (b) The modal class is 30-40. ∴ Class mark = 30 + 40 = 70 = 35 Hence, median class is 60-70. 22 (v) (a) As we know, Mode = 3 Median − 2 Mean 21. (i) (d) Class mark of class 30-34 ∴ 3 Median = Mode + 2 Mean = 30 + 34 2 Subjective Questions = 64 2 Lower limit + Upper limit 1. We know that, Class mark = = 32 (ii) (d) Let’s make the table 2 Class Class marks Frequency (fi ) xifi ∴ Class mark of 15-35 is = 15 + 35 = 50 = 25 (xi ) 22 340 770 Class mark of 45-60 is 45 + 60 = 105 = 52.5 864 22 896 925 Sum of all the observations Σxifi = 3795 2. Arithmetic mean = Number of observations 15-19 17 20 =1+2+K+ n n 20-24 22 35 n [2 × 1 + (n − 1)1] 25-29 27 32 =2 30-34 32 28 n 35-39 37 25 [Q1 + 2 + K + n is an AP series whose first term is a = 1 and common difference is d = 1. We know that, the Total Σfi = 140 sum of n th term of an AP is Sn = n [2 a + (n − 1)d ]] 2 Σxifi = 3795 (iii) (a) Mean time (X) = Σxifi =2 + n −1 = n +1 22 Σfi 3795 3. Since, given data is not continuous, so we subtract 0.5 from = the lower limit and add 0.5 in the upper limit of each class. 140 Now, we first find the class mark xi of each class and then = 27.10 proceed as follows

162 CBSE Term II Mathematics X (Standard) Class Class marks Frequency (fi ) f i xi Given, mean = 14 3.5-7.5 (xi ) 5 27.5 ∴ Σfi xi = 14 ⇒ 10k + 360 = 14 5.5 Σfi k + 24 7.5-11.5 9.5 4 38 ⇒ 10k + 360 = 14(k + 24) ⇒ 10k + 360 = 14k + 336 11.5-15.5 13.5 9 121.5 ⇒ 14k − 10k = 360 − 336 ⇒ 4k = 24 15.5-19.5 17.5 10 175 ∴ k = 24 = 6 Σ fi = 28 Σ fi xi = 362 4 Therefore, (x) mean = Σfi xi = 362 = 12. 93 Hence, the value of k is 6. Σfi 28 7. Table of given data is Hence, mean of the given data is 12.93. Class interval Frequency (fi) Mid-value xi fi 4. We first, find the class mark xi of each class and then (xi) proceed as follows. 11-13 3 36 12 84 144 Class Class marks Frequency f i xi 13-15 6 14 234 (xi ) (fi ) 20f 1-3 2 9 18 15-17 9 16 110 3-5 4 22 88 96 5-7 6 27 162 17-19 13 18 Σfi xi = 7-10 8.5 17 144.5 704 + 20f Σ fi xi = 412.5 19-21 f 20 Σ fi = 75 21-23 5 22 Therefore, mean (x) = Σ fi xi = 412.5 = 5.5 23-25 4 24 Σ fi 75 Total Σfi = 40 + f Hence, mean of the given distribution is 5.5. Q Mean = Σfi xi 5. Since, given data is not continuous, so we subtract 0.5 from Σfi the lower limit and add 0.5 in the upper limit of each class. ∴ 18 = 704 + 20f Class mark Mid-value (xi ) Number of days f i xi 40 + f [Q mean = 18, given] (f i ) 15.5-18.5 17 18.5-21.5 17 1 60 ⇒ 720 + 18f = 704 + 20f 21.5-24.5 92 ⇒ 16 = 2f 24.5-27.5 20 3 234 ⇒ f=8 27.5-30.5 377 Hence, missing frequency in the given data is 8. 23 4 780 8. Total 26 9 29 13 30 Mileage Class marks Number of cars f i xi (kmL −1 ) (xi ) (f i ) 77 Q Mean (x) = Σ fi xi = 780 = 26 Σ fi 30 10-12 11 7 Hence, the mean of pages written per day is 26. 12-14 13 12 156 6. Table for the given data is 14-16 15 18 270 xi fi fixi 16-18 17 13 221 7 35 5 k 10k Total Σ fi = 50 Σ fi xi = 724 10 8 120 15 4 80 Here, Σ fi = 50 20 5 125 25 Σ fi = k + 24 Σ fi xi = 10 k + 360 and Σ fi xi = 724 Total Q Mean (x) = Σ fi xi = 724 = 14.48 Σ fi 50 Hence, mean mileage is 14.48 kmL−1. No, the manufacturer is claiming mileage 1.52 kmL−1 more Here, Σfi = k + 24 and Σfi xi = 10k + 360 than average mileage.

CBSE Term II Mathematics X (Standard) 163 9. 0-20 20-40 40-60 60-80 In the given table, the highest frequency is 53 and Age of patients (in years) 35(f0 ) 315(f1) 120(f2) 50 corresponding class of this frequency is 35-40. Number of patients Thus, 35-40 is a modal class. Here, maximum frequency is 315 and the class Here, l = 35, f1 = 53, f0 = 36, f2 = 40 and h = 5 corresponding to this frequency is 20-40. So, the modal Q Mode = l + f1 − f0 × h class is 20-40. 2f1 − f0 − f2 ∴ l = 20, f1 = 315, f0 = 35, f2 = 120 and h = 20 = 35 + 53 − 36 × 5 2 × 53 − 36 − 40 Now, Mode = l + ⎛ f1 − f0 ⎞ × h = 35 + 17 × 5 = 35 + 85 ⎜⎝ − f0 − f2⎠⎟ 106 − 76 30 2 f1 = 35 + 2. 83 = 37. 83 (approx) 315 − 35 = 20 + ⎜⎝⎛ 2 315 − 35 − 120⎠⎟⎞ × 20 Hence, mode of given data is 37.83. × 12. Given frequency distribution table is = 20 + ⎜⎝⎛ 63028−0155⎞⎠⎟ × 20 Size of items (in cm) Frequency = 20 + 280 × 20 0-4 5 475 4-8 7 = 20 + 11.79 = 31.79 8-12 9 Hence, average age of maximum number of patients is 31.79. 12-16 17 10. Given, distribution table is 16-20 12 Class Frequency 20-24 10 10-15 45 15-20 24-28 6 20-25 30(f0 ) 25-30 75( f1 ) The maximum frequency in the given distribution table is 30-35 20( f2 ) 35-40 17, which lies in the class interval 12-16. 35 15 ∴Modal class = 12 -16 So, l = 12 , f1 = 17, f0 = 9, f2 = 12 and h = 4 Q Mode = l + ⎛ f1 − f0 ⎞ × h ⎝⎜ − f0 − f2⎟⎠ 2 f1 The highest frequency in the given data is 75 and the = 12 + ⎛⎝⎜ 2 × 17 − 9 12 ⎞⎠⎟ × 4 corresponding class is 20-25, which is a modal class. 17 − 9 − Here, l = 20, f1 = 75, f0 = 30, f2 = 20 and h = 5 = 12 + ⎝⎛⎜ 34 8 21⎞⎟⎠ × 4 Q Mode = l + f1 − f0 × h − 2f1 − f0 − f2 = 12 + 32 = 12 + 2. 46 = 14. 46 13 = 20 + 2 × 75 − 30 20 × 5 75 − 30 − Hence, mode of given distribution is 14.46. 13. Given, frequency distribution table is = 20 + 45 × 5 150 − 50 = 20 + 225 = 20 + 2.25 = 22.25 Class Frequency (fi) 100 15-20 3 11. Given, distribution table is 20-25 8 Class Frequency 25-30 9 25-30 20 30-35 36 30-35 10 35-40 53 40-45 40 35-40 3 45-50 28 50-55 14 40-45 2 The maximum frequency in the given distribution table is 10, which lies in the class interval 30-35. ∴ Modal Class = 30-35 So, l = 30, f1 = 10, f0 = 9, f2 = 3 and h = 5

164 CBSE Term II Mathematics X (Standard) Q Mode = l + f1 − f0 × h 17. The cumulative distribution (less than type) table is shown 2f1 − f0 − f2 below = 30 + 2 10 − 9 − 3 × 5 Weight (in kg) Cumulative frequency (cf ) × 10 − 9 Less than 45 4 Less than 50 = 30 + 5 = 30 + 0.625 = 30.625 Less than 55 4+ 4=8 8 Less than 60 8 + 13 = 21 Less than 65 21 + 5 = 26 Hence, mode of given distribution is 30.625. Less than 70 26 + 6 = 32 Less than 75 32 + 5 = 37 14. The cumulative frequency of first observation is the same as Less than 80 37 + 2 = 39 its frequency, since there is no frequency before it. 39 + 1 = 40 Now, the cumulative frequency table is Age Frequency Cumulative frequency (in years) (fi ) (cf) 10 3 3 11 18 3 + 18 = 21 18. Here, we observe that, all 34 students have scored marks more than or equal to 0. Since, 32 students have scored 12 13 21 + 13 = 34 marks more than or equal to 10. So, 34 − 32 = 2 students lies 13 12 34 + 12 = 46 in the interval 0-10 and so on. 14 7 46 + 7 = 53 Now, we construct the frequency distribution table. 15 27 53 + 27 = 80 Class interval Number of candidates (f i ) 15. Here, we observe that, 17 students have scored marks below 0-10 34 − 32 = 2 20 i.e. it lies between class interval 0-20 and 22 students 10-20 32 − 30 = 2 have scored marks below 40, so 22 − 17 = 5 students lies in 20-30 30 − 27 = 3 the class interval 20-40 continuting in the same manner, we 30-40 27 − 23 = 4 get the complete frequency distribution table for given data. 40-50 23 − 17 = 6 50-60 17 − 11 = 6 Marks Number of students 60-70 11 − 6 = 5 0-20 17 70-80 6− 4=2 20-40 80-90 4 40-60 22 − 17 = 5 60-80 29 − 22 = 7 19. The cumulative frequency table for given distribution is 80-100 37 − 29 = 8 50 − 37 = 13 Class Frequency (fi) Cumulative Frequency (cf) 16. Here, we observe that 10 students have scored marks below 500-600 36 10 i.e. it lies between class interval 0-10. Similarly, 50 600-700 32 36 students have scored marks below 20. So, 50 − 10 = 40 700-800 32 (f) students lies in the interval 10 - 20 and so on. The table of a 800-900 20 36 + 32 = 68 frequency distribution for the given data is 900-1000 30 68 + 32 = 100 100 + 20 = 120 Class interval Number of students (f i ) 120 + 30 = 150 0-10 10 10-20 Here, N = 150 = 75, which lies in the cumulative frequency 20-30 50 − 10 = 40 2 2 30-40 130 − 50 = 80 40-50 270 − 130 = 140 100, whose corresponding class is 700-800. Thus, modal class 50-60 440 − 270 = 170 60-70 570 − 440 = 130 is 700-800. 70-80 670 − 570 = 100 80-90 740 − 670 = 70 Here, l = 700, cf = 68, f = 32 and h = 100 90-100 780 − 740 = 40 800 − 780 = 20 N − cf Q Median = l + 2 f ×h = 700 + 75 − 68 × 100 = 700 + 700 32 32 = 700 + 21.88 = 721.88 Hence, median of the given distribution is 721.88.

CBSE Term II Mathematics X (Standard) 165 20. By assumed mean method, Mean (x) = a + Σ fi di Size of agricultural Number of Cumulative Σ fi holdings (in hec) families (f i ) frequency (cf ) = 125 + (− 80) 0-5 50 5-10 10 10 10-15 15 25 = 125 − 1.6 = 123.4 kg 15-20 30 55 20-25 80 (f) 135 23. Here, given mode is 54, which lies between 45-60. 25-30 40 175 Therefore, the modal class is 45-60. 30-35 20 195 5 200 ∴ l = 45, f1 = 16, f0 = f, f2 = 12 and h = 15 Q Mode = l + ⎛ f1 − f0 ⎞ × h ⎝⎜ − f0 − f2⎟⎠ 2 f1 ∴ 54 = 45 + ⎜⎝⎛ 2 × 16 − f 12 ⎠⎞⎟ × 15 16 − f − I. Here, N = 200 Now, N = 200 = 100, which lies in the interval 15-20. ⇒ 9 = 16 − f × 15 22 20 − f Here, l = 15 , h = 5, f = 80 and cf = 55 ⇒ 9(20 − f) = 15(16 − f) ⎛N cf ⎞ ⇒ 180 − 9f = 240 − 15f ⎟ ∴ Median = + ⎜ 2 − ⎟ × h ⇒ 6f = 240 − 180 = 60 l ⎜ f ⇒ f = 10 ⎜⎝ ⎠⎟ Hence, required value of f is 10. = 15 + ⎜⎝⎛ 100 − 55⎞⎟⎠ × 5 24. The given distribution table is 80 = 15 + ⎛⎜⎝ 1465⎟⎠⎞ Class Frequency (f) = 15 + 2.81 = 17. 81 hec 0-20 10 II. In a given table 80 is the highest frequency. 20-40 8 40-60 12 (f0 ) So, the modal class is 15-20. 60-80 16 (f1) 21. Given, median = 137 units and mean= 137.05 units. 80-100 4 (f2) We know that, Mode = 3(Median) − 2(Mean) The highest frequency in the given distribution table is 16, whose corresponding class is 60-80. Thus, 60-80 is the modal = 3 (137) − 2(137.05) class of the given distribution. = 411 − 274.10 Here, l = 60, f1 = 16, f0 = 12, f2 = 4 and h = 20 Q Mode = l + f1 − f0 × h = 136.90 Hence, the value of mode is 136.90 units. 2f1 − f0 − f2 22. We first find the class mark xi, of each class and then = 60 + 16 − 12 × 20 proceed as follows 2 × 16 − 12 − 4 Weight Number of Class Deviations = 60 + 4 × 20 = 60 + 80 (in kg) wrestlers di = xi − a, fi di 32 − 16 16 (fi ) marks ( xi ) a = 125 −80 100-110 = 60 + 5 = 65 4 105 −20 Hence, mode of the given distribution is 65. 110-120 14 115 −10 −140 25. (i) We observe that the number of patients which take medical treatment in a hospital on a particular day less 120-130 21 a = 125 0 0 than 10 is 0. Similarly, less than 20 include the number of patients which take medical treatment from 0-10 as well 130-140 8 135 10 80 as the number of patients which take medical treatment from 10-20. 140-150 3 145 20 60 So, the total number of patients less than 20 is N = Σ fi = 50 Σ fi di = − 80 0 + 60 = 60, we say that the cumulative frequency of the class 10-20 is 60. Similarly, for other classes, which is ∴Assumed mean (a) = 125, shown below the table. Class width (h ) = 10 and total observation (N ) = 50

166 CBSE Term II Mathematics X (Standard) (ii) Also, we observe that all 300 patients which take medical 27. First we construct the cumulative frequency table treatment more than or equal to 10. Since, there are 60 patients which take medical treatment in the interval Speed (in km/h) Number of Cumulative 10-20, this means that there are 300 − 60 = 240 patients players (f i ) frequency (cf ) which take medical treatment more than or equal to 20. 85-100 Continuing in the same manner, which is shown below 100-115 11 11 the table. 115-130 130-145 9 (f) 11 + 9 = 20 (i) Less than type (ii) More than type 20 + 8 = 28 8 28 + 5 = 33 5 Age (in years) Number of Age (in years) Number of It is given that, N = 33 Less than 10 students students More than or ∴ N = 33 = 16. 5 0 equal to 10 300 2 2 Less than 20 60 More than or 240 So, the median class is 100-115. equal to 20 Here, l = 100, f = 9, cf = 11 and h = 15 Less than 30 102 More than or 198 ⎝⎜⎛ N − cf⎞⎠⎟ equal to 30 2 Q Median = l + × h Less than 40 157 More than or 143 f equal to 40 = 100 + (16. 5 − 11) × 15 Less than 50 227 More than or 73 9 equal to 50 = 100 + 5. 5 × 15 9 Less than 60 280 More than or 20 equal to 60 = 100 + 82. 5 9 Less than 70 300 = 100 + 9.17 = 109.17 26. Hence, the median bowling speed is 109.17 km/h. Height (in cm) Frequency Cumulative Cumulative 28. Here, the given data is in ascending order of xi. frequency frequency Cumulative frequency table for the given data is (fi ) (given) (cf ) 150-155 12 a xi fi cf 155-160 b 12 18 8 160-165 10 25 165-170 d 12 + b 2 10 18 170-175 e c 22 + b 175-180 2 22 + b + d 3 11 29 Total 50 43 22 + b + d + e 24 + b + d + e 4 16 45 48 5 20 65 f 6 25 90 7 15 105 On comparing last two tables, we get 8 9 114 a = 12 9 6 120 ∴ 12 + b = 25 ⇒ b = 25 − 12 = 13 Here, n = 120 (even) 22 + b = c ∴Median = 1 ⎣⎢⎡Value of ⎩⎧⎨⎝⎛⎜ n ⎠⎞⎟ th + ⎛⎝⎜ n + 1⎞⎠⎟ th ⎫⎤ observations ⇒ c = 22 + 13 = 35 2 2 2 ⎬⎭⎦⎥ 22 + b + d = 43 = 1 ⎢⎡⎣Value of ⎩⎨⎧⎛⎜⎝ 1220⎞⎟⎠ th + ⎝⎛⎜ 120 + 1⎠⎟⎞ th ⎫⎤ observations ⇒ 22 + 13 + d = 43 2 2 ⎬⎭⎦⎥ ⇒ d = 43 − 35 = 8 = 1 [Value of 60th observation + Value of 61th observation] 22 + b + d + e = 48 2 ⇒ 22 + 13 + 8 + e = 48 ⇒ e = 48 − 43 = 5 Both 60th and 61th observations lie in the cumulative and 24 + b + d + e = f frequency 65 and its corresponding value of xi is 5. ⇒ 24 + 13 + 8 + 5 = f ∴ Median = 1 (5 + 5) = 5 ∴ f = 50 2

CBSE Term II Mathematics X (Standard) 167 29. First we construct a cumulative frequency table. Here, N = 51 ∴ N = 51 = 25.5 Weekly income Number of families Cumulative frequency 22 (in `) (fi ) (cf ) Since, the cumulative frequency just greater than 25.5 is 29 and the corresponding class interval is 145-150. 0-1000 250 250 1000-2000 190 = f 250 + 190 = 440 ∴ Median class = 145-150 = mid class Now, l = 145, f = 18, cf = 11 and h = 5 2000-3000 100 440 + 100 = 540 ⎧N − cf ⎫ f ⎪ ∴ Median = l+ ⎪ 2 ⎬ × h = 145 + ⎧25. 5− 11 ⎫ × 5 ⎨ ⎨⎩ 18 ⎬⎭ 3000-4000 40 540 + 40 = 580 ⎪ ⎪ 4000-5000 15 580 + 15 = 595 ⎩⎭ 5000-6000 5 595 + 5 = 600 = 145 + 72.5 = 145 + 4.03 = 149.03 18 It is given that, N = 600 Hence, the required median height is 149.03 cm. ∴ N = 600 = 300 31. Given, median = 32 22 and N = Σf = 100 Since, cumulative frequency 440 lies in the interval Let f1 and f2 be the frequencies of the class interval 10-20 1000 - 2000. and 40-50, respectively. Here, l = 1000, f = 190, cf = 250 and h = 1000 Since, sum of frequencies = 100 ⎩⎨⎧N2 − cf ⎬⎫ ∴ 10 + f1 + 25 + 30 + f2 + 10 = 100 ⎭ Q Median = l + × h ⇒ f1 + f2 = 100 − 75 ⇒ f1 + f2 = 25 f ⇒ f2 = 25 − f1 …(i) = 1000 + (300 − 250) × 1000 Now, the cumulative frequency table for given distribution is 190 = 1000 + 50 × 1000 Class Frequency (f i ) Cumulative 190 interval frequency (cf ) = 1000 + 5000 0-10 10 10 19 = 1000 + 263.15 10-20 f1 10 + f1 20-30 25 35 + f1 = 1263.15 30-40 30 (f) 65 + f1 40-50 f2 65 + f1 + f2 Hence, the median income is ` 1263.15. 50-60 10 75 + f1 + f2 Total 30. To calculate the median height, we need to convert the given N = f1 + f2 + 75 data in the continuous grouped frequency distribution. Given, distribution is of less than type and 140, 145, 150, …, Here, N = 100 ⇒ N = 50 165 gives the upper limits of the corresponding class 2 intervals. So, the classes should be below 140, 140-145, 145-150, …, 160-165. Clearly, the frequency of class interval below 140 is 4, since Given, median = 32, which belongs to the class 30-40. there are 4 boys with height less than 140. For the frequency of class interval 140-145 subtract the number of boys having So, the median class is 30-40. height less than 140 from the number of boys having height less than 145. Then, l = 30, h = 10, f = 30 and cf = 35 + f1 Thus, the frequency of class interval 140 -145 is 11 − 4 = 7. ⎧N − cf ⎫ Similarly, we can calculate the frequencies of other class f ⎪ Q Median = l + ⎪ 2 ⎬ × h intervals and get the following table ⎨ ⎪ ⎪ ⎩⎭ Class interval Frequency (f i ) Cumulative ∴ 32 = 30 + ⎧ 50 − 35 − f1 ⎫ × 10 frequency (cf ) ⎩⎨ 30 ⎬ ⎭ 4 Below 140 4 11 ⇒ 32 − 30 = 15 − f1 140-145 11 − 4 = 7 29 3 145-150 29 − 11 = 18 = f 40 150-155 40 − 29 = 11 46 ⇒ 2 × 3 = 15 − f1 155-160 46 − 40 = 6 51 ⇒ f1 = 15 − 6 = 9 160-165 51 − 46 = 5 On putting the value of f1 in Eq. (i), we get f2 = 25 − 9 = 16 Hence, the missing frequencies are f1 = 9 and f2 = 16.

168 CBSE Term II Mathematics X (Standard) 32. First, we construct a cumulative frequency table = ` 12.727 (in thousand) = 12.727 × 1000 = ` 12727 Salary (in ` Number of Cumulative frequency Hence, the median and modal salary are ` 13421 and thousand) persons (f i ) (cf ) ` 12727, respectively. 49 (cf) 5-10 49 (f0 ) 33. (i) 10-15 f1 = 133 133 + 49 = 182 15-20 63 (f2) 182 + 63 = 245 Time Number of Class mark f i xi 20-25 245 + 15 = 260 8 × 10 = 80 25-30 15 260 + 6 = 266 (in seconds) students (f i ) (xi ) 30-35 266 + 7 = 273 35-40 6 273 + 4 = 277 0-20 8 10 40-45 277 + 2 = 279 45-50 7 279 + 1 = 280 20-40 10 30 10 × 30 = 300 4 40-60 13 50 13 × 50 = 650 2 60-80 6 70 6 × 70 = 420 1 80-100 3 90 3 × 90 = 270 N = 280 Σfi = 40 Σfixi = 1720 ∴ N = 280 = 140 Mean (x) = Σfi xi = 1720 = 43 22 Σfi 40 (i) Here, median class is 10-15, because 140 lies in it. ∴ Mean time is 43s. ∴l = 10, f = 133, cf = 49 and h = 5 (ii) ⎝⎛⎜ N − cf⎟⎠⎞ Time Number of Cumulative 2 frequency (cf ) Q Median = l + ×h (in seconds) students (f i ) 8 f 0-20 8 = 10 + (140 − 49) × 5 20-40 10 8 + 10 = 18 133 = 10 + 91 × 5 40-60 13 18 + 13 = 31 133 60-80 6 31 + 6 = 37 = 10 + 455 = 10 + 3.421 133 80-100 3 37 + 3 = 40 = ` 13.421 (in thousand) Σfi = 40 = 13.421 × 1000 Modal class is a class having highest frequency. = ` 13421 So, 40-60 is modal class (ii) Here, the highest frequency is 133, which lies in the To find median class, we find cumulative frequency interval 10-15, called modal class. N = 40 = 20 22 ∴l = 10, h = 5, f1 = 133, f0 = 49, and f2 = 63. ∴ 40-60 has cumulative frequency greater than 20. ∴ Mode = l + ⎛ f1 − f0 ⎞ × h Thus, 40-60 is the median class. ⎝⎜ − f0 − f2⎠⎟ 2 f1 ∴Sum of lower limits of median class and modal class = 40 + 40 = 80. = 10 + ⎧ 133 − 49 ⎫ × 5 ⎩⎨2 133 − 49 − ⎬ (iii) Students finished the race within 1 min × 63 ⎭ = Students between 0-20 + Students between 20-40 = 10 + 84 × 5 + Students between 40-60 266 − 112 = 8 + 10 + 13 = 31 = 10 + 84 × 5 = 10 + 2.727 154

Chapter Test Multiple Choice Questions (ii) The modal value of the given data is 1. A survey conducted by a group of students is (a) 150 (b) 150.91 (c) 145.6 (d) 140.9 given as (iii) The median value of the given data is Family Size 1-3 3-5 5-7 7-9 9-11 (a) 140 (b) 146.67 (c) 130 (d) 136.6 Number of families 7 8 2 21 (iv) Assumed mean method is useful in determining the The mean of the data is (a) Mean (b) Median (c) Mode (d) All of these (a) 6.8 (b) 4.2 (c) 5.4 (d) None of these (v) The manufacturer can claim that the mileage for his car is 2. The relationship among mean, median and (a) 144 km/charge (b) 155 km/charge mode for a distribution is (c) 165 km/charge (d) 175 km/charge (a) Mode = Median − 2 mean (b) Mode = 3 Median − 2 mean Short Answer Type Questions (c) Mode = 2 Median − 3 mean (d) Mode = Median − mean 5. Find the median of the first ten prime numbers. 3. For the following distribution 6. An aircraft has 120 passenger seats. The Class 0 - 5 5 -10 10 -15 15 - 20 20 - 25 number of seats occupied during 100 flights is Frequency given in the following table. 10 15 12 20 9 Number of seats 100-104 104-108 108-112 112-116 116-120 The sum of lower limits of the median class and Frequency 15 20 32 18 15 modal class is (a) 15 (b) 25 (c) 30 (d) 35 Determine the mean number of seats occupied over the flights. Case Based MCQs 7. The following distribution gives the daily 4. A Tesla car manufacturing industry wants to income of 50 workers of a factory: declare the mileage of their electric cars. For this, they recorded the mileage (km/charge) of Daily income (in `) 100-120 120-140 140-160 160-180 180-200 100 cars of the same model. Details of which are given in the following table. Number of workers 12 14 8 6 10 Mileage 100-120 120-140 140-160 160-180 Write the above distribution as ‘less than type’ (km/charge) 14 24 36 26 cumulative frequency distribution. [CBSE 2015] Number of Cars 8. Find the mode of the following frequency distribution. [CBSE 2019] Class 0-10 10-20 20-30 30-40 40-50 50-60 60-70 Frequency 8 10 10 16 12 6 7 Based on the above information, answer the Long Answer Type Questions following questions. 9. Find the mean of the following frequency (i) The average mileage is distribution using assumed mean method. (a) 140 km/charge (b) 150 km/charge Class 2-8 8-14 14-20 20-26 26-32 (c) 130 km/charge Frequency 6 3 12 11 8 (d) 144.8 km/charge Answers For Detailed Solutions Scan the code 1. (b) 2. (b) 3. (b) 4. (i) (d) (ii) (b) (iii) (b) (iv) (a) (v) (a) 5. 12 6. 109.92 8. 36, 9. 18.8

170Mathematics (Standard) CBSE Term II Mathematics X (Standard) Class 10th (Term II) Practice Paper 1* (Solved) Instructions Time : 2 Hr Max. Marks : 40 1. The question paper contains three sections A, B and C. 2. Section A has 5 questions with 3 internal choices. 3. Section B has 4 questions with 3 internal choices. 4. Section C has 1 Case Based MCQs comprises of 5 MCQs. 5. There is no negative marking. * As exact Blue-print and Pattern for CBSE Term II exams is not released yet. So the pattern of this paper is designed by the author on the basis of trend of past CBSE Papers. Students are advised not to consider the pattern of this paper as official, it is just for practice purpose. Section A (3 Marks Each) This section consists of 5 questions of Short Answer Type. 1. Find the value of k for which the quadratic equation (3k + 1)x 2 + 2(k + 1)x + 1 = 0, has equal roots . Also find these roots. 2. Write the expression an − ak for the AP a, a + d, a + 2d, … Hence, find the common difference of the AP for which 25th term is 10 more than the 23rd term. Or If two towers of heights x m and y m subtend angles of 45° and 60°, respectively at the centre of a line joining their feet, then find the ratio of (x + y): y. 3. The length of common chord of two intersecting circles is 30 cm. If the diameters of these two circles are 50 cm and 34 cm, then calculate the distance between their centres. Or Given, a line segment AB. Divide it in the ratio m : n by construction, where both m and n are positive integers and let m = 4 and n = 3. 4. From a solid cube of side 7 cm, a conical cavity of height 7 cm and radius 3 cm is hollowed out. Find the volume of the remaining solid. 5. The mode of the following series is 36. Find the missing (x) frequency in it. Class interval 0-10 10-20 20-30 30-40 40-50 50-60 60-70 Frequency 8 10 x 16 12 6 7 Or The 8th term of an AP is 17 and its 14th term is 29. Find its common difference.

CBSE Term II Mathematics X (Standard) 171 Section B (5 Marks Each) This section consists of 4 questions of Long Answer Type. 6. Construct a tangent to a circle of radius 4 cm from a point on the concentric circle of radius 6 cm and measure its length. Also, verify the measurement by actual calculation. Or A decorative block as shown in figure is made of two solids, a cube and a hemisphere. 4.2 cm 5 cm 5 cm 5 cm The base of the block is the cube with edge of 5 cm and the hemisphere attached on the top has a diameter of 4.2 cm. If the block is to be painted, then find the total area to be painted. [take, π = 22 / 7] 7. From the point, 36 m above the surface of a lake, the angle of elevation of a bird is observed to be 30° and angle of depression of its image in the water of the lake is observed to be 60°. Find the actual height of the bird above the surface of the lake. Or Prove that the intercept of a tangent between two parallel tangents to a circle subtends a right angle at the centre. 8. Find the solution of the equation x − 1 + 2x +1 = 5, x ≠ − 1 ,1 by factorisation method. 2x +1 x −1 2 2 Or Find the median for the following frequency distribution. Height (in cm) Frequency 160-162 15 163-165 117 166-168 136 169-171 118 172-174 14 9. A girl of height 120 cm is walking away from the base of a lamp-post at a speed of 1.2 m/s. If the lamp is 4.8 m above the ground, then find the length of her shadow after 6 s. Section C (1 Mark Each) This section consists of 1 Case Based comprises of 5 MCQs. 10. In a mathematic class, a teacher explain the concept for determine the mean by defining the formula x = Σf i x i . Σf i

172 CBSE Term II Mathematics X (Standard) Further, a teacher give one example for explaining the above (ii) Find the cumulative frequency value in the concepts. interval (40-55). The marks obtained by 30 students of class X of a certain school in a mathematics paper consisting of 100 marks are (a) 5 (b) 12 presented in table below (c) 2 (d) 18 Class interval 10-25 25-40 40-55 55-70 70-85 85-100 (iii) Through cumulative frequency table, which central measurement can be determined. Number of 2 3 7 6 6 6 Students (a) mean (b) mode (c) median (d) None of these (i) Find the average marks obtained by the students. (a) 61 (iv) Find the lower limit of the median class. (b) 62 (c) 63 (a) 55 (b) 40 (d) 64 (c) 70 (d) 25 (e) Find the upper limit of modal class. (a) 40 (b) 55 (c) 70 (d) 25 Solutions 1. Given quadratic equation is On subtracting Eq. (ii) from Eq. (i), we get (3k + 1)x2 + 2(k + 1)x + 1 = 0 an − ak = [a + (n − 1) d] − [a + (k − 1) d] On comparing with ax2 + bx + c = 0, we get ⇒ an − ak = a + (n − 1) d − a − (k − 1) d a = 3k + 1, b = 2(k + 1), c = 1 ⇒ an − ak = (n − 1 − k + 1) d Since, the roots are equal, so b2 − 4ac = 0 ⇒ an − ak = (n − k) d ∴ [2(k + 1)]2 − 4(3k + 1) (1) = 0 Now, an − ak = (n − k) d ⇒ 4(k + 1)2 − 4(3k + 1) = 0 ⇒ a 25 − a 23 = (25 − 23) d [put n = 25and k = 23] ⇒ 4(k2 + 2k + 1) − 4(3k + 1) = 0 ⇒ 10 = 2d [Qa 25 − a 23 = 10, given] ⇒ 4(k2 + 2k + 1 − 3k − 1) = 0 ⇒ d=5 ⇒ k2 + 2k + 1 − 3k − 1 = 0 Hence, the common difference is 5. Or ⇒ k2 + 2k + 1 − 3k − 1 = 0 [Q 4 ≠ 0] Let AB = x m be the height of a tower and CD = y m be the height of other tower and ∠AEB = 45° and ∠CED = 60°. ⇒ k2 − k = 0 D ⇒ k(k − 1) = 0 B ⇒ k = 0 or k = 1 y x We know that, if roots are equal, then roots will be the form of − b , − b. 45° 60° 2a 2a A aE a C Thus, roots are − b , − b. Let E be the point (centre) on the line AC. 2a 2a i.e. AE = EC = a m ∴ Equal roots are − (k + 1) , − (k + 1) (3k + 1) (3k + 1) In right angled ΔBAE, perpendicular tan 45°= AB base When k = 0, equal roots are − (0 + 1) and − (0 + 1) AE ⎢⎡⎣Q tan θ = ⎤ ⎥⎦ 0+1 0+1 ⇒ x =1 a i.e. − 1 and − 1. [Q tan 45°= 1] ⇒ x=a When k = 1, equal roots are − (1 + 1) and − (1 + 1) …(i) 3+1 3+1 Again, in right angled ΔDCE, i.e. − 1 and − 1. 22 tan 60° = DC CE 2. Given, first term = a and common difference = d ⇒ 3=y ∴ an = a + (n − 1) d …(i) a [Q tan 60°= 3] and ak = a + (k − 1) d …(ii) ⇒ y = 3a …(i)

CBSE Term II Mathematics X (Standard) 173 ∴ (x + y): y = (a + 3a): a 3 = (1 + 3): 3 [given] (iv) Now, through the point A 4(m = 4), draw a line parallel to A7B by making an angle equal to ∠AA 7 B at A4 Hence, the requred ratio (x + y): y is (1 + 3): 3. intersecting AB at a point C. 3. Let, PQ be the length of the common chord of two Then, AC : BC = 4 : 3 intersecting circles. Alternative Method ∴ PQ = 30cm (i) Draw any ray AX making an acute angle with AB. Diameters of two circles are 50 cm and 34 cm. (ii) Draw a ray BY parallel to AX by making Join AB. ∠ABY = ∠BAX. P (iii) Locate the points A1, A 2, A 3, A 4 (for m = 4) on AX and A OB similarly B 1,B 2,B3 (for n = 3) on BY, such that A A 1 = A 1A 2 = A 2A 3 = A 3A 4 Q = BB1 = B1B2 = B2B3 X Since, AB bisects the common chord PQ perpendicularly. A4 A3 ∴ OP = OQ = 1 × PQ = 1 × 30 = 15 cm A2 22 A1 C Radius, AP 1 × 50 = 25 cm AB =2 B1 and radius, PB = 1 × 34 = 17 cm B2 2 B3 In right angled ΔAOP, Y OA = (AP)2 − (OP)2 (iv) Join A 4B3. Let it intersects AB at point C. Then, AC = 4 [by using Pythagoras theorem] BC 3 = (25)2 − (15)2 4. Given, side of a solid cube, a = 7 cm Height of conical cavity, i.e. cone, h = 7 cm = 625 − 225 = 400 = 20 cm Radius of conical cavity, r = 3 cm In right angled ΔPOB, 7 cm OB = (PB)2 − (OP)2 = (17)2 − (15)2 3 cm [by using Pythagoras theorem] Now, volume of cube = a3 = (7)3 = 343 cm 3 and volume of conical cavity = 1 π × r2 × h = 289 − 225 3 = 64 = 8 cm = 1 × 22 × 3 × 3 × 7 = 66 cm 3 ∴ Distance between centres, 37 AB = OA + OB = 20 cm + 8 cm = 28 cm According to the question, Volume of remaining solid Or = Volume of cube − Volume of conical cavity Given A line segment AB, m = 4 and n = 3. = 343 − 66 = 277 cm 3 Hence, the required volume of solid is 277 cm 3. Steps of construction 5. Since, the mode of the given series is 36, which lies in the class 30-40. (i) Draw any ray AX making an acute angle with AB. So, the modal class is 30-40. (ii) Locate 7 (i.e. m + n ) points A1, A2 , ...,A7 on AX, Then, l = 30, f1 = 16, f0 = x, f2 = 12 and h = 10 such that AA1 = A1A 2 = ... = A 6A 7. (iii) Join BA 7. X A7 A6 A5 A4 A3 A2 A1 A CB

174 CBSE Term II Mathematics X (Standard) Also, mode = 36 ∴ (PM)2 = (OP)2 − (OM)2 [by Pythagoras theorem] Q Mode = l + ⎨⎧h × f1 − f0 ⎫ ⇒ (PM)2 = (6)2 − (4)2 = 36 − 16 = 20 ∴ ⎩ f1 − f0 − ⎬ ⇒ 2 f2 ⎭ ⇒ PM = 20 = 4.47 ≈ 4.5 ⇒ 36 = 30 + ⎨⎧10 × × 16 − x 12 ⎫ Hence, the length of tangent is 4.5 cm. ⎩ 2 16 − x − ⎬ Or ⎭ 36 = 30+ 10 (16 − x) Given, edge of cube = 5 cm (20 − x) and diameter of hemisphere = 4.2 cm radius of hemisphere = 4.2 cm = 2.1 cm 36 − 30 = 10 (16 − x) ⇒ 6 = 10(16 − x) (20 − x) 1 (20 − x) 2 ⇒ 10 (16 − x) = 6 (20 − x) Clearly, total surface area of the cube = 6 (Edge)2 ⇒ 160 − 10 x = 120 − 6x = 6 × 5 × 5 = 150 cm2 ⇒ − 10x + 6x = 120 − 160 ⇒ − 4x = − 40 Now, area to be painted on the cube ∴ x = − 40 = 10 = Total surface area of cube −4 − Base area of hemisphere Hence, the missing frequency is 10. = 150 − πr2 = 150 − 22 × 4.2 × 4.2 72 2 Or = 150 − 13.86 = 136.14 cm 2 Given, Area to be painted on the hemisphere ε8 = 17, ε14 = 29 Let be a first term a = Curved surface area of hemisphere = 2πr2 = 2 × 22 × 4.2 × 4.2 = 27.72 cm 2 εn = a + (n − 1)d 17 = 9 + (8 − 1)d 72 2 ⇒ 17 = a + 7d …(i) ∴ Total area to be painted = Area to be painted on the cube Similarly, 29 = a + 13d …(ii) + Area to be painted on the hemisphere = 136.14 + 27.72 = 163.86 cm 2 Subtract Eq. (i) from Eq. (ii), 7. Let QR be the surface of the lake and P be point above the a + 13d − a − 7d = 29 − 17 = d = 2 surface such that PQ = 36 m. Let B represents the bird and 6. Given, two concentric circles of radii 4 cm and 6 cm with Bʹ be its image in the lake. common centre O. Here, we have to draw two tangents to inner circle C1 from a B point of outer circle C2. x Steps of construction S (i) Draw two concentric circles C1 and C2 with common P 30° 36 m centre O and radii 4 cm and 6 cm, respectively. 60° R 36 m (ii) Take any point P on outer circle C2 and join OP. Q (iii) Now, bisect OP. Let Mʹ be the mid-point of OP. (iv) Taking Mʹ as centre and OMʹ as radius, draw a dotted (36+x) m circle which cuts the inner circle C1 at two points M and Pʹ. (v) Join PM and PPʹ. Thus, PM and PPʹ are required Bʹ tangents. ∴ ∠BPS = 30° and ∠BʹPS = 60° M Also, SR = PQ = 36 m Let BS = x P Mʹ O C1 C2 ⇒ BʹR = BR = (36 + x) m Pʹ ∴ BʹS = SR + BR On measuring PM and PPʹ, we get = 36 + 36 + x = (72 + x) m PM = PPʹ = 4.5 cm In right angled ΔPSB, Calculation In right angled ΔOMP, ∠ PMO = 90° BS = tan 30° PS

CBSE Term II Mathematics X (Standard) 175 x1 ⎢⎣⎡Q tan 30° = 1⎤ = 1 (∠1 + ∠1 + ∠3 + ∠3) ∴ PS = 3 3 ⎥⎦ 2 ⇒ PS = 3x …(i) = 1 (∠1 + ∠2 + ∠3 + ∠4) 2 Again, in right angled ΔBʹSP, BʹS = tan 60°⇒ 72 + x = 3 [from Eqs. (i) and (ii)] PS PS = 1 (180°) = 90° [Q B’S = (72+x) m and tan 60° = 3 ] 2 ⇒ PS = 72 + x [Q QR is a straight line, therefore 3 …(ii) ∠1 + ∠2 + ∠3 + ∠4 = 180°] From Eqs. (i) and (ii), we get Hence proved. 3x = 72 + x 8. We have, x −1 + 2x + 1 = 5 3 2x + 1 x −1 2 ⇒ 3 ⋅ 3x = 72 + x Let y = x −1 , then given equation becomes 2x +1 ⇒ 3x − x = 72 ⇒ 2x = 72 ⇒ x = 72 ⇒ x = 36 y + 1 = 5 ⇒ y 2 + 1 = 5 ⇒ 2y 2 − 5y + 2 = 0 y2 y2 2 ∴ Height of bird above surface of the lake, This is a quadratic equation. BR = BS + SR = 36 + 36 = 72 m Or By using factorisation method, 2y 2 − 4y − y + 2 = 0 Let, AB and CD are two tangents to a circle and AB || CD. ⇒ 2y(y − 2) − 1 (y − 2) = 0 Tangent BD subtends ∠BOD at the centre. ⇒ (2y − 1)(y − 2) = 0 To prove ∠BOD = 90° ⇒ 2y − 1 = 0 or y − 2 = 0 Construction Join OP, OQ and OR. ⇒ y = 1 or y = 2 2 AQ B Put y = x −1 , we get 2x +1 2 x − 1 = 1 or x − 1 = 2 O1 2x +1 2 2x +1 3 P ⇒ 2x − 2 = 2x + 1 or x − 1 = 4x + 2 4 ⇒ −2 = 1, which is not true. Consider, x − 1 = 4x + 2 C RD ⇒ 3x = − 3 ⇒ x = − 1 Or Proof Here, OP ⊥ BD The given series is in inclusive form. Converting it to exclusive form and preparing the cumulative frequency table [Q a tangent at any point of a circle is perpendicular to the is given below radius through the point of contact] In right angled ΔOQB and Δ OPB, Class interval Frequency (fi ) Cumulative frequency BQ = BP 159.5-162.5 15 162.5-165.5 117 15 [Q the lengths of tangents drawn from an 165.5-168.5 136 132 external point are equal] 168.5-171.5 118 268 171.5-174.5 14 386 OQ = OP [radii] N = Σ f i = 400 400 OB = OB [common] Total ∴ ΔOQB ≅ Δ OPB [by SSS congruency] Then, ∠1 = ∠2 [by CPCT] …(i) Similarly, in right angled ΔOPD and ΔORD, …(ii) Here, N = 400 ∠3 = ∠4 Now, N = 400 = 200 22 ∴ ∠BOD = ∠1 + ∠3 = 1 (2∠1 + 2∠3) 2

176 CBSE Term II Mathematics X (Standard) The cumulative frequency just greater than 200 is 268 and the ⇒ 3x = 7.2 ⇒ x 7.2 = 2.4 m corresponding class is 165.5-168.5. =3 Thus, the median class is 165.5-168.5. Hence, the length of her shadow after 6 s is 2.4 m. 10. (i) (b) Let us make the following table for the given data. ∴ l = 165.5, h = 3 and f = 136 and C = 132 Cumulative frequency ⎧ N − C. f ⎫ Class Frequency Class marks f i xi ⎨⎪h 2 ⎪ Interval (cf ) ∴ Median = l + ⎪ × ⎬ (1/2) (xi ) 2 ⎪ f 10 + 25 = 17.5 5 2 ⎩⎭ 10-25 2 35.0 12 25-40 = 165.5 + ⎧⎨3 × (200 − 132) ⎬⎫ 40-55 25 + 40 = 32.5 18 ⎩ 136 ⎭ 55-70 2 70-85 3 97.5 24 = 165.5 + 3 × 68 = 165.5 + 1.5 = 167 136 85-100 7 40 + 55 = 47.5 332.5 30 2 Hence, the median height is 167 cm. 9. Let AB be the lamp-post, CD be the girl and D be the 6 55 + 70 = 62.5 375.0 position of girl after 6 s. 2 Again, let DE = x m be the length of shadow of the girl. 6 70 + 85 = 77.5 465.0 2 A 85 + 100 = 92. 5 2 6 555.0 4.8 m C Σfixi = 1860.0 E Total Σfi = 30 B Dx Given, CD = 120 cm = 1.2 m, AB = 4.8 m Here, Σfi = 30 and Σfixi = 1860.0 Q Average, x = Σfixi = 1860.0 = 62 and speed of the girl = 1.2 m/s Σfi 30 ∴ Distance of the girl from lamp-post after 6 s. Hence, average marks obtained by student is 62. BD = 1.2 × 6 = 7.2 m [Q distance = speed × time] (ii) (b) The cumulative frequency value in the interval In ΔABE and ΔCDE, 40-55 is 12. ∠B = ∠D [each 90°] (iii) (c) Through cumulative frequency table, median can be determined. ∠E = ∠E [common angle] (iv) (a) Here, N = 30 = 15, which lies in the cumulative ∴ ΔABE ~ ΔCDE [by AA similarity criterion] 22 ⇒ BE = AB …(i) frequency 18, whose corresponding frequency is 55-70. DE CD Hence, lower limit of the median class is 55. On substituting all the values in Eq. (i), we get 7.2 + x = 4.8 [Q BE = BD + DE = 7.2 + x] (v) (b) In the given data, the highest frequency is 7, whose x 1.2 corresponding interval is 40-55. ⇒ 7.2 + x = 4 Hence, upper limit of the modal class is 55. x ⇒ 7.2 + x = 4x

CBSEMTeramtIhI eMmathaemtiactiscs (XS(tSatannddaarrdd)) 177 Class 10th (Term II) Practice Paper 2* (Unsolved) Instructions Time : 2 Hr Max. Marks : 40 1. The question paper contains three sections A, B and C. 2. Section A has 5 questions with 3 internal choices. 3. Section B has 4 questions with 3 internal choices. 4. Section C has 1 Case Based MCQs comprises of 5 MCQs. 5. There is no negative marking. * As exact Blue-print and Pattern for CBSE Term II exams is not released yet. So the pattern of this paper is designed by the author on the basis of trend of past CBSE Papers. Students are advised not to consider the pattern of this paper as official, it is just for practice purpose. Section A (3 Marks Each) This section consists of 5 questions of Short Answer Type. 1. Find the roots of the quadratic equation 9x 2 − 9(a + b)x + (2a2 + 5ab + 2b 2 ) = 0. 2. Which term of the progression 19, 18 1 , 17 2, ... is the first negative term? 55 Or From the top of a 10 m high building, the angle of elevation of the top of a tower is 60° and the angle of depression of its foot is 45°. Determine the height of the tower. 3. In the given figure, AB is the diameter of a circle with centre O and QC is a tangent to the circle at C. If ∠CAB = 30°, then find ∠CQA and ∠CBA. C A 30° B Q O Or A copper wire 4 mm in diameter is evenly wound about a cylinder whose length is 24 cm and diameter 20 cm so as to cover the whole surface. Find the length and weight of the wire assuming the specific density to be 8.88 gm/cm 3. 4. Draw a circle of radius 6 cm. Take a point P on it. Without using the centre of the circle, draw a tangent to the circle at point P.

178 CBSE Term II Mathematics X (Standard) 5. The mean of the following frequency table is 50 but the frequencies f1 and f 2 in class intervals 20-40 and 60-80 are missing. Find the missing frequencies. Class interval 0-20 20-40 40-60 60-80 80-100 Total Frequency 17 f1 32 f2 19 120 Or Find the volume area of the largest right circular cone that can be cut out of a cube whose edge is 10 cm. Section B (5 Marks Each) This section consists of 4 questions of Long Answer Type. 6. A cone of maximum size is cut-out from a cube of edge 14 cm. Find the surface area of the remaining solid left out after the cone is cut-out. Or The angle of elevation of a jet plane from a point A on the ground is 60°. After a flight of 15 sec, the angle of elevation changes to 30°. If the jet plane is flying at a constant height of 1500 3 m, find the speed of the jet plane. 7. If α and β are the zeroes of the quadratic polynomial f (x) = 3x 2 − 4x + 1, find a quadratic polynomial whose zeroes are α 2 and β2 . βα Or If m times the mth term of an AP is equal to n times its nth term, then show that (m + n) th term of the AP is zero. 8. Construct a tangent to a circle of radius 1.8 cm from a point on the concentric circle of radius 2.8 cm and measure its length. Also, verify the measurement by actual calculation. 9. In the given figure, PT is a tangent and PAB is a secant. If PT = 6 cm and AB = 5 cm, then find the length of PA. T O P B A Or Find the mean, mode and median of the following data. Class 0-10 10-20 20-30 30-40 40-50 50-60 60-70 Frequency 3 4 7 15 10 7 4 Section C (1 Mark Each) This section consists of 1 Case Based comprises of 5 MCQs. 10. In one corner of the drawing room, a flower basket is kept inside the glass, lies on the table. The basket is designed in such a way that every one pleases to see it.

CBSE Term II Mathematics X (Standard) 179 The shape of flower basket is hemisphere with radius 60 cm and upper shape is conical with height 120 cm from the bottom surface. F PE 120 cm 180 cm O60 cm D 60 cm A Oʹ B (i) Find the capacity of the glass. (a) 14.256 m 3 (b) 12.256 m 3 (c) 142.56 m 3 (d) 14.256 m 3 7 7 7 5 (ii) Find the volume of the cone. (a) 0.54 m3 (b) 0.45 m3 (c) 0.25 m3 (d) 0.52 m3 (iii) Find the curve surface area of hemisphere. (a) 0.201 m2 (b) 0.104 m2 (c) 0.102 m2 (d) 0.401 m2 (iv) The volume of two combined figure is equal to the sum of (a) two individual volumes (b) two individual curve surface area (c) volumes and curve surface area (d) None of these (v) If the cost of painting the glass outside is ` 1.20 per m 2, find the total cost of painting the CSA of the glass. (a) ` 55 (b) ` 55.02 (c) ` 57 (d) ` 57.02 Answers 1. (2a + b) , (a + 2b) 2. 25th term or 10( 3+ )1m 3. ∠CQA = 30° and ∠CBA = 60° or 4.21 kg 33 8. 2.14 cm 5. f1 = 28 and f 2 = 24 or 261.9 cm2 6. 1022 + 154 5 cm2 or 720 km/h 7. k ⎜⎛⎝ x 2 − 28 x + 13⎠⎞⎟ where k is any non-zero real number 9 9. 4 cm or (i) 37.4, (ii) 36.15, (iii) 37.3 10. (i) (a) (ii) (b) (iii) (c) (iv) (a) (v) (d)

180Mathematics (Standard) CBSE Term II Mathematics X (Standard) Class 10th (Term II) Practice Paper 3* (Unsolved) Instructions Time : 2 Hr Max. Marks : 40 1. The question paper contains three sections A, B and C. 2. Section A has 5 questions with 3 internal choices. 3. Section B has 4 questions with 3 internal choices. 4. Section C has 1 Case Based MCQs comprises of 5 MCQs. 5. There is no negative marking. * As exact Blue-print and Pattern for CBSE Term II exams is not released yet. So the pattern of this paper is designed by the author on the basis of trend of past CBSE Papers. Students are advised not to consider the pattern of this paper as official, it is just for practice purpose. Section A (3 Marks Each) This section consists of 5 questions of Short Answer Type. 1. Find the solution of the equation x −3 − x + 3 = 48 , x ≠ 3, x ≠ −3. x +3 x − 3 7 2. Write the expression an − ak for the AP a, a + d, a + 2d, … Hence, find the common difference of the AP for which 25th term is 10 more than the 23rd term. Or A statue 1.6 m tall stands on the top of a pedestal. From a point on the ground, the angle of elevation of the top of the statue is 60° and from the same point the angle of elevation of the top of the pedestal is 45°. Find the height of the pedestal. 3. In the given figure, PQ and QR are tangents to the circle centre O, at P and R, respectively. Find the value of x. R 20° 50° Q SO x° P Or Draw a circle with the help of circular solid ring. Construct a pair of tangents from a point P to the circle. 4. For grouped data, if Σf i = 20, Σf i x i = 2p + 20 and mean of distribution is 12, then find the value of p.

CBSE Term II Mathematics X (Standard) 181 5. A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere is 14 cm and the total height of the vessel is 13 cm. Find the inner surface area of the vessel. Or Two pillars of equal height are on either sides of a road, which is 100 m wide. The angles of the top of the pillars are 60° and 30° at a point on the road between the pillars. Find the position of the point between the pillars. Also, find the height of each pillar. Section B (5 Marks Each) This section consists of 4 questions of Long Answer Type. 6. A solid toy is in the form of a hemisphere surmounted by a right circular cone. Height of the cone is 4 cm and the diameter of the base is 8 cm. If a right circular cylinder circumscribes the solid. Find how much more space it will cover? Or The angle of elevation of the top of a tower from certain point is 30°. If the observer moves 20 m towards the tower, the angle of elevation of the top increases by 15°. Find the height of the tower. 7. Construct a tangent to a circle of radius 1.8 cm from a point on the concentric circle of radius 2.8 cm and measure its length. Also, verify the measurement by actual calculation. Or From an external point P, two tangents PA and PB are drawn to the circle with centre O. Prove that OP is the perpendicular bisector of AB. 8. Solve the following quadratic equation by factorisation method. a + 1 + x = 1 + 1 + 1, a + b ≠ 0. b a b x Or The angles of a triangle are in AP. If the greatest angle equals to the sum of the other two, then find the angles. Also, conclude that find these angles are multiple of which angle. 9. The median of the distribution given below is 14.4. Find the values of x and y, if the total frequency is 20. Class interval 0-6 6-12 12-18 18-24 24-30 Frequency 4x5 y1 Section C (1 Mark Each) This section consists of 1 Case Based comprises of 5 MCQs. 10. Suppose, there are two windows in a house. A window of the house is at a height of 1.5 m above the ground and the other window is 3 m vertically above the lower window. C B 30° h – 4.5 m (Shyam) D 3m 3m (RAM) A E 1.5 m 1.5 m P Q

182 CBSE Term II Mathematics X (Standard) Anil and Sanjeev are sitting in the two windows. At an instant, the angles of elevation of a balloon from these windows are observed as 45° and 30°, respectively. (i) Find the height of the balloon from the ground. (a) 6.8 m (b) 8.6 m (c) 9.4 m (d) 9.6 m (ii) Among Anil and Sanjeev, who is more closer to the balloon? (a) Sanjeev (b) Anil (c) cannot say (d) None of these (iii) If the balloon is moving towards the building, then will both the angles of elevation remain same? (a) cannot say (b) Yes (c) No (d) None of these (iv) If the height of any tower is double and the distance between the observer and foot of the tower is also doubled, then the angle of elevation (a) remain same (b) become double (c) become triple (d) None of these (v) Suppose a tower and a pole is standing on the ground. And the angle of elevation from bottom of pole is θ1 and elevation from top of pole to the top of tower is θ2. θ1 B Choose the correct option. θ1 A (a) θ1 >θ2 (b) θ1 = θ2 (c) θ1 < θ2 (d) None of these Answers 1. −4, 9 2. 5 or 0.8( 3 + 1) m 3. 45° 4. 110 5. 572 cm2 or 25 m, 43.3 m 4 6. 128 π cm3 or 10( 3 +1) m 7. 2.14 cm 8. x = −a or x = −b or ` 30°, 60° and 90° angles are the multiple of 30° 3 9. x = 4 and y = 6 10. (i) (b) (ii) (a) (iii) (c) (iv) (a) (v) (a)