Vedanta Excel in Mathematics Book 9 Final (2080)_compressed

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 301 Vedanta Excel in Mathematics - Book 9 e) Draw a histogram to represent the data given below in the table. Mid-point 12 16 20 24 28 32 36 40 Frequency 5 11 17 13 15 20 12 8 4. a) The heights (in cm) of 30 pupils are given below: 135, 140, 120, 160, 110, 107, 150, 136 102, 113, 116, 109, 129, 124, 131, 147, 142, 123, 118, 152 128, 143, 127, 134, 115, 126, 136, 147, 129, 139 Construct a frequency distribution table of continuous class with length 10 and draw a histogram to represent the data. b) The daily wages (in Rs) of 40 employees of a factory are given below: 415, 440, 405, 412, 402, 418, 446, 406, 435, 422, 425, 430, 428, 416, 445, 421, 449, 437, 441, 419, 432, 426, 443, 436, 424, 438, 423, 448, 414, 427, 443, 412, 448, 447, 425, 415, 445, 434, 440, 428 Construct a frequency distribution table of continuous class with length 10 and draw a histogram to represent the data. Project work and activity section 5. a) Measure the heights of every student in your class. Show the heights in a frequency distribution table of continuous class of length 10. Then, present the data in a histogram. b) Show the marks obtained by the students of your class in mathematics in recently conducted exam in a frequency distribution table of continuous class of length 5. Then, show it in a histogram. 2. Frequency polygon Frequency polygon is an alternative to a histogram that represents the similar information about the data. It is the piecewise linear curve formed by connecting the midpoints of the tops of the adjacent rectangles of the histogram. In other words, it is a graphical representation of continuous data in which the frequencies of the classes are plotted by dots against the mid-points of each class and adjacent dots are joined by straight line segments. The frequency polygon can be constructed by the following two methods. a) Using a histogram Study the following steps to construct a frequency polygon using a histogram. (i) Draw a histogram from the given data. (ii) Mark the mid-point at the top of each rectangle of the histogram drawn. Statistics (I): Classification and Graphical Representation of Data

Vedanta Excel in Mathematics - Book 9 302 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur (iii) Also, mark the mid-point of the class interval preceding the lower class interval and midpoint of the class interval succeeding the highest class interval with zero frequencies. (iv) At last, join the consecutive midpoints by start line segments to obtained the required frequency polygon. Worked-out Examples Example 1: The marks obtained by the students in mathematics are given below. Draw a frequency polygon by using histogram. Marks obtained 10-20 20-30 30-40 40-50 50-60 60-70 70-80 80-90 Number of students 6 10 18 24 20 14 12 5 Solution: Marks obtained 0 5 10 20 30 40 50 60 70 80 90 100 10 15 20 25 Number of students b) Without using a histogram Study the following steps to construct a frequency polygon without using a histogram. (i) Find the mid-value (class-mark) of each interval using the formula, mid-value = lower limit + upper limit 2 (ii) On a graph paper, mark the mid-values along X-axis and their corresponding frequencies along Y-axis. (iii) At last, join the consecutive midpoints by start line segments to obtained the required frequency polygon. Example 2: Draw a frequency polygon for the following data without using data. Weight (in kg) 10 - 20 20 - 30 30 - 40 40 - 50 50 - 60 60 - 70 Number of people 6 12 18 20 15 9 Statistics (I): Classification and Graphical Representation of Data

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 303 Vedanta Excel in Mathematics - Book 9 Solution: Weight (in kg) Mid-value (m) Number of people (f) Points 0 - 10 10 - 20 20 - 30 30 - 40 40 - 50 50 - 60 60 - 70 70 - 80 0 + 10 2 = 5 10 + 20 2 = 15 20 + 30 2 = 25 30 + 40 2 = 35 40 + 50 2 = 45 50 + 60 2 = 55 60 + 70 2 = 65 70 + 80 2 = 75 0 6 12 18 20 15 9 0 (5, 0) (15, 6) (25, 12) (35, 18) (45, 20) (55, 15) (65, 9) (75, 0) Frequency polygon 0 2 4 6 5 15 25 35 45 55 65 75 8 10 12 14 16 18 20 22 Number of people Weight in kg EXERCISE 17.3 General section 1. a) Define frequency polygon. b) Write down the ways of drawing the frequency polygon. 2. a) The marks obtained by 90 students of class 9 in Mathematics are shown in the table given below. Construct a frequency polygon to represent the data by using histogram. Marks obtained 20-30 30-40 40-50 50-60 60-70 70-80 80-90 Number of students 3 8 10 25 20 15 9 Statistics (I): Classification and Graphical Representation of Data

Vedanta Excel in Mathematics - Book 9 304 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur b) From the data given below, draw a histogram and frequency polygon imposed on it. Data 10-20 20-30 30-40 40-50 50-60 60-70 70-80 Frequency 6 10 16 20 15 8 5 3. a) Construct a frequency polygon to represent the data without using histogram. Weight (in kg) 10-20 20-30 30-40 40-50 50-60 60-70 Number of people 10 15 18 20 25 15 b) In a study of dental problem, the following data was recorded. Present the data in a frequency polygon without using histogram. Age (in years) 0-10 10-20 20-30 30-40 40-50 50-60 60-70 No. of people 5 14 25 30 35 40 20 Creative section 4. a) The marks obtained by 40 students in mathematics in the final examination of class 9 are given below. 65, 40, 24, 18, 36, 45, 75, 89, 60, 70, 15, 99, 52, 85, 64, 44, 28, 23, 37, 58, 46, 55, 27, 16, 49, 73, 90, 67, 95, 80, 34, 94, 65, 26, 41, 63, 58, 25, 75, 81 (i) Make a frequency distribution table with class interval of length 10. (ii) Construct the frequency polygon for the data. b) The following are the ages (in years) of patients admitted in a month in a hospital. 25, 18, 61, 23, 35, 60, 46, 57, 70, 15, 12, 72, 39, 51, 30, 20, 32, 71, 42, 59, 29, 19, 38, 24, 48, 40, 32, 54, 49, 28, 68, 45 (i) Construct a frequency distribution table of continuous class with length 10. (ii) Draw a frequency polygon to represent the data. Project work and activity section 5. a) Collect the ages of family members of your 10 classmates. Show the ages in a frequency distribution table of class of length 10. Then, construct the frequency polygon. b) Show the marks obtained by the students of your class in mathematics in recently conducted exam in a frequency distribution table of continuous class of length 5. Then, represent the data in the frequency polygon. 3. Ogive (Cumulative frequency curve) The sum of the frequencies of all the values up to a given value is known as cumulative frequency. It is denoted by c.f. Ogive is a graphic presentation of the cumulative frequency distribution of continuous series. In the case of a continuous series, if the upper limit (or the lower limit) of each class interval is taken as x-coordinate and its corresponding c.f. as y-coordinate and the points are plotted in the graph, we obtain a curve by joining the points with freehand. Such curve is known as ogive or cumulative frequency curve. Statistics (I): Classification and Graphical Representation of Data

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 305 Vedanta Excel in Mathematics - Book 9 Since ‘less than’ c.f. and ‘more than’ c.f. are the two types of cumulative frequency distributions, there are two types of ogive. They are less than ogive and more than ogive. (i) Less than ogive (or less than cumulative frequency curve) When the upper limit of each class interval is taken as x-coordinate and its corresponding frequency as y-coordinate, the ogive so obtained is known as less than ogive (or less than cumulative frequency curve). Obviously, less than ogive is an increasing curve, sloping upwards from left to right and has the shape of an elongated S. (ii) More than ogive (or more than cumulative frequency curve) When the lower limit of each class interval is taken as x-coordinate and its corresponding frequency as y-coordinate, the ogive so obtained is known as more than ogive (or more than cumulative frequency curve). More than ogive is a decreasing curve sloping downwards from left to right and has the shape of an elongated S, upside down. 17.8 Construction of less than ogive and more than ogive Study the following steps to construct a less than ogive. (i) Make a less than cumulative frequency table. (ii) Choose the suitable scale and mark the upper class limits of each class interval along X-axis and cumulative frequencies along Y-axis. (iii) Plot the coordinates of (the upper limit, less than c.f.) on the graph. (iv) Join the point by freehand and obtain a less than ogive. In the case of more than ogive, we should prepare the more than cumulative frequency table. The lower class limits of each class interval are marked on x-axis. Then, the process is similar to the construction of less than ogive. Worked-out Examples Example 1: Look at the given ‘less than’ ogive and answer the following questions. (i) How many students are there in all? (ii) How many students have obtained less than 20 marks? (iii) In which class interval of obtained marks, there is the maximum number of students? Solution: (i) Total number of students is 40 (ii) Number of students who obtained less than 20 marks is 10. (iii) The class interval of marks obtained by maximum number of students is (20 - 30). Cumulative No. of students Marks obtained 10 20 30 10 20 30 40 50 40 50 Statistics (I): Classification and Graphical Representation of Data

Vedanta Excel in Mathematics - Book 9 306 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Example 2: The table given below shows the marks obtained by 50 students in Mathematics. Construct (i) less than ogive and (ii) more than ogive. Marks 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60 60 – 70 No. of students 3 8 10 17 6 2 4 Solution: Less than cumulative frequency table. Marks No. of students (f) Upper limit less than c.f. Coordinates (x, y) 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60 60 - 70 3 8 10 17 6 2 4 10 20 30 40 50 60 70 3 (less than 10) 11 (less than 20) 21 (less than 30) 38 (less than 40) 44 (less than 50) 46 (less than 60) 50 (less than 70) (10, 3) (20, 11) (30, 21) (40, 38) (50, 44) (60, 46) (70, 50) The less than ogive Upper limits 0 10 10 20 30 40 50 60 70 20 30 40 50 60 Less than cumulative frequencies (10, 3) (20, 11) (30, 21) (40, 38) (50, 44) (60, 46) (70, 50) More than cumulative frequency table Marks No. of students (f) Lower limit more than c.f. Coordinates 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60 60 - 70 3 8 10 17 6 2 4 0 10 20 30 40 50 60 47 + 3 = 50 (more than 0) 39 + 8 = 47 (more than 10) 29 + 10 = 39 (more than 20) 12 + 17 = 29 (more than 30) 6 + 6 = 12 (more than 40) 4 + 2 = 6 (more than 50) 4 (more than 60) (0, 50) (10, 47) (20, 39) (30, 29) (40, 12) (50, 6) (60, 4) Statistics (I): Classification and Graphical Representation of Data

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 307 Vedanta Excel in Mathematics - Book 9 The more than ogive If we draw both the less than ogive and more than ogive of a distribution on the same graph paper, they intersect at a point. The foot of the perpendicular drawn from the point of intersection of two ogives to the x-axis gives the value of median of the distribution. For example, Marks No. of students (f) Upper limit Less than c.f. Lower limit More than c.f. 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60 60 – 70 3 8 10 17 6 2 4 10 20 30 40 50 60 70 3 11 21 38 44 46 50 0 10 20 30 40 50 60 50 47 39 29 12 6 4 Lower limits 0 10 10 20 30 40 50 60 70 20 30 40 50 60 More than cumulative frequencies (40, 12) (50, 6) (60, 4) (30, 29) (20, 39) (10, 47) (0, 50) Upper limits 0 10 10 20 30 40 50 60 70 20 30 40 50 60 Less than cumulative frequencies (10, 3) (20, 11) (40, 38) (30, 21) (40, 12) (50, 6)(60, 4) (50, 44) (60, 46) (0, 50) (10, 47) (20, 39) (30, 29) (70, 50) Statistics (I): Classification and Graphical Representation of Data

Vedanta Excel in Mathematics - Book 9 308 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur From the graph, the perpendicular drawn from the point of intersection of two ogives meets x-axis at 32.65 (approx.) units from the origin. So, the required median of the given distribution is 32.65. EXERCISE 17.4 General section 1. a) Study the given less than cumulative frequency curve, answer the following questions. (i) How many people are participated in the survey? (ii) How many people are there, who are less than 30 years? (iii)How many people are there in the age group (30 - 40) years? b) Study the given less than ogive curve, answer the following questions. (i) How many students are there who obtained less than 50 marks? (ii) In which class interval do the marks obtained by the maximum member of students fall? (iii)Find the number of students who obtained marks in the interval (10 - 20)? 2. a) Study the more than ogive curve given alongside, answer the following questions. (i) How many students are there? (ii) How many students are there who obtained more than 20 marks? Cumulative numbers of people Ages (in years) 10 20 30 10 20 30 40 50 60 40 50 60 Numbers of students Marks obtained less than 5 10 Y 15 10 20 30 40 50 60 70 20 25 30 35 Numbers of students 4 8 12 10 20 30 40 50 60 16 20 24 Marks obtained (more than) Statistics (I): Classification and Graphical Representation of Data

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 309 Vedanta Excel in Mathematics - Book 9 b) Study the given more than ogive and answer the following questions. (i) How many workers are there altogether? (ii) How many workers have monthly salary more than Rs 15,000? Creative section 3. a) Draw a ‘less than’ ogive from the data given below. Data 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60 Frequency 5 8 15 10 6 3 b) Construct a ‘more than’ ogive from the data given below. Marks 20–30 30–40 40–50 50–60 60–70 70–80 80–90 90–100 No. of students 6 10 16 15 10 8 4 3 c) The table given below shows the marks obtained by 60 students in mathematics. Construct (i) less than ogive and (ii) more than ogive. Marks 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60 No. of students 4 10 20 15 6 5 4. a) Draw ‘less than’ ogive and ‘more than’ ogive on the same coordinate axes and find the median of the distribution. Data 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60 60 – 70 Frequency 6 12 18 20 15 9 b) The table given below shows the daily wages 40 workers of a factory. Wages (in Rs) 50 – 55 55 – 60 60 – 65 65 – 70 70 – 75 75 – 80 No. of workers 4 6 10 12 5 3 Draw ‘less than’ ogive and ‘more than’ ogive on the same coordinate axes and find the median of the distribution. 5. a) The following are the marks obtained by the students in mathematics in an examination. 49, 23, 64, 35, 44, 60, 79, 20, 41, 50 32, 34, 25, 57, 40, 54, 23, 60, 63, 68 (i) Make a frequency distribution table of class interval 10. (ii) Construct a less than ogive from the above data. Numbers of workers 15 30 45 5 10 15 20 25 30 60 75 90 Monthly salary (in Rs 1,000) 35 Statistics (I): Classification and Graphical Representation of Data

Vedanta Excel in Mathematics - Book 9 310 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur b) The following are the ages (in years) of patients admitted in a month in a hospital. 15, 12, 23, 35, 46, 57, 18, 12, 39, 51 32, 42, 59, 18, 38, 48, 40, 32, 54, 49 (i) Make a frequency distribution table of class size 10. (ii) Construct a more than ogive from the above data. OBJECTIVE QUESTIONS Let’s tick (√) the correct alternative. 1. The correction factor of a data with classes 10 – 19, 20 – 29, 30 – 39, … is (A) 0.5 (B) 1 (C) 1.5 (D) 2 2. The number of times an observation occurs in the given data is called (A) tally marks (B) data (C) frequency (D) cumulative frequency 3. The data that can take values between the class intervals is (A) individual (B) discrete (C) continuous (D) frequency 4. In a frequency distribution, if the lower limit of a class is 20 and width of the class is 4, what is its upper limit? (A) 12 (B) 18 (C) 22 (D) 24 5. The mid-value of a class interval 40-60 is (A) 45 (B) 50 (C) 55 (D) 57 6. Which of the following is the graphical representation of ungrouped data? (A) histogram (B) frequency polygon (C) pie chart (D) ogive 7. Which of the following is the graphical representation of grouped data? (A) histogram (B) pie chart (C) bar graph (D) pictograph 8. Histogram is a graph of a …………… frequency distribution. (A) continuous (B) discontinuous (C) discrete (D) individual 9. A line graph for the continuous frequency distribution is (A) frequency polygon (B) histogram (C) less than ogive (D) more than ogive 10. The frequency polygon can be drawn by (A) using histogram (B) without using histogram (C) only (A) (D) both (A) and (B) https://www.geogebra.org/m/xev3naw3 Vedanta ICT Corner Please! Scan this QR code or browse the link given below: Statistics (I): Classification and Graphical Representation of Data

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 311 Vedanta Excel in Mathematics - Book 9 18.1 Central tendency – Looking back Classwork-Exercise 1. Let’s say and write the answers as quickly as possible. a) Average of 2 and 4 is ................. b) Average of 1, 7 and 13 is ................. c) Average of 4, 6, 8, 10 is ................. d) Average of 3 and x is 5, x = ................. e) Average of m and 8 is 6, m = ................. 2. a) Σfx = 50, n = 5, x = .............. b) Σfx = 85, n = 17, x = .............. c) Σfx = 120, x =15, n = .............. d) Σfx = 200, x =20, n = .............. e) x =12, n = 15, Σfx =.............. f) x =18, n = 20, Σfx =.............. 3. a) Median of 4, 6, 9 is ..................... b) Median of 5, 8, 10, is ..................... c) Median of 5, 1, 4, 3, 7 is ............. d) Median of 10, 8, 15, 17, 12 is ............. 18.2 Measures of central tendency The measure of central tendency gives a single central value that represents the characteristics of entire data. A single central value is the best representative of the given data towards which the values of all other data are approaching. Average of the given data is the measure of central tendency. There are three types of averages which are commonly used as the measure of central tendency. They are: mean, median and mode. 18.3 Arithmetic mean Arithmetic mean is the most common type of average. It is the number obtained by dividing the sum of all the items by the number of items. i.e., mean = sum of all the items the number of items (i) Mean of non–repeated data If x represents all the items and n be the number of items, then mean ( x ) = Σx n Unit 18 Statistics (II): Measures of Central Tendency

Vedanta Excel in Mathematics - Book 9 312 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Worked-out Examples Example 1: Calculate the average of the following marks obtained by 7 students of a class in mathematics. 87, 63, 45, 72, 95, 35, 79 Solution: Here, Σx = 87 + 63 + 45 + 72 + 95 + 35 + 79 = 476 and n = 7 Now, mean ( x ) = Σx n = 476 7 = 68 Example 2: If the average of the following wages received by 5 workers is Rs 400, find the value of p. 300, 380, p, 420, 460. Solution: Here, Σx = 300 + 380 + p + 420 + 460 = 1560 + p and n = 5 Now, average = Σx n or, 400 = 1560 + p 5 or, 1560 + p = 2000 or, p = 440 Hence, the required value of p is Rs 440. Example 3: In a discrete data, if Σfx = 400 + 20a and Σf = 20 + a, find the mean ( x ) . Solution: Here, Σfx = 400 + 20a and Σf = 20 + a, x = ? Now, x = Σfx Σf = 400 + 20a 20 + a = 20(20 + a) 20 + a = 20 Example 4: In a discrete series, the mean is 50, Σfx = 100p + 150 and N = 4p – 15, find the values of p and N. Solution: Here, mean ( x ) = 50, Σfx = 100 p + 150 and N = 4p – 15 Now, x = Σfx N or, 50 = 100p + 150 4p – 15 or, 50 (4p – 15) = 100p + 150 or, 200p – 750 = 100p + 150 or, 100p = 900 ∴ p = 9 Also, N = 4p – 15 = 4 × 9 – 15 = 21 Hence, the required values of p is 9 and N is 21. (ii) Mean of individual repeated data (Mean of a frequency distribution) In the case of repeated data, follow the steps given below to calculate the mean. – Draw a table with 3 columns. – Write down the items (x) in ascending or descending order in the first column and the corresponding frequencies in the second column. – Find the product of each item and its frequency (fx) and write in the third column. Statistics (II): Measures of Central Tendency

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 313 Vedanta Excel in Mathematics - Book 9 – Find the total of f column and fx column. – Divide the sum of fx by the sum of f (total number of items), the quotient is the required mean. Example 5: From the table given below, calculate the mean mark. Marks 14 17 28 35 40 No. of students 5 10 15 8 12 Solution: Calculation of mean marks: Marks (x) No. of students (f) fx 14 17 28 35 40 5 10 15 8 12 70 170 420 280 480 Total N = 50 Σfx =1420 Now, mean mark ( x ) = Σfx N = 1420 50 = 28.4 Hence, the required mean mark is 28.4. Example 6: If the mean of the data given below be 32, find the value of p. x 10 20 30 40 50 60 f 5 8 10 9 p 1 Solution: x f fx 10 20 30 40 50 60 5 8 10 9 p 1 50 160 300 360 50p 60 Total N = 33 + p Σfx = 930 + 50p Now, mean ( x ) = Σfx N or, 32 = 930 + 50p 33 + p or, 1056 + 32p = 930 + 50p or, 18p = 126 or, p = 7 Hence, the required value of p is 7. EXERCISE 18.1 General section 1. a) The marks obtained by 6 students out of 50 full marks are given below. Find the average marks. 48, 26, 36, 14, 42, 38 Statistics (II): Measures of Central Tendency

Vedanta Excel in Mathematics - Book 9 314 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur b) The daily wages (in Rs) of 5 workers are given below. Calculate the average wage. 360, 500, 450, 540, 400 c) The ages of Anupa, Bimlesh, Chandra, Dipika and Elina are 12, 18, 13, 16 and 6 years respectively. Find their average age. d) Find the mean from the data given below. 72, 46, 54, 80, 99, 62, 56 2. a) The average age of 5 students is 9 years. Out of them the ages of 4 students are 5, 7, 8 and 15 years. What is the age of the remaining student? b) If 7 is the mean of 3, 6, a, 9 and 10, find the value of a. c) Find x if the mean of 2, 3, 4, 6, x, and 8 is 5. 3. a) In a discrete series, Σfx = 15p + 800, Σf = p + 10 and mean ( x ) = 20, find the value of p. b) In a discrete data, if Σfx = 12 + 34n, N = 12 + 3n and x = 10, find the n and N. c) In a discrete series, if Σfx = 45 + 60k and N = 9 + 12k, find the mean ( x ). d) In a discrete series, if the sum of frequencies (Σf) = a2 + 4a and Σfx = 25a2 + 100a, find the mean ( x ). Creative section 4. a) Find the mean of the data given in the table. Marks obtained 40 50 60 70 80 90 No. of students 2 7 4 8 6 3 b) The ages of the students of a school are given below. Find the average age. Age (in years) 5 8 10 12 14 16 No. of students 20 16 24 18 25 15 c) The weight of the teachers of a school are given below. Find the average weight. Weight (in kg) 50 55 60 65 70 75 80 No. of teachers 2 5 7 10 4 3 1 d) Compute the arithmetic mean from the following frequency distribution table. Height (in cm) 58 60 62 64 66 68 No. of teachers 12 14 20 13 8 5 5. a) Find the missing frequency p for the following distribution whose mean is 50. x 10 30 50 70 90 f 17 p 32 24 19 b) The mean of the data given below is 17. Determine the value of m. x 5 10 15 20 25 30 f 2 5 10 m 4 2 Statistics (II): Measures of Central Tendency

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 315 Vedanta Excel in Mathematics - Book 9 c) The heights of plants planted by the students of class IX in the afforestation program are given below. The average height of the plant is 36 cm. Height (in cm) 5 15 25 35 45 55 Number of plants 8 5 12 p 24 16 (i) Find the value of p. (ii) Find the total number of plants. d) The speeds of vehicles recorded in a highway during 15 minutes on a day is given in the following table. The average speed of the vehicles was 54 km per hour. Speed (km/hr) 10 30 50 70 90 No. of vehicles 7 k 10 9 13 (i) Find the value of k. (ii) Find the total number of vehicles counted during the time. 6. a) Find the value of p, for the following distribution whose mean is 10. x 3 5 8 p 16 21 f 8 4 9 10 3 6 b) Find the missing value of k if the mean of the following distribution is 21. x 9 14 17 k 28 34 f 12 15 16 25 13 19 c) If the mean of the following data is 14, find the value of m. x m 10 15 20 25 f 7 6 8 4 m d) If the mean of the following data is 55, determine the value of ‘a’. x 10 30 50 70 89 f 7 8 a 15 a 18. 4 Median Let’s take a series: 25, 13, 7, 10, 16, 22, 19 Now, arranging the series in ascending order. 7 10 13 16 19 22 25 Thus, when we arrange the given series in ascending order, the fourth item 16 falls exactly at the middle of the series. Therefore, 16 is called the median of the series. In this way, median is the value that falls exactly at the middle of an array data. i.e., Median = the middle value of a set of arry data (i) Median of ungrouped data To find the median of an ungrouped data, arrange them in ascending or descending order. Let the total number of observation be n. Statistics (II): Measures of Central Tendency

Vedanta Excel in Mathematics - Book 9 316 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur ⇒ If n is odd, the median is the value of the n + 1 2 th observation. ⇒ If n is even, the median is the average of th n 2 and th n 2 + 1 the observations. Worked-out Examples Example 1: The heights of 7 students (in cm) are given below. Find the median height. 115, 119, 125, 110, 140, 135, 128 Solution: Arranging the heights in ascending order, we have, 110, 115, 119, 125, 128, 135, 140 Here, n= 7 Now, the position of median = n + 1 2 th term = 7 + 1 2 th term = 4th term Thus, median lies at the 4th position. ∴ Median = 125. Example 2: The daily wages (in Rs) of 12 workers are given below. Calculate the median wage. 400, 350, 450, 520, 250, 375, 480, 550, 380, 555, 600, 580 Solution: Arranging the wages in ascending order, we have, 250, 350, 375, 380, 400, 450, 480, 520, 550, 555, 580, 600 Here, n = 12 Now, the position of median = n + 1 2 th term = 12 + 1 2 th term = 6.5th term 6.5th term is the average of 6th and 7th terms. ∴ Median = 450 + 480 2 = 930 2 = 465 Hence, median wage is Rs 465. Example 3: If the given data is in ascending order and median is 34, find the value of x. 10, 15, 26, 3x + 1, 39, 44, 53 Solution: Here, The given data in ascending order is 10, 15, 26, (3x + 1), 39, 44, 53 Number of observations (n) = 7 Now, the position of median = n + 1 2 th term = 7 + 1 2 th term = 4th term ∴ Median =3x + 1 But, by question median = 34 or, 3x + 1 = 34 or, 3x = 33 ∴ x = 11 Hence, the required value of x is 11. Statistics (II): Measures of Central Tendency

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 317 Vedanta Excel in Mathematics - Book 9 (ii) Median of Discrete series To compute the median of a discrete series of frequency distribution, we should display the data in ascending or descending order in a cumulative frequency table. Then, the median is obtained by using the formula, Median = value of N + 1 2 th observation. Example 4: Compute the median from the table given below, Weight (in kg) 25 30 35 40 45 50 No. of students 4 7 10 4 3 2 Solution: Cumulative frequency table: Weight in kg (x) No. of students (f) c.f. 25 30 35 40 45 50 4 7 10 4 3 2 4 11 21 25 28 30 Total N = 30 Now, position of median = N + 1 2 th term = 30 + 1 2 th term = 15.5th term In c.f. column, the c.f. just greater than 15.5 is 21 and its corresponding values is 35. ∴ Median = 35 kg. 18.5 Quartiles Quartiles are the values that divide the data arranged in ascending or descending order into four equal parts. Hence, there are three quartiles that divide a distribution into four equal part. x1 x2 x3 x4 x5 x6 x7 1st quartile (Q1 ) 2nd quartile (Q2 ) 3rd quartile (Q3 ) ⇒ The first or lower quartile (Q1 ) is the point below which 25 % of the observations lie and above which 75 % of the observations lie. ⇒ The second quartile (Q2 ) is the point below which 50 % of the observations lie and above which 50 % of the observations lie. Of course, the second quartile is the median. ⇒ The third or upper quartile (Q3 ) is the point below which 75 % of the observations lie and above which 25 % of the observations lie. If N be the number of observations in ascending (or in descending) order of a distribution, then in the case of discrete data The position of the first quartile (Q1 ) = N + 1 4 th term Statistics (II): Measures of Central Tendency

Vedanta Excel in Mathematics - Book 9 318 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur The position of the second quartile (Q2 ) = 2 (N + 1)th 4 term = N + 1 2 th term The position of the third quartile (Q3 ) = 3 (N + 1) th 4 term. After finding the positions of a quartile, we apply the similar process of computing the required quartile as like the process of computing median. Example 5: Find the first quartile (Q1 ) and the third quartile (Q3 ) from the data given below. 18, 24, 20, 19, 23, 34, 28 Solution: Arranging the data in ascending order, 18, 19, 20, 23, 24, 28, 34 Here, N = 7 The position of the first quartile (Q1 ) = N + 1 4 th term = 7 + 1 4 th term = 2nd item Thus Q1 lies at the 2nd position. ∴ The first quartile (Q1 ) = 19. Again, the position of the third quartile (Q3 ) = 3(N + 1) 4 th term = 6th term Thus, Q3 lies at the 6th position. ∴ The third quartile (Q3 ) = 28 Example 6: The observations 2x + 1, 3x – 1, 3x + 5, 5x – 1, 38, 7x – 4 and 50 are in ascending order. If the first quartile is 20, what is the value of x? Also, find the third quartile of the observations. Solution: Here, The given data in ascending order is 2x + 1, 3x – 1, 3x + 5, 5x – 1, 38, 7x – 4, 50 No. of items (n) = 7 Now, the position of the first quartile (Q1 ) = N + 1 4 th term = 7 + 1 4 th = 2nd term The value of 2nd term is 3x – 1. So, the first quartile (Q1 ) = 3x – 1. According to question, Q1 = 20 or, 3x – 1= 20 or, 3x = 21 ∴ x = 7 Again, the position of the third quartile (Q3 ) = 3 n + 1 4 th term = 3 7 + 1 4 th term = 6th term The value of 6th term is 7x – 4. ∴ The third quartile (Q3 ) = 7x – 4 = 7 × 7 – 4 = 45. Statistics (II): Measures of Central Tendency

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 319 Vedanta Excel in Mathematics - Book 9 Example 7: Compute the first and the third quartiles from the table given below. Marks 40 50 60 70 80 90 No. of students 4 6 10 12 5 2 Solution: Cumulative frequency distribution table, Marks (x) No. of students (f) c.f. 40 50 60 70 80 90 4 6 10 12 5 2 4 10 20 32 37 39 Total N = 39 Now, the position of the first quartile (Q1 ) = N + 1 4 th term = 39 + 1 4 th term = 10th term In c.f. column, the corresponding value of the c.f. 10 is 50. ∴ The first quartile (Q1 ) = 50 Again, the position of the third quartile (Q3 ) = 3 (N + 1) th 4 term = 30th term In c.f. column, the c.f. just greater than 30 is 32 and its corresponding value is 70. ∴ The third quartile (Q3 ) = 70 18.6 Mode Let’s take a set of ordered observations. 2, 5, 5, 7, 7, 7, 7, 10, 10, 10, 15 Here, the observation ‘7’ appears maximum number of times. Therefore, 7 is the mode of the given observations. Thus, the mode of a set of data is the value with the highest frequency. A distribution that has two modes is called bimodal. The mode of a set of data is denoted by Mo . (i) Mode of discrete data In the case of discrete data, mode can be found just by inspection, i.e., just by taking an item with highest frequency. Example 8: Find the mode for the following distribution, 12, 10, 18, 12, 10, 12, 15, 12, 10 Solution: Arranging the data in ascending order 10, 10, 10, 12, 12, 12, 12, 15, 18 Here, 12 has the highest frequency. ∴ Mode = 12 Example 9: Find the mode of the given distribution: Marks 27 30 36 40 45 50 No. of students 4 10 16 9 6 5 Statistics (II): Measures of Central Tendency

Vedanta Excel in Mathematics - Book 9 320 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Solution: Here, the marks 36 has the highest frequency, i.e., 16. ∴ Mode = 36. (ii) Mode of grouped and continuous data In the case of grouped and continuous data, the class with highest frequency is observed and it is taken as the model class. Then, by using the following formula, mode can be computed. Mode (M0 ) = L + f 1 – f0 2f1 –f0 – f2 × c Where, L = the lower limit of the model class f 0 = the frequency of the class preceding the model class f 1 = the frequency of the model class f 2 = the frequency of the class succeeding the model class c = the width of the class interval Alternatively, mode can also be computed by the following empirical relation. Mode = 3 Median – 2 Mean Example 10: Find the model class from the data given below: x 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50 f 6 8 12 7 3 Solution: Here, 12 is the highest frequency and its corresponding class is 20 – 30. ∴ Model class is 20 – 30. Example 11: Compute the mode from the data given below: Age in years 3 – 5 5 – 7 7 – 9 9 – 11 11 – 13 13 – 15 No. of pupils 10 35 70 35 12 6 Solution: Here, 70 is the highest frequency and its corresponding class is 7 – 9. ∴ Model class is 7 – 9. Now, L = 7, f1 = 70, f0 = 35, f2 = 35 and c = 2 ∴ Mode (M0 ) = L + f 1 – f0 2f1 –f0 – f2 × c = 7 + 70 – 35 140 – 35 – 35 × 2 = 7 + 35 70 × 2 = 8 So, the required mode is 8 years. 18.7 Range The difference between the largest and the smallest score is called range. ∴ Range = Largest score – Smallest score Statistics (II): Measures of Central Tendency

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 321 Vedanta Excel in Mathematics - Book 9 Example 12: The marks obtained by 10 students of class 9 in Mathematics are given below. Find the range. 78 36 27 95 43 15 69 84 72 51 Here, the highest marks = 95 The lowest marks = 15 ∴ Range = highest score – lowest score = 95 – 15 = 80 EXERCISE 18.2 General section 1. a) Define median of a set of observations. b) Define the first quartile (Q1 ) and the third quartile (Q3 ) of set of observations. c) Write the formula to find the position of median of a descrete series. d) Write the formulae to find Q1 and Q3 of a descrete series. e) Define mode. Write the formula to find mode of a grouped and continuous data. f) Define range. 2. a) Find the medians of the following sets of data. (i) 18, 16, 27, 20, 25 (ii) 21, 28, 14, 42, 35 (iii) 15, 30, 35, 25, 20, 45, 40 (iv) 22, 16, 14, 26, 32, 30 (v) 16, 13, 10, 14, 11, 12, 15 b) The weights of five high school students are given below. Find their median weight. 50 kg, 54 kg, 45 kg, 63 kg, 48 kg c) Find the median age of a group of 7 people whose ages in years are as follows. 47, 61, 13, 34, 56, 22, 8 d) What is the position of the median marks in the table given below? Marks 18 27 32 40 46 No. of students 2 3 10 9 5 3. a) Find the first quartiles (Q1 ) of the following sets of data. (i) 14, 12, 17, 23, 20, 16, 10 (ii) 16, 25, 10, 30, 35, 8, 12 (iii) 40, 20, 30, 10, 16, 12, 8 b) Find the third quartiles (Q3 ) of the following sets of data. (i) 15, 9, 21, 33, 27, 39, 45 (ii) 18, 26, 14, 22, 30, 38, 34 (iii) 30, 20, 50, 80, 40, 60, 70 Statistics (II): Measures of Central Tendency

Vedanta Excel in Mathematics - Book 9 322 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 4. a) If the given data is in ascending order and the median is 40, find the value of x. 20, 30, 3x + 5 2 , 50, 60 b) If the given data is in ascending order and the median is 200, find the value of p. 100, 150, 5p + 10 4 , 250, 300 c) a + 1, 2a – 1, a + 7 and 3a + 4 are in ascending order. If the median is 12, find the value of a. d) 5, k + 3, 2k + 1, 3k – 2, 26 and 31 are in ascending order. If its median is 17, find the value of k. 5. a) 20, x + 6, 35, 40, 47, 51, 58 are in ascending order. If the first quartile of the data is 24, find the value of x. b) 110, 2x + 1 3 , 120, 125, 130, 135, and 140 are in ascending order. If the first quartile of the data is 115, find the value of x. c) 9, 11, 12, 14, 18, 8m + 1, 28 are in ascending order. If the third quartile is 25, find the value of m. d) x + 1, x + 3, x + 5, 2x, 3x – 5, 4x – 10 and 3x – 1 are in ascending order. If the upper quartile of the data is 18, find the value of x. 6. a) Find the modes of the following distributions. (i) 7, 9, 5, 7, 10, 9, 7, 12 (ii) 15 kg, 21 kg, 17 kg, 21 kg, 28 kg, 21 kg, 15 kg, 21 kg b) In a class, there are 15 students of 16 years, 14 students of 17 years, and 16 students of 18 years. What is the modal age of the class? c) In a factory, the daily wages of 30 labourers is Rs 25, 35 labourers is Rs 36, and 20 labourers is Rs 45. What is the modal wage of the factory? d) The age of 5000 students of a school are shown in the table. What is its modal class? Age in years No. of students 4 to 8 8 to 12 12 to 16 1050 2856 1094 e) In a factory, number of labourers and their remuneration are as follows. Find the modal class. Remuneration (in Rs) 1500 – 2000 2000 – 2500 2500 – 3000 No. of labourers 220 215 120 f) Find the mode of the following distribution. (i) a, c, b, b, c, a, c, a, b, c, a, c, (ii) d, f, e, f, e, d, f, d, e, f, d, f g) Find the modal size of the shoes from the data given below. Size of shoe 5 6 7 8 9 10 No. of men 10 15 30 25 18 12 Statistics (II): Measures of Central Tendency

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 323 Vedanta Excel in Mathematics - Book 9 h) Find the mode from the following data: Daily wages (in Rs) 70 90 110 130 150 170 No. of workers 4 12 15 18 20 12 7. a) The heights of 10 students of class 9 are given below. 95 cm, 110 cm, 120 cm 90 cm, 100 cm, 105 cm, 98 cm, 115 cm,1 12 cm, 116 cm (i) What is the height of the tallest student? (ii) What is the height of the shortest student? (iii) Find the range of the height. b) Find the range of the following data. 35, 40, 58, 70, 45, 27, 86, 65 c) The marks obtained by 20 students in Mathematics are given below. Find the range. Marks 40 50 60 70 80 90 No. of students 2 3 5 7 2 1 Creative section 8. Find the mode from the following distribution. a) Age in years 20 – 30 30 – 40 40 – 50 50 – 60 60 – 70 70 – 80 No. of people 150 200 140 90 50 40 b) Daily wages (in Rs) 400 – 500 500 – 600 600 – 700 700 – 800 800 – 900 No. of workers 35 50 30 25 18 9. Find the median marks from the data given below: a) Marks 24 36 50 65 78 No. of students 2 4 12 11 6 b) Marks 10 15 20 25 30 35 No. of students 3 7 15 12 7 3 c) d) 10. a) Calculate the first quartile (Q1 ) from the data given below: Marks obtained 32 36 40 44 48 52 No. of students 2 5 9 6 3 2 b) Compute the third quartile (Q3 ) from the following table: Wages per hour (in Rs) 50 60 70 80 90 100 No. of workers 6 10 15 13 8 3 Wage per hour (in Rs) 45 55 65 75 85 95 No. of workers 20 25 24 18 15 7 Weight (in kg) 36 40 25 44 30 49 No. of students 8 10 3 6 5 4 Statistics (II): Measures of Central Tendency

Vedanta Excel in Mathematics - Book 9 324 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur c) Find the first quartile (Q1 ) and the third quartile (Q3 ) from the following distribution: Ages (in years) 22 27 32 37 42 No. of people 35 42 40 30 24 d) Find the quartiles from the following distribution: Height (in cm) 90 100 110 120 130 140 No. of students 20 28 24 40 35 18 Project work and activity section 11. a) Measure the height of your any 10 friends and compute the mean, median, and mode height. b) Write the number of students in class 4 to class 10 of your school. Collect the data of monthly fees of each students of these classes. Show these information in table. Then compute the mean, median and quartiles of the data. OBJECTIVE QUESTIONS Let’s tick (√) the correct alternative. 1. In a discrete data, if Σfx = 66a + 99b and Σf = 6a + 9b, what is the mean ( x ) ? (A) 6 (B) 9 (C) 11 (D) 15 2. In a data having 9 observations, if the 5th term is 20, what is the value of median? (A) 9 (B) 20 (C) 5 (D) 34 3. A data has 7 observations. If its 2nd term is 4x + 1, and the value of Q1 is 45, what is the value of x? (A) 2 (B) 4 (C) 9 (D) 11 4. Of all mathematics textbooks from different publications in a stationery, a student has to purchase the textbook which is most liked by students and teachers. What measure of central tendency would be most appropriate, if the data is provided to her? (A) Mean (B) Mode C) Median (D) Any of the three 5. The mean of three different natural numbers is 20. If lowest number is 7, what could be highest possible number of remaining two numbers? (A) 45 (B) 40 (C) 48 (D) 52 6. In a discrete data, the mean of 100 observations was 60. If the frequency of the observation 60 is mistakenly written as 25 instead of 15 then what will be exact mean? (A) 60 (B) 56 (C) 54 (D) 49 7. The mean of a data with 11 observations is 42. If the mean of first 6 observations is 39 and that of last 6 observations is 44, what is the sixth observation? (A) 35 (B) 36 (C) 38 (D) 43 8. What is the name of the quartile which divides the data below 25%? (A) Q1 (B) Q2 (C) Q3 (D) both (A) and (C) 9. In what ratio does the upper quartile divide the data arranged in ascending order from bottom? (A) 1:2 (B) 1:3 (C) 3:1 (D) 2:1 10. What percent of observations are there below Q3 of a data? (A) 25% (B) 50% (C) 60% (D) 75% Statistics (II): Measures of Central Tendency

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 325 Vedanta Excel in Mathematics - Book 9 19.1 Probability –Looking back Classwork-Exercise 1. Answer the following questions as quickly as possible. a) (i) How many faces are there in a coin? (ii) Which faces can be surely placed when the coin is tossed? (iii) Can both the faces be shown at once? (iv) What is the possibility of getting head? (v) What is the possibility of getting tail? b) (i) How many faces are there in a die? (ii) Which faces can be surely placed when a die is rolled? (iii) What is the possibility of getting 1? 2. Choose the words from the table given below that best describes an event that has a possibility of: a) 0% b) 50% c) 100% d) 120% Certain impossible probable improbable 19.2 Probability – Introduction In our real-life situations, we often talk about the probability of happening of any events. For example, probability of raining, probability of increasing the price of petroleum, probability of winning the games or election and so on are a few cases. Thus, probability refers something likely to occur; however, it is not certain to occur. Facts to remember Probability is the numerical measurement of the degree of certainty of the occurrence of events. For example, when a coin is tossed, it is 50/50 chance that the head or tail occurs. So, the probability of occurrence of head is 1 2 and that of tail is also 1 2. Unit 19 Probability

Vedanta Excel in Mathematics - Book 9 326 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur The first foundation of modern mathematical theory of probability was laid in the seventeenth century by French mathematicians Pierre de Fermat (1601-1665) and Blaise Pascal (1623-1662) while contemplating a gambling problem posed by Chevalier de Mere, a gambler and a member of aristocracy in 1654. James Bernoulli (1654 – 1705), A. de Moivre (1667 – 1754), and Pierre Simon Laplace are among those who made significant contributions to this field. Laplace’s Theorie Analytique des Probabilités, 1812, is considered to be the greatest contribution by a single person to the theory of probability. Following are a few important terms which are frequently used in probability. It is essentially important to have the proper concept of these terms. Experiments and outcomes: An action by which an observation is made is called the experiment. Any experiment whose outcome cannot be predicated or determined in advance is called a random experiment. A few examples of random experiments a) Tossing a coin b) Rolling a die c) Drawing a card from a well shuffled pack of 52 playing cards The results of a random experiment are called outcomes. For example, while tossing a coin, the occurrence of head or tail is the outcome. Sample space: Each performance in a random experiment is called a trial and an outcome of a trial is called a sample point. The set of all possible outcomes (i.e., sample points) of a random experiment is known as sample space. Usually, a sample space is denoted by S. For example, (i) The possible outcomes of a random experiment of throwing a die are 1, 2, 3, 4, 5 or 6. ∴ The sample space, S = {1, 2, 3, 4, 5, 6} (ii) The possible outcomes of a random experiment of tossing a coin are head (H) or tail (T). ∴ The sample space, S = {H, T} (iii) When two coins are tossed simultaneously, the sample space, S = {HH, HT, TH, TT} Event: A sample space S of a random experiment is a universal set. Every non-empty subset of the sample space S is called an event. For example, while tossing a coin, the sample space, S = {H, T}. Here, the subsets {H}, {T} and {H, T} are the events of S. The empty set I is also an event of S, but it is the ‘impossible event’. S is called the ‘sure event’. An event containing only one element of S is called a simple or elementary event. Probability

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 327 Vedanta Excel in Mathematics - Book 9 For example, If S = {H, T}, then {H} and {T} are the elementary events. If S = {1, 2, 3, 4, 5, 6}, then {1}, {2}, {3}, {4}, {5}, and {6} are the elementary events. Exhaustive cases: The total number of all possible outcomes of a random experiment is known as exhaustive cases. For example, while tossing a coin, S = {H, T}. So, exhaustive cases = 2 While tossing two coins simultaneously, S = {HH, HT, TH, TT} So, exhaustive cases = 4. Similarly, the exhaustive cases of tossing three coins simultaneously is 8. Thus, if n denotes the number of tossing coins simultaneously, then the exhaustive cases can be obtained as 2n. Equally likely events Two or more events are said to be equally likely if the chance of occurring any one event is equal to the chance of occurring other cases. For example, while throwing a die, the chance of coming up the numbers 1 to 6 is equal. similarly, while tossing a coin, head (H) or tail (T) are equally likely events. Favourable and unfavourable cases The outcomes in an random experiment which are desirable (or expected) to us are called favourable cases and all other cases are unfavourable cases. For example: While tossing a coin, S = {H, T} Here, the favourable number of case of head is 1 and tail is also 1. While tossing two coins, S = {HH, HT, TH, TT} The favourable number of case of both of them head is 1 and tail is also 1. The favourable number of cases of at least one head is 3 and at least tail is also 3. Some useful facts about playing cards, coin and die a) Playing cards There are 52 cards in a packet of playing cards. Probability

Vedanta Excel in Mathematics - Book 9 328 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur There are 26 red and 26 black coloured cards in the packet. 13 Hearts (♥) and 13 diamonds (♦) are the red coloured cards. 13 Spades (♠) and 13 clubs (♣) are the black coloured cards. There are 4 suits (Heart, Club, Diamond, Spade), each with 13 cards (Ace, 2, 3, 4, 5, 6, 7, 8, 9, 10, Jack, Queen, King). There are 12 face cards in the packet. Among them 4 are Jacks (heart, diamond, spade and club), 4 are Queens and 4 are Kings. b) Coin There are two faces of a coin, head (H) and tail (T). c) Die There are six faces in a die which are numbered from 1 to 6. In a fair die, the sum of the numbers turning on the opposite sides (1 and 6, 2 and 5, 3 and 4) are always equal to 7. 19.3 Probability of an event Let n(S) be the total number of all possible outcomes (Exhaustive number of cases) of an experiment and n(E) be the favourable number of cases of the sample space S, then the probability of happening the event (E) is defined as, P(E) = Favourable number of cases Exhaustive number of cases = n(E) n(S) Also, the probability of non–happening of the event is given by P'(E) = 1 – P(E) 19.4 Probability scale Probability is measured on a scale from 0 to 1. A 0 (zero) probability means there is no chance of an event happening. A probability of 1 means that it is certain the event will happen. In the adjoining spinner, the probability of the pointer to stop on green is 1 4. The probability of the pointer to stop on blue is 1 4. The probability of the pointer to stop on yellow is 1 4. The probability of the pointer to stop on green or yellow is 1 2. The probability of the pointer to stop on black is 0, which means not possible. The probability of the pointer to stop on green, blue, red or yellow is 1. Similarly, there are 6 blue, 9 red and 5 green marbles of same shape and size in a bag. When a marble is randomly taken out, then the probability of getting a blue marble, P (B) is 6 20 = 0.3 Probability

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 329 Vedanta Excel in Mathematics - Book 9 the probability of getting a red marble, P (R) is 9 20 = 0.45 and the probability of getting a green marble, P (G) is 5 20 = 0.25 Facts to remember 1. The probability of an event always lies between 0 to 1, but it can never be less than 0 and more than 1. i.e., for an event E, 0 ≤ P (E) ≤ 1. 2. The probability of a sure event is 1. 3. The probability of impossible event is 0. Worked-out Examples Example 1: An unbiased normal die is rolled once. Answer the following questions. (a) Write down the sample space for this experiment. (b) Find the probability of getting an even number. (c) Find the probability of getting the face with the number 5. (d) Are the events of getting numbers from 1 to 6 equally likely? Give reason. Solution: (a) Sample space (S) = {1, 2, 3, 4, 5, 6} (b) n (S) = 6, Even numbers (E) = {2, 4, 6} ∴The favourable number of cases, n (E) = 3 We have, P (E) = n(E) n(S) = 3 6 = 1 2 (c) n (S) = 6, E = {5} The favourable number of cases, n (E) = 1 ∴ P (E) = n(E) n(S) = 1 6 (d) While rolling a die, the chance of turning up the numbers from 1 to 6 are the same. Thus, the events are equally likely. Example 2: The flash cards numbered from 1 to 20 are shuffled well and a card is drawn at random. (a) Define sample space. (b) If n (E) denotes the favourable events and n (S) denotes total number of events, write the formula to find the probability P (E). (c) Find the probability of getting a card having number which is exactly divisible by 5. (d) Find the probability of not getting a composite numbered card. Solution: (a) The set of all possible outcomes in a random experiment is called sample space. (b) P (E) = n(E) n(S) Probability

Vedanta Excel in Mathematics - Book 9 330 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur (c) S = {1, 2, 3, …, 20} ∴n (S) = 20 Numbers exactly divisible by 5, E = {5, 10, 15, 20} ∴n (E) =4 ∴ P (E) = n(E) n(S) = 4 20 = 1 5 (d) Composite numbers (E) = {4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20} ∴n (E) =11 P (E) = n(E) n(S) = 11 20 ∴The probability of not getting a composite numbered card, P(E) = 1 – P (E) = 1 – 11 20 = 9 20 Example 3: From a well shuffled pack of 52 playing cards, a card is randomly drawn. Answer the following questions. (a) What is the probability of a sure event? (b) What is the probability of getting a king? (c) What is the probability of getting a red-faced card? (d) What is the probability of not getting a king of spade? Solution: (a) The probability of sure event is 1. (b) n (S) =52 The favourable number of cases, n (E) = 4 ∴ P (E) = n(E) n(S) = 4 52 = 1 13 (c) The favourable number of cases, n (E) = 6 P (E) = n(E) n(S) = 6 52 = 3 26 (d) The favourable number of cases, n (E) = 1 ∴ P (E) = n(E) n(S) = 1 52 ∴The probability of not getting a king of spade, P(E) = 1 – P (E) = 1 – 1 52 = 51 52 Example 4: Three unbiased coins are tossed simultaneously. Write down the sample space and find the probability of getting (a) all heads (b) at most two heads (c) one tail (d) at most one tail. Solution: To find the sample space S, Here, sample space of tossing the first coin = {H, T} Sample space of tossing the first and the second coins = H T H HH HT T TH TT = (HH, HT, TH, TT} E = {king of spade, king of heart, king of diamond, king of club} ∴n (E) = 4 E = {faced-cards of heart and diamonds} ∴n (E) = 6 E = {king of spade} ∴n (E) = 1 Probability

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 331 Vedanta Excel in Mathematics - Book 9 Sample space of tossing all the three coins = HH HT TH TT H HHH HHT HTH HTT T THH THT TTH TTT = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT} ∴ n(S) = 8 (a) The number of favourable cases of all heads, n(E) = 1 ∴ P(E) = n(E) n(S) = 1 8 (b) The number of favourable cases of at most two heads, n(E) = 7 ∴ P(E) = n(E) n(S) = 7 8 (c) The number of favourable cases of one tail, n(E) = 3 ∴P (E) = n(E) n(S) = 3 8 (d) The number of favourable cases of at most one tail, n(E) = 4 ∴P (E) = n(E) n(S) = 4 8 = 1 2 19.5 Empirical probability (or Experimental probability) When a coin is tossed, theoretically the probability of getting head is 1 2 . The probability obtained in this way is called theoretical probability. On the other hand, if a coin is tossed 20 times, theoretically the head should occur 1 2 × 20 times, i.e., 10 times. However, in real experiment it may not happen, i.e., the occurrence of head may be 6, 9, 11, 15, or any other number of times. The probability of any event which is estimated on the basis of the number of actual experiments is known as empirical (or experimental) probability. If n(S) be the total number of times, an experiment is repeated and n(E) be number of observed outcomes, the empirical probability is defined as, P(E) = Number of observed outcomes Total number of times an experiment is repeated = n(E) n(S) Example 5: When a coin is tossed 200 times, head occurs 80 times. Find the probability of (a) head and (b) tail. Solution: Here, total number of trials n(S) = 200 n(E) = n(H) = 80 (a) Now, P(E) = P(H) = n(E) n(S) = 80 200 = 0.4 (b) Again, the probability of tail, P(T) = 1 – 0.4 = 0.6 Example 6: The number of match–sticks in each of 20 boxes were counted. The results are shown in the table given below: No. of match–sticks 39 40 41 42 Frequency 5 8 4 3 If one of these boxes is selected at random, what is the probability that (a) it contains 40 sticks? (b) it contains more than 40 sticks? Alternative process P(E) = P(T) = 200 – 80 = 120 ∴P(T) = P(E) P(S) = 120 200 = 0.6 Probability

Vedanta Excel in Mathematics - Book 9 332 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Solution: (a) Here, n(S) = 20 n(E) = 8 ∴ P(E) = n(E) n(S) = 8 20 = 2 5 Example 7: If the probability of germinating a seed of a pea plant is 0.85, how many seeds out of 1000 will germinate? Solution: Here, n(S) = 1000 P(E) = 0.85 n(E) = ? Hence, the required number of geminated seeds is 850. Example 8: Bulbs are packed in cartons, each containing 60 bulbs. 500 cartons were examined for defective bulbs and the results are given in following table. No. of defective bulbs 0 1 2 3 4 5 More than 5 No. of cartons 280 120 45 31 16 6 2 If one carton is selected at random, what is the probability that: (a) it has no defective bulb? (b) it has defective bulbs less than 3? (c) it has defective bulbs more than 4? Solution: Here, total number of cartons, n(S) = 500 (a) The number of cartons that have no defective bulbs = n(E1 ) = 280 ∴ P(E1 ) = n(E1 ) n(S) = 280 500 = 0.56 (b) The number of cartons that have less than 3 defective bulbs n(E2 ) = 280 + 120 + 45 = 445 ∴ P(E2 ) = n(E2 ) n(S) = 445 500 = 0.89 (c) The number of cartons that have more than 4 defective bulbs = n(E3 ) = 6 + 2 = 8 ∴ P(E3 ) = n(E3 ) n(S) = 8 500 = 0.016 EXERCISE 19.1 General section 1. Answer the following questions. a) Define probability? b) If the number of favourable outcomes and possible outcomes are n(E) and n(S) respectively, find the probability of the event ‘E’. c) What is the probability of an event if it is certain? d) If the event is impossible, what is its probability? (b) Again, n(E) = 4 + 3 = 7 ∴ P(E) = n(E) n(S) = 7 20 . Now, P(E) = n(E) n(S) or, 0.85 = n(E) 1000 n(E) = 850 Probability

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 333 Vedanta Excel in Mathematics - Book 9 e) In what case, the probability of an event is maximum? f) What is the probability scale of any event ‘E’ in a random experiment? 2. a) What do you mean by sample space in probability? b) Write the sample space of the following random experiments. (i) Tossing of two coins simultaneously (ii) Tossing of three coins simultaneously (iii) Rolling a die (iv) Rolling two dice simultaneously c) What do you mean by equally likely event? Write with examples. g) Define empirical probability. In what way is it different from theoretical probability? 3. a) What is the probability of getting 4 when a die is rolled once? b) A bag contains 5 identical balls of green, yellow, blue, red and purple. If a ball is drawn randomly from the bag, what is the probability of getting blue ball? c) A marble is drawn from a box containing 6 blue and 9 yellow marbles. What is the probability of getting blue ball? d) There are 15 black, 5 green, 10 red and 10 yellow balls in a bag. If a ball is drawn randomly, find the probability that the ball is green. Creative section 4. a) An unbiased coin is tossed once. Answer the following questions. (i) Write down the sample space for this experiment. (ii) Find the probability of getting head. (iii) Find the probability of getting tail. (iv) Are the events of getting head and tail equally likely? Give reason. b) An unbiased die is rolled once. Answer the following questions. (i) Write down the sample space for this experiment. (ii) Find the probability of getting an odd number. (iii) Find the probability of getting the face with the number 4. (iv) Are the events of getting numbers from 1 to 6 equally likely? Give reason. c) A normal die is thrown once. Answer the following questions. (i) Define sample space. (ii) Find the probability of getting a prime numbered face. (iii) Find the probability of getting a composite numbered face. (iv) Write down the case under which the probability of an event is maximum. 5. a) The flash cards numbered from 1 to 10 are shuffled well and a card is drawn at random. (i) If n (E) denotes the favourable events and n (S) denotes total number of events, write the formula to find the probability P (E). (ii) What is the probability of getting a card with prime number? (iii) What is the probability of getting card with number 9? (iv) Find the probability of not getting card with number 7. Probability

Vedanta Excel in Mathematics - Book 9 334 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur b) From a pack of flash cards numbered from 1 to 20, a card is drawn at random. (i) What is the probability of getting the card with odd number? (ii) What is the probability of getting card with a number which is exactly divisible by 3? (iii) What is the probability of getting card with a number which is exactly divisible by 5? (iv) What is the probability of getting card with a number which is not exactly divisible by 8? c) From the number cards numbered from 10 to 25, a card is drawn at random. (i) What is the probability of getting the card with square number? (ii) What is the probability of getting card with cube number? (iii) What is the probability of not getting card with odd number? (iv) Write a necessary condition for the probability of an event to be 0. 6. a) From a well shuffled pack of 52 playing cards, a card is randomly drawn. Answer the following questions. (i) What is the probability of an impossible event? (ii) What is the probability of getting a queen? (iii) What is the probability of getting a card of heart? (iv) What is the probability of not getting an ace? b) From a well shuffled pack of 52 playing cards, a card is drawn randomly. Answer the following questions. (i) Find the probability of getting a black king. (ii) Find the probability of getting a red queen. (iii) Find the probability of getting a faced card. (iv) Find the probability of not getting a card of club. 7. a) In a class of 40 students, 25 are boys and the rest are girls. Among them 3 boys and 5 girls wear spectacles. If a teacher wants to choose one of the students randomly for class monitor, find the probability that the students so chosen is (i) a boy (ii) a girl (iii) not a girl (iv) wearing the spectacles b) A box contains a dozen of marbles of the same shape and size. Among them, 3 marbles are red, 5 marbles are black and the rest are white marbles. If a marble is drawn at random, find the probability that the marble so drawn is (i) black marble (ii) white marble (iii) not red marble c) Find the probability of touching the following letters of the word ‘PROBABILITY’ by closing the eyes. (i) touching ‘P’ (ii) touching ‘B’ (iii) touching ‘I’ (iv) not touching ‘R’ (v) not touching ‘A’ (vi) not touching ‘B’ d) In the adjoining spinner, find the probability of the pointer to stop on (i) the number 7 (ii) the sectors of odd numbers only (iii) the sectors of numbers exactly divisible by 3 Probability

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 335 Vedanta Excel in Mathematics - Book 9 (iv) the sectors of numbers greater than 5 (v) the sectors of numbers except 7 and 8 (vi) the sectors of numbers whose sum is 10 8. a) Out of 1000 newly born babies, 562 are girls. What will be the empirical probability that the newly born baby is a girl? b) In a survey of 100000 people who are chain–smokers, 100 people were found suffering from lungs–cancer. What is the probability of the people suffering from cancer? c) If the probability of germinating a seed of a flower is 0.87, how many seeds out of 1000 will germinate? d) A dice is thrown 300 times and the record of outcomes is given in the table. Outcomes 1 2 3 4 5 6 Frequency 45 36 50 55 60 54 Calculate the empirical probability of getting the numbers (i) less than 3 and (ii) greater than 4. e) Glass tumblers are packed in cartons, each containing 12 tumblers. 200 cartons were examined for broken glasses and the results are given in the table below: No. of broken glasses 0 1 2 3 4 More than 4 Frequency 164 20 9 4 2 1 If one cartoon is selected at random, what is the probability that: (i) it has no broken glass? (ii) it has broken glasses less than 3? (iii) it has broken glasses more than 1? (iv) it has broken glasses more than 1 and less than 4. Project work and activity section 9. a) Make a few number of peer groups of students in your class. Each group will take a coin. One student will throw the coin and another student will record the outcome as head (H) or tail (T) in each peer group. (i) Throw the coin 10 times and find the empirical probability of getting H or T. (ii) Throw the coin 20 times and find the empirical probability of getting H or T. (iii) Throw the coin 50 times and find the empirical probability of getting H or T. b) Now, compare the empirical probabilities obtained by each group and discuss in the class. Probability

Vedanta Excel in Mathematics - Book 9 336 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur OBJECTIVE QUESTIONS Let’s tick (√) the correct alternative. 1. Which of the following is not an example of random experiment? (A) Tossing a coin (B) Rolling a die (C) Drawing a card from well shuffled deck of 52 cards (D) Throwing a stone from the roof of a house 2. If the number of favourable cases and exhaustive cases of an experiment are n (E) and n (S), the probability of event E is (A) n(E) n(S) (B) n(S) n(E) (C) n(E) n(S) + n(E) (D) n(S) – n(E) n(S) 3. The probability of sure event is (A) 0 (B) 0.5 (C) 1 (D) None of these 4. The probability scale of any event E is (A) 0 < P (E) < 1 (B) 1 < P (E) < 0 (C) 0 ≤ P (E) ≤ 1 (D) 1 ≤ P (E) ≤ 0 5. For any event E, P'(E) is equal to (A) - P (E) (B) 1 – P (E) (C) P (E) – 1 (D) P (E) 6. In a class of 40 students, 3 boys and 5 girls wear spectacles. If the principal called one of the students at random in the office, the probability of this student wearing spectacles is (A) 3 40 (B) 1 8 (C) 4 5 (D) 1 5 7. A bag contains 20 identical balls out of which 5 are red, 7 are white and the rest are green. If a ball is drawn at random, the probability of getting a green ball is (A) 1 4 (B) 7 20 (C) 3 5 (D) 2 5 8. A coin is flipped to decide which team starts the game. What is the probability that your team will start the game? A) 0 (B) 1 (C) 0.5 (D) 1 9. A card is drawn from a well shuffled deck of 52 cards. What is the probability that the card drawn will be a king? (A) 1 52 (B) 1 13 (C) 3 13 (D) 12 13 10. When a coin is tossed 50 times, head occurs 20 times. What is the probability of getting tail? (A) 0.4 (B) 0.6 (C) 0.8 (D) 0.9 https://www.geogebra.org/m/nutvyzce Vedanta ICT Corner Please! Scan this QR code or browse the link given below:

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 337 Vedanta Excel in Mathematics - Book 9 Assessment -VI 1. The marks obtained by the students of a school in mathematics test are given below. 25, 36, 44, 15, 52, 33, 68, 37, 50, 29, 34, 46, 16, 43, 24, 56, 70, 32, 22, 40, 58, 48, 27, 43, 36 Answer the following questions. (a) Make a frequency distribution table of class interval of 10. (b) Draw the histogram of the data. (c) Construct the frequency polygon by using the histogram. (d) How is the area under the histogram equal to the area under frequency polygon of the same data? 2. The following table shows the marks obtained by the students of grade IX in mathematics test. Marks obtained 15 30 45 60 75 90 No. of students 2 3 7 10 8 4 Answer the following questions. (a) What type of data are shown in the table? (b) Construct a cumulative frequency distribution table? (c) Calculate the median mark. (d) Calculate the mean mark of the students who secured more than median mark. 3. The table given below represents the height of the students of class-IX. Height (in cm) 125 130 138 143 148 155 No. of students 4 7 13 9 5 2 Answer the following questions. (a) What type of data are in the table? (b) Construct a cumulative frequency distribution table. (c) Calculate the lower quartile. (d) Find the percentage of the students who is shorter than the Q1 height. 4. A woman is pregnant. It is not sure that on which day she gives birth for a child. (a) What is the probability of a sure event? (b) Find the probability of giving birth of the child on Wednesday. (c) Find the probability of giving birth of the child on a day which starts with ‘S’. (d) Find the probability of not giving birth on Friday. 5. The flash cards numbered from 1 to 10 are shuffled well and a card is drawn at random. (a) If n (E) denotes the favourable events and n (S) denotes total number of events, write the formula to find the probability P (E). (b) What is the probability of getting a card with prime number? (c) What is the probability of getting a card with number 2? (d) Find the probability of not getting a card with number 6. From a well shuffled pack of 52 playing cards, a card is randomly.

Vedanta Excel in Mathematics - Book 9 338 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 20.1 Trigonometry - Introduction Trigonometry is a branch of mathematics that studies relationships between sides and angles of triangles. The word Trigonometry is derived from Greek, where trigonon means ‘triangle’ and metron means ‘measure’. Throughout history, trigonometry has it’s broad applications in many area such as surveying, celestial mechanics, navigation, and so on. 20.2 Trigonometric ratios In the given right–angled triangle, ∠B is a right angle. The longest side, which is the opposite side of right angle is the hypotenuse (h). When we take ∠C as the reference angle, the side opposite to it is perpendicular and the side adjacent to it is the base. So, AB is perpendicular and BC is base. When we take ∠A as the reference angle, BC is the perpendicular and AB is the base. The ratios of any two sides of a right–angled triangle with respect to one of the two acute angles are known as Trigonometric Ratios. The acute angle about which the ratio is formed is called the angle of reference. Sine, cosine and tangent are the name of the three main trigonometric ratios. (i) The sine of an angle A, written as sinA, is the ratio of opposite side (perpendicular) and hypotenuse. i.e., sinA = opposite hypotenuse = perpendicular hypotenuse = p h = BC AC (ii) The cosine of an angle A, written as cosA, is the ratio of adjacent side (base) and hypotenuse. i.e., cosA = adjacent hypotenuse = base hypotenuse = b h = AB AC (iii) The tangent of an angle A, written as tanA, is the ratio of opposite side (perpendicular) and adjacent side (base). i.e., tanA = opposite adjacent = perpendicular base = p b = BC AB In the above cases, the acute angle A is taken as the angle of reference. Now, let’s find the trigonometric ratios of taking another angle of reference C. (i) sinC = perpendicular hypotenuse = p h = AB AC A B C hypotenuse perpendicular (opposite) base (adjacent) A B C hypotenuse perpendicular (opposite) base (adjacent) C B A h b p C B A h b p Unit 20 Trigonometry

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 339 Vedanta Excel in Mathematics - Book 9 (ii) cosC = base hypotenuse = b h = BC AC (iii) tanC = perpendicular base = p b = AB BC Thus, sine, cosine, and tangent are known as the fundamental trigonometric ratios. The reference angles are denoted by different Greek letters such as a (alpha), b (beta), y (gamma), q (theta), f (phi), etc. However, the reference angles are also denoted by English capital letters like A, B, C, ... etc. 20.3 Relation between trigonometric ratios We know that sinA = p h and cosA = b h ∴ sin2 A = p2 h2 and cos2 A = b2 h2 Now, sin2 A + cos2 A = p2 h2 + b2 h2 = p2 + b2 h2 = h2 h2 = 1 Also, we have, tanA = p b . Dividing the numerator and the denominator by h, we get tanA = p/h b/h = sinA cosA Thus, tan A = sinA cosA . Example 1: In the right angled triangle ABC given alongside, ∠ABC = 90o and ∠ACB =θ. Answer the following questions. (a) Express tanθ in terms of the sides of the triangle. (b) Which trigonometric ratio is represented by BC AC? (c) If AC = 20 cm and sinθ = 4 5 , find the length of AB. (d) What is the value of AB AC 2 + BC AC 2 ? Solution: Here, for reference angle θ, hypotenuse (h) = AC, perpendicular (p) = AB and base (b) = BC. (a) tanθ = p b = AB BC (b) BC AC = b h = cosθ Thus, cosθ is represented by BC AC Thus, sin2 A + cos2 A = 1 (i) sin2 A = 1 – cos2 A and sinA = 1 – cos2 A (ii) cos2 A = 1 – sin2 A and cosA = 1 – sin2 A https://www.geogebra.org/m/uffpj4tr Vedanta ICT Corner Please! Scan this QR code or browse the link given below: Worked-out examples Trigonometry A B C q

Vedanta Excel in Mathematics - Book 9 340 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur (c) AC = 20 cm and sinθ = 4 5 We have, sinθ = 4 5 or, AB(p) AC(h) = 4 5 or, AB 20 cm = 4 5 ∴ AB = 16 cm d) AB AC 2 + BC AC 2 = p h 2 + b h 2 = sin2 θ + cos2 θ = 1 Example 2: In the given triangle ABC; AD ⊥ BC, AC = 20 cm, BC = 21 cm, DC = 16 cm and ∠BAD = θ . (a) Find the length of AD. (b) Find the value of tanθ (c) Find the length of AB. (d) Show that AB AD 2 – BD AD 2 = 1 Solution: Here, in ∆ABC; AD ⊥ BC, AC = 20 cm, BC = 21 cm, DC = 16 cm and ∠BAD = θ. (a) In rt ∠ed ∆ ADC, using Pythagoras theorem, AD = AC2 – DC2 = 202 – 162 = 400 – 256 = 144 = 12 cm (b) In rt ∠ed ∆ ABD, for reference angle θ, hypotenuse (h) = AB, perpendicular (p) = BD = 21 cm – 16 cm = 5 cm, and base (b) = AD = 12 cm ∴ tanθ = p b = BD AD = 5 12 (c) In rt ∠ed ∆ ABD, using Pythagoras theorem, AB = BD2 + AD2 = 52 + 122 = 25 + 144 = 169 = 13 cm (d) L.H.S.= AB AD 2 – BD AD 2 = 13 12 2 _ 5 12 2 = 169 144 – 25 144 = 144 144 = 1 = R.H.S. Proved Example 3: In the adjoining figure, ∠ABC = 90°, AB = 3 cm, BC = 4 cm, BD⊥ AC and ∠DBC = θ. (a) Show that ∠BAC = ∠DBC = θ (b) Find the length of AC. (c) Find the value of cosθ. (d) Prove that tanθ.cosθ = sinθ Alternative process sinθ = 4 5 or, p h = 4 5 ∴p = 4x and h = 5x Now, AC (h) = 20 cm or, 5x = 20 cm ∴ x = 4 cm Also, AB (p) = 4x = 4 × 4 cm = 16 cm q A B C D 16 cm 20 cm 21 cm A B C D q 4 cm 3 cm Trigonometry

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 341 Vedanta Excel in Mathematics - Book 9 Trigonometry Solution: Here, in ∆ABC; BD ⊥ AC, AB = 3 cm, BC = 4 cm and ∠DBC = θ. Suppose, ∠ACB = α (a) In rt ∠ed ∆ ABC, ∠BAC = 90o – ∠ACB = 90o – α … (i) In rt ∠ed ∆ BCD, ∠DBC = 90o – ∠BCD = 90o – α … (ii) From (i) and (ii), we get, ∠BAC = ∠DBC = θ (b) In rt ∠ed ∆ ABC, using Pythagoras theorem, AC = AB2 + BC2 = 32 + 42 = 9 + 16 = 25 = 5 cm (c) In rt ∠ed ∆ ABC, for reference angle (∠BAC) = θ, hypotenuse (h) = AC = 5 cm, perpendicular (p) = BC = 4 cm, and base (b) = AB = 3 cm ∴ cosθ = b h = 3 5 (d) L.H.S. = tanθ.cosθ = sinq cosq × cosθ = sinθ = R.H.S. Proved EXERCISE 20.1 General section 1. Write the trigonometric ratios (sine, cosine, and tangent) with respect to the given angle of reference in terms of the ratios of sides of the following right angled triangles. 2. Write the values of the trigonometric ratios (sine, cosine, and tangent) to the given angle of reference. 3. a) From the given figure, find the trigonometric ratios of sinα and tanθ. b) In the given triangle, find the trigonometric ratios of sinβ and cosθ. 9 5 13

Vedanta Excel in Mathematics - Book 9 342 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur c) Find the trigonometric ratios of tanα and cosθ from the figure. d) Find the trigonometric ratios of sina and cosq from the figure alongside. Creative section 4. a) In the given right angled triangle ABC, ∠ABC = 90o and ∠ACB =θ. Answer the following questions. (i) Express sinθ in terms of the sides of the triangle. (ii) Which trigonometric ratio is represented by AB BC ? (iii) What is the value of AB AC 2 + BC AC 2 ? (iv) If BC = 12 cm and cosθ = 4 5 , find the length of AC. b) In the right angled triangle PQR given alongside, ∠PQR = 90o and ∠QPR =α. Answer the following questions. (i) Express cosα as the ratio of sides of the triangle. (ii) Which trigonometric ratio is represented by QR PR ? (iii) What is the value of PQ PR 2 + QR PR 2 ? (iv) If PQ = 16 cm and tanα = 3 4 , find the length of QR. 5. a) In the adjoining figure; BA ⊥ AC, AD ⊥ DC, BC = 13 cm, AD = 3 cm, DC = 4 cm, ∠ABC = θ and ∠ACD = α. (i) Find the length of AC. (ii) Find the value of sinα. (iii) Find the length of AB. (iv) Show that AB BC 2 + AC BC 2 = 1 b) In the figure given alongside; AB ⊥ BC, AC ⊥ CD, BC = 3 cm, CD = 12 cm, AD = 13 cm, ∠CAD = θ and ∠ACB = α. (i) Find the length of AC. (ii) Find the value of tanθ. (iii) Find the length of AB. (iv) Show that AB AC 2 + BC AC 2 = 1 P Q T R q a A B C q P R Q a 3 cm 13 cm 4 cm 13 cm 3 cm 12 cm Trigonometry

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 343 Vedanta Excel in Mathematics - Book 9 Trigonometry c) In the given triangle ABC; AD ⊥ BC, AB = 25 cm, BD = 7 cm, BC = 17 cm and ∠ACB = θ. (i) Find the length of AD. (ii) Find the value of tanθ. (iii) Find the length of AC. (iv) Show that AC AD 2 – CD AD 2 = 1 6. a) In the adjoining figure, ABCD is a rhombus in which AC = 6 cm, BD = 8 cm and ∠ADB = θ. (i) What type of triangle is DAOD. (ii) What is the length of AD? (iii) What is the value of sinθ? (iv) Show that AD OD 2 – AO OD 2 = 1 b) In the figure alongside, O is the center of circle, P is the mid-point of the chord XY. If ∠POY = α, XY = 18 cm and OY = 15 cm, answer the following questions. (i) Name the special type of triangle POY. (ii) What is the length of OP? (iii) What is the value of cosα? (iv) Show that OP OY 2 + PY OY 2 = 1 7. a) In the adjoining figure, AB = 6 cm, BC = 8 cm, ∠ABC = 90°, BD ⊥ AC and ∠ABD = θ. (i) Show that ∠ABD = ∠BCD = θ (ii) Find the length of AC. (iii) Find the value of sinθ. (iv) Prove that cosθ.tanθ = sinθ. b) In the given figure, AB = 12 cm, BC = 5 cm, ∠ABC = 90°, BD ⊥ AC and ∠DBC = α. (i) Show that ∠BAC = ∠DBC = α (ii) Find the length of AC. (iii) Find the value of cos α. (iv) Prove that (1 – sin2 α) . tan2 α = sin2 α. q A B D C 25 cm 7 cm 17 cm A D C O q B O X P Y a 6 cm 8 cm 12 cm 5 cm

Vedanta Excel in Mathematics - Book 9 344 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 20.4 Values of trigonometric ratios of some standard angles The angles like 0°, 30°, 45°, 60°, and 90° are commonly known as the standard angles. The values of trigonometric ratios of these standard angles can also be obtained geometrically without using the table or a calculator. (i) Values of trigonometric ratios of the angles 45° Let PQR be an isosceles right–angled triangle right angled at Q. Here, PQ = QR = a (suppose) ∠QRP = ∠QPR Now, ∠PQR + ∠QRP + ∠QPR = 180° or, 90° + ∠QRP + ∠QRP = 180° ∴ ∠QRP = 45° Again, in rt. ∠ed ∆ PQR, using Pythagoras theorem, PR = PQ2 + QR2 = a2 + a2 = 2a2 = a 2 Now, in rt. ∠ ed ∆ PQR, sinR = p h = PQ PR = a a 2 = 1 2 , cosR = b h = QR PR = a a 2 = 1 2 , tanR = p b = PQ QR = a a = 1 (ii) Values of trigonometric ratios of the angle 30° and 60° Let PQR be an equilateral triangle, where PQ = QR = RP = 2a (suppose) and ∠P = ∠Q = ∠R = 60°. PM ⊥ QR is drawn. As the perpendicular drawn from a vertex to the opposite side bisects the opposite side as well as the vertical angle in an equilateral triangle, QM = MR = 1 2 × 2a = a ∠QPM = ∠MPR = 1 2 × 60° = 30° In rt. ∠ed ∆PQM, using Pythagoras Theorem, PM = PQ2 – QM2 = (2a)2 – a2 = 3a2 = a 3 Now, in rt. ∠ed ∆ PQM, sin (QPM) = sin30° = p h = QM PQ = a 2a = 1 2 cos (QPM) = cos30° = b h = PM PQ = a 3 2a = 3 2 tan (QPM) = tan30° = p b = QM PM = a a 3 = 1 3 Trigonometry

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 345 Vedanta Excel in Mathematics - Book 9 Trigonometry Now, in rt. ∠ed ∆ PQM, sin (PQM) = sin60° = p h = PM PQ = a 3 2a = 3 2 cos (PQM) = cos60° = b h = QM PQ = a 2a = 1 2 tan (PQM) = tan60° = p b = PM QM = a 3 a = 3 (iii) Values of trigonometric ratios of the angle 0° Let PQR be a right-angled triangle in which ∠RPQ is a right angle. Suppose, ∠PQR = θ. Here, when θ tends to be 0°, PR also tends to be 0 and QR tends to be equal to PQ. ∴ When θ = 0 then, PR = 0 and QR = PQ Now, in rt. ∠ed ∆ PQR, sinθ = p h = PR QR cosθ = b h = PQ QR tanθ = p b = PR PQ ∴ sin0° = 0 PQ = 0 ∴ cos0° = PQ PQ = 1 ∴ tan0° = 0 PQ = 0 (iv) Values of trigonometric ratios of the angle 90° Let PQR be a right-angled triangle in which ∠RPQ is a right angle. Suppose, ∠PQR = θ. Here, when θ tends to be 90°, PQ tends to be 0 and QR tends to be equal to PR. ∴When θ = 90°, then PQ = 0 and QR = PR Now, in rt. ∠ed ∆ PQR, sinθ = p h = PR QR cosθ = b h = PQ QR tanθ = p b = PR PQ ∴ sin90° = PR PR = 1 ∴ cos90° = 0 PR = 0 ∴ tan90° = PR 0 = ∞ The values of trigonometric ratios of the standard angles are given in the following table. Trigonometric ratios Angles 0° 30° 45° 60° 90° sin 0 1 2 1 2 3 2 1 cos 1 3 2 1 2 1 2 0 tan 0 1 3 1 3 ∞

Vedanta Excel in Mathematics - Book 9 346 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Example 3: In the given figure, ABC is a right angled triangle and ACDE is a rectangle. Find the size of BD. Solution: Here, in rectangle ACDE , AE = CD = 2 m and ED = AC = 50 m In rt. ∠ed ∆ ABC, tanA = p b = BC AC = BC 50 or, tan30° = BC 50 or, 1 3 = BC 50 or, BC = 50 3 = 50 3 × 3 3 = 50 × 1.732 3 = 28.87 m. Now, BD = BC + CD = 28.87 m + 2m = 30.87 m. Example 4: In the given figure, a ladder rest against a vertical wall making an angle of 60o with the ground. If the height of the wall at which the upper end of the ladder is supported is 30 ft, find the length of the ladder and the distance of the foot of the ladder from the wall. Solution: Here, height of the wall (AC) = 30 ft, angle made by the ladder (θ) = 60o Length of ladder (AB) =? Distance between the foot of the ladder and the wall (BC) =? https://www.geogebra.org/m/j8wckkch Vedanta ICT Corner Please! Scan this QR code or browse the link given below: A B C 60° 30ft. Worked-out examples Example 1: Find the value of sin2 30° + sin2 45° + sin2 60° . Solution: Here, sin2 30° + sin2 45° + sin2 60° = 1 2 1 2 + + 3 2 = 1 4 + 1 2 + 3 4 = 1 + 2 + 3 4 = 6 4 = 1 1 2 2 2 2 Example 2: Prove that 1 + tan30° 1 – tan30° = 1 + sin60° 1 – sin30° Solution: L.H.S. = 1 + tan30° 1 – tan30° = 1 3 1 + 1 3 1 – = 3 + 1 3 – 1 = 3 + 1 3 – 1 × 3 + 1 3 + 1 = ( 3 + 1)2 ( 3) 2 – (1)2 = 2 + 3 R.H.S. = 1 + sin60° 1 – sin30° = 3 2 1 + 1 2 1 – = 2 + 3 2 – 1 = 2 + 3 ∴ L.H.S. = R.H.S. proved. Trigonometry

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 347 Vedanta Excel in Mathematics - Book 9 Trigonometry In right angled triangle ABC, sin60° = p h = AC AB tan60° = p b = AC BC or, 3 2 = 30 AB or, 3 = 30 BC or, 3AB = 60 or, 3 BC = 30 or, AB = 60 3 × 3 3 = 3 3 60 = 20 3 or, BC = 30 3 × 3 3 = 3 3 30 = 10 3 Hence, the length of the ladder is 20 3 ft. and the distance between the foot of the ladder from the wall is 10 3 ft. Example 5: A tree of 18 m height is broken by the wind so that its top touches the ground. If the height of the remaining part of the tree is 6 m, find the angle made by the broken part of the tree with the ground. Solution: Let, AB be the height of the tree before it was broken, CD be the broken part of the tree. Here, AB = 18 m, BD = 6 m ∴ CD = AD = 18 m – 6 m = 12 m Let, ∠BCD = θ In right angled triangle BCD, sinθ = p h = BD CD or, sinθ = 6 m 12 m or, sinθ = 1 2 or, sinθ = sin30° ∴ q = 30° Hence, the angle made by the broken part of the tree makes an angle of 30o with the ground. EXERCISE 20.2 General section 1. Evaluate: a) sin60°.cos30° b) sin30°.tan45° c) 2cos45°.tan60° d) sin30° – cos30°.tan30° e) tan2 45° + cos2 45° f) sin2 30° – cos2 60° g) 1 2 × sin30° × cos45° h) 2tan30°.cos30° i) 2 3 tan60°.sin60° j) tan2 60° + 4 cos2 45° k) 2 2 sin45°.cos60° l) sin45° × cos45° × tan45° m) sin2 30° + cos2 60° n) tan30°.tan60°.cos45° o) sin30°.cos60°.tan45° p) sin60°.cos30°.tan30° q) sin90°.cos45°.tan0° r) tan0° + cos0° + sin0° s) sin30° (cos30° + sin60°) t) 2 3sin30°.tan30° u) 3.cos60° sin60° v) tan60° 2cos30° w) sin30° + sin60° cos30° + cos60° x) 2tan30° 1 – tan2 30° y) sin60°.cos30° + cos60°.sin30° sin60°.cos30° – cos60°.sin30° z) cos60°.cos30° – sin60°.sin30° cos60°.cos30° + sin60°.sin30° A C q B 6m D 18m

Vedanta Excel in Mathematics - Book 9 348 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Creative section - A 2. Prove that a) cos60°.tan60° = sin60° b) sin30°.tan60° = cos30° c) cos2 30° + cos2 60° = 1 d) sin2 30° + sin2 60° = 1 e) 2tan30° 1 – tan2 30° = tan60° f) 2tan30° 1 + tan2 30° = sin60° g) 2sin60°.cos60° = cos30° h) 1 + tan30° 1 – tan30° = 1 + sin60° 1 – sin30° 3. In the following right angled triangles, find the unknown sizes of the sides. 4. In the given figure, calculate the length of the side AC. (sin28° = 0.46) 5. In the following right angled triangles, find the unknown sizes of the acute angles. 7. a) An electric pole 15 m high is supported by a wire fixing its one end on the ground at some distance from the pole. If the wire joining the top of the pole is inclined to the ground at an angle of 60°, find the length of the wire. b) A ladder rests on the top of the wall in such a way that it makes an angle of 45° with the ground. If the height of the wall is 25 ft, find the length of the ladder. Also, find the distance of the foot of the ladder and the wall. P Z X Y R 6. a) In the given figure, ABC is a right angled triangle and ACDE is a rectangle. Find the size of BD. Creative section - B b) In the figure, PQR is a right angled triangle and PRST is a rectangle. Find the measure of TS. F P Q R G E Trigonometry

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 349 Vedanta Excel in Mathematics - Book 9 Trigonometry 8. a) A ladder 24 meter long is resting against a wall. If the distance between the foot of the ladder from the wall is 12 m, find the angle made by the ladder with the floor. b) The length of the shadow of a 30 3m high tree is 90 m at 2:30 p.m., find the altitude of sun. Project work and activity section 9. a) Draw two right-angled triangles using two different sets of Pythagorean triplets. Then write the trigonometric ratios of sine, cosine and tangent in each triangle taking both the acute angles of the triangles as reference angles. b) Draw three right-angled triangles of your own measurements with one of the acute angles being 30°, 60° and 45° respectively in three triangles. Measure the lengths of sides of each triangle. Then find the value of the following trigonometric ratios upto three decimal places. a) sin30°, cos30° , tan30° b) sin60°, cos60° , tan60° c) sin45°, cos45° , tan45° OBJECTIVE QUESTIONS Let’s tick (√) the correct alternative. 1. In a right angled triangle, if θ be a reference angle, the side opposite to θ is (A) base (b) (B) perpendicular (p) (C) hypotenuse (h) (D) altitude 2. In a right angled triangle, tangent of an angle is the ratio of: (A) perpendicular and hypotenuse (B) base and hypotenuse (C) perpendicular and base (D) base and perpendicular 3. In ∆ABC given alongside; what is the trigonometric ratio of sinα? (A) 3 4 (B) 3 5 (C) 4 5 (D) 5 3 4. What is the value of sin2 A + cos2 A? (A) -1 (B) 0 (C) 1 (D) 2 5. Which one of the following relations is NOT true? (A) sin2 A×cos2 A = 1 (B) sin2 A = 1 – cos2 A (C) cos2 A = 1 – sin2 A (D) sin2 A+cos2 A = 1 6. The value of sinA in terms of cosA is: (A) 1 – cosA (B) 1 – cos2 A (C) 1 – cos2 A (D) 1+ cos2 A 7. The value of the trigonometric ratio sin45° is (A) 0 (B) 1 2 (C) 1 2 (D) 3 2 8. The value of the expression sin30° + cos60° ? (A) 1 (B) 0 (C) 3 2 (D) 1 2 9. The maximum value of the expression 3sinA + 4cosA is (A) 7 (B) 4 (C) 3 (D) 0 10. If 2sinθ – 1 = 0 (00 ≤ θ ≤ 00 ), then the value of θ is (A) 00 (B) 300 (C) 450 (D) 600 B 4 cm C a A3 cm

Vedanta Excel in Mathematics - Book 9 350 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Assessment –VII 1. In the given right angled triangle ABC; ∠ABC = 90o and ∠ACB =θ. Answer the following questions. a) Which trigonometric ratio is represented by AB BC ? b) What is the value of AB AC 2 + BC AC 2 c) If AC = 20 cm and cosθ = 4 5 , find the length of BC. d) If the length of AC were twice the length of AB, what would be the value of θ? 2. In the adjoining figure, ∠QPR = 90°, PQ = 8 cm, QR = 10 cm, PS ⊥ QR and ∠PQR = θ. (a) Show that ∠SPR = ∠PQR = θ (b) Find the length of PR. (c) Find the value of sin θ. (d) Prove that tan θ. cos θ = sin θ 3. In the figure alongside, O is the centre of circle; P is the mid-point of the chord XY. If ∠POY = α, XY = 24 cm and OY = 13 cm, answer the following questions. (a) Name the special type of triangle POY. (b) What is the length of OP? (c) What is the value of tanα? (d) If OP and PY were equal in length, what would be the value of α? 4. In the adjoining figure, AB is the height of the tree and BC is the distance between the tree and the car. If BC = 15 m and ∠ACB = 30o , answer the following questions. (a) What is the value of sin 30o ? (b) Write the trigonometric ratio that is applicable to find the height of the tree. (c) Find the height of the tree. (d) What would be the value of ∠ACB when the height of the tree AB and the distance between the tree and the car BC were equal? A B C q q 8 cm 10 cm P Q R S a P O X Y A B 15 m C 30°