347 Solution: (a) Continuous series. (b) Constructing the histogram for the given data, (c) 780 students attended the lest. (d) 200 of them did not achieve any grade. 6. The 140 families are asked about the daily wages and the result obtained as follows: Wages (Rs.) 150-200 200-250 250-300 300-350 350-400 Frequency 5 18 32 45 20 (a) Which graphical presentation is the best for the data given? (b) Represent the data in histogram. (c) If the family with daily income below Rs. 200 is considered as absolute poverty line, find the number of family below the absolute poverty line. (d) How much more money should they earn daily to rise above the absolute poverty line? Solution: (a) As the data given is continuous, Histogram is best for it. (b) Constructing the histogram for the given data, (c) 5 families are below absolute poverty line. (d) They have to earn upto Rs. 50 to rise above the absolute poverty line. 350 300 250 200 150 100 50 0 10 20 30 40 50 60 Marks 70 80 90 100 50 45 40 35 30 25 20 15 10 5 0 150 200 250 300 350 400 Wages
348 17.3 Frequency Polygon PRACTICE 17.3 Read / Understand / Think / Do Keeping Skill Sharp 1. A cargo company is preparing to deliver the boxes. Different boxes must deliver in different van according to their weight. The boxes are separated as following: Weight in kg 10-20 20-30 30-40 40-50 50-60 No. of boxes 5 10 18 12 8 (a) With which value should we mark on y-axis corresponding to frequency to construct the frequency polygon? (b) Present the given data into frequency polygon. (c) Which group has the highest number of boxes? (d) If Rs. 50 is charged for each box above the weight of 40 kg in toll checkpoint, how much money should be paid as toll charge? Solution: (a) No. of boxes should be marked in y-axis. (b) Constructing the histogram for the given data, (c) In the group of 30 - 40 kg has highest number of boxes. (d) 12 + 8 = 20 boxes are above 40 kg so that toll charge = 20 × Rs. 50 = Rs. 1000 should be paid. 2. A farmer recorded the height of the plants in his her farm as follows: Height in cm 0-5 5-10 10-15 15-20 20-25 No. of Plants 5 10 18 12 8 (a) In which axis the height of plants need to mark to construct the frequency polygon? (b) Present the given data into frequency polygon. (c) How many plants are there in tallest plants group? (d) If each plant costs Rs. 15 for the plants of height 10 cm and above, how much money will the farmer get? Solution: (a) The height of plants marks in the x-axis to construct the frequency polygon. (b) Constructing the histogram for the given data, 20 18 16 14 12 10 8 6 4 2 0 10 30 40 50 60 70 Weight (in kg) 20
349 (c) There are 8 plants in tallest plant group (d) 18 + 12 + 8 = 38 plants are above 10 cm, so the farmer gets Rs. 15 × 38 = Rs. 570 for those plants. 3. The height of students of grade nine of a particular school is as given in table: Height in cm 100-110 110-120 120-130 130-140 140-150 150-160 No. of Students 4 16 20 24 10 5 (a) In which axis the Number of Students need to mark to construct the frequency polygon? (b) Present the following data into frequency polygon. (c) How many total students are there altogether? (d) If only students above the height of 120 cm can swim in pool, how much money should be paid at Rs. 300 per each ticket? Solution: (a) The Number of Students marks in the y-axis to construct the frequency polygon. (b) Constructing the histogram for the given data, (c) There are 79 student in total. (d) 20 + 27 + 10 + 5 = 59 students are above 120 cm, so the total amount to pay = Rs. 300 × 59 = Rs. 17700. 4. In an exam of in full marks 50, the marks obtained by the students of class nine are as follows: Marks 2.5-7.5 7.5-12.5 12.5-17.5 17.5-22.5 22.5-27.5 27.5-32.5 No. of Students 7 5 12 8 4 2 (a) What is another name for frequency polygon? (b) Present the following data into frequency polygon. (c) How many total students are there altogether? (d) If the marks below 35% is considered as non-graded, find the number of students whose certificate will not be issued? Solution: (a) The another name of frequency polygon is line graph. 20 18 16 14 12 10 8 6 4 2 0 5 15 20 25 30 Height (in cm) 10 30 25 20 15 10 5 0 100 120 130 140 150 160 Height (in cm) 110 y-axis
350 (b) Constructing the histogram for the given data, (c) 7 + 17 + 8 + 4 = 2 = 38 students are there altogether. (d) The 35% of 50 = 17.5 marks is pass marks. So 7 + 5 + 12 = 24 students failed or non-graded. Hence 24 student's certificate won't be issued. 17.4 Cumulative Frequency Curve (Ogive) PRACTICE 17.4 Read / Understand / Think / Do Keeping Skill Sharp 1. The given ogive shows the records of the marks of students, (a) What is cumulative frequency? (b) What is the total number of students? (b) How many students are there who obtained less than 75 mark? (c) What is the number of students who obtained the marks between 20 to 40? Solution: (a) The accumulated frequencies with each of its previous frequencies is called the cumulative frequency. (b) The total number of students = 30. (c) There are 30 students who scored less than 75 marks. (d) The number of students who scored between 20 and 40 is 10 – 4 = 6 students. 2. Study the given cumulative frequency curve and answer the following questions: (a) What does the adjoining graph represent? (b) What is the number of students who secured 70 to 80 marks? (c) How many students secured the marks more than 80? (d) If the pass mark is 40, how many students passed the test? Solution: (a) The adjoining figure represents less than cumulative frequency curve of test marks. 14 12 10 8 6 4 2 0 5 15 20 25 30 Marks 10 35
351 (b) The number of students who secured 70 to 80 marks is 20 – 13 = 7. (c) 30 – 20 = 10 students secured more than 80 marks. (d) No students secured less than 40. Hence all 30 students pass the test. 3. Study the given cumulative frequency curve and answer the following questions: (a) What does the adjoining graph represent? (b) What is the number of students who secured the mark more than 65? (c) Find the number of students who secured the marks between 5 and 45? (d) If the pass mark is 25, how many students failed the exam? Solution: (a) The graph adjoining represents more than cumulative frequency curve of marks obtained. (b) 25 students secured more than 65 marks. (c) 60 – 40 = 20 students secured between 5 and 45. (d) 49 students secured marks more than 25 hence 60 – 49 = 11 students failed. 4. The given ogive shows the record of weights of some girls. Answer the following questions: (a) What does the adjoining graph represent? (b) Find the number of girls whose weight is between 20 kg and 40 kg. (c) How many girls are less than 20kg? (d) What it the total number of girls surveyed? Solution: (a) The adjoining figure represents more than cumulative curve of weight of girls. (b) 72 – 18 = 54 girls are there whose weight is between 20 kg and 40 kg. (c) 98 – 72 = 26 girls are less than 20 kg. (d) All together 98 girls were surveyed. 5. Here is given a data of some survey done by grade nine students on the topic of pocket money in class 8: Pocket Money (Rs.) 0-10 10-20 20-30 30-40 40-50 No of students 5 11 18 10 7 (a) What is less than ogive? (b) Construct the less than type cumulative curve (ogive) of the following data: (c) How many students are there in class 8? (d) If cost of a burger is Rs. 30, how many students cannot afford a burger? Solution: (a) The graphical representation of less than value of class and its corresponding cumulative frequency is called less than ogive. (b) Constructing the less than type cumulative curve (ogive) from the given data, Class f Less then c.f. 0 - 10 5 10 5 10 - 20 11 20 16
352 20 - 30 18 30 34 30 - 40 10 40 44 40 - 50 7 50 51 (c) There are 51 students in class 8. (d) 20 + 7 = 17 students can afford burger of cost Rs. 30. 6. The number of students of a government school who created a Google account at following age for the first time are given below. Age in years 10-11 11-12 12-13 13-14 14-15 Frequency 15 27 35 23 10 (a) What is the first limit to construct the less than ogive for the above data? (b) Construct the less than type cumulative curve (ogive) of the given data: (c) How many students are involved in survey? (d) If the age limit is 12 years to create Google account, how many of them used fake date of birth? Solution: (a) The first limit to construct the less than ogive for the above data is 11. (b) Constructing the less than type cumulative curve (ogive) from the given data, Class f Less then c.f. 10 - 11 15 11 15 11 - 12 27 12 42 12 - 13 35 13 77 13 - 14 23 14 100 14 - 15 10 15 110 55 50 45 40 35 30 25 20 15 10 5 0 10 30 40 50 Pocket Money 20 120 110 100 90 80 70 60 50 40 30 20 10 0 11 13 14 15 Age (in years) 12
353 (c) 110 students are involved in the survey. (d) 15 + 27 = 42 students used fake date of birth to creat an account in google. 7. A hotel recorded the number of costumers for 40 days. Number of costumer 20-40 40-60 60-80 80-100 100-120 Number of days 4 8 6 12 10 (a) What is more than ogive? Define it. (b) Construct the more than type cumulative curve (ogive) of the above data: (c) How many days the hotel served more than 100 costumers? (d) If the cook of the hotel get bonus of Rs. 1500 per day when the number of costumer is more than 80, how much bonus did he get in those 40 days? Solution: (a) The graphical representation of curve made by more than limit and its corresponding cumulative frequencies is called more than ogive. (b) Constructing the less than type cumulative curve (ogive) from the given data, Class f More than c.f. 20 - 40 4 20 40 40 - 60 8 40 36 60 - 80 6 60 28 80 - 100 12 80 22 100 - 120 10 100 10 (c) The hotel served more than 100 customers for 10 days. (d) 12 + 10 = 22 days, there were more than 80 costumers. Hence Rs. 1500 × 22 = Rs. 33,000 as bonus in those 40 days. 8. The administration of a fun park recorded the ages of the visitors in a certain day as follows: Age in years 5-10 10-15 15-20 20-25 25-30 30-35 Frequency 5 10 20 15 25 15 (a) What is the first limit to construct the more than ogive for the above table? (b) Construct the more than type cumulative curve (ogive) of the given data: (c) Which age group has maximum number of visitors? (d) If the rates of tickets are as follows: for below 15 –free and for more than 15 years Rs. 20 per person, find the total income of that day. Solution: (a) The first limit to construct the more than ogive for the above table is 5. (b) Constructing the less than type cumulative curve (ogive) from the given data, 50 45 40 35 30 25 20 15 10 5 0 20 60 80 100 No. of costumers 40
354 Class f More than c.f. 5 - 10 5 5 90 10 - 15 10 10 65 15 - 20 20 15 75 20 - 25 15 20 55 25 - 30 25 25 40 30 - 35 15 30 15 (c) The maximum number of visitors are of age 25 – 30. (d) There are 75 visitors of age more than 15 yrs. Hence, the total income is Rs. 20 × 75 = Rs. 1500. 9. The marks obtained by some students are as following: Marks 0-10 10-20 20-30 30-40 40-50 No. of Students 5 11 18 10 7 (a) Define ogive. (b) Prepare the more than cf and less than cf table from the given data: (c) Construct the less than and more than type cumulative curve (ogives) of the following data in the same graph: (d) Determine the median marks of the students from the graph. [md=25.28] Solution: (a) The curve obtained by plotting the class sizes with their corresponding cumulative frequencies is called ogive. (b) Preparing more than cf and less than cf table from the given data, Class f More than c.f. Less than c.f. 0 - 10 5 0 51 10 5 10 - 20 11 10 46 20 16 20 - 30 18 20 35 30 34 30 - 40 10 30 17 40 44 40 - 50 7 40 7 50 51 100 90 80 70 60 50 40 30 20 10 0 5 15 20 25 30 Age (in years) 10 35
355 (c) Constructing the less than and more than type cumulative curve (ogives) from the given data, (d) The median of data is 25. [25.28 actual] 10. A medical recorded the data of ages of people who were infected by viral infection last week. Age in years 4-8 8-12 12-16 16-20 20-24 24-28 Frequency 3 4 7 3 5 6 (a) What is the last limit to construct the less than and the more than ogive for the above table? (b) Prepare the more than cf and less than cf table from the given data: (c) Construct the less than and more than type cumulative curve (ogives) of the following data in the same graph: (d) Determine the median age of people suffering from the viral infection from the graph. [md=17] Solution: (a) For above the last limit of less than ogive is 28 and that of more than ogive is 24. (b) Preparing more than cf and less than cf table from the given data, Class f More than c.f. Less than c.f. 4 - 8 3 4 28 8 3 8 - 12 4 8 25 12 7 12 - 16 7 12 21 16 14 16 - 20 3 16 14 20 17 20 - 24 5 20 11 24 22 24 - 28 6 24 6 28 28 (c) Constructing the less than and more than type cumulative curve (ogives) from the given data, (d) The median from ogive is 16. [17 in calculation]. 55 50 45 40 35 30 25 20 15 10 5 0 10 30 40 50 Marks 20 25 30 25 20 15 10 5 0 4 12 16 20 24 28 Ages (in years) 8
356 CHAPTER MEASUREMENT OF CENTRAL TENDENCY 18 18.1 Arithmetic Mean PRACTICE 18.1 Read / Understand / Think / Do Keeping Skill Sharp 1. (a) Define arithmetic mean. (b) Write the formula to find the mean of an individual data. (c) Write the formula to calculate the mean of a discrete series. (d) In a data, the sum of N observations is Σfx. What is the mean of the data? Solution: (a) The value obtained by dividing the total sum of all observation in data by the number of all observation is called mean. (b) The formula to find the mean of an individual data is x = x n , n = number of items. (c) The formula to find the mean of a descrete data is x = fx N , N = sum of frequency of number of items. (d) x = fx N 2. (a) In an individual data, the sum of x of 26 observations is 390. What is its mean? (b) If the sum of wages of 35 labours is Rs. 20125, what is their average wages? (c) In a discrete data, the sum of fx of 36 observations is 1620. What is the mean? (d) In a discrete data, the number of observations is 30 and Σfx is 750. What is its AM? Solution: (a) Here, x = 390, n = 26 So, x = x n = 390 26 = 15 (b) Here, sum of wages (x) = Rs. 20,125 No. of workers (n) = 35 average wage = x n = Rs. 20125 35 = Rs. 575 (c) Here, fx = 1620, N = f = 36 x = fx N = 1620 36 = 45 (d) Here, fx = 750, N = 30 AM = fx N = 750 30 = 25
357 3. Find the mean of each of the following sets of scores: (a) 2, 3, 2, 4, 6 (b) 1, 2, 2, 4, 5, 5, 7, 6, 3, 4 (c) 1, 0, 1, 0, 1, 0 (d) 1, 3, 5, 7, 5, 3, 1 Solution: (a) Here, x : 2, 3, 2, 4, 6 x = 2 + 3 + 2 + 4 + 6 = 17 n = 5 Hence, x = x n = 17 5 = 3.4 (b) Here, x : 1, 2, 2, 4, 5, 5, 7, 6, 3, 4 x = 1 + 2 + 2 + 4 + 5 + 5 + 7 + 6 + 3 + 4 = 39 and n = 10 Hence, x = x n = 39 10 = 3.9 (c) Here, x : 1, 0, 1, 0, 1, 0 x = 1 + 0 + 1 + 0 + 1 + 0 = 3 n = 6 Hence, x = x n = 3 6 = 0.5 (d) Here, x : 1 , 3, 5, 7, 5, 3, 1 x = 1 + 3 + 5 + 7 + 5 + 3 + 1 = 25 n = 7 Hence, x = x n = 25 7 = 3.57 4. (a) If the mean of 1, 3, 5, 7 and x is 10, find the value of x. (b) If the mean of 10, 30, 40, 50, 60, p and 90 is 50, find the value of p. Solution: (a) Here, x : 1, 3, 5, 7, x x = 1 + 3 + 5 + 7 + x = 16 + x n = 5, x = 10 Hence, x = x n or, 10 = 16 + x 5 or, 16 + x = 50 x = 34 (b) Here, x : 10, 30, 40, 50, 60, p, 90 x = 10 + 30 + 40 + 50 + 60 + p + 90 = 280 + p n = 7, x = 50 Hence, we know x = x n or, 50 = 280 + p 7 or, 280 + p = 350 p = 70 5. (a) If the mean of an individual data is 14 in which N = 24 + a and Σx = 360 + 2a, find the value of a. (b) If the mean of a discrete data is 56 in which N = 32 + k and Σfx = 2567 + 5k, find the value of k. Solution: (a) Here, x = 14, N = 24 + a, x = 360 + 2a
358 Hence, x = x N or, 14 = 360 + 2a 24 + a or, 336 + 14a = 360 + 2a or, 14a – 2a = 360 – 336 or, 12a = 24 a = 2 (b) Here, x = 56, N = 32 + k, fx = 2567 + 5k Hence, x = fx N or, 56 = 2567 + 5k 32 + k or, 1792 + 56k = 2567 + 5k or, 56k – 5k = 2567 – 1792 or, 51k = 775 or, k = 15.2 6. (a) If the following table gives the marks obtained by 50 students, find the mean. Marks (x) 20 25 30 35 40 No. of Students (f) 10 6 10 15 9 (b) If the following table gives the marks obtained by 50 students, find the mean. Marks (x) 14 15 16 17 18 No. of Students (f) 8 9 12 16 12 Solution: (a) Here, calculation of mean x f fx 20 10 200 25 6 150 30 10 300 35 15 525 40 9 360 N = 50 fx = 1535 Now, x = fx N = 1535 50 = 30.7 (b) Here, computation of mean x f fx 14 8 112 15 9 135 16 12 192 17 16 272 18 12 216 N = 57 fx = 927 Now, x = fx N = 927 57 = 16.26
359 7. (a) If the arithmetic mean of the given data is 7, find the value of p. x 5 6 7 8 9 f 6 4 3 6 p (b) Find the value of k, if the arithmetic mean of the given data is 29.625. Marks (x) 20 25 30 35 40 No. of Students (f) 10 5 k 12 5 Solution: (a) Here, x = 7 To find: p = ? Compulation of mean x f fx 5 6 30 6 4 24 7 3 21 8 6 48 9 p 9p N = 19 + p fx = 123 + 9p Now, x = fx N or, 7 = 123 + 9p 19 + p or, 133 + 7p = 123 + 9p or, 9p – 7p = 133 – 123 or, 2p = 10 p = 5 (b) Here, x = 29.625 To find: k = ? Computation of mean x f fx 20 10 200 25 5 125 30 k 30k 35 12 420 40 5 200 N = 32 + k fx = 945 + 30k Now, x = fx N or, 29.625 = 945 + 30k 32 + k or, 948 + 29.625k = 945 + 30k or, 948 – 945 = 30k – 29.625 or, 3 = 0.375k or, k = 3 0.375 k = 8.
360 18.2 Median PRACTICE 18.2 Read / Understand / Think / Do Keeping Skill Sharp 1. (a) Define median. (b) In which percent does the median divide the data? (c) Which is the median term of a data having N observations? (d) If a data has 12 observations, what is the position of the median in it? Solution: (a) Median: The value or item or observation which divides the data into two equal halves being one above that value and another below that value. (b) Median divide the data into 50%. (c) The median term of a data having N observation is N 2 . (d) If the data has 12 observations the position of median is 12 + 1 2 th item = 6.5th item. 2. Write the median in each of the following cases: (a) 3, 2, 7, 1, 4 (b) 10, 8, 6, 5, 4, 1, 2 Solution: (a) Here, x : 3, 2, 7, 1, 4 Arranging in ascending order, 1, 2, 3, 4, 7 N = 5 Median (Md) = N + 1 2 th item = 5 + 1 2 th item = 6 2 th item = 3rd item = 3. (b) Here, x : 10, 8, 6, 5, 4, 1, 2 Arranging in ascending order, 1, 2, 4, 5, 6, 8, 10 N = 7 Median (Md) = N + 1 2 th item = 7 + 1 2 th item = 8 2 th item = 4th item = 5. 3. Find the median in each of the following cases: (a) 3, 4, 7, 7, 7, 1, 2, 2, 7, 1, 9 (b) 9, 9, 9, 8, 8, 8, 8, 4, 4, 1, 1, 1, 1, 1, 6, 8, 5 Solution: (a) Here, x : 3, 4, 7, 7, 7, 1, 2, 2, 7, 1, 9 Arranging in ascending order, 1, 1, 2, 2, 3, 4, 7, 7, 7, 7, 9 N = 11 Now, Md = N + 1 2 th item = 11 + 1 2 th item = 12 2 th item = 6th item = 4. (b) Here, 9, 9, 9, 8, 8, 8, 8, 4, 4, 1, 1, 1, 1, 1, 6, 8, 5 Arranging in ascending order; 1, 1, 1, 1, 1, 4, 4, 5, 6, 8, 8, 8, 8, 8, 9, 9, 9 N = 17 Now, Md = N + 1 2 th item = 17 + 1 2 th item = 18 2 th item = 9th item = 6.
361 4. (a) What is the value of x if the data 40, 50, 3x + 4 2 , 80, 90 are in ascending order and the median is 14? (b) If the data x – 1, x + 2, 2x – 1, 2x + 3, 3x + 2 are in ascending order and the median is 17, what is the value of x? (c) If the data x, 2x, 3x + 1, 4x – 1, 6x, 7x – 1 are in ascending order and its median is 35, find the value of x. Solution: (a) Here, x : 40, 50, 3x + 4 2 , 80, 90 are in ascending order. N = 5, Md = 14 To find: x = ? We know that, Md = N + 1 2 th item or, 14 = 5 + 1 2 th item or, 14 = 3rd item or, 14 = 3x + 4 2 or, 3x + 4 = 28 or, 3x = 24 or, x = 24 3 x = 8 (b) Here, x – 1, x + 2, 2x – 1, 2x + 3, 3x + 2 are in ascending order. N = 5, Md = 17 To find: x = ? We know that, Md = N + 1 2 th item or, 17 = 5 + 1 2 th item or, 17 = 3rd item or, 17 = 2x – 1 or, 2x = 18 x = 9 (c) Here, x, 2x, 3x + 1, 4x – 1, 6x, 7x – 1 are in ascending order. N = 6, Md = 35 To find: x = ? We known that, Md = N + 1 2 th item or, 35 = 6 + 1 2 th item or, 35 = 3.5th item or, 35 = 3rd item + 4th item 2 or, 35 = 3x + 1 + 4x – 1 2 or, 70 = 7x x = 10.
362 5. Find the median from the data in each of the following tables: (a) x 50 60 70 80 90 99 f 15 11 14 5 3 1 (b) Wages 45 55 65 75 85 No. of workers 24 23 31 12 23 (c) Marks 10 20 30 40 50 No. of workers 1 4 8 6 2 (d) x 5 15 25 35 45 55 f 7 8 9 2 1 4 Solution: (a) Here, computation of median x f cf 50 15 15 60 11 26 70 14 40 80 5 45 90 3 48 99 1 49 N = 49 Now, median (Md) = N + 1 2 th item = 49 + 1 2 th item = 25th item The nearest greater value of 25 in cf column is 26 and corresponding to it is 60. Hence, Md = 60. (b) Here, computation of median x f cf 45 24 24 55 23 47 65 31 78 75 12 90 85 23 113 N = 113 Now, median (Md) = N + 1 2 th item = 113 + 1 2 th item = 57th item The nearest greater value of 57 is 78 whose corresponding item is 65. Hence, Md = 65. (c) Here, computation of median x f cf 10 1 1 20 4 5 30 8 13 40 6 19 50 2 21 N = 21
363 Now, median (Md) = N + 1 2 th item = 21 + 1 2 th item = 11th item The nearest greater value of 11 in cf column is 13 and its corresponding item is 30. Hence, Md = 30. (d) Here, computation of median x f cf 5 7 7 15 8 15 25 9 24 35 2 26 45 1 27 55 4 31 N = 31 Now, median (Md) = N + 1 2 th item = 31 + 1 2 th item = 16th item The nearest greater value of 16 in cf column is 24 and its corresponding item is 25. Hence, Md = 25. 6. (a) Determine the median size of shoes from the following data of 51 pairs: Size 4 5 6 7 8 9 No. of pairs 8 14 k 10 3 1 (b) Calculate the median from the data given below: Marks 28 24 32 35 48 45 Total No. of Students 24 23 28 a 59 54 225 Solution: (a) Here, N = 51 Computation of median, x f 4 8 5 14 6 15 7 10 8 3 9 1 36 + k Since, N = 51 i.e. f = 51 or, 36 + k = 51 k = 15 Representing the table with actual frequencies x f cf 4 8 8 5 14 22 6 15 37 7 10 47 8 3 50 9 1 51 N = 51
364 Now, Md = N + 1 2 th item = 51 + 1 2 th item = 26th item The nearest greater value of 26 is 37 and its corresponding item is 6. Hence, Md = 6. (b) Here, computation of Md x f 28 24 24 23 32 28 35 a 48 59 45 54 N = 188 + a Since, N = 225 f = 225 or, 188 + a = 225 a = 37 Representing the data will actual frequencies and arranging in ascending order Computation of median. x f cf 28 24 23 24 23 47 32 28 75 35 37 112 48 59 166 45 54 225 N = 225 Now, Md = N + 1 2 th item = 225 + 1 2 th item = 113th item The nearest greater value of 113 in cf column is 166 and its corresponding item is 45. Hence, Md = 45.
365 18.3 Quartiles PRACTICE 18.3 Read / Understand / Think / Do Keeping Skill Sharp 1. (a) Define quartiles. (b) What is the first quartile? (c) What is the third quartile? (d) What is the formula to calculate the third quartile of a data with N items? Solution: (a) Quartiles: the values or items or observations that divides the data into four equal parts are called quartiles. (b) First quartile: The value of item or observation that divides the data into 25% below and 75% above it is called first quartile. (c) Third quartile: The value or item or observation that divides the data into 75% below and 25 above it is called third quartile. (d) Q3 = 3(N + 1) 4 th item 2. (a) If a data has 7 items, what is the position of Q1 in it? (b) If a data has 11 observations, what is the position of the third quartile in it? Solution: (a) Here, N = 7 Q1 = N + 1 4 th item = 7 + 1 4 th item = 2nd item Hence, Q1 is 2nd item if data has 7 observation. (b) Here, N = 11 Q3 = 3(N + 1) 4 th item = 3(11 + 1) 4 th item = 36 4 th item = 9th item Hence, the 9th item is Q3 of data has 11 observations. 3. (a) Write the first quartile in each of the following data: (i) 1, 1, 2, 3, 4, 7, 9 (ii) 12, 10, 5, 4, 4, 3, 1 (b) Write the third quartile of the following data: (i) 12, 14, 15, 16, 17, 18, 21 (ii) 9, 9, 9, 8, 8, 8, 8, 4, 4, 3, 2 Solution: (a) (i) Here, x : 1, 1, 2, 3, 4, 7, 9 Since the given data is given in ascending order. We can proceed. N = 7 Now, Q1 = N + 1 4 th item = 7 + 1 4 th item = 2nd item Hence, Q1 = 1.
366 (ii) Here, x : 12, 10, 5, 4, 4, 3, 1 Arranging the data into ascending order, x : 1, 3, 4, 4, 5, 10, 12 N = 7 Now, Q1 = N + 1 4 th item = 7 + 1 4 th item = 2nd item Q1 = 3 (b) (i) Here, x : 12, 14, 15, 16, 17, 18, 21 Since the data is in ascending order, we can proceed. N = 7 Now, Q3 = 3(N + 1) 4 th item = 3(7 + 1) 4 th item = 24 4 th item = 6th item Q3 = 18. (ii) Here, x : 9, 9, 9, 8, 8, 8, 8, 4, 4, 3, 2 Arranging the data in ascending order. x : 2, 3, 4, 4, 8, 8, 8, 8, 9, 9, 9 N = 11 Now, Q3 = 3(N + 1) 4 th item = 3(11 + 1) 4 th item = 36 4 th item = 9th item Q3 = 4. 4. Find the first quartile from the data given below: (a) 18, 15, 5, 17, 20, 21, 23 (b) 15, 27, 18, 13, 12, 16, 22, 21, 25 Solution: (a) Here, x : 18, 15, 5, 17, 20, 21, 23 Arranging the given data into ascending order, x : 5, 15, 17, 18, 20, 21, 23 N = 7 Now, Q1 = N + 1 4 th item = 7 + 1 4 th item = 2nd item Q1 = 15. (b) Here, x : 15, 27, 18, 13, 12, 16, 22, 21, 25 Arranging the given data in ascending order; x : 12, 13, 15, 16, 18, 21, 22, 25, 27 N = 9 Now, Q1 = N + 1 4 th item = 9 + 1 4 th item = 2.5th item Q1 = 2nd item + 0.5(3rd item – 2nd item) = 13 + 0.5(15 – 13) = 13 + 0.5 × 2 = 14. 5. Find the third quartiles from the data given below: (a) 12, 13, 10, 15, 8, 17, 6, 12, 14, 19, 25 (b) 16, 32, 10, 25, 22, 28, 34 Solution: (a) Here, x : 12, 13, 10, 15, 8, 17, 6, 12, 14, 19, 25 Arranging the given data in ascending order, x : 6, 8, 10, 12, 12, 13, 14, 15, 17, 19, 25 N = 11
367 Now, Q3 = 3(N + 1) 4 th item = 3(11 + 1) 4 th item = 36 4 th item = 9th item Q3 = 17. (b) Here, x : 16, 32, 10, 25, 22, 28, 34 Arranging the given data in ascending order, x : 10, 16, 22, 25, 28, 32, 34 N = 7 Now, Q3 = 3(N + 1) 4 th item = 3(7 + 1) 4 th item = 24 4 th item = 6th item Q3 = 32. 6. (a) If the first quartile of the data given below is 41, find the value of x: 23, 2x – 3, 4x – 5, 29, 30, 34, 38. (b) If the third quartile of the data given below is 21, find the value of y. 12, 14, 15, 17, 2y – 1, 2y + 3, 27 Solution: (a) Here, x : 23, 2x – 3, 4x – 5, 29, 30, 34, 38 (ascending order) N = 7, Q1 = 41 To find x = ? We know that, Q1 = N + 1 4 th item or, 41 = 7 + 1 4 th item or, 41 = 2nd item or, 41 = 2x – 3 or, 2x = 44 x = 22. (b) Here, x : 12, 14, 15, 17, 2y – 1, 2y + 3, 27 N = 7, Q3 = 21 To find: y = ? Now, Q3 = 3(N + 1) 4 th item or, 21 = 3(7 + 1) 4 th item or, 21 = 6th item or, 21 = 2y + 3 or, 2y = 18 y = 9. 7. Find the first quartile from the data given below: (a) x 10 20 30 40 50 f 7 8 12 5 6 (b) Weight (kg) 32 35 40 42 50 No. of students 5 15 18 13 10 (c) Age (yrs.) 12 13 14 15 16 Frequency 6 12 19 11 6 (d) Weight (kg.) 38 42 44 48 50 No. of students 3 4 8 6 2 Solution: (a) Here, computation of Q1
368 x f cf 10 7 7 20 8 15 30 12 27 40 5 32 50 6 38 N = 38 Now, Q1 = N + 1 4 th item = 38 + 1 4 th item = 9.75th item The nearest greater value than 9.75 in cf column is 15 and the item corresponding to it is 20. Hence, Q1 = 20. (b) Here, computation of Q1 Weight (kg) (x) No. of students (f) cf 32 5 5 35 15 20 40 18 38 42 13 51 50 10 61 N = 61 Now, Q1 = N + 1 4 th item = 61 + 1 4 th item = 15.25th item The nearest greater value of 15.25 in cf column is 20 and the item corresponding to it is 35. Hence, Q1 = 35. (c) Here, computation of Q1 Age (yrs) (x) Frequency (f) cf 12 6 6 13 12 18 14 19 37 15 11 218 16 6 54 N = 54 Now, Q1 = N + 1 4 th item = 54 + 1 4 th item = 13.75th item The nearest greater value than 13.75 in cf column is 18 and the item corresponding to it is 13. Hence, Q1 = 18. (d) Here, computation of Q1 Weight (kg) (x) No. of students (f) cf 38 3 3 42 4 7 44 8 15 48 6 21 50 2 23 N = 23 Now, Q1 = N + 1 4 th item = 23 + 1 4 th item = 6th item
369 The nearest greater value of 6 in cf column is 7 and the item corresponding to it is 42. Hence, Q1 = 42. 8. Find the third quartile from the data given below: (a) Age (yrs.) 12 13 14 15 16 Frequency 6 12 19 11 6 (b) Weight (kg.) 38 42 44 48 50 No. of students 3 4 8 6 2 Solution: (a) Here, computation of Q3 Age (yrs.) (x) Frequency (f) cf 12 6 6 13 12 18 14 19 37 15 11 41 16 6 47 N = 47 Now, Q3 = 3(N + 1) 4 th item = 3(47 + 1) 4 th item = 36th item The nearest greater value of 36 in cf column is 37 and the item corresponding to it is 14. Hence, Q3 = 14. (b) Here, computation of Q3 Weight (kg.) (x) No. of students (f) cf 38 3 3 42 4 7 44 8 15 48 6 21 50 2 23 N = 23 Now, Q3 = 3(N + 1) 4 th item = 3(23 + 1) 4 th item = 18th item The nearest greater value of 18 in cf column is 21 and the item corresponding to it is 48. Hence, Q3 = 48. 9. Find the quartiles (Q1 and Q3) from the data given below: (a) x 50 60 70 80 90 99 f 15 11 14 5 3 1 (b) Wages 45 55 65 75 85 No. of workers 24 23 31 12 23 Solution: (a) Here, computation of Q1 and Q3
370 x f cf 50 15 15 60 11 26 70 14 40 80 5 45 90 3 48 99 1 49 N = 49 Now, Q1 = N + 1 4 th item = 49 + 1 4 th item = 12.5th item The nearest greater value of 12.5 in cf column is 15 and its corresponding item is 50. Q1 = 50. Again, Q3 = 3(N + 1) 4 th item = 3(49 + 1) 4 th item = 37.5th item The nearest greater value of 37.5 in cf column is 30 and its corresponding item is 70. Hence, Q3 = 70. (b) Here, computation of Q1 and Q3 x f cf 45 24 24 55 23 47 65 31 78 75 12 90 85 23 113 N = 113 Now, Q1 = N + 1 4 th item = 113 + 1 4 th item = 28.5th item The nearest greater value of 28.5 in cf column is 47 and its corresponding item is 55. Q1 = 55. Again, Q3 = 3(N + 1) 4 th item = 3(113 + 1) 4 th item = 85.5th item The nearest greater value of 85.5 in cf column is 90 and its corresponding item is 75. Hence, Q3 = 75.
371 18.4 Mode PRACTICE 18.4 Read / Understand / Think / Do Keeping Skill Sharp 1. (a) Define mode. (b) How is the mode of an individual data found? (c) How is the mode of a discrete data found? Solution: (a) Mode: The item or value of observation which occurs mostly in data is called Mode. (b) The mostly repeated item is regarded as mode in individual series/data. (c) The item corresponding to the highest frequency is regarded as mode in descrete series/data. 2. (a) What is the mode of the data: 3, 3, 4, 3, 4, 6, 7? (b) What is the mode of the data: 1, 1, 1, 1, 1, 0, 0, 0? Solution: (a) Here, 3, 3, 4, 3, 4, 6, 7 Since 3 has repeated for most times (3 times). Hence, Mo = 3. (b) Here, 1, 1, 1, 1, 1, 0, 0, 0 Since 1 has repeated for most time (5 times). Hence, Mo = 1. 3. Find the mode for the following data: (a) 34, 35, 26, 35, 34, 38, 34, 39, 34, 39, 23, 31, 30, 33, 34, 35, 34 (b) 56, 65, 75, 56, 67 56, 43, 46, 37, 83, 56, 56, 34, 29, 45, 56, 67, 75, 83 Solution: (a) Here, 34, 35, 26, 35, 34, 38, 34, 39, 34, 39, 23, 32, 30, 33, 34, 35, 34 23 has repeated for 1 time 26 has repeated for 1 time 30 has repeated for 1 time 31 has repeated for 1 time 33 has repeated for 1 time 34 has repeated for 6 times 35 has repeated for 3 times 38 has repeated for 1 time 39 has repeated for 2 times Hence, most repeated item is 34 so Mo = 34. (b) Here, 56, 65, 75, 56, 67 56, 43, 46, 37, 83, 56, 56, 34, 29, 45, 56, 67, 75, 83 29 has repeated for 1 time 34 has repeated for 1 time 37 has repeated for 1 time 43 has repeated for 1 time 45 has repeated for 1 time 46 has repeated for 1 time
372 56 has repeated for 6 times 65 has repeated for 1 time 67 has repeated for 2 times 75 has repeated for 2 times 83 has repeated for 2 times Hence, 56 has repeated for most time. So, M0 = 56. 4. (a) The mean of the given data 1, x – 3, 2x + 1, 3x + 5, 2, 2, 5, 7 is 7. Find its mode. (b) If the mean of the given data x, x + 3, 2x + 15, 3x – 1, 13, 3x + 10 is 10, find its mode. (c) The median of the given data 12, 15, 2x + 2, 2x + 3, 3x – 3, 18, 25, 27, 30 in ascending order is 18. Find its mode. (d) The data 25, 30, 35, 40, x + 10, 2x – 20, 55, 60, 65, 70 in ascending order. If its median is 40, find its mode. Solution: (a) Here, x : 1, x – 3, 2x + 1, 3x + 5, 2, 2, 5, 7 x = 7, N = 8 Now, x = x N = 1 + x – 3 + 2x + 1 + 3x + 5 + 2 + 2 + 5 + 7 8 or, 56 = 6x + 20 or, 6x = 56 – 20 or, 6x = 36 x = 6 Hence the actual items are 1, 6 – 3, 2 × 6 + 1, 3 × 6 + 5, 2, 2, 5, 7 = 1, 3, 13, 23, 2, 2, 5, 7 Which shows the most repeated item is 2. Hence, Mo = 2. (b) Here, x : x, x + 3, 2x + 15, 3x – 1, 13, 3x + 10 x = 10, N = 6. We know, x = x N = x + x + 3 + 2x + 15 + 3x – 1 + 13 + 3x + 10 6 or, 10 = 10x + 40 6 or, 10x + 40 = 60 or, 10x = 20 x = 2 The actual data is 2, 2 + 3, 2 × 2 + 15, 3 × 2 – 1, 13, 3 × 2 + 10 = 2, 5, 19, 5, 13, 16 Which shows the most repeated item is 5. Hence, Mo = 5. (c) Here, x : 12, 15, 2x + 2, 2x + 3, 3x – 3, 18, 25, 27, 30 (ascending order) Md = 18, N = 9, Mo = ? We know that, Md = N + 1 2 th item or, 18 = 9 + 1 2 th item or, 18 = 5th item or, 18 = 3x – 3 or, 3x = 21 x = 7
373 Rewriting the data in actual value: 12, 15, 2 × 7 + 2, 2 × 7 + 3, 3 × 7 – 3, 18, 25, 27, 30 i.e. 12, 15, 16, 17, 18, 18, 25, 27, 30 which shows the most repeated item as 18 Hence, Mo = 18. (d) Here, x : 25, 30, 35, 40, x + 10, 2x – 20, 55, 60, 65, 70 (ascending order) N = 10, Md = 40, Mo = ? We know that, Md = N + 1 2 th item or, 40 = 10 + 1 4 th item or, 40 = 5.5th item or, 40 = 5th item + 6th item 2 or, 40 = x + 10 + 2x – 20 2 or, 80 = 3x – 10 or, 90 = 3x x = 30 Rewriting the items with their actual values 25, 30, 35, 40, 30 + 10, 2 × 30 – 20, 55, 60, 65, 70 i.e. 25, 30, 35, 40, 40, 40, 55, 60, 65, 70 which shows the most repeated item is 40. Hence, Mo = 40. 5. (a) Find the mode for the following data: Marks 50 60 70 80 90 99 No. of students 15 11 14 5 3 1 (b) Determine the modal weight of children: Size 14 15 16 17 18 19 No. of pairs 7 5 4 9 7 8 Solution: (a) Since the highest frequency is 15 and its corresponding item is 50. Mo = 50 (b) Since the highest frequency is 9 and its corresponding item is 17. Mo = 17. 6. Find the mean and median of data. Estimate the approximate mode using Mo = 3 Median – 2 Mean. (a) 3, 4, 4, 4, 5, 2, 3, 2, 5, 6, 3, 4, 5, 3, 2, 6, 3, 4, 4, 5, 2, 1, 3, 3, 2, 3 (b) 15, 10, 12, 19, 10, 11, 18, 16, 15, 14, 15, 15, 14, 13, 15, 15, 16, 18, 13, 12 Solution: (a) 3, 4, 4, 4, 5, 2, 3, 2, 5, 6, 3, 4, 5, 3, 2, 6, 3, 4, 4, 5, 2, 1, 3, 3, 2, 3 preparing frequency table for the data. x f cf fx 1 1 1 1 2 5 6 10 3 8 14 24 4 6 20 24 5 4 24 20 6 2 26 12 N = 26 fx = 91
374 Now, x = fx N = 91 26 = 3.5 Md = N + 1 2 th item = 26 + 1 2 th item = 13.5th item The nearest greater value of 13.5 in cf column is 24 and its corresponding item is 3. Hence Md = 3 Finally estimated approximate Mo = 3Md – 2 x = 3 × 3 – 2 × 3.5 = 9 – 7 = 2 (b) 15, 10, 12, 19, 10, 11, 18, 16, 15, 14, 15, 15, 14, 13, 15, 15, 16, 18, 13, 12 preparing the frequency table for data. x f cf fx 10 2 2 20 11 1 3 11 12 2 5 24 13 2 7 26 14 2 9 28 15 6 15 90 16 2 17 32 18 2 19 36 19 1 20 19 N = 20 fx = 286 Now, x = fx N = 286 20 = 14.3 Md = N + 1 2 th item = 20 + 1 2 th item = 10.5th item The nearest greater value of 10.5 in cf column is 15 and its corresponding item is 15. Hence Md = 15 Finally estimated approximate Mo = 3Md – 2 x = 3 × 15 – 2 × 14.3 = 45 – 28.6 = 16.4 7. Find the mean and median and estimate approximate Mo using Mo = 3Md – 2 x . (a) Heights (inc.) 59 61 62 63 64 No. of persons 12 13 7 7 1 (b) Length (in cm) 12 15 18 21 24 No. of copies 3 4 9 8 1 Solution: (a) Here, computation mean and median Heights (inc.) (x) No. of persons (f) cf fx 59 12 12 708 61 13 25 793 62 7 32 434 63 7 39 441 64 1 40 14 N = 40 fx = 2440 Now, x = fx N = 2440 40 = 61
375 Md = N + 1 2 th item = 40 + 1 2 th item = 20.5th item The nearest greater value of 20.5 in cf column is 25 and its corresponding item is 61. Hence, Md = 61 Finally estimated approximate Mo = 3Md – 2 x = 3 × 61 – 2 × 61 = 183 – 122 = 61 (b) Here, computation mean and median Length (in cm) (x) No. of copies (f) cf fx 12 3 3 36 15 4 7 60 18 9 16 162 21 8 24 168 24 1 25 24 N = 25 fx = 450 Now, x = fx N = 450 25 = 18 Md = N + 1 2 th item = 25 + 1 2 th item = 13th item The nearest greater value in of column of 13 is 16 and its corresponding item is 18. Hence, Md = 18 Finally estimated approximate Mo = 3Md – 2 x = 3 × 18 – 2 × 18 = 54 – 36 = 18
376 CHAPTER PROBABILITY 19 19.1 Introduction to Probability PRACTICE 19.1 Read / Understand / Think / Do Keeping Skill Sharp 1. State whether the following is an experiment or not: (a) The earth is round. (b) Acid turns blue litmus paper into red. (c) The number of rotten apples in a box of apples. Solution: (a) No (b) Yes (c) Yes 2. Which of the following are random experiments? (a) Drawing a red ball from an urn containing 5 red balls and 4 blue balls. (b) Choosing a vowel from the set of the English alphabet. (c) Electrolysis of acidified water. Solution: (a) Yes (b) Yes (c) No. 3. In a single throw of a die, the sample space is S = {1, 2 ,3, 4, 5, 6}. What are the sets representing the events: (a) The event of getting a prime number? (b) The event of getting multiples of 3? (c) The sure event? Solution: (a) {2, 3, 5} (b) {3, 6} (c) {1, 2, 3, 4, 5, 6} 4. Suppose a biased or unfair die is rolled. Will all the outcomes be equally likely? Solution: Yes. 5. What is the number of favourable cases for? (a) an event E1 ={2, 4, 6) of getting an even number in a single throw of a die ? (b) an event E of drawing a black card from a pack of 52 cards? (c) an event E of getting a total of 10 in a simultaneous throw of two dice? Solution: (a) n(E1) = 3 (b) n(E) = 26 (c) n(E) = 3 6. In an experiment of throwing a single die, the sample space is S = {1, 2, 3, 4, 5, 6}. State whether the following pairs of sets E ={1, 2, 3) and F = {4, 5, 6} are true or false. (a) Complementary to each other; (b) Mutually exclusive; (c) Both complementary and mutually exclusive. Solution: (a) Yes (b) Yes (c) Yes
377 19.2 Empirical and Classical Probabilities PRACTICE 19.2 Read / Understand / Think / Do Keeping Skill Sharp 1. A coin is tossed twice. List the sample space. What is the probability that; (a) Both are tails? (b) Getting one head and one tail (c) Getting head at first and tail at second (d) Getting no head (e) At least one of the two is a head? Solution: Here, S = {HH, HT, TH, TT} n(S) = 4 (a) E = Both are tails = {TT} n(E) = 1 P(E) = n(E) n(S) = 1 4 (b) E = Getting one Head and one Tail = {HT, TH} n(E) = 2 P(E) = n(E) n(S) = 2 4 = 1 2 (c) E = Getting Head a first and Tail of second = {HT} n(E) = 1 P(E) = n(E) n(S) = 1 4 (d) E = Getting No Head = {TT} n(E) = 1 P(E) = n(E) n(S) = 1 4 (e) E = At least one of two is a head = {HH, HT, TH} n(E) = 3 P(E) = n(E) n(S) = 3 4 2. Three coins are thrown. List the sample space. What is the probability of getting; (a) 2 heads and 1 tail? (b) exactly 1 head and 2 tails? (c) at least 1 head and 1 tail? Solution: Here, S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT} n(S) = 8 (a) E = 2 Heads and 1 tail = {HHT, HTH, THH} n(E) = 3 P(E) = n(E) n(S) = 3 8 (b) E = exactly 1 Head and 2 Tail = {TTH, THT, HTT} n(E) = 3 P(E) = n(E) n(S) = 3 8 (c) E = at least one head and 1 tail = {HHT, HTH, HTT, THH, THT, TTH} n(E) = 6 P(E) = n(E) n(S) = 6 8 = 3 4 3. A fair dice is tossed. List the sample space. Find the probability of the event; (a) getting 2 (b) getting 2 or 6 (c) getting an even number (d) getting 1 or 4 or 6
378 (e) getting more then 6 (f) getting more than 4 (g) a number greater or equal to 2 (h) not getting 6 (i) factors of 6 (j) prime numbers Solution: Here, S = {1, 2, 3, 4, 5, 6} n(S) = 6 (a) E = getting 2 = {2} n(E) = 1 P(E) = n(E) n(S) = 1 6 (b) E = getting 2 or 6 = {2, 6} n(E) = 2 P(E) = n(E) n(S) = 2 6 = 1 3 (c) E = getting an even number = {2, 4, 6} n(E) = 3 P(E) = n(E) n(S) = 3 6 = 1 2 (d) E = getting 1 or 4 or 6 = {1, 4, 6} n(E) = 3 P(E) = n(E) n(S) = 3 6 = 1 2 (e) E = getting more than 6 = { } n(E) = 0 P(E) = n(E) n(S) = 0 6 = 0 (f) E = getting more than 4 = {5, 6} n(E) = 2 P(E) = n(E) n(S) = 2 6 = 1 3 (g) E = a number greater or equal 2 = {2, 3, 4, 5, 6} n(E) = 5 P(E) = n(E) n(S) = 5 6 (h) E = not getting 6 = {1, 2, 3, 4, 5} n(E) = 5 P(E) = n(E) n(S) = 5 6 (i) E = Factors of 6 = {1, 2, 3, 6} n(E) = 4 P(E) = n(E) n(S) = 4 6 = 2 3 (j) E = Prime numbers = {2, 3, 5} n(E) = 3 P(E) = n(E) n(S) = 3 6 = 1 2 4. Two dice are thrown simultaneously. List the sample space. Find the probability of; (a) 1 on the first die (b) showing both die the same number, (c) getting a total of less than 7 (d) getting a total of 7 Solution: Here, S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)} n(s) = 36 (a) E = 1 on first die = {(1, 1) (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)}, n(E) = 6 P(E) = n(E) n(S) = 6 36 = 1 6
379 (b) E = both die the same number = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)} n(E) = 6 P(E) = n(E) n(S) = 6 36 = 1 6 (c) E = getting a total of less than 7 = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (2, 1), (2, 2), (2, 3), (2, 4), (3, 1), (3, 2), (3, 3), (4, 1), (4, 2), (5, 1)} n(E) = 15 P(E) = n(E) n(S) = 15 36 = 5 12 (d) E = getting a total of 7 = {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)} n(E) = 6 P(E) = n(E) n(S) = 6 36 = 1 6 5. One card is drawn from a pack of 52 cards. Find the probability of getting; (a) a black card, (b) a spade, (c) a Queen, (d) a card with 8, (e) a face card, (f) a black coloured king, (g) a Jack of heart, (h) no red ace, (i) a queen or a king, (j) a ace or red king. Solution: Here, S = {playing cards} n(S) = 52 (a) B = a black card n(B) = 26 P(B) = n(B) n(S) = 26 52 = 1 2 (b) E = a spade n(E) = 13 P(E) = n(E) n(S) = 13 52 = 1 4 (c) Q = a queen n(Q) = 4 P(Q) = n(Q) n(S) = 4 52 = 1 13 (d) E = a card with 8 n(E) = 4 P(E) = n(E) n(S) = 4 52 = 1 13 (e) F = a face card n(F) = 12 P(F) = n(F) n(S) = 12 52 = 3 13 (f) BK = a black coloured king n(Bk) = 2 P(Bk) = n(Bk) n(S) = 2 52 = 1 26 (g) JH = a Jack of Heart n(JH) = 1 P(JH) = n(JH) n(S) = 1 52 (h) R = No red card n( R ) = 26
380 P( R ) = n( R ) n(S) = 26 52 = 1 2 (i) Q K = a queen or a king = n(Q K) = 8 P(Q K) = n(Q K) n(S) = 8 52 = 2 13 (j) A RK = a ace or Red King n(A RK) = 6 P(A RK) = n(A RK) n(S) = 6 52 = 3 26 6. (a) What is the probability that a pregnant women giving birth to a child is on Thursday of a week ? (b) Find the probability of giving birth to a baby by a pregnant woman on Tuesday or Saturday. (c) What is the probability of born two daughters when a married woman gives birth on planning of two children? Find it. (d) When a couple gives birth to three children, find the probability of giving birth the same sex. Solution: (a) S = {Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday} n(S) = 7 E = giving birth to a child on Thursday n(E) = 1 P(E) = n(E) n(S) = 1 7 (b) S = {Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday} E = {giving birth to a chid on Tuesday or Saturday} n(E) = 2 P(E) = n(E) n(S) = 2 7 (c) S = {SS, SD, DS, DD} n(S) = 4 DD = {both daughter} N9DD) = 1 P(DD) = n(DD) n(S) = 1 4 (d) S = {SSS, SSD, SDS, SDD, DSS, DSD, DDS, DD} n(S) = 8 E = same sex = {SSS, DDD} n(E) = 2 P(E) = n(F) n(S) = 2 8 = 2 8 = 1 4 7. A bag contains 5 white balls, 4 black balls and 3 red balls. What is the probability that the ball drawn is; (a) white? (b) black? (c) red? (d) black or red? Solution: A bag contains 5 white balls, 4 black balls, 3 red balls i.e. n(S) = 12 (a) W = white balls n(W) = 5 P(W) = n(W) n(S) = 5 12 (b) B = black balls n(B) = 4
381 P(B) = n(B) n(S) = 4 12 = 1 3 (c) R = red balls n(R) = 3 P(R) = n(R) n(S) = 3 12 = 1 4 (d) B R = black or red balls n(B R) = 7 P(B R) = n(B R) n(S) = 7 12 8. Tickets numbered form 1 to 30 are mixed up together and then a ticket is drawn at random. What is the probability that the ticket has a number; (a) divisible by 4, (b) a prime number. Solution: Here, {1, 2, 3, ..., 30} n(S) = 30 (a) D4 = divisible by 4 = {4, 6, 12, 16, 20, 24, 28} 4(D4) = 7 P(D4) = n(D4) n(S) = 7 30 (b) PR = prime number = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29} n(PR) = 10 P(PR) = n(PR) n(S) = 10 30 = 1 3 9. While rotating the needle on the given spinner as in the figure, find the probability of the needle stopping at 1. (a) 1 (b) odd number Solution: Here, S = {1, 2, 3, 4} n(S) = 4 (a) E = {1} n(E) = 1 P(E) = n(E) n(S) = 1 4 (b) O = odd number = {1, 3} n(O) = 2 P(O) = n(O) n(S) = 2 4 = 1 2 10. (a) A stamping machine turns out 500 parts per hour. Experience has shows that there are approximately 12 defective parts per hour. Find the probability that a part picked at random will be defective. (b) The students of a school planted 1500 saplings of mango. If the probability of survival is 0.8, find the total number of saplings that survived. Solution: (a) Here, S = sample space n(S) = 500 D = defective part n(D) = 12 P(D) = n(D) n(S) = 12 500 = 3 125 (b) Here, S = sample space = total sapling n(S) = 1500 E = survival of sapling P(E) = 0.8 n(E) = ?
382 We know that, P(E) = n(E) n(S) or, 0.8 = n(E) 1500 or, n(E) = 0.8 × 1500 or, n(E) = 1200 11. A die is thrown 500 times. Suppose E denotes the occurrence of one of numbers 1, 2, 3, 4, 5, 6 representing the dots on the face of the die. The result is shown in the following table: Events (E) 1 2 3 4 5 6 Frequency n(E) 75 83 80 92 90 80 Find the probability of getting; (a) an odd number (b) a number less than 3 (c) a number greater than 3. Solution: Here, n(S) = 500 (a) O = odd numbers n(O) = 75 + 80 + 90 = 245 P(O) = n(O) n(S) = 245 500 = 49 100 (b) T = a number less than 3 n(T) = 75 + 83 = 158 P(T) = n(T) n(S) = 158 500 = 79 250 (c) T = a number greater than 3 n(T) = 92 + 90 + 80 = 262 P(T) = n(T) n(S) = 162 500 = 131 250 12. (a) Out of 1000 mangoes the probability of selecting of rotten mango is 0.15. What is the number of good mangoes? Find. (b) Out of 1500 students in a school, the probability of success of girl student is 0.45. Find the number of boys students. Solution: Here, n(S) = 1500 Probability of success of girl (P(G)) = 0.45 Probability of success of boy (P(B)) = ? Since P(B) + P(G) = 1 as complementary. or, P(B) + 0.45 = 1 or, P(B) = 0.55 Now, P(B) = n(B) n(S) 0.55 = n(B) 1500 n(B) = 825 1850 "Alternative method" n(S) = 1500, P(G) = 0.45, n(G) = ? We know, P(G) = n(G) n(S) or, 0.45 = n(G) 1500 or, n(G) = 675 Hence, n(B) = 1500 – 675 = 825
383 13. (a) If P(A) = 1 3 and P(B) = 2 5 , find, (i) P( A ) (ii) P( B ) (iii) P(A) × P( B ) (iv) P( A ) × P(B) (b) If P(A) = x 2 and P( A ) = 3 4 , find the value of x. Solution: (a) P(A) = 1 3 and P(B) = 2 5 (i) P( A ) = 1 – P(A) = 1 – 1 3 = 2 3 (ii) P( B ) = 1 – P(B) = 1 – 2 5 = 3 5 (iii) P(A) × P( B ) = 1 3 × 3 5 = 1 5 (iv) P( A ) × P(B) = 2 3 × 2 5 = 4 15 (b) Here, P(A) = x 2 , P( A ) = 3 4 To find: x = ? We know that, P(A) + P( A ) = 1 or, x 2 + 3 4 = 1 or, x 2 = 1 – 3 4 or, x 2 = 1 4 or, x = 2 4 x = 1 2
Add 1. T ( ( ( ( Soluti ( ( ( ( 2. T ( ( ( ( Soluti ( ( ( ditional Prac The marks obtai Marks No. of students (a) In which se (b) Construct a (c) What is the (d) What is the ion: (a) The given d (b) Cumulative (c) Median (Md The nearest Md = 5 (d) The average 60 × 4 4 The given data s (a) How many in class inte (b) How many Write in cla (c) Plot the ge frequency data. (d) How much Calculate it ion: (a) 5 students ar (b) 20 students (c) Representin x f ctice – VI ined by 20 stude 20 3 eries, the given d a cumulative fre e median marks e average marks ata is presented i frequency table, x 20 30 40 50 60 70 d) = N + 1 2 th item greater value of 50. e marks of the stu + 70 × 4 + 4 = 240 + 8 shows the marks students are get erval. y students are g ass interval. etting marks an polygon by re h is the averag t.? re getting most m are getting low m g the continuous 10 - 15 20 384 ents in full mark 30 4 data is presented equency table fro of the class 9 stu of the students in discrete series. N m = 22 + 1 2 th ite 11.5 in cf column udents getting mo + 280 8 = 65 s of 105 students tting the most m getting in the lo nd number of st epresenting the ge marks of t marks between 35 marks between 10 data and cf table 15 - 20 20 25 3 ks 75 is given bel 40 50 3 2 d? om the given dat udents? Find it. getting more th f 3 4 3 2 4 6 N = 22 em = 11.5th item n is 12 and its cor ore than median s s recorded in the marks? Write ow interval? tudents on a e continuous he students. 5-40. 0-15. e, - 25 25 - 30 30 15 low. 0 60 4 ta. an median score 2 rresponding item score (50) is e form of a table 0 30 - 35 10 70 4 e? Calculate it. cf 3 7 10 12 16 22 m = 50 e. 35 - 40 5
385 Constricting the ogive curve, (d) Computation of average marks. (Taking mid value). x f fx 12.5 20 250 17.5 25 437.5 22.5 30 675 27.5 15 412.5 32.5 10 325 37.5 9 187.5 N = 105 fx = 2287.5 Average ( x ) = fx N = 2287.5 105 = 21.79. 3. The time spent on the studying at home by 20 students is given in the table below. Times in hour 1 2 3 4 5 No. of students 8 6 2 3 1 (a) Write the model value of the data. (b) Find the average time spent of the students. (c) Find the median of the given data. (d) If the students who study either 1 hour or 2 hours are removed from the above data, what would be the average study time? Calculate. Solution: (a) Since the highest frequency is 8 and corresponding item to it is 1. Modal value = 1 hour. (b) Calculating the average x f fx cf 1 8 8 8 2 6 12 14 3 2 6 16 4 3 12 19 5 1 5 20 N = 20 fx = 43 Now, Average ( x ) = fx N = 43 20 = 2.15 hours. 35 30 25 20 15 10 5 0 10 20 25 30 35 40 Class interval Marks obtained 15
386 (c) (Md) = N + 1 2 th item = 20 + 1 2 th item = 10.5th item The nearest greater value of 10.5 in cf column is 14 and its corresponding item is 2. Md = 2 hours. (d) After removing the students studying 1 hr. and 2 hrs. x f fx 3 2 6 4 3 12 5 1 5 N = 6 fx = 23 Average ( x ) = fx N = 23 6 = 3.83 hours. 4. The heights of trees in a jungle are given below. Height (meter) 0 - 10 10 - 20 20 - 30 30 - 40 40 - 50 50 - 60 No. of plants 5 15 20 18 12 7 (a) In which class intervals is the greatest frequency? (b) Construct a cumulative frequency table from the given data. (c) Construct the histogram of the given data. (d) Prove that the probability of getting the height of tree more than 50 meters is 1 11 . Solution: (a) In 20 - 30 class interval the greatest frequency is (b) Cumulative frequency table height (m) (x) No. of plants (f) cf 0 - 10 5 5 10 - 20 15 20 20 - 30 20 40 30 - 40 18 58 40 - 50 12 70 50 - 60 7 77 N = 77 (c) Constructing the histogram from the above data, 22 20 18 16 14 12 10 8 6 4 2 0 10 30 40 50 60 70 Height (m) 20
( 5. H w ( ( ( ( Soluti ( ( ( ( 6. T ( ( ( ( Soluti ( (d) Here, n(S) = E = get n(E) = P(E) = Here are given written from 1 to (a) Write the fo probability (b) Find is th getting the (c) What is the (d) In which co ion: (a) P(E) = n(E) n(S) (b) E = getting t Also, n(S) = P(E) = (c) E = gettin = {1, 2 Also, n(S) = P(E) = Alternatively E = = Also, n(S) = P(E) = Hence P( E (d) In the certai number of fa i.e. n(E) = So, P(E) = The marks of 60 Marks No. of students (a) Write the m (b) Construct a (c) Construct t (d) Find the pr ion: (a) The model c = 77 tting the height m 7 n(E) n(S) = 7 77 = 1 11 , 20 number car o 20. formula for findi P(E). he probability number card wh e probability of g onditions can get , n(E) = cardinal the number card = 20 n(E) n(S) = 6 20 = 3 10 ng the number ca 2, 3, 4, 6, 7, 8, 9, = 20 n(E) n(S) = 16 20 = 4 5 ly, = getting the numb = {5, 10, 15, 20} = 20 n(E) n(S) = 4 20 = 1 5 ) = P(not divisib in case we can g favourable cases i n(S) n(E) n(S) = n(E) n(F) = 1 0 students out of 0 - 10 5 model class of th a cumulative fre the less than ogiv robability of gett class of given dat 387 more than 50 m proved. rds ing of hich is exactly d getting the numb t maximum valu lity of event and n exactly divisible ard not divisible b 11, 12, 13, 14, 16 ber card divisible n(E) = 4 ble by 5) = 1 – P( get maximum va is equal to numbe f full marks 75 ar 10 - 20 2 15 e given data. equency table fro ve from the give ting the marks le ta is 20 - 30 divisible by 3. ber card NOT d ue of probability n(S) = sample ca by 3 = {3, 6, 9, 1 by 5 6, 17, 18 19} n e by 5 (E) = 1 – 1 5 = 4 5 alue of probabilit er of sample case re as given below 0 - 30 30 - 4 20 18 om the given dat en data. ess than 20. divisible by 5? Fi y? Write with re se 12, 15, 18} n(E n(E) = 16 ty as 1, because es. w. 40 40 - 50 12 ta. ind it. ason. E) = 6 in such situation 50 - 60 7
388 (b) Cumulative frequency table, x f cf 0 - 10 5 5 10 - 20 15 20 20 - 30 20 40 30 - 40 18 58 40 - 50 12 70 50 - 60 7 77 N = 77 (c) Construction less than cf table, x f less than cf 0 - 10 5 10 5 10 - 20 15 20 20 20 - 30 20 30 40 30 - 40 18 40 58 40 - 50 12 50 70 50 - 60 7 60 77 N = 77 Constructing ogive from the above table, (d) Here, E = getting marks less than 20 n(E) = 20 n(S) = 77 Hence, P(E) = n(E) n(S) = 20 77 7. In the figure, a cubical die with its six faces is shown. (a) Write down the sample space for a die rolled. (b) Find the probability of getting 3 on the face. (c) Find the probability of NOT getting odd number on the faces of die. (d) In which conditions can get minimum value of probability? Write with example. Solution: (a) S = {1, 2, 3, 4, 5, 6} n(S) (b) E = getting 3 = {3} n(E) = 1 P(E) = n(E) n(S) = 1 6 (c) E = getting not odd number = {2, 4, 6} n(E) = 3 80 70 60 50 40 30 20 10 0 10 30 40 50 60 70 Marks 20
389 P(E) = n(E) n(S) = 3 6 = 1 2 (d) In impossible case, the minimum value (i.e. 0) can be get. For example: Probability of getting more than 8 in dice rolling is impossible and P(E) = 0. 8. Three coins are tossed simultaneously at random. (a) Write down its sample space. (b) Find the probability of getting all the heads. (c) Find the probability of getting at least one head. (d) Is the probability of getting all the heads and all the tails equal? Justify. Solution: (a) S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT} n(S) = 8 (b) E = getting all heads = {HHH} n(E) = 1 P(E) = n(E) n(S) = 1 8 (c) E = at least one Head = {HHH, HHT, HTH, HTT, THH, THT, TTH} n(E) = 7 P(E) = n(E) n(S) = 7 8 (d) Since P(HHH) = 1 8 Also, P(TTT) = n(TTT) n(S) = 1 8 Hence, They are equal. 9. In a pack of cards, there are 52 playing cards as shown in the figure. (a) Write the probability scale of the events. (b) A card is drawn randomly from the well shuffled cards. What is the probability of getting the face card? (c) One card is drawn randomly from the well shuffled cards. What is the probability of NOT getting a spade? (d) In which conditions can we get maximum value of probability? Write with reason. Solution: (a) 1 13 1 4 1 2 3 4 0 1
390 (b) F = face card n(F) = 12 n(S) = 52 Hence P(F) = n(F) n(S) = 12 52 = 3 13 (c) E = getting spade n(F) = 13 n(S) = 52 Hence, P(E) = n(E) n(S) = 13 52 = 1 4 So, P( E ) = probability of not getting spade = 1 – P(E) = 1 – 1 4 = 3 4 (d) In sure case we get maximum value of probability as 1 because in sure case favourable cases and sample cases will be same. 10. A bag contains 5 red TT balls, 7 white TT balls and 3 black TT balls having the same size. (a) Write the formula for finding the probability of the events. (b) A ball is drawn randomly from the bag. Find the probability getting white balls. (c) A ball is drawn randomly from the bag. What is the probability NOT getting black balls? Calculate it. (d) A marble is drawn randomly from the bag. What is the probability of getting either black or red or white balls? Is it the maximum probability of such events? Solution: (a) n(E) = number of favourable cases to event E. n(S) = number of sample cases. the probability of event E = P(E) = n(E) n(S) (b) W = white ball n(W) = 7 and n(S) = 15 Hence, P(W) = n(W) n(S) = 7 15 (c) B = black balls n(B) = 3 and n(S) = 15 P(B) = n(B) n(S) = 3 15 = 1 5 Hence, P( B ) = probability of not getting black balls = 1 – P(B) = 1 – 1 5 = 4 5 (d) E = either black or red or white n(E) = 15 and n(S) = 15 P(E) = n(E) n(S) = 15 15 = 1 Yes, it is the maximum probability as it is sure/certain case.
391 UNIT VII TRIGONOMETRY CHAPTER TRIGONOMETRY 20 20.1 Trigonometric Ratios PRACTICE 20.1 Read / Understand / Think / Do Keeping Skill Sharp 1. Given the triangle ABC, B = 90°, (a) Which side is the base for the angle A? (b) Which side is the base for the angle C? (c) Which side is the base for the angle B? Solution: (a) AB (b) BC (c) None 2. (i) ΔPQR is a right-angled triangle at Q. In each of the following, write down the following trigonometric ratios: sin P, cos R and tan P. (a) PQ = QR = 1, PR = 2 (b) PQ = 5, QR = 12, PR = 13 (c) PQ = 2, PR = 4, QR = 2 3 (ii) Given ΔLMN, ∠ M = 90°, (a) sin N = 15 17 , h = 5; find p. (b) cos N = 8 17 , b = 3; find h. (c) tan L = 8 15 , p = 4; find b. Solution: (i) (a) Here, PQ = 1 unit QR = 1 unit PR = 2 unit Now, sin P = p h = QR PR = 1 2 cos R = b h = PQ PR = 1 2
392 tan P = p b = QR PQ = 1 1 = 1 (b) Here, PQ = 5 units QR = 12 units PR = 13 units Now, sin P = p h = PQ PR = 12 13 cos R = b h = QR PR = 12 13 tan P = p b = PQ QR = 12 5 (c) Here, PQ = 2 units QR = 4 units PR = 2 3 units Now, sin P = p h = QR PR = 2 3 4 = 3 2 cos R = b h = PQ PR = 2 4 = 1 2 tan P = p b = QR PQ = 2 3 2 = 3 (ii) (a) Here, sin N = 15 17 , h = 5 To find p = ? we know that, sin N = p h or, 15 17 = p 5 or, p = 75 17 (b) Here, cos N = 8 17 , b = 3 To find: h = ? we know that, cos N = b h or, 8 17 = 3 h or, h = 3 × 17 8 = 51 8 (c) Here, tan L = 8 15 , p = 4 To find b = ? we know that, tan L = p b or, 8 15 = 4 b or, b = 4 × 15 8 or, b = 15 2
393 3. From the following figures, find the values of sin θ, sin , cos and tan θ: (a) (b) (c) (d) Solution: (a) For reference angle For reference angle p = DA, b = AB, h = BD p = BD, b = CD, h = BC Hence, sin = p h = DA BD Hence, sin = p h = BD BC tan = p b = DA AB cos = b h = CD BC (b) For reference angle For reference angle p = QR, b = PQ, h = PR p = PR, b = SR, h = PS Hence, sin = p h = QR PR Hence, sin = p h = PR PS tan = p b = QR PQ cos = b h = SR PS (c) For reference angle For reference angle p = KN, b = MK, h = NM p = LM, b = KL, h = KM Hence, sin = p h = KN NM Hence, sin = p h = LM KM tan = p b = KN MK cos = b h = KL KM (d) For reference angle For reference angle p = YZ, b = ZX, h = XY p = WX, b = YW, h = XY Hence, sin = p h = YZ XY Hence, sin = p h = WX XY tan = p b = YZ ZX cos = b h = YW XY 4. (a) If sin θ = 4 5 , find the values of cos and tan . (b) If cos A = 5 13 , find the values of sin A and tan A. (c) If tan A = 12 5 , find the values of sin A and cos A. Solution: (a) Here, sin = 4 5 i.e. p h = 4 5 p = 4, h = 5 by Pythagoras theorem, h2 = p2 + b2
394 or, 52 = 42 + b2 or, 25 – 16 = b2 or, b2 = 9 or, b = 3 Hence, cos = b h = 3 5 tan = p b = 4 3 (b) Here, cos A = 5 13 i.e. b h = 5 13 b = 5, h = 13 by Pythagoras theorem, h2 = p2 + b2 or, 132 = p2 + 52 or, 169 = p2 + 25 or, 144 = p2 or, p = 12 Hence, sin A = p h = 12 13 tan A = p b = 12 5 (c) Here, tan A = 12 5 i.e. p b = 12 5 p = 12, b = 5 by Pythagoras theorem, h2 = p2 + b2 or, h2 = 122 + 52 = 144 + 25 = 169 h = 13 Hence, sin A = p h = 12 13 cos A = b h = 5 13 5. Find the value of; (a) sin2 A + cos2 A (b) 1 – cos2 θ (c) sin A ÷ cosA (d) sec2 θ – tan2 θ (e) cosec2 A – cot2 A (f) 1 + tan2 θ Solution: (a) sin2 A + cos2 A = p h 2 + b h 2 = p2 h2 + b2 h2 = p2 + b2 h2 = h2 h2 [ h2 = p2 + b2 ] = 1
395 (b) 1 – cos2 θ = 1 – b h 2 = 1 – b2 h2 = h2 – b2 h2 = p2 h2 [ h2 = p2 + b2 ] = p h 2 = sin2 (c) sin A ÷ cosA = p h ÷ b h = p h × h b = p b = tan A (d) sec2 θ – tan2 θ = h b 2 – p b 2 = h2 b2 – p2 b2 = h2 –p2 b2 = b2 b2 [ h2 = p2 + b2 ] = 1 (e) cosec2 A – cot2 A = h p 2 – b p 2 = h2 p2 – b2 p2 = h2 – b2 p2 = p2 p2 [ h2 = p2 + b2 ] = 1 (f) 1 + tan2 θ = 1 + p b 2 = 1 + p2 b2 = b2 + p2 b2 = h2 b2 [h2 = p2 + b2 ] = h b 2 = sec2 θ 6. Prove that: (a) sec2 A – tan2 A = 1 (b) 1 – sin2 θ = cos2 θ (c) cot A = cos A sin A (d) sec2 θ – 1 = tan2 θ (e) cosec A . sin A = 1 (f) cos A = 1 sec A Solution: (a) Here, sec2 A – tan2 A = 1 LHS = sec2 A – tan2 A = h b 2 – p b 2 = h2 b2 – p2 b2 = h2 – p2 b2 = b2 b2 [ h2 = p2 + b2 ] = 1 = RHS, proved. (b) Here, 1 – sin2 θ = cos2 θ LHS = 1 – sin2 θ = 1 – p h 2 = 1 – p2 h2 = h2 – p2 h2 = b2 h2 [ h2 = p2 + b2 ] = b h 2 = (cos) 2 = cos2 = RHS, proved.
396 (c) Here, cot A = cos A sin A LHS = cos A = b p = b h p h = cos A sin A = RHS, proved. (d) Here, sec2 θ – 1 = tan2 θ LHS = sec2 θ – 1 = h b 2 – 1 = h2 b2 – 1 = h2 – b2 b2 = p2 b2 [ h2 = p2 + b2 ] = p b 2 = (tan) 2 = tan2 = RHS, proved. (e) Here, cosec A . sin A = 1 LHS = cosec A . sin A = h p × p h = 1 = RHS, proved. (f) Here, cos A = 1 sec A LHS = cos A = b h = b b h b = 1 sec A = RHS, proved. 7. If sin A = 4 5 , prove that: (a) cosec2 A – cot2 A = 1 (b) 1 – cos2 A = sin2 A (c) 1 + tan2 A = sec2 A Solution: Here, sin A = 4 5 i.e. p h = 4 5 p = 4, h = 5 by Pythagoras theorem, h2 = p2 + b2 or, 52 = 42 + b2 or, 25 = 16 + b2 or, b2 = 9 or, b = 3