Oasis School Mathematics-8 247 Example 3 Write in three digits bearing. (a) N60° E (b) S 30° W Solution: From the figure, S30°W = (180°+ 30°) N 60° E = 060° = 210° Example 4 If the bearing of B from A is 60°, what is the bearing of A from B? Solution: Here, The bearing of B from A is 060°. i.e. ∠NAB = 60° Since, NA||N'B ∠NAB + ∠N'BA = 180° [∵ Sum of co-interior angles is 180°] 60° + ∠N'BA = 180° or, ∠N'BA = 180° – 60° = 120° ∴ The bearing of A from B = reflex ∠N'BA = (360° – 120°) = 240° Exercise 21.1 1. Write down the compass bearing of the following figure: 2. Draw the diagrams to show each of the following bearings. (a) N50°E (b) S 30°E (c) S 60°W (d) N 80°W N S W O 300 N A O 600 A 600 B N' N 300 N S A B S A 450 O 600 O N S W SW S N E NW N E S W (a) (b) (c) (d) (e) C O O O
248 Oasis School Mathematics-8 3. From the given figure, write down the three digit bearing of NE, E, SE, S, SW, W, and NW. 4. Write the three digit bearing of given points from O. 5. Sketch the following bearing. (a) 075° (b) 145° (c) 215° (d) 305° 6. Write the three digit bearing of: (a) N60°E (b) S70°W (c) S50°E (d) N30°W 7. In each ofthe given figures, bearing ofBfrom A is given. Find the bearing of A from B. 8. From the given map of Nepal, using three digit bearing, write down the bearing of: (a) Kathmandu from Pokhara. (b) Gorkha from Nepalganj. (c) Jumla from Dhangadi. (d) Biratnagar from Janakpur. O E SW SE NW NE S N W N E O 1500 O N D 400 O N C O N B 1600 A O (a) N (b) (c) (d) (e) 800 N A N N N N A B A B B B A 2300 3100 400 1150 N' N' N' (a) (b) (c) (d)
Oasis School Mathematics-8 249 Answer 1. (a) S300 E (b) S450 W (c) N600 W (d) S450 W (e) N450 W 2. Consult your teacher. 3. 0450 , 0900 , 1350 , 1800 , 2250 , 2700 , 3150 4. (a) 0800 , (b) 2000 , (c) 3200 (d) 2700 (e) 1500 5. Consult your teacher. 6. (a) 0600 , (b) 2500 (c) 1300 , (d) 3300 7. (a) 2950 (b) 2200 (c) 500 (d) 1300 8. Consult your teacher 21.2 Scale Drawing The figures or maps can be drawn by using different scales. It is impossible to show lengths and distance which are too great or too small in a sheet of paper. In this case, we have to enlarge smaller figure and reduce the larger figure by taking a suitable scale. Scale factor In the given fig. (i) and (ii), we can see that each side of second figure is twice that of first figure. Here each side of first figure is enlarged twice. ∴ AB A'B' = 3 cm 6 cm = 1 2 or, BC B'C' = 3 cm 6 cm = 1 2 or, AC A'C' = 3 cm 6 cm = 1 2 Thus, first figure is the scale drawing of second figure to the scale 1:2. Thus, the ratio 1:2 is called the scale factor. Note : The scale factor 1:100 means 1 unit length in drawing is 100 units of length in actual size. i.e. Length in drawing Actual length = 1 100 = scale factor. A A C 3 cm Figure (i) 3 cm 3 cm Figure (ii) 6 cm 6 cm 6 cm C' A' B'
250 Oasis School Mathematics-8 Worked Out Examples Example 1 What is the actual distance between two places which is represented by 1.5 cm on a map which is drawn to the scale 1:60,00,000? Solution: Here, Scale = 1: 6000000 i.e. 1 cm on a map represents 6000000 cm. = 6000000 1000 m = 60000 100 m. = 60 km. ∴ 1.5 cm on a map represents = 60 × 1.5 km = 90 km. ∴ Distance between the two places = 90 km Example 2 The actual length and breadth of a rectangular field is 100 m and 80 m respectively, which is drawn on the scale 1:800. Find the length and breadth of the field on the drawing. Solution: Here, 800 cm is represented by 1 cm i.e. 800 100 m is represented by 1 cm or, 8 m is represented by 1 cm length of the field = 100 m 8 m is represented by 1 cm 1 m is represented by 1 8 cm 80 m is represented by 1 8 × 80 cm = 10 cm 100 m is represented by 1 8 × 100 cm = 12.5 cm. Alternative method We have, scale = length in map actual length 1:800 = length in map/100 m or, length in map = 100 800 m = 1 8 m = 1 8 × 100 cm = 12.5 cm. Again, Scale = Length in map Actual length 1:800 = Length in map 80 m ∴ Length in map = 80 800 m = 1 10 m = 1 10 ×100 cm = 10 cm.
Oasis School Mathematics-8 251 Example 3 If an aeroplane flies 600 km in the bearing of 030° and 800 km in the bearing of 120°, find the distance between the two places and the bearing of the starting place from the last place using the scale 1 cm = 100 km. Solution: In the given figure, A is the starting place, last place is B, AB = 10 cm. (By measurement) therefore, distance between the places (AB)=10 cm = 1000 km. [∵ 1 cm = 100 km] From the given figure ∠ABN2 = 113° (By measurement). Hence, bearing of the starting place from the last place = 360° – 113° = 247°. Exercise 21.2 1. Given figure shows the position of students in a class. If the scale is 1:150. Find the actual distance between: (a) Ramesh and Shyam. (b) Goma and Bidhya. (c) Ramesh and Goma. (d) Shyam and Bidhya. 2. If the scale in a map is 1:1500. Find the actual distance between the given places if the distance in the map is: (a) 4 cm (b) 8 cm (c) 6 cm (d) 15 cm (e) 29 cm 3. The scale in a map is 1:5000. Find the distance between two places in the map if the actual distance between the two places is: (a) 0.8 km (b) 600 m (c) 1500 m (d) 3 km (e) 4 km 4. (a) Find the actual height of a tree whose height in a map is 5 cm and is represented by a scale 1: 600. (b) What is the actual distance between two places which is represented by 25 mm. on a map which is drawn to the scale of 1 mm = 2 km? 5. The figure alongside shows the outline sketch of a house plan. Draw a scale of 1 cm to represent 1 m. (a) Find the actual length and breadth of bedroom, kitchen and toilet. (b) Find the area of the garage. 8 cm 300 1200 N A B N N2 1 6 cm Ramesh Goma Shyam Bidhya
252 Oasis School Mathematics-8 Answer 1. Consult your teacher. 2. (a) 60m (b) 120m (c) 90m (d) 225m (e) 435m 3. (a) 16cm (b) 12cm (c) 30cm (d) 60cm (e) 80cm 4. (a) 30m (b) 50km 5. (a) Bedroom 4m, 3m Kitchen - 4m, 2m, Toilet-4m, 1m. (b) 9m2 6. Consult your teacher 7. (a) 152km (b) 88km (c) 152 km. (d) 72km 8. 500km and 2470 9. (a) 100km (b) 0900 6. Find the actual distance between the given places from the given scale. (a) Kathmandu and Dhangadi (b) Pokhara and Jumla (c) Gorkha and Janakpur 7. The figure or map of Nepal is drawn on the scale of 1 cm is equal to 40 km showing the major trade centers. Find out the actual distance between: (a) Dhangadhi and Nepalgunj. (b) Bhaktapur and Birgunj. (c) Jiri and Biratnagar. (d) Birgunj and Biratnagar. 8. If an aeroplane flies 400 km in the bearing of 030° and then 300 km in the bearing of 120°, find the distance between the two places and the bearings of the starting place from the last place using the scale 1 cm = 100 km. 9. A motor is at S, a place 200 km east to T. Travelling a distance of 300 km to the due west of S, it reached at U. Express this information by the drawing and find (a) the distance from T to U. (b) the bearing of T from U. T U N 2cm S 3cm Using scale : 1cm = 100km
Oasis School Mathematics-8 253 Estimated Teaching Hours 10 Contents: • Pie chart • Mean • Median • Mode Expected Learning Outcomes At the end of this unit, students will be able to develop to following competencies: • Take the informations from the Pie chart • Make the pie-chart from the given data • Calculate the arithmetic mean from the given individual and discrete data • Calculate the median from the given individual and discrete data • Calculate mode from individual and discrete data • Calculate range from given individual data Teaching Materials • Graph sheet, A4 size paper, etc. Statistics
254 Oasis School Mathematics-8 Unit 22 Statistics You know! • Information from line graph • Information from bar graph. Pie- chart Study the given figure and answer the questions given below: • Which coloured is the biggest one? • Which coloured sector is the smallest one? • If red coloured sector represents the number of students in VIII and blue coloured sector represents the number of students in class IX, which class has more students? • Discuss what is this diagram called? • Red coloured sector is the biggest one. • Blue coloured sector is the smallest one. • Class VIII has more students than in class IX. • This diagram is called Pie-chart.
Oasis School Mathematics-8 255 22.1 Pie–chart or Pie–diagram or Circular Diagram A pie–chart is a diagrammetical representation of statistical data in circular form with several subdivisions. It is also known as a circular diagram or pie–diagram or angular diagram. The radial lines divide the pie–diagram into different sectors in such a way that the area of the sectors are proportional to the different components of the data. As there is 360° at the centre, the sum of the angles representing different items is equal to 360°. Worked Out Examples Example 1 Represent the following data for the amount spent on the construction of a house by a pie–chart. Item Cement Timber Bricks Labour Steel Misc. Total Expenditure (in Rs. thousands) 180 90 135 225 135 135 900 Solution: Here, Total value = 900, total angle = 360° ∴ Angle at the centre by an item = Value of an item Total value × 360° The following pie–diagram shows the amount spent in the construction of a house. Calculation for construction of a pie diagram. Items of Expenditure Expenditure in Rs. thousand Angle at centre Cement 180 360 900 × 180 = 720 Timber 90 360 900 × 90 = 360 Bricks 135 360 900 × 135 = 540 Labour 225 360 900 × 225 = 900 Angle at the centre by an item = Value of an item Total value × 360° 54° 54° 72° 36° 54° 90°
256 Oasis School Mathematics-8 Steel 135 360 900 × 135 = 540 Misc. 135 360 900 × 135 = 540 Total 900 3600 Example 2 The total monthly income of a family is Rs. 7,200. The sources of income are shown in the given pie chart. What is the income of the family from each title? Solution: Here, 360° represents the total income Rs. 7,200. 1° represents the total income Rs. 7,200 360 85° represents the total income = Rs. 7,200 360 × 85 = Rs. 1,700 Income from business = Rs 1,700. Similarly, Income from house rent = Rs. 7,200 360 × 75 = Rs. 1,500 Income from salary = Rs. 7,200 360 × 125 = Rs. 2,500. Income from agriculture = Rs. 7,200 360 × 25 = Rs. 500 Income from others = Rs. 7,200 360 × 50 = Rs. 1,000. Value of an item = Total value 3600 × angle of that item Remember ! 54° 25° 75° 85° 125°
Oasis School Mathematics-8 257 Exercise 22.1 1. (a) The following table gives the details of monthly expenditure of a family. Draw a pie–chart to present the following information. Item Food Clothing Rent Eduction Lighting Miscellaneous Expenditure (in Rs.) 6000 2000 2000 2500 2500 3000 (b) The number of students in different levels of school is given in the table. Represent it in a pie chart. Level Pre primary Primary Lower Sec. Secondary Higher Sec. Number of students 240 540 360 600 420 (c) Represent the monthly expenses of a household in a pie-chart. Heading Food Eduction House Rent Transport Others Expenditure 6000 2000 2000 2500 2500 2. Circular graph given below shows the number of students enrolled in classes V, VI, VII and VIII of a school. Study the graph carefully and answer the following questions. (a) Which class has the maximum number of students? (b) If 1% equals 2 students, how many students are there in class VII ? (c) What is the total number of students enrolled? 3. Given pie chart shows the votes obtained by four candidates A, B, C and D in an election. If the total number of vote cast is 5,400. Find the number of votes obtained by each candidate. Answer 1. Consult your teacher 2. (a) Class VI (b) 20 (c) 720 3. A = 1275 B = 1425 C = 1575 and D = 1125 VI V VII, 10% VIII 850 750 1050 950 B C A D
258 Oasis School Mathematics-8 Project Work • Collect the informations about the monthly expense of your family in different titles. Show the information in the pie-chart. • From the school record, collect the number of students in the school in last 10 years. Show the data in line graph. 22.2 Measures of Central Tendency The single value which represents the given set of variables is called the volume central tendency. Measure of central tendency is also called average. There are mainly three types of measurement of central tendency. They are (i) Mean (ii) Median (ii) Mode. I. Arithmetic Mean The most popular and widely used measure for representing the entire data by single value is arithmetic mean. The arithmetic mean or simply mean or average is defined as the total sum of observation divided by total number of observations. That is, Arithmetic mean = Total sum of observations Total number of observations It is denoted by X. It is simply called Mean. (i) Mean of ungrouped data when none is repeated: Mean of ungrouped data is obtained by adding all the observations and dividing the sum by the total number of observations. ∴ Arithmetic Mean (X) = ∑X n . Where ∑X = sum of n items, n = number of items. Note : The Greek letter ∑(sigma) means "Summation of". Calculation of Arithmetic Mean in repeated data (discrete series) If the variable x1 , x2 , x3 ……. xn are repeated f1 ,f2 f 3 …….fn times, respectively, then mean of this data. (x) = ∑fx ∑f where, ∑fx = f1 x1 + f2 x2 + f3 x3 + ……….+ fnxn and ∑f = f1 +f2 + f3 + ……+ fn Remember ! N = ∑f
Oasis School Mathematics-8 259 Mean of grouped and continuous data In the case of grouped and continuous data, we should find the mid -value of each class interval. Mid -value of a class (m) = upper limit + lower limit 2 Mid value represents the variable of that class. ∴ In continuous series, Mean (x) = ∑fm ∑f Where, m = mid – value. Worked Out Examples Example 1 The heights of 5 students are 145 cm, 150 cm, 149 cm, 150 cm and 151 cm respectively. Find the mean height per student. Solution: Given, height of 5 students 145cm, 150 cm, 149cm, 155cm, 151cm. Now, ∑X = (145 + 150 + 149 + 155 + 151)cm = 750 cm Number of items (n) = 5. We have, Mean (X) = ∑X n = 750 5 = 150 cm. Example 2 The marks obtained by 10 students of a class are 46, 51, 62, 70, 35, x, 50, 83, 65, 52. If the average marks is 55, find x. Solution: Here, Mean = 55, n = 10, ∑X = (46 + 51 + 62 + 70 + 35 + x + 50 + 83 + 65 + 52) = 514 + x Now, mean (X) = ∑X n or, 55 = 514 + x 10 or, 550 = 514 + x ∴ x = 550 – 514 = 36
260 Oasis School Mathematics-8 Example 3 The marks obtained by 50 students in an examination are given below: Marks 10 15 20 25 30 No. of students 5 10 18 9 8 Find the average marks. Solution: Let marks be denoted by X and no. of students by ƒ. Calculation of average marks x f fx 10 5 50 15 10 150 20 18 360 25 9 225 30 8 240 N=50 ∑fX= 1025 Here, N = 50, ∑ƒx = 1025, (x) = ? Now, Mean (x) = ∑fx N = 1025 50 = 20.5 Hence, their average marks = 20.5 Example 4 Find the arithmetic mean from the following distribution: Class-interval 10-20 20-30 30-40 40-50 50-60 Frequency 3 6 8 5 2 Solution: Let mid -value of classes be denoted by m and frequency by ƒ.
Oasis School Mathematics-8 261 Calculation of mean: Class-interval f Mid value (m) fm 10-20 3 15 45 20-30 6 25 150 30-40 8 35 280 40-50 5 45 225 50-60 2 55 110 N=24 ∑fm = 810 Here, N = 24, ∑ƒm = 810, X = ? Now, X = ∑fm N = 810 24 = 33.75 Hence, the arithmetic mean is 33.75. Example 5 The average age of 30 boys in a class is 14 years and that of 20 girls is 12 years. Find average age of the students in the class. Solution: Here, n1 = 30, x1 = 14 yrs., n2 = 20, x2 = 12 years. Combined average age of boys and girls (x2 ) = ? Now, x 12 = n1 x1 + n2 x2 n1 +n2 = 30 × 14 + 20 × 12 30 + 20 = 420 + 240 50 = 660 50 = 13.2 years. Hence, the average age of the students in the class is 13.2 years. Exercise 22.2 1. Find the arithmetic mean for the following data. (a) Rs. 5, Rs. 10, Rs. 15, Rs. 20, Rs. 25, Rs. 30, Rs. 35 (b) 50 kg, 52 kg, 53 kg, 54 kg, 60 kg, 65 kg, 71 kg, 75 kg (c) 10 cm, 12 cm, 14 cm, 16 cm, 20 cm, 24 cm.
262 Oasis School Mathematics-8 2. Calculate the average marks of : 12, 15, 18, 17, 16, 20. 3. (a) If the arithmetic mean of 5, 7, 9, x and 15 is 10, find the value of x. (b) If the mean of 3, 7, 10, 15 and x is 12, find the value of x. 4. (a) From each of the following frequency distribution table, find the mean. x 10 20 30 40 50 f 5 10 15 10 9 (b) Calculate the mean mark from the following frequency distribution table. Marks obtained 10 20 30 40 50 No. of students 5 2 3 6 4 (c) x 5 10 15 20 25 30 f 2 5 10 7 4 2 5. From each of the following frequency distribution table, find the mean. (a) Class 0-10 10-20 20-30 30-40 40-50 No. of students 5 10 15 10 9 (b) Marks obtained 10-20 20-30 30-40 40-50 50-60 Frequency 4 6 8 5 3 (c) x: 0-10 10-20 20-30 30-40 40-50 50-60 f: 5 4 6 2 8 5 6. Find the combined mean (x12 ) of the following two series. (a) n1 = 70, x1 = 75, n2 = 30, x2 = 65 (b) n1 = 40, x1 = 55, n2 = 60, x2 = 65 7. (a) The mean weight of 30 boys in a class is 45 kg. And that of 20 girls is 40 kg. Find the mean weight of the students in the class.
Oasis School Mathematics-8 263 (b) The mean marks in Maths of 100 students in a class was 72. The mean mark of 70 boys was 75. Find out the mean marks of girls in the class. Answer 1. (a) 20 (b) 60 (c) 16 2. 16.33 3. (a) 14 (b) 25 4. (a) 31.63 (b) 31 (c) 17 5. (a) 26.63 (b) 33.84 (c) 31.33 6. (a) 72 (b) 61 7. (a) 43 kg (b) 65 22.3 Median The data of the variables are generally collected randomly. One must arrange them systematically in order. Suppose the data are arranged in ascending order of magnitude. The data which lies exactly in the middle is called median. It is denoted by Md. Hence, there will be almost same number of variables on either side of median. Calculation of median for individual series Let there be n observations of variables. Then positional value of median is given by n +1 2 th ( ( where 'n' is number of observations. From this, we can easily find the median value. Calculation of median for discrete distribution If the variables are given in terms of their respective frequencies, we will find the cumulative frequency of each variable. The median position will be obtained by using the formula, Position of Median = N+1 2 th ( ( item Where, N is the number of observation. Worked Out Examples Example 1 Calculate the median of the given data. 9, 13, 12, 16, 15, 20, 23. Solution: Arranging the above data in ascending order 9, 12, 13, 15, 16, 20, 23 Number of items (n) = 7 We have Median (Md) = N+1 2 th ( ( item
264 Oasis School Mathematics-8 = 7+1 2 th ( ( item = 4th item 4th item of the given data is 15. Hence, median (Md) = 15. Example 2 The height of the 6 students of class VIII is given below. 150 cm, 151 cm, 148 cm, 154 cm, 156 cm, 157 cm. Find their median height. Solution: Arranging data in ascending order 148 cm, 150 cm, 151 cm, 154 cm, 156 cm, 157 cm Number of terms (n) = 6 We have Median (Md) = N+1 2 th ( ( term = 6+1 2 th ( ( term = 3.5th term = 3rd term + 4th term 2 = 151cm + 154 cm 2 = 152.5 cm Example 3 If the median of the data 14, 18, x + 2, x + 4, 30, 32 arranged in ascending order is 24, find the value of x. Solution: Given data 14, 18, x + 2, x + 4, 30, 32 Number of item (n) = 6 We have Median (Md) = N+1 2 th ( ( item = 6+1 2 th ( ( item = 3.5th item = (x+2) + (x+4) 2 = 2x + 6 2 = x + 3 Since, the median is 24, x + 3 = 24 or, x = 21 Hence, median (Md) = 21.
Oasis School Mathematics-8 265 Exercise 22.3 1. Answer the following: (a) In how many equal parts does the median divide the whole data? (b) If 'n' be the number of item, find which item is the median? 2. Calculate the median of the given data (a) 4, 7, 9, 12, 15 (b) 6, 17, 19, 23, 24, 28 (c) 120 cm, 124 cm, 125 cm, 130 cm, 115 cm, 135 cm, (d) 60 ft, 50 ft, 40 ft, 25 ft, 80 ft, 70 ft, 85 ft, 65 ft. (e) 12, 22, 32, 41, 26, 30, 14, 11, 18, 35 (f) 15, 29, 43, 13, 50, 30, 20, 15, 55, 28, 35 3. (a) If the numbers 6, 8, 10, x + 2, 14, 16, 18 are in ascending order and the median is 12, find the value of x. (b) If the data 50, 60, 3x + 10 4 , 160, 180 are in ascending order and the median is 100, find the value of x. (c) If x-1, 2x + 1, x + 5 and 3x + 1 are in ascending order, the median of the data is 18 find the value of x. (d) x + 1, 2x –1, x + 2, 3x + 4 are in ascending order, if the median is 12, find the value of x. Answer 1. Consult your teacher. 2. (a) 9 (b) 21 (c) 124.5 (d) 62.5 ft (e) 24 (f) 29 3. (a) 10 (b) 130 (c) 10 (d) 6 22.4 Mode The mode or the modal value is that value in a distribution which occurs most frequently. It is the most common of value found in a distribution. Thus, the most typical, repeated value of a distribution is called the mode. It is denoted by M0 . Calculation of Mode for ungrouped data a. Individual series: In case of individual series, mode can often be found by inspection. To find the mode, count the number of times various values repeat themselves. Then, the value occurring maximum number of times is the modal value. For example, the modal
266 Oasis School Mathematics-8 value of the given series of individual observation is as follows. Calculation of Mode: 5, 6, 2, 0, 2, 1, 1, 4, 2, 2, 5, 4 Variable No. of occurrence 0 1 1 2 2 4 5 2 6 1 Here, a variate value 2 repeats maximum number of times 4 times. ∴ Mode = 2 b. Discrete series (or ungrouped frequency distribution) In case of discrete series also, mode can be obtained by inspection only. Mode occurs with maximum frequency. Variable values having greatest frequency is the mode. Worked Out Examples Example 1 Find the modal value for the following data: Sizes of shoes 5 6 7 8 9 10 11 Number of persons 15 20 25 35 27 18 10 Solution: Here, the maximum frequency is 35. So corresponding modal size is 8. ∴ Mode = 8 Exercise 22.4 1. Find the mode for each of the following data: (i ) 65, 6, 1, 0, 2, 1, 2, 5, 2, 6, 4, 4, 2, 4 (ii) Rs. 3, Rs. 4, Rs. 2, Rs. 5, Rs. 3, Rs. 4, Rs. 5, Rs. 5, Rs. 4, Rs. 6, Rs. 5, Rs. 6, Rs. 6, Rs. 3, Rs. 7, Rs. 5, Rs. 7
Oasis School Mathematics-8 267 2. A boy scored the following marks in various class tests during term exams, each test being marked out of 20. 15, 17, 16, 7, 10, 12, 14, 16, 19, 12, 16. What are his modal marks ? 3. Find the mode for the following data. x 10 15 20 25 30 35 40 ƒ 3 6 10 12 7 5 2 4. Calculate the mode for the following data. Age in years 10 11 12 13 14 15 16 No. of students 10 15 20 25 18 12 7 Answer 1. (i) 2 (ii) 5 2. 16 3. 25 4. 13 22.5 Range Marks obtained by 20 students of class VIII are given below: 72, 78, 42, 90, 63, 48, 30, 25, 82, 92, 46, 89, 97, 29, 53, 76, 58, 43, 39, 75 If we arrange the given data in ascending order, 25, 29, 30, 39, 42, 43, 46, 48, 53, 58, 63, 72, 75, 76, 78, 82, 89, 90, 92, 97 It can be observed that, The lowest marks of the class = 25 The highest marks of the class = 97 The marks of other students are spread between 25 and 97. Difference between the highest marks and the lowest marks = 97 – 25 = 72 This value is called the range. Hence the difference between the highest and the lowest values of the observation is called the range. Range = highest value of the observation – lowest value of the observation . i.e. Range = Largest item – Smallest item = L – S
268 Oasis School Mathematics-8 Worked Out Examples Example 1 Find the range of the following data: 20, 25, 31, 32, 18, 15 Solution: Given data: 15, 18, 20, 25, 31, 32 Largest item (L) = 32 Smallest item (S) = 15 We have, Range = L – S = 32 – 15 = 17 Exercise 22.5 1. Find the range of the following data: (a) 3, 7, 9, 12, 6, 15 (b) 19, 12, 8, 17, 23, 25, 30 (c) 65, 60, 52, 35, 80, 68, 92, 31, 56. 2. The height (in cm) of 20 students in a class is given below: 140, 141, 151, 152, 162, 157, 136, 145, 151, 150, 165, 135, 142, 146, 148, 140, 141, 152, 156, 160 (i) What is the height of the tallest student? (ii) What is the height of the shortest student? (iii) Find the range. Answer 1. (a) 12 (b) 22 (c) 61 2. (i) 165 cm (ii) 135cm (iii) 30 Project Work • Collect the marks obtained by the students of your class in Mathematics in an examination. Find (i) mean (ii) median (iii) mode (iv) Range and present the result in your classroom. • Collect the information about the ages of the teachers teaching in your school and find (i) mean age (ii) median (iii) mode (iv) range and present the result in your classroom.
Oasis School Mathematics-8 269 Miscellaneous Exercise 1. The pie-diagram shows the votes secured by 4 competitive candidates in the election of Rural municipality Study the given chart and answer the questions given below. (a) How many candidates are there? (b) If 7200 voters cast their votes, find the number of votes obtained by all candidates. (c) Who won the election and by what majority? [Ans: (a) 4, (b) A :200, B : 2160, C:1400, D:1440, (c) A won the election by 40 votes] 2. Daily wages of 12 workers is given below. Rs. 120, Rs. 100, Rs. 110, Rs. 115, Rs. 108, Rs. 115, Rs. 102, Rs. 118, Rs. 125, Rs. 127, Rs. 130, Rs. 115 (a) What is the highest and lowest wage of the worker? (b) Calculate the highest and lowest wage of the worker? (c) Find the average wage of the worker. (d) Find the median. (e) Find the mode of the data. [Ans: (a) Rs. 130 and Rs. 100, (b) Rs. 30 (c) Rs. 116.25, (d) 115, (e) 115] 3. The monthly income of a family is given below. Source Salary Business Agriculture Rent Income Rs. 5000 Rs. 3400 Rs. 3000 Rs. 3000 (a) What is the total income of the family? (b) If total income is represented 3600 by how much degree Rs. 5000, Rs. 3400, Rs. 3000 and Rs. 3000 are presented by? (c) Draw the pie-chart to show the above information. [Ans: (a) Rs. 14400, (b) 1250 , 850 , 750 , 750 , (c) Consult your teacher.] 1100 1080 720 700
270 Oasis School Mathematics-8 Objective Questions Choose the correct alternatives. 1. Given pie-chart shows the expenditure of a household. (a) On which title there is large expenditure? (i) Food (ii) Cloths (iii) Education (b) If the total expenditure is Rs. 72,000 then 10 is equivalent to (i) Rs. 200 (ii) Rs. 2000 (iii) Rs. 20 (c) The expenditure on food is (i) Rs. 3000 (ii) Rs. 30,000 (iii) Rs. 300 (d) The expenditure of the household on education is (i) Rs. 15,00 (ii) Rs. 150 (iii) Rs. 15,000 2. The average of first n1 data is x1 that of next n2 data is x2 then the mean of whole data is (i) n1 n1 n2 n2 + + x2 x1 (ii) n1 n1 n2 n2 + + x1 x2 (iii) n1 n1 n2 n2 – – x2 x1 3. Median divides the whole data into (i) three equal parts (ii) four equal parts (iii) two equal parts 4. The term which is repeated the maximum number of times is (i) the median (ii) the mean (iii) the mode 5. The median of the given data 22, 27, 26, 28, 23, 29, 34, 32, is (i) 27 (ii) 27.5 (iii) 28 6. The difference between the largest item and the smallest item is (i) mode (ii) range (iii) median Food Cloths 1500 1300 Education Miscellaneous 500 300
Oasis School Mathematics-8 271 Attempt all the questions. 1. Study the given figures that shows the number of students in different classes of school. (a) What information does the given Pie-chart give? (1) (b) If 10 = 4 students, find the number of students in each class. (2) (c) Which class has the highest number of students and by how much? (1) 2. Given Pie-chart shows the favourite subject of the students of class VIII. If there are 180 students, (a) How many students does 10 represent? (1) (b) How many student's favourite subject is Science? (1) (c) How many more students like Maths than Nepali? (1) (d) Which is the least favourite subject among all? (1) 3. Marks obtained by 50 students of class VIII is given below. Marks 25 32 42 50 68 75 80 Number of students 5 10 12 3 6 7 7 (a) Calculate the average marks obtained by the students. (2) (b) Calculate the median marks obtained by the students. (2) (a) What is the gap between the highest and the lowest marks? What is this gap called? (2) Assessment Test Paper Full marks - 14 Class V 80% Class IV 50% Class IX 100% Class VII 60% Class VIII 70% Maths 1040 1000 760 800 English Nepali Science
272 Oasis School Mathematics-8 S.N. Areas Total working hours Knowledge Understanding Application Higher ability Total number of items Total number of questions Total Marks Number of items Marks Number of items Marks Number of items Marks Number of items Marks 1. Sets 10 1 1 1 1 2 3 1 1 5 2 3 2. Statistics 10 3 3. Arithmetic 45 2 2 3 4 3 5 2 3 10 3 14 4. Mensuration 15 1 1 1 1 1 2 1 1 4 1 5 5. Algebra 30 2 2 1 2 2 4 1 2 6 3 10 6. Geometry 50 2 2 2 4 2 6 2 3 8 3 15 Total 160 8 8 8 12 10 20 7 10 33 12 50 b|:6JoM • k|Zgkq lgdf{0f ubf{ k|To]s If]qdf / ;du|df 1fg, af]w, k|of]u / pRr bIftfsf nflu tf]lsPcg';f/sf ef/ ldn]sf] x'g'kb{5 . t/ ;+1fgfTds txdf @ c+s;Dd 36a9 x'g ;Sg]5 . • ;Gbe{ lbP/ k|Zg lgdf{0f ug'{kg]{5 . k|To]s k|Zgdf PseGbf a9L ;+1fgfTds txsf pkk|Zg ;dfj]z ug{ ;lsg]5 . • Application / higher ability txsf k|Zg lgdf{0f ubf{ ;DalGwt If]qsf cnfjf cGo If]qsf ljifoj:t';+u ;DalGwt k|Zg klg /xg ;Sg]5g\ . • x/]s If]qcGtu{t /x]sf ;a} pkIf]qsf ljifoj:t' ;dfg'kflts ?kdf ;dfj]z x'g] u/L k|Zg lgdf{0f ug'{kg]{5 . Subject: Compulsory Mathematics -8 Time : 2 hrs. Full marks : 50 Specification grid for class 6 to 8 Prescribed by CDC, Nepal