{"id":79697,"date":"2021-10-18T17:00:44","date_gmt":"2021-10-18T11:30:44","guid":{"rendered":"https:\/\/www.cbselabs.com\/?p=79697"},"modified":"2021-10-20T16:45:05","modified_gmt":"2021-10-20T11:15:05","slug":"data-handling-ncert-extra-questions-for-class-8-maths","status":"publish","type":"post","link":"https:\/\/www.cbselabs.com\/data-handling-ncert-extra-questions-for-class-8-maths\/","title":{"rendered":"Data Handling Class 8 Extra Questions Maths Chapter 5"},"content":{"rendered":"
Extra Questions for Class 8 Maths Chapter 5 Data Handling<\/strong><\/p>\n Data Handling Class 8 Extra Questions Question 1.<\/strong> Class 8 Data Handling Extra Questions Question 2.<\/strong> Class 8 Maths Chapter 5 Extra Questions Question 3.<\/strong> Data Handling Class 8 Worksheet Question 4.<\/strong> Data Handling Class 8 Extra Questions With Answers Question 5.<\/strong> Data Handling Class 8 Worksheets With Answers Question 6.<\/strong> Short Answer (SA) Questions<\/strong> Data Handling Extra Questions Class 8 Question 8.<\/strong> Question 9. Question 10. Question 11. Question 12. Question 13. Question 14. Question 15. Data Handling Class 8 Extra Questions Maths Chapter 5 Extra Questions for Class 8 Maths Chapter 5 Data Handling Data Handling Class 8 Extra Questions Very Short Answer Type Data Handling Class 8 Extra Questions Question 1. In the class interval 5-10, find the (i) lower limit (ii) upper limit (iii) class mark (iv) class …<\/p>\n","protected":false},"author":27,"featured_media":159550,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[2],"tags":[],"yoast_head":"\nData Handling Class 8 Extra Questions Very Short Answer Type<\/h3>\n
\nIn the class interval 5-10, find the
\n(i) lower limit
\n(ii) upper limit
\n(iii) class mark
\n(iv) class size
\nSolution:
\n(i) lower limit = 5
\n(ii) upper limit = 10
\n(iii) Class mark = \\(\\frac { 5+10 }{ 2 }\\) = \\(\\frac { 15 }{ 2 }\\) = 7.5
\n(iv) Class size = 10 – 5 = 5<\/p>\n
\nA group of 20 students recorded their heights (in cm). The data received were as given below. What is the range?
\n150, 120, 112, 160, 155, 151, 158, 142, 148, 149, 161, 165, 140, 157, 156, 146, 148, 153, 138, 135
\nSolution:
\nThe minimum height =112 cm
\nMaximum height = 165 cm
\nRange = Maximum height – Minimum height = 165 cm – 112 cm = 47 cm<\/p>\n
\nIn the given pie chart, which colour is most popular? Which colour is the least popular?
\n
\nSolution:
\nRed colour is the most popular and the blue colour is the least popular.<\/p>\n
\nA die is thrown once. Find the probability of getting a number greater than 4.
\nSolution:
\nNumber greater than 4 = 5, 6
\nn(E) = 2
\nSample space n(S) = 6
\nProbability of getting a number greater than 4
\n= \\(\\frac { n(E) }{ n(S}\\) = \\(\\frac { 2 }{ 6 }\\) = \\(\\frac { 1 }{ 3 }\\)
\nWhere re(E): Number of favourable outcomes
\nn(S): Total number of outcomes<\/p>\n
\nA class consists of 21 boys and 9 girls. A student is to be selected for social work. Find the probability that
\n(i) a girl is selected
\n(ii) a boy is selected
\nSolution:
\nSample space n(S) = 21 + 9 = 30
\nNumber of girls n(E) = 9
\n(i) Probability of selecting a girl
\n= \\(\\frac { n(E) }{ n(S}\\) = \\(\\frac { 9 }{ 30 }\\) = \\(\\frac { 3 }{ 10 }\\)
\n(ii) Probability of selecting a boy
\n= \\(\\frac { n(E) }{ n(S}\\) = \\(\\frac { 21 }{ 30 }\\) = \\(\\frac { 7 }{ 10 }\\)<\/p>\n
\nThe following pie chart depicts the percentage of students, nationwide. What is the percentage of
\n(i) Indian students
\n(ii) African students?
\n
\nSolution:
\n(i) Percentage of Indian students = \\(\\frac { 180\\times 100 }{ 360 }\\) = 50%
\n(ii) Percentage of African students = \\(\\frac { 45\\times 100 }{ 360 }\\) = 12\\(\\frac { 1 }{ 2 }\\)%<\/p>\n
\nData Handling Class 8 Questions With Answers Question 7.<\/strong>
\nFill in the blanks:
\n
\nSolution:
\nClass-marks are
\nClass-mark
\n
\n<\/p>\n
\nConstruct a frequency table for the following marks obtained by 50 students using equal Intervals taking 16-24 (24 not included) as one of the class-intervals.
\n52, 16, 18, 20, 42, 48, 39, 38, 54, 58, 47, 37, 25, 16, 42, 49, 36, 35, 53, 21, 30, 43, 56, 34, 33, 17, 22, 24, 37, 41, 40, 50, 54, 56, 54, 36, 38, 42, 44, 56, 17, 18, 22, 24, 17, 48, 58, 23, 29, 58
\nSolution:
\n<\/p>\n
\nThe double bar graph shows the average monthly temperatures of two cities over 4 months period. Read the graph carefully and answer the questions given below:
\n(i) What does each 1 cm block on the vertical axis represent?
\n
\n(ii) What was the average monthly temperature in Dehradun in
\n(a) March
\n(b) April
\n(c) May
\n(d) June?
\n(iii) What was the average monthly temperature in Delhi for the whole 4 months?
\n(iv) In which month was the difference between the temperature of Delhi and Dehradun maximum and how much?
\nSolution:
\n(i) 1 cm block on vertical axis = 10\u00b0C
\n(ii) The average monthly temperature in Dehradun in the month of
\n(a) March was 25\u00b0C
\n(b) April was 34\u00b0C
\n(c) May was 40\u00b0C
\n(d) June was 36\u00b0C
\n(iii) The average monthly temperature in Delhi in the 4 months
\n
\n(iv) Difference between the average monthly temperature of Delhi and Dehradun was maximum in the month of June, i.e. (50\u00b0 – 36\u00b0) = 14\u00b0C.<\/p>\n
\nThe following table represents the number of students in a school playing six different games.
\n
\nPresent the above information on a bar graph.
\nSolution:
\n<\/p>\n
\nPrepare a grouped frequency table for the given histogram.
\n
\nSolution:
\n<\/p>\n
\nA bag contains 144 coloured balls represented by the following table. Draw a pie chart to show this information.
\n
\nSolution:
\n<\/p>\n
\nMrs Verma spends her allowance in the following way.
\n
\nRepresent the above information by a pie chart.
\nSolution:
\n
\n<\/p>\n
\nWhat is the probability of getting a marble which is not red from a bag containing 3 black, 8 yellow, 2 red and 5 white marbles?
\nSolution:
\nTotal number of balls = 3 black + 8 yellow + 2 red + 5 white = 18
\nn( S) = 18
\nNumber of the balls which are not red = 3 + 8 + 5 = 16
\nn(E) = 16
\nProbability = \\(\\frac { n(E) }{ n(S) }\\) = \\(\\frac { 16 }{ 18 }\\) = \\(\\frac { 8 }{ 9 }\\)<\/p>\n
\nFrom a well shuffled deck of 52 playing cards, a card is selected at random. Find the probability of getting
\n(i) a black card
\n(ii) a black king
\n(iii) an ace
\n(iv) a card of diamond
\nSolution:
\nHere, n(S) = 52
\n(i) Total number of black card = 26
\nn(E) = 26
\nProbability of getting a black card = \\(\\frac { n(E) }{ n(S) }\\) = \\(\\frac { 26 }{ 52 }\\) = \\(\\frac { 1 }{ 2 }\\)
\n(ii) Number of black king = 2
\nn(E) = 2
\nProbability of getting a black king = \\(\\frac { n(E) }{ n(S) }\\) = \\(\\frac { 2 }{ 52 }\\) = \\(\\frac { 1 }{ 26 }\\)
\n(iii) Number of aces = 4
\nn(E) = 4
\nProbability of getting an ace = \\(\\frac { n(E) }{ n(S) }\\) = \\(\\frac { 4 }{ 52 }\\) = \\(\\frac { 1 }{ 13 }\\)
\n(iv) Number of diamond cards = 13
\nn(E) = 13
\nProbability of getting a card of diamond = \\(\\frac { n(E) }{ n(S) }\\) = \\(\\frac { 13 }{ 52 }\\) = \\(\\frac { 1 }{ 4 }\\)<\/p>\n
\n
\n
\n
\n
\n
\n
\n
\n
\n
\n
\n
\n
\n
\n
\n
\n
\n
\n
\n
\n
\n
\n<\/p>\nExtra Questions for Class 8 Maths<\/a><\/h4>\n
NCERT Solutions for Class 8 Maths<\/a><\/h4>\n","protected":false},"excerpt":{"rendered":"