# Data handling-NCERT Solutions

Class VIII Math
NCERT Solution for Data Handling
EXERCISE – 5.1
1.    For which of these would you use a histogram to show the data?
(a) The number of letters for different areas in a postman�s bag.
(b) The height of competitors in an athletics meet.
(c) The number of cassettes produced by 5 companies.
(d) The number of passengers boarding trains from 7:00 am to 7:00 pm at a station.
Give reasons for each.
Ans. In (b) and (d), data can be divided into class intervals. So, their histograms can be drawn.
2.    The shoppers who come to a departmental store are marked as : man (M), woman (W), boy (B) or girl (G). The following list gives the shoppers who came during the first hour in the morning.
W W W G B W W M G G M M W W W W G B M W B G G M W W M M W W W M W B W G M W W W W G W M M W W M W G W M G W M M B G G W
Ans. We arrange the data in a table using tally marks as
The bar graph of the above data is as under
3.    The weekly wages (in Rs) of 30 workers in a factory are:
830, 835, 890, 810, 835, 836, 869, 845, 898, 890, 820, 860, 832, 833, 855, 845, 804, 808, 812, 840, 885, 835, 835, 836, 840, 868, 890, 806, 840.
Using tally marks a frequency table with intervals as 800 � 810, 810 � 820 and so on.
Ans. Let us make a grouped frequency table as under:
4.    Draw a histogram for the frequency table made for the data in question 3, and answer the following questions.
(i)   Which group has the maximum number of workers?
(ii)  How many workers earn Rs. 850 and more?
(iii) How many workers earn less than the 850?
Ans.
(i)   830-840 group has max.no. of workers.
(ii)  10
(iii) 20
The number of hours for which students of a particular class watched television during holidays is shown through the given graph (on previous page).
(i)   For how many hours did the maximum number of students watch TV ?
(ii)  How many students watched TV for less than 4 hours?
(iii) How many students spend more than 5 hours in watching TV?
Ans. Clearly from the given histogram, we find that
(i)   The maximum number of students watch TV for 4-5 hours.
(ii)  The number of students who watch TV for less than 4 hours are 4 + 8 + 22 = 34.
(iii) The number of students who spend more than 5 hours in watching TV are 8 + 6 = 14.
EXERCISE – 5.2
1.    A survey was made to find the type of music that a certain group of young people likes in a city. Following pie chart shows the findings of this survey. From this pie chart answer the following questions:
(i)   If 20 people like classical music, how many young people were surveyed?
(ii)  Which type of music is liked by the maximum number of people?
(iii) If a cassette company were to make 1000 CD's, how many of each type would it make?
Ans. (i) Let x be the number of young people surveyed

(ii)  light music is liked by the maximum number of people.
(iii) Number of CD's in respect of

2.    A group of 360 people were asked to vote for their favourite season from the three seasons-Rainy, winter and summer.
(i)   Which season got the most votes?
(ii)  Find the central angle of each sector.
(iii) Draw a ple chart to show this information.
Ans. (i) The winter season got the maximum votes.
(ii)  The proportion of sectors (winter, summer and rainy seasons) are as

Now, the various components may be shown by the adjoining pie chart.
3.    Draw a pie chart showing the following information, table shows the colours preferred by a group of people.
Ans.
4.    The adjoining pie chart gives the marks scored in an examination bV a student in Hindi, Eng,ish, Mathemahcs, Social Science and Science. If the total marks obtained by the students were 540 answer the following questions.
(i)   In which subject did the student score 105 marks?
(ii)  How many more marks were obtained by the student in Mathematics than in Hindi?
(iii) Examine whether the sum of the marks obtained in Social Science and Mathematics is more than that in Science and Hindi. (Hint: Just study the central angles.)
Ans. (i) For 540 marks, the central angle = 360°
∴ For 1 mark, the central angle

Hence, from the given pie chart, the subject is Hindi.
(i)   Difference between the central angles made by the subject of Mathematics and Hindi
= 90° – 70° – 20°.
For the central angle of 360°, marks obtained
= 540
For the central angle of 1°, marks obtained

For the central angle of 20°, marks obtained
Hence, 30 more marks were obtained by the student in Mathematics than Hindi.
(iii) Sum of the central angles made by Social Science and Mathematics = 65° + 90° = 155°.
And, the sum of the central angles made by Science and Hindi = 80° + 70° = 150°
Science 155° > 150°, therefore, the sum of the marks obtained in Social Science and Mathematics is more than that in Science and Hindi.
5.    The number of students in a hostel, speaking different languages is given below. Display the data in a pie chart.
Ans.

EXERCISE – 5.3
1.    List the outcomes you can see in these experiments.
(i)   Spinning a wheel,
(ii)  Tossing two coins together
Ans. (i) List of outcomes of spinning the given wheel are A, B, C and D.
(ii)  When two coins are tossed together, the possible outcomes of the experiment are HH, HT, TH and TT.
2.    When a die is thrown, list the outcomes of an event of getting.
(i)   (a) A prime number,
(b) Not a prime number
(ii)  (a) A number greater than 5,
(b) A number not greater than 5.
Ans. (i) In a throw of die, list of the outcome of an event 5 of getting
(a) A prime number are 2,3 and 5.
(b) Not a prime number are 1, 4 and 6.
(ii)  In a throw of die, list of the outcomes of an event of getting
(a) A number greater than 5 is 6.
(b) A number not greater than 5 are 1,2,3 and 4.
3.    Find the
(i)   Probability of the pointer stopping on D in (Q. 1 (a)).
(ii)  Probability of getting an ace from a well shuffled deck of 52 playing cards.
(iii) Probability of getting a red apple (see adjoining figure).
Ans. (i) Out of 5 sectors, the pointer can stop at any of sectors in 5 ways.
∴ Total number of elementary events = 5.                There is only one 'D( on the spinning wheel,
∴ Favourable number of outcomes = 1
∴ Required probability =1/5
(ii)  Out of 52 cards, one card can be drawn in 52 ways.
∴ Total number of outcomes = 52
There are 4 aces in a pack of 52 cards, out of which one ace can be drawn in 4 ways.
∴ Favourable number of cases = 4

(iii) Out of 7 apples, one apple can be drawn in 7 ways.
∴ Total number of outcomes = 7
There are 4 red apples, in a bag of 7 apples, out of which 1 red apple can be drawn in 4 ways.
∴ Favourable number of cases = 4
So, the required probability = 4/7
4.    Numbers 1 to 10 are written on ten separate slips (one number on one slip), kept in a box and mixed well. One slip is chosen from the box without looking into it. What is the probability of
(i)   Gettting a number 6?
(ii)  Getting a number less than 6?
(ii)  Getting a number greater than 6?
(iv) Getting a 1-digit number?
Ans. Out of 10 slips, 1 slip can be drawn In 10 ways.
So, the total numlier of outcomes = 10
(i)   An event of getting a number 6, i.e., if we obtain a slip having number 6 as an outcome.
So favourable number of outcomes = 1
∴ Required probability = 1/10
(ii)  An event of getting a number less than 6, i.e., If we obtain a slip having any of numbers 1, 2, 3, 4, 5 as an outcome.
So, favourable number of cases = 5

(iii) An event of getting a number greater than 6, i.e., if we obtain a slip having any of numbers 7, 8, 9, 10 as an outcome.
So, favourable number of cases = 4

(iv) An event of getting a one-digit number, i.e., if we obtain a slip having any of numbers 1,2, 3, 4, 5, 6, 7, 8, 9 as an outcome.
So, favourable number of cases = 9.
∴ Required probability = 9/10.
5.    If you have a spinning wheel with 3 green sectors, 1 blue sector and 1 red sector, what is the probability of getting a green sector? What is the probability of getting a non-blue sector?
Ans. Out of 5 sectors, the pointer can stop at any of sectors in 5 ways.,
∴ Total number of outcomes = 5
There are 3 green sectors in the spinning wheel, out of which one can be obtained in 3 ways.
∴ Favourable number of outcomes = 3
So, the required probability = 3/5
Further, there are 4 non-blue sectors in the spinning wheel, out of which one can be obtained in 4 ways.
So, the required probability = 4/5.
6.    Find the probabilities of the events given in question 2.
Ans. In a single throw of a die, we can get any one of the six numbers 1, 2, ...., 6 marked on its six faces.
Therefore, the total number of outcomes = 6.
(i)   Let A denote the event “getting a prime number”. Clearly, event A occurs if we obtain 2, 3, 5 as an outcome
∴ Favourable number of outcomes = 3.

(ii)  Let A denote the event “not getting a prime number”. Clearly, event A occurs if we obtain 1, 4, 6 as an outcomes
∴ Favourable number of outcomes = 3.

(iii) The event “getting a number greater than 5” wil occur if we obtain the number 6.
∴ Favourable number of outcomes = 1. Hence, required probability « 1/6
(iv) The event “getting a number not greater than 5” will occur 'if we obtain one of the numbers 1, 2, 3, 4, 5.
∴ Favourable number of outcomes = 5.
Hence, required probability = 5/6