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 a.m. to 7.00 p.m. at a station. Give a reason for each.
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 a.m. to 7.00 p.m. at a station. Give a reason for each.
Explanation:
a) The number of areas cannot be split into class intervals so it cannot be represented using a histogram.
b) Since this information can be divided into class interval, a histogram can be used to display this data.
c) Since, companies cannot be divided into intervals, we cannot use histogram to represent this data.
d) Time for boarding the train can be divided into intervals. So, we can use histogram here.
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 of 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 M W G W M G W M M B G G W.
Make a frequency distribution table using tally marks. Draw a bar graph to illustrate it.
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 of 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 M W G W M G W M M B G G W.
Make a frequency distribution table using tally marks. Draw a bar graph to illustrate it.
Explanation:
3. The weekly wages (in ₹) 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, 878, 840, 868, 890, 806, 840.
Using tally marks, make a frequency table with intervals as 800 – 810, 810 – 820 and so on.
3. The weekly wages (in ₹) 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, 878, 840, 868, 890, 806, 840.
Using tally marks, make a frequency table with intervals as 800 – 810, 810 – 820 and so on.
Explanation:
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 ₹ 850 and more?
(iii) How many workers earn less than ₹ 850?
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 ₹ 850 and more?
(iii) How many workers earn less than ₹ 850?
Explanation:
i) 830-840 has the maximum number of workers, i.e., 9
ii) Workers earning more than 850 is 10
iii) Workers earning less than 850 is 20
5. The number of hours for which students of a particular class watched television during holidays is shown in the given graph.
Answer the following:
(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 spent more than 5 hours watching TV?
5. The number of hours for which students of a particular class watched television during holidays is shown in the given graph.
Answer the following:
(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 spent more than 5 hours watching TV?
Explanation:
(i) 32 students watched TV for 4-5 hours.
(ii) The number of students who watched TV less than 4 hours= 22+8+4=34
(iii) The number of students who spent more than 5 hours watching TV
=8+6=14
6. A survey was made to find the type of music that a certain group of young people liked in a city.
An adjoining pie chart shows the findings of this survey. From this pie chart, answer the following:
(i) If 20 people liked 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 CDs, how many of each type would they
make?
6. A survey was made to find the type of music that a certain group of young people liked in a city.
An adjoining pie chart shows the findings of this survey. From this pie chart, answer the following:
(i) If 20 people liked 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 CDs, how many of each type would they
make?
Explanation:
(i) 10% represents 100 people.
⟹20% represents = (100×20)/10 = 200
∴ Number of people surveyed= 200
(ii) Since 40% of respondents liked light music and no other song form was preferred, we can conclude that light music is the most popular.
(iii) CDs of classical music = (10 × 1000)/100 = 100
CDs of semi-classical music = (20 × 1000)/100 = 200
CDs of light music = (40 × 1000)/100 = 400
CDs of folk music = (30 × 1000)/100 = 300
7. 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 pie chart to show this information
7. 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 pie chart to show this information
Explanation:
(i) According to the table given in the question, the winter season got the most votes.
(ii) Central angle of summer season= (90×360)/360= 90degree
Central angle of rainy season= (120×360)/360= 120degree
Central angle of winter season= (150×360)/360= 150degree
iii)
8. Draw a pie chart showing the following information. The table shows the colours preferred by a group of people.
8. Draw a pie chart showing the following information. The table shows the colours preferred by a group of people.
Explanation:
Here, central angle = 360degree Total number of people = 36
9. The adjoining pie chart gives the marks scored in an examination by a student in Hindi, English, Mathematics, 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?
(Hint: for 540 marks, the central angle = 360°. So, for 105 marks, what is the central angle?)
(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).
9. The adjoining pie chart gives the marks scored in an examination by a student in Hindi, English, Mathematics, 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?
(Hint: for 540 marks, the central angle = 360°. So, for 105 marks, what is the central angle?)
(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).
Explanation:
(i) The student scored 105 marks in Hindi.
(ii) Marks obtained in Mathematics = 135 Marks obtained in Hindi = 105 Difference = 135 – 105 = 30
Thus, 30 more marks were obtained by the student in Mathematics than in Hindi.
(iii) The sum of marks in Social Science and Mathematics = 97.5 + 135 = 232.5 The sum of marks in Science and Hindi = 120 + 105 = 225
∴ the sum of the marks in Social Science and Mathematics is more than in Science and Hindi.
10. The number of students in a hostel speaking different languages is given below. Display the data in a pie chart.
10. The number of students in a hostel speaking different languages is given below. Display the data in a pie chart.
Explanation:
11. List the outcomes you can see in these experiments.
(a) Spinning a wheel (b) Tossing two coins together
11. List the outcomes you can see in these experiments.
(a) Spinning a wheel (b) Tossing two coins together
Explanation:
(a) There are 4 outcomes as there are 4 letters in the wheel, A, B, C, D.
(b) There are four possible outcomes HH, HT, TH, and TT.
12. 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
12. 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
Explanation:
(i)
(a) Outcomes of the event of getting a prime number are 2, 3 and 5.
(b) Outcomes of the event of not getting a prime number are 1, 4 and 6.
(ii)
(a) Outcomes of the event of getting a number greater than 5 is 6.
(b) Outcomes of the event of not getting a number greater than 5 are 1, 2, 3, 4 and 5.
13. Find the.
(a) Probability of the pointer stopping on D in (Question 1-(a)).
(b) Probability of getting an ace from a well-shuffled deck of 52 playing cards.
(c) Probability of getting a red apple. (See figure below)
13. Find the.
(a) Probability of the pointer stopping on D in (Question 1-(a)).
(b) Probability of getting an ace from a well-shuffled deck of 52 playing cards.
(c) Probability of getting a red apple. (See figure below)
Explanation:
(a) In the spinning wheel, there are five pointers A, A, B, C, D.
Thus, number of total outcome= 5
So, the probability of the pointer stopping on D = 1/5
(b) There are 4 aces in a deck of 52 playing cards. So, there are four events for getting an ace.
Therefore, the probability of getting an ace = 4/52 = 1/13
(c) Total number of apples = 7
Number of red apples = 4
Probability of getting a red apple = 4/7
14. 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) getting a number 6?
(ii) getting a number less than 6?
(iii) getting a number greater than 6?
(iv)getting a 1-digit number?
14. 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) getting a number 6?
(ii) getting a number less than 6?
(iii) getting a number greater than 6?
(iv)getting a 1-digit number?
Explanation:
(i)Total number of outcome= 10
The result of obtaining number 6 from ten separate slips is one.
Therefore, probability of getting a number 6 = 1/10
(ii) Numbers less than 6 are 1, 2, 3, 4 and 5, which are five
So, number of possible outcome= 5
Therefore, probability of getting a number less than 6 =5/10 =1/2
(iii) Number greater than 6 out of ten that are 7, 8, 9, 10
Thus, number of possible outcome= 4
Therefore, probability of getting a number greater than 6 = 4/10 = 2/5
(iv) One-digit numbers are 1, 2, 3, 4, 5, 6, 7, 8, 9 out of ten.
Therefore, probability of getting a 1-digit number = 9/10
15. 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?
15. 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?
Explanation:
No. of total sectors present= 5
No. of green sectors= 3
∴ Probability of getting a green sector = 3/5
No. of blue sector present= 1
Hence, non-blue sectors = 5 – 1 = 4 sectors
∴ Probability of getting a non-blue sector= 4/5
16. Find the probabilities of the events given in Question 2.
16. Find the probabilities of the events given in Question 2.
Explanation:
Number of total outcome in a die= 6
a) No of prime numbers in a die= 3
So, there are 3 outcome out of 6
Therefore, probability of getting a prime number= 3/6 or ½
b) No of non prime numbers in a die= 3
So, there are 3 outcome out of 6
Therefore, probability of getting a non prime number= 3/6 or ½
c) Number of possible output=1 (Since, 6 is the only number in a die greater than 5)
Therefore, probability of getting a number greater than 5= 1/6
d) Numbers not greater than 5 are 1, 2, 3, 4 and 5. So there are 5 outcomes out of 6.
Therefore. probability of not getting a number greater than 5= 5/6
Also Read: Data Handling Class 8 Extra Questions
