ML Aggarwal Solutions for Class 8 Maths Chapter 19 Data Handling help students with the technique of handling the given set of data. The collection of information in the form of numerical is known as data. The solutions are well structured by experienced teachers at BYJUâ€™S, as per the studentsâ€™ understanding capacity. The solutions in simple language, not only boost exam preparation but also helps them to score high marks in academics. Practising these solutions regularly, students obtain problem solving and time management skills, which are vital from the exam point of view. For more conceptual knowledge, students can download ML Aggarwal Solutions for Class 8 Maths Chapter 19 Data Handling PDF, from the links available here.
Chapter 19 provides information based on the data in the form of a frequency distribution table, pie chart and bar graph to represent the given set of data. These solutions are the best study tools, as they clear the studentsâ€™ doubts within a short duration.
ML Aggarwal Solutions for Class 8 Maths Chapter 19 Data Handling Download PDF
Access ML Aggarwal Solutions for Class 8 Maths Chapter 19 Data Handling
Exercise 19.1
1. The result of a survey of 200 people about their favourite fruit is given below:
Fruit  Apple  Orange  Banana  Grapes  Guava  Pineapple  Papaya 
Number of people  45  30  20  50  15  25  15 
Represent the above data by a bar graph.
Solution:
Fruit  Apple  Orange  Banana  Grapes  Guava  Pineapple  Papaya 
Number of people  45  30  20  50  15  25  15 
The bar graph is shown below
2. Mr Khurana has two kitchen appliance stores. He compares the sales of two stores during a month and recovered as given below
Item  Number of items sold  
Store A  Store B  
Grill
Toaster Oven Blender Coffee maker 
40
35 30 40 35 
20
15 30 30 40 
Represent the above data by a double bar graph.
Solution:
Item  Number of items sold  
Store A  Store B  
Grill
Toaster Oven Blender Coffee maker 
40
35 30 40 35 
20
15 30 30 40 
The double bar graph of the given data is shown below
3. The number of goals scored by a football team in different matches is given below:
3, 1, 0, 4, 6, 0, 0, 1, 1, 2, 2, 3, 5, 1, 2, 0, 1, 0, 2, 3, 9, 2, 0, 1, 0, 1, 4, 1, 0, 2, 5, 1, 2, 2, 3, 1, 0, 0, 0, 1, 1, 0, 2, 3, 0, 1, 5, 2, 0
Make a frequency distribution table using tally marks
Solution:
Frequency table for the given data is as follows:
Number of goals scored  Tally Marks  Frequency of matches 
0
1 2 3 4 5 6 9 
    
14
13 10 5 2 3 1 1 
Total  49 
4. Given below a bar graph:
Read the bar graph carefully and answer the following questions:
(i) What is the information given by the bar graph?
(ii) On which item the expenditure is maximum?
(iii) On which item the expenditure is minimum?
(iv) State whether true or false:
Expenditure on education is twice the expenditure on clothing
Solution:
(i) Representation of the expenditure of monthly salary on different heads is the information given in the bar graph
(ii) The expenditure on food is maximum
(iii) The expenditure on clothing is minimum
(iv) Yes, the expenditure on education is twice the expenditure on clothing
5. Given below a double bar graph
Read the double bar graph carefully and answer the following questions:
(i) What is the information given by the double graph?
(ii) Which mode of transport girls using more than the boys?
(iii) Which mode of transport boys using the most?
(iv) In which mode of transport number of girls is half the number of boys?
Solution:
From the double bar graph:
(i) The bar graph represents number of boys and girls, going to school using different modes of transport
(ii) The mode of transport, girls using more than the boys is school bus
(iii) The mode of transport, boys using the most is bicycle
(iv) The mode of transport, number of girls is half the number of boys is walking
6. Using class intervals 05, 510, construct the frequency distribution table for the following data:
13, 6, 12, 9, 11, 14, 2, 8, 18, 16, 9, 13, 17, 11, 19, 6, 7, 12, 22, 21, 18, 1, 8, 12, 18
Solution:
The frequency table is as follows:
Class Intervals  Tally Marks  Frequency 
05
510 1015 1520 2025 

 
2
7 8 6 2 
Total  25 
7. Given below are the marks secured by 35 students in a surprise test:
41, 32, 35, 21, 11, 47, 42, 00, 05, 18, 25, 24, 29, 38, 30, 04, 14, 24, 34, 44, 48, 33, 36, 38, 41, 48, 08, 34, 39, 11, 13, 27, 26, 43, 03.
Taking class intervals 010, 1020 â€¦â€¦. Construct frequency distribution table. Find the number of students obtaining below 20 marks.
Solution:
The frequency table of the given data is shown below:
Class  Tally Marks  Frequency 
010
1020 2030 3040 4050 
5
5 7 10 8 

Total  35 
Number of students obtaining below 20 marks = 5 + 5
= 10
Hence, 10 students are getting below 20 marks
8. The electricity bills (in ?) of 40 houses in a locality are given below:
78, 87, 81, 52, 59, 65, 101, 108, 115, 95, 98, 65, 62, 121, 128, 63, 76, 84, 89, 91, 65, 101, 95, 81, 87, 105, 129, 92, 75, 105, 78, 72, 107, 116, 127, 100, 80, 82, 61, 118. Construct a grouped frequency distribution table of class size 10.
Solution:
The frequency distribution table for the given data is as follows:
Class Intervals
(Electricity bill in Rs) 
Tally Marks  Frequency
(Number of houses) 
5060
6070 7080 8090 90100 100110 110120 120130 

  
2
6 5 8 5 7 3 4 
Total  40 
9. Draw a histogram for the frequency table made for data in Question 8, and answer the following questions:
(i) Which group has the maximum number of houses?
(ii) How many houses pay less than Rs 100?
(iii) How many houses pay Rs 100 or more?
Solution:
Histogram of the given data in Question 8 is as follows:
(i) Group 8090 has maximum number of house
(ii) Number of houses who pay less than Rs 100 = 2 + 6 + 5 + 8 + 5
= 26
Hence, 26 houses pay less than Rs 100
(iii) Number of houses who pay Rs 100 or more = 7 + 3 + 4
= 14
Hence, 14 houses pay Rs 100 or more
10. The weights of 29 patients in a hospital were recorded as follows:
Weight (in kg)  5055  5560  6065  6570  7075  7580 
Number of patients  7  4  4  9  2  3 
Draw a histogram to represent this data visually
Solution:
Weight (in kg)  5055  5560  6065  6570  7075  7580 
Number of patients  7  4  4  9  2  3 
The histogram for the given data is shown below:
11. In a study of diabetic patients, the following data was obtained:
Age (in years)  1020  2030  3040  4050  5060  6070  7080 
Number of patients  3  8  30  36  27  15  6 
Represent the above data by a histogram
Solution:
Age (in years)  1020  2030  3040  4050  5060  6070  7080 
Number of patients  3  8  30  36  27  15  6 
The histogram representing the above given data is as follows:
12. The histogram showing the weekly wages (in Rs) of workers in a factory is given alongside:
Answer the following:
(i) What is the frequency of class 400425?
(ii) What is the class having a minimum frequency?
(iii) How many workers get more than Rs 425?
(iv) How many workers get less than Rs 475?
(v) Number of workers whose weekly wages are more than or equal to Rs 400 but less than Rs 450
Solution:
The weekly wages of workers in a factory is shown in the given histogram:
(i) The frequency of class 400425 is 18
(ii) The class having the minimum frequency is 475500
(iii) 34 workers are getting more than Rs 425
(iv) 54 workers are getting less than Rs 475
(v) Number of workers whose weekly wages are more than or equal to Rs 400 but less than Rs 450 is 28
13. The number of hours for which students of a particular class watched television during holidays is shown in the histogram below.
Answer the following:
(i) For how many hours did the maximum number of students watch T.V.?
(ii) How many students watched T.V. for less than 4 hours?
(iii) How many students spent more than 5 hours in watching T.V.?
(iv) How many students spent more than 2 hours but less than 4 hours in watching T.V.?
Solution:
From the given histogram,
(i) Maximum number of students watch T.V for 45 hours
(ii) 34 students watch T.V. for less than 4 hours
(iii) 14 students spent more than 5 hours in watching T.V.
(iv) 30 students spent more than 2 hours but less than 4 hours in watching T.V.
14. The number of literate females in the age group of 10 to 40 years in a town is shown in the histogram alongside.
Answer the following questions:
(i) Write the classes assuming all the classes are of equal width.
(ii) What is the class size?
(iii) In which age group are the literate females the least?
(iv) In which age group is the number of literate females the highest?
Solution:
(i) From the given histogram, the classes having equal width are 1015, 1520, 2025, 2530, 3035, 3540
(ii) In the given histogram, the class size is 5
(iii) In 1015 age group, the literate females are the least
(iv) In 1520 age group, the literate females are the highest
Exercise 19.2
1. The following data represents the different number of animals in a zoo. Prepare a pie chart for the given data.
Animals  Deer  Tiger  Elephant  Giraffe  Reptiles 
Number of animals  40  10  30  15  25 
Solution:
Animals  Number of animals  Central degree 
Deer
Tiger Elephant Giraffe Reptiles 
40
10 30 15 25 
(360^{0} Ã— 40) / 120 = 120^{0}
(360^{0} Ã— 10) / 120 = 30^{0} (360^{0} Ã— 30) / 120 = 90^{0} (360^{0} Ã— 15) / 120 = 45^{0} (360^{0} Ã— 25) / 120 = 75^{0} 
Total  120  360^{0} 
Pie chart for the given data is shown below
2. The following data represents the monthly expenditure of a family (in T) on various items. Draw a pie chart to represent this data.
Items  Food  House rent  Education  Savings  Health  Others 
Expenditure (in Rs)  12500  5000  7500  10000  5000  10000 
Solution:
Items  Expenditure (in Rs)  Central angles 
Food
House rent Education Savings Health Others 
12500
5000 7500 10000 5000 10000 
(12500 Ã— 360^{0}) / 50000 = 90^{0}
(5000 Ã— 360^{0}) / 50000 = 36^{0} (7500 Ã— 360^{0}) / 50000 = 54^{0} (10000 Ã— 360^{0}) / 50000 = 72^{0} (5000 Ã— 360^{0}) / 50000 = 36^{0} (10000 Ã— 360^{0}) / 50000 = 72^{0} 
Total  50000  360^{0} 
Pie chart for the given data is shown below
3. The following data represents the percentage distribution of the expenditure incurred in publishing a book.
Items  Paper cost  Printing cost  Binding  Royality  Transportation
cost 
Promotion cost 
Expenditure (in %)  25%  20%  20%  10%  15%  10% 
Solution:
Items  Expenditure  Central angles 
Paper cost
Printing cost Binding Royality Transportation cost Promotion cost 
25%
20% 20% 10% 15% 10% 
(360^{0} Ã— 25) / 100 = 90^{0}
(360^{0} Ã— 20) / 100 = 72^{0} (360^{0} Ã— 20) / 100 = 72^{0} (360^{0} Ã— 10) / 100 = 36^{0} (360^{0} Ã— 15) / 100 = 54^{0} (360^{0} Ã— 10) / 100 = 36^{0} 
Total  100%  360^{0} 
Pie chart representing the given data is as follows:
4. The following data represents the number of students got admission in different streams of a college:
Stream  Science  Arts  Commerce  Law  Management 
Number of students  400  300  500  250  350 
Draw a pie chart to represent this data
Solution:
Stream  Number of students  Central angle 
Science
Arts Commerce Law Management 
400
300 500 250 350 
(400 Ã— 360^{0}) / 1800 = 80^{0}
(300 Ã— 360^{0}) / 1800 = 60^{0} (500 Ã— 360^{0}) / 1800 = 100^{0} (250 Ã— 360^{0}) / 1800 = 50^{0} (350 Ã— 360^{0}) / 1800 = 70^{0} 
Total  1800  360^{0} 
Pie chart representing the above given data is shown below:
5. The adjoining pie chart shows the expenditure of a country on various sports during year 2012. Study the pie chart carefully and answer the following questions:
(i) What percent of total expenditure is spent on cricket?
(ii) How much percent more is spent on hockey than that on tennis?
(iii) If the total amount spent on sports in 2012 is Rs 1, 80, 00, 000 then find the amount spent on Badminton
(iv) If the total amount spent on sports in 2012 is Rs 2, 40, 00, 000 then find the amount spent on cricket and hockey together.
Solution:
The given pie chart represents the expenditure of a country on various sports during year 2012
(i) Given that expenditure on a cricket = 90^{0}
So,
(90 / 360^{0}) Ã— 100% = 25%
Therefore, 25% expenditure is spent on cricket
(ii) Given that expenditure on a hockey = 75^{0}
So,
(70 / 360^{0}) Ã— 100% = (125 / 6)% =
%
Expenditure on tennis = 50^{0}
Hence, (50 / 360^{0}) Ã— 100% = (125 / 9)% = 13.9%
So,
(125 / 6) â€“ (125 / 9) = 125 / 18
= (375 â€“ 250) / 18
= 125 / 18
= 6.95% more
Therefore, 6.95% more is spent on hockey than that on tennis
(iii) Total amount spent on sports = Rs 1, 80, 00, 000
Total amount spent on Badminton = Rs 1, 80, 00,000 Ã— (60 / 360^{0})
= Rs 30,00,000
(iv) If the total amount spent on sports = 2,40,00,000
Total amount spent on cricket and hockey together = 90^{0} + 75^{0}
= 165^{0}
= (165^{0} / 360^{0}) Ã— 2, 40, 00,000
= 1, 10, 00, 000
6. The adjoining pie chart shows the number of students enrolled in class VI to class X of a school.
If 1440 students are enrolled from VI to X, then answer the following questions:
(i) How many students are enrolled in class VIII?
(ii) How many students are more in class IX than in class X?
(iii) What is the sum of students enrolled in VII and VIII?
(iv) Find the ratio of students enrolled in VI to students enrolled in X
Solution:
The given pie chart represents the enrolment of students from class VI to class X in a school.
Total number of students enrolled from VI to X = 1440 students
(i) Enrolment of class VIII = (85 / 360^{0}) Ã— 1440 = 340 students
(ii) Difference in X and IX class enrolment = 75^{0} – 50^{0}
= 25^{0}
(25 / 360^{0}) Ã— 1440 = 100 students
Therefore, 100 students are more in class IX than in class X
(iii) Sum of students enrolled in VII and VIII classes = 70^{0} + 85^{0}
= 155^{0}
(155 / 360^{0}) Ã— 1440 = 620 students
Therefore, the sum of students enrolled in class VI and VIII = 620 students
(iv) Ratio between the students enrolled in VI to students enrolled in X classes = 80^{0}: 50^{0}
= 8: 5
Therefore, the ratio between the students enrolled in VI to students enrolled in X classes is 8: 5
Exercise 19.3
1. List the outcomes you can see in these experiments
Solution:
(i) The outcomes in spinning wheel = A, A, A, B, C, D
(ii) The outcomes in drawing a ball from a bag containing 5 identical balls of different colours = White, Red, Blue, Green, Yellow
2. A die is rolled once. Find the probability of getting
(i) an even number
(ii) a multiple of 3
(iii) not a multiple of 3
Solution:
Total outcomes of a die when rolled once:
1, 2, 3, 4, 5, 6 = 6
(i) An even number: 2, 4, 6
i.e, Favourable outcomes = 3
Therefore,
Probability P(E) = 3 / 6
= 1 / 2
(ii) Multiple of 3 = 3, 6
i.e, Favourable outcomes = 2
Therefore,
Probability P(E) = 2 / 6
= 1 / 3
(iii) Not a multiple of 3 = 1, 2, 4, 5
i.e Favourable outcomes = 4
Therefore,
Probability P(E) = 4 / 6
= 2 / 3
3. Two coins are tossed together. Find the probability of getting
(i) two tails
(ii) atleast one tail
(iii) no tail
Solution:
The total outcomes, when two coins are tossed together = 2 Ã— 2
= 4
Therefore, outcomes are,
HH, HT, TH, TT
(i) Favourable outcomes of getting two tails = 1
Hence,
Probability P(E) = 1 / 4
(ii) Favourable outcomes of getting atleast one tail = TH, HT, TT
= 3
Hence,
Probability P(E) = 3 / 4
(iii) Favourable outcomes of getting no tail = HH
= 1
Hence,
Probability P(E) = 1 / 4
4. Three coins are tossed together. Find the probability of getting
(i) atleast two heads
(ii) atleast one tail
(iii) atmost one tail
Solution:
Three coins are tossed together
Hence,
Total outcomes = 8
= HHH, HHT, HTH, THH, HTT, TTH, TTT, THT
(i) Favourable outcomes of getting atleast two heads = HHH, HHT, HTH, THH
= 4 in numbers
Therefore,
Probability P(E) = (Number of favourable outcome) / (Number of possible outcome)
= 4 / 8
= 1 / 2
(ii) Favourable outcomes of getting atleast one tail = HHT, HTH, HTT, TTT, THH, THT, TTH
= 7 in numbers
Therefore,
Probability P(E) = (Number of favourable outcome) / (Number of possible outcome)
= 7 / 8
(iii) Favorable outcomes of getting atmost one tail = HHH, HHT, HTH, THH
= 4 in numbers
Therefore,
Probability P(E) = (Number of favourable outcome) / (Number of possible outcome)
= 4 / 8
= 1 / 2
5. Two dice are rolled simultaneaously. Find the probability of getting
(i) the sum as 7
(ii) the sum as 3 or 4
(iii) prime numbers on both the dice.
Solution:
Two dice are rolled simultaneously, then
Total outcomes = 6 Ã— 6
= 36
(i) Sum as 7 = (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)
= 6
Therefore,
Probability P(E) = (Favourable outcome) / (Total outcome)
= 6 / 36
= 1 / 6
(ii) The sum as 3 or 4 = (1, 2), (1, 3), (2, 1), (2, 2), (3, 1)
= 5
Therefore,
Probability P(E) = (Favourable outcome) / (Total outcome)
= 5 / 36
(iii) Prime numbers on both the side = (2, 2), (2, 3), (2, 5), (3, 2), (3, 3), (3, 5), (5, 2), (5, 3), (5, 5)
= 9
Therefore,
Probability P(E) = (Favourable outcome) / (Total outcome)
= 9 / 36
= 1 / 4
6. A box contains 600 screws, one tenth are rusted. One screw is taken out at random from the box. Find the probability that it is
(i) a rusted screw
(ii) not a rusted screw
Solution:
Given
Rusted screw = (1 / 10) of 600
= (1 / 10) Ã— 600
= 60 seconds
(i) Favourable outcomes of picking rusted screw = 60
Therefore,
Probability P(E) = 60 / 600
= 1 / 10
(ii) Probability (of not rusted screw) = 1 â€“ Probability (of rusted screw)
= 1 â€“ 1 / 10
= (10 â€“ 1) / 10
= 9 / 10
7. A letter is chosen from the word â€˜TRIANGLEâ€™. What is the probability that it is a vowel?
Solution:
Given word,
â€˜TRIANGLEâ€™
Total number of outcomes = 8
Vowels = I, A, E = 3
Therefore,
Probability P(E) = 3 / 8
Hence, the probability of vowel in â€˜TRIANGLEâ€™ is 3 / 8
8. A bag contains 5 red, 6 black and 4 white balls. A ball is drawn at random from the bag, find the probability the ball is drawn is
(i) white
(ii) not black
(iii) red or black
(iv) neither red nor black
Solution:
Given
In a bag, there are 5 red, 6 black and 4 white balls.
Then, total number of outcomes = 5 + 6 + 4
= 15
(i) Probability of white ball = 4 / 15
(ii) Probability of not black = 5 + 4
= 9 balls
Therefore,
Probability of not black = 9 / 15 = 3 / 5
(iii) Probability of red or black = 5 + 6
= 11
Therefore,
Probability of red or black = 11 / 15
(iv) Probability of ball which is neither red nor black, i.e, white ball = 4
Therefore,
Probability of ball which is neither red nor black = 4 / 15
9. A box contains 17 cards numbered 1, 2, 3â€¦â€¦.., 17 and are mixed thoroughly. A card is drawn at random from the box. Find the probability that the number on that card is
(i) odd
(ii) even
(iii) prime
(iv) divisible by 3
(v) divisible by 2 and 3 both
Solution:
Given
A box contains 17 cards numbered 1 to 17
So, total number of outcomes = 17
(i) Card bearing odd number
(1, 3, 5, 7, 9, 11, 13, 15, 17) = 9
Therefore,
Probability P(E) = 9 / 17
(ii) Even number
(2, 4, 6, 8, 10, 12, 14, 16) = 8
Therefore,
Probability P(E) = 8 / 17
(iii) Prime numbers
(2, 3, 5, 7, 11, 13, 17) = 7
Therefore,
Probability P(E) = 7 / 17
(iv) Numbers divisible by 3
3, 6, 9, 12, 15 = 5
Therefore,
Probability P(E) = 5 / 17
(v) Numbers divisible by 2 and 3 both
6, 12 = 2
Therefore,
Probability P(E) = 2 / 17
10. A card is drawn from a wellshuffled pack of 52 cards. Find the probability that the card drawn is:
(i) an ace
(ii) a red card
(iii) neither a king nor a queen
(iv) a red face card or an ace
(v) a card of spade
(vi) nonface card of red colour
Solution:
Total number of playing cards = 52
One card is drawn
(i) An ace = 4
Therefore,
Probability P(E) = 4 / 52
= 1 / 13
(ii) A red card = 13 + 13 = 26
Therefore,
Probability P(E) = 26 / 52
= 1 / 2
(iii) Neither a king nor a queen
Number of cards = 52 â€“ (4 + 4)
= 52 â€“ 8
= 44
Therefore,
Probability P(E) = 44 / 52
= 11 / 13
(iv) A red face card = 6
Therefore,
Probability P(E) = 6 / 52
= 3 / 26
(v) A card of spade or an ace = 13 + 3
= 16
Therefore,
Probability P(E) = 16 / 52
= 4 / 13
(vi) Nonface card of red colour = 26 â€“ 6
= 20
Therefore,
Probability P(E) = 20 / 52
= 5 / 13
11. In a lottery, there are 5 prized tickets and 995 blank tickets. A person buys a lottery ticket. Find the probability of his winning a prize.
Solution:
Given
Number of prized tickets = 5
Number of blank tickets = 995
So, total number of tickets = 5 + 995
= 1000
Probability of prized ticket P(E) = (Number of favourable outcome) / (Number of possible outcome)
= 5 / 1000
= 1 / 200
Therefore, the probability of his winning prize is 1 / 200