{"id":602348,"date":"2022-05-13T11:25:39","date_gmt":"2022-05-13T11:25:39","guid":{"rendered":"https:\/\/www.indcareer.com\/schools\/?p=602348"},"modified":"2022-05-14T06:13:54","modified_gmt":"2022-05-14T06:13:54","slug":"ml-aggarwal-solutions-for-class-8-maths-chapter-19-data-handling","status":"publish","type":"post","link":"https:\/\/www.indcareer.com\/schools\/ml-aggarwal-solutions-for-class-8-maths-chapter-19-data-handling\/","title":{"rendered":"ML Aggarwal Solutions for Class 8 Maths Chapter 19- Data Handling"},"content":{"rendered":"\n

Class 8: Maths Chapter 19 solutions. Complete Class 8 Maths Chapter 19 Notes.<\/p>\n\n\n\n

ML Aggarwal Solutions for Class 8 Maths Chapter 19- Data Handling<\/h2>\n\n\n\n

ML Aggarwal 8th Maths Chapter 19, Class 8 Maths Chapter 19 solutions<\/p>\n\n\n\n

Exercise 19.1<\/h4>\n\n\n\n

1. The result of a survey of 200 people about their favourite fruit is given below:<\/strong><\/p>\n\n\n\n

Fruit<\/strong><\/td>Apple<\/strong><\/td>Orange<\/strong><\/td>Banana<\/strong><\/td>Grapes<\/strong><\/td>Guava<\/strong><\/td>Pineapple<\/strong><\/td>Papaya<\/strong><\/td><\/tr>
Number of people<\/strong><\/td>45<\/strong><\/td>30<\/strong><\/td>20<\/strong><\/td>50<\/strong><\/td>15<\/strong><\/td>25<\/strong><\/td>15<\/strong><\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

Represent the above data by a bar graph.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Fruit<\/td>Apple<\/td>Orange<\/td>Banana<\/td>Grapes<\/td>Guava<\/td>Pineapple<\/td>Papaya<\/td><\/tr>
Number of people<\/td>45<\/td>30<\/td>20<\/td>50<\/td>15<\/td>25<\/td>15<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

The bar graph is shown below<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

2. Mr Khurana has two kitchen appliance stores. He compares the sales of two stores during a month and recovered as given below<\/strong><\/p>\n\n\n\n

Item<\/strong><\/td>Number of items sold<\/strong><\/td><\/tr>
Store A<\/strong><\/td>Store B<\/strong><\/td><\/tr>
Grill<\/strong>Toaster<\/strong>Oven<\/strong>Blender<\/strong>Coffee maker<\/strong><\/td>40<\/strong>35<\/strong>30<\/strong>40<\/strong>35<\/strong><\/td>20<\/strong>15<\/strong>30<\/strong>30<\/strong>40<\/strong><\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

Represent the above data by a double bar graph.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Item<\/td>Number of items sold<\/td><\/tr>
Store A<\/td>Store B<\/td><\/tr>
GrillToasterOvenBlenderCoffee maker<\/td>4035304035<\/td>2015303040<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

The double bar graph of the given data is shown below<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

3. The number of goals scored by a football team in different matches is given below:<\/strong><\/p>\n\n\n\n

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<\/strong><\/p>\n\n\n\n

Make a frequency distribution table using tally marks<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Frequency table for the given data is as follows:<\/p>\n\n\n\n

Number of goals scored<\/td>Tally Marks<\/td>Frequency of matches<\/td><\/tr>
01234569<\/td>\"ML\"ML\"ML\"ML|||||||<\/td>14131052311<\/td><\/tr>
Total<\/td><\/td>49<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

4. Given below a bar graph:<\/strong><\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

Read the bar graph carefully and answer the following questions:<\/strong><\/p>\n\n\n\n

(i) What is the information given by the bar graph?<\/strong><\/p>\n\n\n\n

(ii) On which item the expenditure is maximum?<\/strong><\/p>\n\n\n\n

(iii) On which item the expenditure is minimum?<\/strong><\/p>\n\n\n\n

(iv) State whether true or false:<\/strong><\/p>\n\n\n\n

Expenditure on education is twice the expenditure on clothing<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) Representation of the expenditure of monthly salary on different heads is the information given in the bar graph<\/p>\n\n\n\n

(ii) The expenditure on food is maximum<\/p>\n\n\n\n

(iii) The expenditure on clothing is minimum<\/p>\n\n\n\n

(iv) Yes, the expenditure on education is twice the expenditure on clothing<\/p>\n\n\n\n

5. Given below a double bar graph<\/strong><\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

Read the double bar graph carefully and answer the following questions:<\/strong><\/p>\n\n\n\n

(i) What is the information given by the double graph?<\/strong><\/p>\n\n\n\n

(ii) Which mode of transport girls using more than the boys?<\/strong><\/p>\n\n\n\n

(iii) Which mode of transport boys using the most?<\/strong><\/p>\n\n\n\n

(iv) In which mode of transport number of girls is half the number of boys?<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

From the double bar graph:<\/p>\n\n\n\n

(i) The bar graph represents number of boys and girls, going to school using different modes of transport<\/p>\n\n\n\n

(ii) The mode of transport, girls using more than the boys is school bus<\/p>\n\n\n\n

(iii) The mode of transport, boys using the most is bicycle<\/p>\n\n\n\n

(iv) The mode of transport, number of girls is half the number of boys is walking<\/p>\n\n\n\n

6. Using class intervals 0-5, 5-10, construct the frequency distribution table for the following data:<\/strong><\/p>\n\n\n\n

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<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

The frequency table is as follows:<\/p>\n\n\n\n

Class Intervals<\/td>Tally Marks<\/td>Frequency<\/td><\/tr>
0-55-1010-1515-2020-25<\/td>||\"ML\"ML\"ML||<\/td>27862<\/td><\/tr>
Total<\/td><\/td>25<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

7. Given below are the marks secured by 35 students in a surprise test:<\/strong><\/p>\n\n\n\n

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.<\/strong><\/p>\n\n\n\n

Taking class intervals 0-10, 10-20 \u2026\u2026. Construct frequency distribution table. Find the number of students obtaining below 20 marks.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

The frequency table of the given data is shown below:<\/p>\n\n\n\n

Class<\/td>Tally Marks<\/td>Frequency<\/td><\/tr>
0-1010-2020-3030-4040-50<\/td>\"ML\"ML\"ML\"ML\"ML<\/td>557108<\/td><\/tr>
Total<\/td><\/td>35<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

Number of students obtaining below 20 marks = 5 + 5<\/p>\n\n\n\n

= 10<\/p>\n\n\n\n

Hence, 10 students are getting below 20 marks<\/p>\n\n\n\n

8. The electricity bills (in ?) of 40 houses in a locality are given below:<\/strong><\/p>\n\n\n\n

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.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

The frequency distribution table for the given data is as follows:<\/p>\n\n\n\n

Class Intervals(Electricity bill in Rs)<\/td>Tally Marks<\/td>Frequency(Number of houses)<\/td><\/tr>
50-6060-7070-8080-9090-100100-110110-120120-130<\/td>||\"ML\"ML\"ML\"ML\"ML|||||||<\/td>26585734<\/td><\/tr>
Total<\/td><\/td>40<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

9. Draw a histogram for the frequency table made for data in Question 8, and answer the following questions:<\/strong><\/p>\n\n\n\n

(i) Which group has the maximum number of houses?<\/strong><\/p>\n\n\n\n

(ii) How many houses pay less than Rs 100?<\/strong><\/p>\n\n\n\n

(iii) How many houses pay Rs 100 or more?<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Histogram of the given data in Question 8 is as follows:<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

(i) Group 80-90 has maximum number of house<\/p>\n\n\n\n

(ii) Number of houses who pay less than Rs 100 = 2 + 6 + 5 + 8 + 5<\/p>\n\n\n\n

= 26<\/p>\n\n\n\n

Hence, 26 houses pay less than Rs 100<\/p>\n\n\n\n

(iii) Number of houses who pay Rs 100 or more = 7 + 3 + 4<\/p>\n\n\n\n

= 14<\/p>\n\n\n\n

Hence, 14 houses pay Rs 100 or more<\/p>\n\n\n\n

10. The weights of 29 patients in a hospital were recorded as follows:<\/strong><\/p>\n\n\n\n

Weight (in kg)<\/strong><\/td>50-55<\/strong><\/td>55-60<\/strong><\/td>60-65<\/strong><\/td>65-70<\/strong><\/td>70-75<\/strong><\/td>75-80<\/strong><\/td><\/tr>
Number of patients<\/strong><\/td>7<\/strong><\/td>4<\/strong><\/td>4<\/strong><\/td>9<\/strong><\/td>2<\/strong><\/td>3<\/strong><\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

Draw a histogram to represent this data visually<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Weight (in kg)<\/td>50-55<\/td>55-60<\/td>60-65<\/td>65-70<\/td>70-75<\/td>75-80<\/td><\/tr>
Number of patients<\/td>7<\/td>4<\/td>4<\/td>9<\/td>2<\/td>3<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

The histogram for the given data is shown below:<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

11. In a study of diabetic patients, the following data was obtained:<\/strong><\/p>\n\n\n\n

Age (in years)<\/strong><\/td>10-20<\/strong><\/td>20-30<\/strong><\/td>30-40<\/strong><\/td>40-50<\/strong><\/td>50-60<\/strong><\/td>60-70<\/strong><\/td>70-80<\/strong><\/td><\/tr>
Number of patients<\/strong><\/td>3<\/strong><\/td>8<\/strong><\/td>30<\/strong><\/td>36<\/strong><\/td>27<\/strong><\/td>15<\/strong><\/td>6<\/strong><\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

Represent the above data by a histogram<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Age (in years)<\/td>10-20<\/td>20-30<\/td>30-40<\/td>40-50<\/td>50-60<\/td>60-70<\/td>70-80<\/td><\/tr>
Number of patients<\/td>3<\/td>8<\/td>30<\/td>36<\/td>27<\/td>15<\/td>6<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

The histogram representing the above given data is as follows:<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

12. The histogram showing the weekly wages (in Rs) of workers in a factory is given alongside:<\/strong><\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

Answer the following:<\/strong><\/p>\n\n\n\n

(i) What is the frequency of class 400-425?<\/strong><\/p>\n\n\n\n

(ii) What is the class having a minimum frequency?<\/strong><\/p>\n\n\n\n

(iii) How many workers get more than Rs 425?<\/strong><\/p>\n\n\n\n

(iv) How many workers get less than Rs 475?<\/strong><\/p>\n\n\n\n

(v) Number of workers whose weekly wages are more than or equal to Rs 400 but less than Rs 450<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

The weekly wages of workers in a factory is shown in the given histogram:<\/p>\n\n\n\n

(i) The frequency of class 400-425 is 18<\/p>\n\n\n\n

(ii) The class having the minimum frequency is 475-500<\/p>\n\n\n\n

(iii) 34 workers are getting more than Rs 425<\/p>\n\n\n\n

(iv) 54 workers are getting less than Rs 475<\/p>\n\n\n\n

(v) Number of workers whose weekly wages are more than or equal to Rs 400 but less than Rs 450 is 28<\/p>\n\n\n\n

13. The number of hours for which students of a particular class watched television during holidays is shown in the histogram below.<\/strong><\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

Answer the following:<\/strong><\/p>\n\n\n\n

(i) For how many hours did the maximum number of students watch T.V.?<\/strong><\/p>\n\n\n\n

(ii) How many students watched T.V. for less than 4 hours?<\/strong><\/p>\n\n\n\n

(iii) How many students spent more than 5 hours in watching T.V.?<\/strong><\/p>\n\n\n\n

(iv) How many students spent more than 2 hours but less than 4 hours in watching T.V.?<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

From the given histogram,<\/p>\n\n\n\n

(i) Maximum number of students watch T.V for 4-5 hours<\/p>\n\n\n\n

(ii) 34 students watch T.V. for less than 4 hours<\/p>\n\n\n\n

(iii) 14 students spent more than 5 hours in watching T.V.<\/p>\n\n\n\n

(iv) 30 students spent more than 2 hours but less than 4 hours in watching T.V.<\/p>\n\n\n\n

14. The number of literate females in the age group of 10 to 40 years in a town is shown in the histogram alongside.<\/strong><\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

Answer the following questions:<\/strong><\/p>\n\n\n\n

(i) Write the classes assuming all the classes are of equal width.<\/strong><\/p>\n\n\n\n

(ii) What is the class size?<\/strong><\/p>\n\n\n\n

(iii) In which age group are the literate females the least?<\/strong><\/p>\n\n\n\n

(iv) In which age group is the number of literate females the highest?<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) From the given histogram, the classes having equal width are 10-15, 15-20, 20-25, 25-30, 30-35, 35-40<\/p>\n\n\n\n

(ii) In the given histogram, the class size is 5<\/p>\n\n\n\n

(iii) In 10-15 age group, the literate females are the least<\/p>\n\n\n\n

(iv) In 15-20 age group, the literate females are the highest<\/p>\n\n\n\n

Exercise 19.2<\/h4>\n\n\n\n

1. The following data represents the different number of animals in a zoo. Prepare a pie chart for the given data.<\/strong><\/p>\n\n\n\n

Animals<\/strong><\/td>Deer<\/strong><\/td>Tiger<\/strong><\/td>Elephant<\/strong><\/td>Giraffe<\/strong><\/td>Reptiles<\/strong><\/td><\/tr>
Number of animals<\/strong><\/td>40<\/strong><\/td>10<\/strong><\/td>30<\/strong><\/td>15<\/strong><\/td>25<\/strong><\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Animals<\/td>Number of animals<\/td>Central degree<\/td><\/tr>
DeerTigerElephantGiraffeReptiles<\/td>4010301525<\/td>(3600<\/sup> \u00d7 40) \/ 120 = 1200<\/sup>(3600<\/sup> \u00d7 10) \/ 120 = 300<\/sup>(3600<\/sup> \u00d7 30) \/ 120 = 900<\/sup>(3600<\/sup> \u00d7 15) \/ 120 = 450<\/sup>(3600<\/sup> \u00d7 25) \/ 120 = 750<\/sup><\/td><\/tr>
Total<\/td>120<\/td>3600<\/sup><\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

Pie chart for the given data is shown below<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

2. The following data represents the monthly expenditure of a family (in T) on various items. Draw a pie chart to represent this data.<\/strong><\/p>\n\n\n\n

Items<\/strong><\/td>Food<\/strong><\/td>House rent<\/strong><\/td>Education<\/strong><\/td>Savings<\/strong><\/td>Health<\/strong><\/td>Others<\/strong><\/td><\/tr>
Expenditure (in Rs)<\/strong><\/td>12500<\/strong><\/td>5000<\/strong><\/td>7500<\/strong><\/td>10000<\/strong><\/td>5000<\/strong><\/td>10000<\/strong><\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Items<\/td>Expenditure (in Rs)<\/td>Central angles<\/td><\/tr>
FoodHouse rentEducationSavingsHealthOthers<\/td>125005000750010000500010000<\/td>(12500 \u00d7 3600<\/sup>) \/ 50000 = 900<\/sup>(5000 \u00d7 3600<\/sup>) \/ 50000 = 360<\/sup>(7500 \u00d7 3600<\/sup>) \/ 50000 = 540<\/sup>(10000 \u00d7 3600<\/sup>) \/ 50000 = 720<\/sup>(5000 \u00d7 3600<\/sup>) \/ 50000 = 360<\/sup>(10000 \u00d7 3600<\/sup>) \/ 50000 = 720<\/sup><\/td><\/tr>
Total<\/td>50000<\/td>3600<\/sup><\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

Pie chart for the given data is shown below<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

3. The following data represents the percentage distribution of the expenditure incurred in publishing a book.<\/strong><\/p>\n\n\n\n

Items<\/strong><\/td>Paper cost<\/strong><\/td>Printing cost<\/strong><\/td>Binding<\/strong><\/td>Royality<\/strong><\/td>Transportation<\/strong>cost<\/strong><\/td>Promotion cost<\/strong><\/td><\/tr>
Expenditure (in %)<\/strong><\/td>25%<\/strong><\/td>20%<\/strong><\/td>20%<\/strong><\/td>10%<\/strong><\/td>15%<\/strong><\/td>10%<\/strong><\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Items<\/td>Expenditure<\/td>Central angles<\/td><\/tr>
Paper costPrinting costBindingRoyalityTransportation costPromotion cost<\/td>25%20%20%10%15%10%<\/td>(3600<\/sup> \u00d7 25) \/ 100 = 900<\/sup>(3600<\/sup> \u00d7 20) \/ 100 = 720<\/sup>(3600<\/sup> \u00d7 20) \/ 100 = 720<\/sup>(3600<\/sup> \u00d7 10) \/ 100 = 360<\/sup>(3600<\/sup> \u00d7 15) \/ 100 = 540<\/sup>(3600<\/sup> \u00d7 10) \/ 100 = 360<\/sup><\/td><\/tr>
Total<\/td>100%<\/td>3600<\/sup><\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

Pie chart representing the given data is as follows:<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

4. The following data represents the number of students got admission in different streams of a college:<\/strong><\/p>\n\n\n\n

Stream<\/strong><\/td>Science<\/strong><\/td>Arts<\/strong><\/td>Commerce<\/strong><\/td>Law<\/strong><\/td>Management<\/strong><\/td><\/tr>
Number of students<\/strong><\/td>400<\/strong><\/td>300<\/strong><\/td>500<\/strong><\/td>250<\/strong><\/td>350<\/strong><\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

Draw a pie chart to represent this data<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Stream<\/td>Number of students<\/td>Central angle<\/td><\/tr>
ScienceArtsCommerceLawManagement<\/td>400300500250350<\/td>(400 \u00d7 3600<\/sup>) \/ 1800 = 800<\/sup>(300 \u00d7 3600<\/sup>) \/ 1800 = 600<\/sup>(500 \u00d7 3600<\/sup>) \/ 1800 = 1000<\/sup>(250 \u00d7 3600<\/sup>) \/ 1800 = 500<\/sup>(350 \u00d7 3600<\/sup>) \/ 1800 = 700<\/sup><\/td><\/tr>
Total<\/td>1800<\/td>3600<\/sup><\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

Pie chart representing the above given data is shown below:<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

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:<\/strong><\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

(i) What percent of total expenditure is spent on cricket?<\/strong><\/p>\n\n\n\n

(ii) How much percent more is spent on hockey than that on tennis?<\/strong><\/p>\n\n\n\n

(iii) If the total amount spent on sports in 2012 is Rs 1, 80, 00, 000 then find the amount spent on Badminton<\/strong><\/p>\n\n\n\n

(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.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

The given pie chart represents the expenditure of a country on various sports during year 2012<\/p>\n\n\n\n

(i) Given that expenditure on a cricket = 900<\/sup><\/p>\n\n\n\n

So,<\/p>\n\n\n\n

(90 \/ 3600<\/sup>) \u00d7 100% = 25%<\/p>\n\n\n\n

Therefore, 25% expenditure is spent on cricket<\/p>\n\n\n\n

(ii) Given that expenditure on a hockey = 750<\/sup><\/p>\n\n\n\n

So,<\/p>\n\n\n\n

(70 \/ 3600<\/sup>) \u00d7 100% = (125 \/ 6)% =
\"ML%<\/p>\n\n\n\n

Expenditure on tennis = 500<\/sup><\/p>\n\n\n\n

Hence, (50 \/ 3600<\/sup>) \u00d7 100% = (125 \/ 9)% = 13.9%<\/p>\n\n\n\n

So,<\/p>\n\n\n\n

(125 \/ 6) \u2013 (125 \/ 9) = 125 \/ 18<\/p>\n\n\n\n

= (375 \u2013 250) \/ 18<\/p>\n\n\n\n

= 125 \/ 18<\/p>\n\n\n\n

= 6.95% more<\/p>\n\n\n\n

Therefore, 6.95% more is spent on hockey than that on tennis<\/p>\n\n\n\n

(iii) Total amount spent on sports = Rs 1, 80, 00, 000<\/p>\n\n\n\n

Total amount spent on Badminton = Rs 1, 80, 00,000 \u00d7 (60 \/ 3600<\/sup>)<\/p>\n\n\n\n

= Rs 30,00,000<\/p>\n\n\n\n

(iv) If the total amount spent on sports = 2,40,00,000<\/p>\n\n\n\n

Total amount spent on cricket and hockey together = 900<\/sup> + 750<\/sup><\/p>\n\n\n\n

= 1650<\/sup><\/p>\n\n\n\n

= (1650<\/sup> \/ 3600<\/sup>) \u00d7 2, 40, 00,000<\/p>\n\n\n\n

= 1, 10, 00, 000<\/p>\n\n\n\n

6. The adjoining pie chart shows the number of students enrolled in class VI to class X of a school.<\/strong><\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

If 1440 students are enrolled from VI to X, then answer the following questions:<\/strong><\/p>\n\n\n\n

(i) How many students are enrolled in class VIII?<\/strong><\/p>\n\n\n\n

(ii) How many students are more in class IX than in class X?<\/strong><\/p>\n\n\n\n

(iii) What is the sum of students enrolled in VII and VIII?<\/strong><\/p>\n\n\n\n

(iv) Find the ratio of students enrolled in VI to students enrolled in X<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

The given pie chart represents the enrolment of students from class VI to class X in a school.<\/p>\n\n\n\n

Total number of students enrolled from VI to X = 1440 students<\/p>\n\n\n\n

(i) Enrolment of class VIII = (85 \/ 3600<\/sup>) \u00d7 1440 = 340 students<\/p>\n\n\n\n

(ii) Difference in X and IX class enrolment = 750<\/sup> \u2013 500<\/sup><\/p>\n\n\n\n

= 250<\/sup><\/p>\n\n\n\n

(25 \/ 3600<\/sup>) \u00d7 1440 = 100 students<\/p>\n\n\n\n

Therefore, 100 students are more in class IX than in class X<\/p>\n\n\n\n

(iii) Sum of students enrolled in VII and VIII classes = 700<\/sup> + 850<\/sup><\/p>\n\n\n\n

= 1550<\/sup><\/p>\n\n\n\n

(155 \/ 3600<\/sup>) \u00d7 1440 = 620 students<\/p>\n\n\n\n

Therefore, the sum of students enrolled in class VI and VIII = 620 students<\/p>\n\n\n\n

(iv) Ratio between the students enrolled in VI to students enrolled in X classes = 800<\/sup>: 500<\/sup><\/p>\n\n\n\n

= 8: 5<\/p>\n\n\n\n

Therefore, the ratio between the students enrolled in VI to students enrolled in X classes is 8: 5<\/p>\n\n\n\n

Exercise 19.3<\/h4>\n\n\n\n

1. List the outcomes you can see in these experiments<\/strong><\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) The outcomes in spinning wheel = A, A, A, B, C, D<\/p>\n\n\n\n

(ii) The outcomes in drawing a ball from a bag containing 5 identical balls of different colours = White, Red, Blue, Green, Yellow<\/p>\n\n\n\n

2. A die is rolled once. Find the probability of getting<\/strong><\/p>\n\n\n\n

(i) an even number<\/strong><\/p>\n\n\n\n

(ii) a multiple of 3<\/strong><\/p>\n\n\n\n

(iii) not a multiple of 3<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Total outcomes of a die when rolled once:<\/p>\n\n\n\n

1, 2, 3, 4, 5, 6 = 6<\/p>\n\n\n\n

(i) An even number: 2, 4, 6<\/p>\n\n\n\n

i.e, Favourable outcomes = 3<\/p>\n\n\n\n

Therefore,<\/p>\n\n\n\n

Probability P(E) = 3 \/ 6<\/p>\n\n\n\n

= 1 \/ 2<\/p>\n\n\n\n

(ii) Multiple of 3 = 3, 6<\/p>\n\n\n\n

i.e, Favourable outcomes = 2<\/p>\n\n\n\n

Therefore,<\/p>\n\n\n\n

Probability P(E) = 2 \/ 6<\/p>\n\n\n\n

= 1 \/ 3<\/p>\n\n\n\n

(iii) Not a multiple of 3 = 1, 2, 4, 5<\/p>\n\n\n\n

i.e Favourable outcomes = 4<\/p>\n\n\n\n

Therefore,<\/p>\n\n\n\n

Probability P(E) = 4 \/ 6<\/p>\n\n\n\n

= 2 \/ 3<\/p>\n\n\n\n

3. Two coins are tossed together. Find the probability of getting<\/strong><\/p>\n\n\n\n

(i) two tails<\/strong><\/p>\n\n\n\n

(ii) atleast one tail<\/strong><\/p>\n\n\n\n

(iii) no tail<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

The total outcomes, when two coins are tossed together = 2 \u00d7 2<\/p>\n\n\n\n

= 4<\/p>\n\n\n\n

Therefore, outcomes are,<\/p>\n\n\n\n

HH, HT, TH, TT<\/p>\n\n\n\n

(i) Favourable outcomes of getting two tails = 1<\/p>\n\n\n\n

Hence,<\/p>\n\n\n\n

Probability P(E) = 1 \/ 4<\/p>\n\n\n\n

(ii) Favourable outcomes of getting atleast one tail = TH, HT, TT<\/p>\n\n\n\n

= 3<\/p>\n\n\n\n

Hence,<\/p>\n\n\n\n

Probability P(E) = 3 \/ 4<\/p>\n\n\n\n

(iii) Favourable outcomes of getting no tail = HH<\/p>\n\n\n\n

= 1<\/p>\n\n\n\n

Hence,<\/p>\n\n\n\n

Probability P(E) = 1 \/ 4<\/p>\n\n\n\n

4. Three coins are tossed together. Find the probability of getting<\/strong><\/p>\n\n\n\n

(i) atleast two heads<\/strong><\/p>\n\n\n\n

(ii) atleast one tail<\/strong><\/p>\n\n\n\n

(iii) atmost one tail<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Three coins are tossed together<\/p>\n\n\n\n

Hence,<\/p>\n\n\n\n

Total outcomes = 8<\/p>\n\n\n\n

= HHH, HHT, HTH, THH, HTT, TTH, TTT, THT<\/p>\n\n\n\n

(i) Favourable outcomes of getting atleast two heads = HHH, HHT, HTH, THH<\/p>\n\n\n\n

= 4 in numbers<\/p>\n\n\n\n

Therefore,<\/p>\n\n\n\n

Probability P(E) = (Number of favourable outcome) \/ (Number of possible outcome)<\/p>\n\n\n\n

= 4 \/ 8<\/p>\n\n\n\n

= 1 \/ 2<\/p>\n\n\n\n

(ii) Favourable outcomes of getting atleast one tail = HHT, HTH, HTT, TTT, THH, THT, TTH<\/p>\n\n\n\n

= 7 in numbers<\/p>\n\n\n\n

Therefore,<\/p>\n\n\n\n

Probability P(E) = (Number of favourable outcome) \/ (Number of possible outcome)<\/p>\n\n\n\n

= 7 \/ 8<\/p>\n\n\n\n

(iii) Favorable outcomes of getting atmost one tail = HHH, HHT, HTH, THH<\/p>\n\n\n\n

= 4 in numbers<\/p>\n\n\n\n

Therefore,<\/p>\n\n\n\n

Probability P(E) = (Number of favourable outcome) \/ (Number of possible outcome)<\/p>\n\n\n\n

= 4 \/ 8<\/p>\n\n\n\n

= 1 \/ 2<\/p>\n\n\n\n

5. Two dice are rolled simultaneaously. Find the probability of getting<\/strong><\/p>\n\n\n\n

(i) the sum as 7<\/strong><\/p>\n\n\n\n

(ii) the sum as 3 or 4<\/strong><\/p>\n\n\n\n

(iii) prime numbers on both the dice.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Two dice are rolled simultaneously, then<\/p>\n\n\n\n

Total outcomes = 6 \u00d7 6<\/p>\n\n\n\n

= 36<\/p>\n\n\n\n

(i) Sum as 7 = (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)<\/p>\n\n\n\n

= 6<\/p>\n\n\n\n

Therefore,<\/p>\n\n\n\n

Probability P(E) = (Favourable outcome) \/ (Total outcome)<\/p>\n\n\n\n

= 6 \/ 36<\/p>\n\n\n\n

= 1 \/ 6<\/p>\n\n\n\n

(ii) The sum as 3 or 4 = (1, 2), (1, 3), (2, 1), (2, 2), (3, 1)<\/p>\n\n\n\n

= 5<\/p>\n\n\n\n

Therefore,<\/p>\n\n\n\n

Probability P(E) = (Favourable outcome) \/ (Total outcome)<\/p>\n\n\n\n

= 5 \/ 36<\/p>\n\n\n\n

(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)<\/p>\n\n\n\n

= 9<\/p>\n\n\n\n

Therefore,<\/p>\n\n\n\n

Probability P(E) = (Favourable outcome) \/ (Total outcome)<\/p>\n\n\n\n

= 9 \/ 36<\/p>\n\n\n\n

= 1 \/ 4<\/p>\n\n\n\n

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<\/strong><\/p>\n\n\n\n

(i) a rusted screw<\/strong><\/p>\n\n\n\n

(ii) not a rusted screw<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given<\/p>\n\n\n\n

Rusted screw = (1 \/ 10) of 600<\/p>\n\n\n\n

= (1 \/ 10) \u00d7 600<\/p>\n\n\n\n

= 60 seconds<\/p>\n\n\n\n

(i) Favourable outcomes of picking rusted screw = 60<\/p>\n\n\n\n

Therefore,<\/p>\n\n\n\n

Probability P(E) = 60 \/ 600<\/p>\n\n\n\n

= 1 \/ 10<\/p>\n\n\n\n

(ii) Probability (of not rusted screw) = 1 \u2013 Probability (of rusted screw)<\/p>\n\n\n\n

= 1 \u2013 1 \/ 10<\/p>\n\n\n\n

= (10 \u2013 1) \/ 10<\/p>\n\n\n\n

= 9 \/ 10<\/p>\n\n\n\n

7. A letter is chosen from the word \u2018TRIANGLE\u2019. What is the probability that it is a vowel?<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given word,<\/p>\n\n\n\n

\u2018TRIANGLE\u2019<\/p>\n\n\n\n

Total number of outcomes = 8<\/p>\n\n\n\n

Vowels = I, A, E = 3<\/p>\n\n\n\n

Therefore,<\/p>\n\n\n\n

Probability P(E) = 3 \/ 8<\/p>\n\n\n\n

Hence, the probability of vowel in \u2018TRIANGLE\u2019 is 3 \/ 8<\/p>\n\n\n\n

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<\/strong><\/p>\n\n\n\n

(i) white<\/strong><\/p>\n\n\n\n

(ii) not black<\/strong><\/p>\n\n\n\n

(iii) red or black<\/strong><\/p>\n\n\n\n

(iv) neither red nor black<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given<\/p>\n\n\n\n

In a bag, there are 5 red, 6 black and 4 white balls.<\/p>\n\n\n\n

Then, total number of outcomes = 5 + 6 + 4<\/p>\n\n\n\n

= 15<\/p>\n\n\n\n

(i) Probability of white ball = 4 \/ 15<\/p>\n\n\n\n

(ii) Probability of not black = 5 + 4<\/p>\n\n\n\n

= 9 balls<\/p>\n\n\n\n

Therefore,<\/p>\n\n\n\n

Probability of not black = 9 \/ 15 = 3 \/ 5<\/p>\n\n\n\n

(iii) Probability of red or black = 5 + 6<\/p>\n\n\n\n

= 11<\/p>\n\n\n\n

Therefore,<\/p>\n\n\n\n

Probability of red or black = 11 \/ 15<\/p>\n\n\n\n

(iv) Probability of ball which is neither red nor black, i.e, white ball = 4<\/p>\n\n\n\n

Therefore,<\/p>\n\n\n\n

Probability of ball which is neither red nor black = 4 \/ 15<\/p>\n\n\n\n

9. A box contains 17 cards numbered 1, 2, 3\u2026\u2026.., 17 and are mixed thoroughly. A card is drawn at random from the box. Find the probability that the number on that card is<\/strong><\/p>\n\n\n\n

(i) odd<\/strong><\/p>\n\n\n\n

(ii) even<\/strong><\/p>\n\n\n\n

(iii) prime<\/strong><\/p>\n\n\n\n

(iv) divisible by 3<\/strong><\/p>\n\n\n\n

(v) divisible by 2 and 3 both<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given<\/p>\n\n\n\n

A box contains 17 cards numbered 1 to 17<\/p>\n\n\n\n

So, total number of outcomes = 17<\/p>\n\n\n\n

(i) Card bearing odd number<\/p>\n\n\n\n

(1, 3, 5, 7, 9, 11, 13, 15, 17) = 9<\/p>\n\n\n\n

Therefore,<\/p>\n\n\n\n

Probability P(E) = 9 \/ 17<\/p>\n\n\n\n

(ii) Even number<\/p>\n\n\n\n

(2, 4, 6, 8, 10, 12, 14, 16) = 8<\/p>\n\n\n\n

Therefore,<\/p>\n\n\n\n

Probability P(E) = 8 \/ 17<\/p>\n\n\n\n

(iii) Prime numbers<\/p>\n\n\n\n

(2, 3, 5, 7, 11, 13, 17) = 7<\/p>\n\n\n\n

Therefore,<\/p>\n\n\n\n

Probability P(E) = 7 \/ 17<\/p>\n\n\n\n

(iv) Numbers divisible by 3<\/p>\n\n\n\n

3, 6, 9, 12, 15 = 5<\/p>\n\n\n\n

Therefore,<\/p>\n\n\n\n

Probability P(E) = 5 \/ 17<\/p>\n\n\n\n

(v) Numbers divisible by 2 and 3 both<\/p>\n\n\n\n

6, 12 = 2<\/p>\n\n\n\n

Therefore,<\/p>\n\n\n\n

Probability P(E) = 2 \/ 17<\/p>\n\n\n\n

10. A card is drawn from a well-shuffled pack of 52 cards. Find the probability that the card drawn is:<\/strong><\/p>\n\n\n\n

(i) an ace<\/strong><\/p>\n\n\n\n

(ii) a red card<\/strong><\/p>\n\n\n\n

(iii) neither a king nor a queen<\/strong><\/p>\n\n\n\n

(iv) a red face card or an ace<\/strong><\/p>\n\n\n\n

(v) a card of spade<\/strong><\/p>\n\n\n\n

(vi) non-face card of red colour<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Total number of playing cards = 52<\/p>\n\n\n\n

One card is drawn<\/p>\n\n\n\n

(i) An ace = 4<\/p>\n\n\n\n

Therefore,<\/p>\n\n\n\n

Probability P(E) = 4 \/ 52<\/p>\n\n\n\n

= 1 \/ 13<\/p>\n\n\n\n

(ii) A red card = 13 + 13 = 26<\/p>\n\n\n\n

Therefore,<\/p>\n\n\n\n

Probability P(E) = 26 \/ 52<\/p>\n\n\n\n

= 1 \/ 2<\/p>\n\n\n\n

(iii) Neither a king nor a queen<\/p>\n\n\n\n

Number of cards = 52 \u2013 (4 + 4)<\/p>\n\n\n\n

= 52 \u2013 8<\/p>\n\n\n\n

= 44<\/p>\n\n\n\n

Therefore,<\/p>\n\n\n\n

Probability P(E) = 44 \/ 52<\/p>\n\n\n\n

= 11 \/ 13<\/p>\n\n\n\n

(iv) A red face card = 6<\/p>\n\n\n\n

Therefore,<\/p>\n\n\n\n

Probability P(E) = 6 \/ 52<\/p>\n\n\n\n

= 3 \/ 26<\/p>\n\n\n\n

(v) A card of spade or an ace = 13 + 3<\/p>\n\n\n\n

= 16<\/p>\n\n\n\n

Therefore,<\/p>\n\n\n\n

Probability P(E) = 16 \/ 52<\/p>\n\n\n\n

= 4 \/ 13<\/p>\n\n\n\n

(vi) Non-face card of red colour = 26 \u2013 6<\/p>\n\n\n\n

= 20<\/p>\n\n\n\n

Therefore,<\/p>\n\n\n\n

Probability P(E) = 20 \/ 52<\/p>\n\n\n\n

= 5 \/ 13<\/p>\n\n\n\n

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.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given<\/p>\n\n\n\n

Number of prized tickets = 5<\/p>\n\n\n\n

Number of blank tickets = 995<\/p>\n\n\n\n

So, total number of tickets = 5 + 995<\/p>\n\n\n\n

= 1000<\/p>\n\n\n\n

Probability of prized ticket P(E) = (Number of favourable outcome) \/ (Number of possible outcome)<\/p>\n\n\n\n

= 5 \/ 1000<\/p>\n\n\n\n

= 1 \/ 200<\/p>\n\n\n\n

Therefore, the probability of his winning prize is 1 \/ 200<\/p>\n\n\n\n

Download PDF<\/strong><\/h2>\n\n\n\n

ML Aggarwal Solutions for Class 8 Maths Chapter 19- Data Handling<\/p>\n\n\n\n

Download PDF<\/strong>: ML Aggarwal Solutions for Class 8 Maths Chapter 19- Data Handling PDF<\/a><\/p>\n\n\n\n

Chapterwise ML Aggarwal Solutions for Class 8 Maths :<\/strong><\/h2>\n\n\n\n