ML Aggarwal Class 8 Maths Solutions Chapter 18 Data Handling

 

Exercise 19.1

 

Question 1. The result of a survey of 200 people about their favourite fruit is given below: Represent the above data by a bar graph.

Answer: A bar graph shows the number of people who like each type of fruit. The x-axis lists the different fruits, while the y-axis shows how many people chose each one. Apple was liked by 45 people, Orange by 30, Banana by 20, Grapes by 50 (the highest), Guava by 15, Pineapple by 25, and Papaya by 15. Each fruit gets its own vertical bar whose height matches the number of people.
In simple words: Draw tall bars to show which fruits more people like. Grapes has the tallest bar because the most people picked it.

Exam Tip: Always label the axes clearly - one axis shows the fruit names and the other shows the number of people. Make the bars the same width and space them evenly.

 

Question 2. Mr Khurana has two kitchen appliance stores. He compares the sales of two stores during a month and recorded as given below: Represent the above data by a double bar graph.

Answer: A double bar graph displays the sales from both stores side by side for each kitchen item. For every appliance, two bars are drawn next to each other - one bar (in one color) shows Store A's sales and another bar (in a different color) shows Store B's sales. This makes it easy to compare which store sold more of each item. For example, Store A sold more Grills and Toasters, Store B sold more Coffee makers, and both stores sold an equal number of Ovens.
In simple words: Draw two bars next to each other for each item. One bar shows how many Store A sold, the other shows how many Store B sold. Different colors help you tell them apart.

Exam Tip: Use a legend (color key) to clearly mark which color stands for Store A and which for Store B. Place the two bars for each item very close together for easy comparison.

 

Question 3. The number of goals scored by a football team in different matches is given below: Make a frequency distribution table using tally marks.
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
Answer:

Total 49
In simple words: Count how many times each number appears in the list. Use tally marks (small lines) to keep track - every fifth mark is shown as a line through four others. The total count for each number is the frequency.

Exam Tip: Make sure your tally marks are neat and clearly grouped in fives. Always verify that the sum of all frequencies equals the total number of matches played.

 

Question 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
Answer:
(i) The bar graph displays how a monthly salary gets divided among different spending areas - Food, Clothing, Education, Miscellaneous expenses, and Savings.
(ii) Food requires the maximum spending.
(iii) Clothing requires the minimum spending.
(iv) True. The spending on education is exactly twice what is spent on clothing.
In simple words: The graph shows where money goes each month. Food takes the most money, and clothing takes the least. Education spending is two times clothing spending.

Exam Tip: When comparing bar heights, look at where each bar reaches on the y-axis. To check if one is twice another, see if the taller bar's height is exactly double the shorter one's height.

 

Question 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?
Answer:
(i) The double bar graph shows how many boys and girls travel to school by different methods - School bus, Walking, Bicycle, and Other vehicles.
(ii) Girls use School bus more often than boys.
(iii) Boys travel to school mostly by Bicycle.
(iv) Walking is the mode where the number of girls equals half the number of boys.
In simple words: This graph compares boys and girls. More girls take the school bus. Most boys ride bicycles. For walking, there are half as many girls as boys.

Exam Tip: Compare the bar pairs carefully. When one value is half another, the smaller bar should reach exactly halfway up the taller bar's height on the graph.

 

Question 6. Using class intervals 0-5, 5-10, 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
Answer:

Total 25
In simple words: Sort each number into the right range bucket. Numbers 0 through 4 go in the first group, 5 through 9 in the next, and so on. Count how many land in each bucket.

Exam Tip: Remember that in the interval 0-5, the number 5 itself goes into the next interval (5-10). The class intervals never overlap - each number fits into exactly one group.

 

Question 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 0-10, 10-20 ...... Construct frequency distribution table. Find the number of students obtaining below 20 marks.
Answer:

Total 35
Students scoring below 20 marks = 5 + 5 = 10
Hence, 10 students achieved below 20 marks.
In simple words: Group the marks into bundles of 10. Count how many students fall in each group. To find students below 20, add the first two groups together.

Exam Tip: "Below 20 marks" means scores in the 0-10 and 10-20 ranges only. Do not include any students from the 20-30 range or higher.

 

Question 8. The electricity bills (in Rs) 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.
Answer:

Total 40
In simple words: Sort all the electricity bills into groups of 10 rupees each. Count how many houses fall in each price range using tally marks.

Exam Tip: Always verify that the class size is consistent (10 rupees in every interval) and that the sum of all frequencies equals the total number of houses (40).

 

Question 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?
Answer:
(i) The group 80-90 has the maximum number of houses (8 houses).
(ii) Houses that pay less than Rs 100 = 2 + 6 + 5 + 8 + 5 = 26 houses.
(iii) Houses that pay Rs 100 or more = 7 + 3 + 4 = 14 houses.
In simple words: Look at the tallest bar in the histogram - that group has the most houses. To count "less than 100", add up all the frequencies from the lower groups before reaching 100. To count "100 or more", add the last three groups.

Exam Tip: In a histogram, the height of each bar shows the frequency. Always check that your addition is correct by verifying that all groups sum to 40 (the total number of houses).

 

Question 10. The weights of 29 patients in a hospital were recorded as follows: Draw a histogram to represent this data visually.

Answer: A histogram is drawn with weight ranges on the horizontal axis and the count of patients on the vertical axis. Each weight group is shown as a rectangular bar, with the bar's height corresponding to how many patients fall within that weight range. The heaviest concentration of patients (9 people) is in the 65-70 kg range, while the fewest patients (2 people) weigh between 70-75 kg.
In simple words: Draw bars for each weight group. Make each bar's height match the number of patients in that group. The 65-70 kg group has the tallest bar.

Exam Tip: In a histogram, bars must be drawn adjacent to each other with no gaps between them. The width of each bar represents the class interval (5 kg in this case), and the height represents the frequency.

 

Question 11. In a study of diabetic patients, the following data was obtained: Represent the above data by a histogram.

Answer: The histogram displays the distribution of diabetic patients across different age groups. Age ranges appear on the horizontal axis while the number of patients is shown on the vertical axis. The 40-50 age group shows the highest frequency with 36 patients, indicating that middle-aged people make up the largest portion of the diabetic patient population. The 10-20 age group has the lowest frequency with only 3 patients.
In simple words: Each bar shows how many patients are in each age group. The tallest bar is for ages 40-50 because most diabetic patients are in that age range.

Exam Tip: Notice that the histogram shows a clear pattern - the number of diabetic patients increases from ages 10-50, then decreases for ages 50 and above. This type of observation is important when analyzing histograms.

 

Question 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 400-425? (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
Answer:
(i) The frequency of class 400-425 is 18 workers.
(ii) The class having the minimum frequency is 475-500 (frequency = 4 workers).
(iii) Workers earning more than Rs 425 = 10 + 20 + 4 = 34 workers.
(iv) Workers earning less than Rs 475 = 6 + 18 + 10 + 20 = 54 workers.
(v) Workers with wages from Rs 400 to less than Rs 450 = 18 + 10 = 28 workers.
In simple words: Read the height of each bar from the graph. For "more than 425", include all wage groups above 425. For "less than 475", add up all groups below 475. For a range like "400 to less than 450", add only those two groups.

Exam Tip: Pay close attention to the wording - "more than", "less than", and "more than or equal to" have different meanings. Check whether the boundaries are included or excluded in each question.

 

Question 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.?
Answer:
(i) The maximum number of students watched T.V. for 4-5 hours.
(ii) Students who watched T.V. for less than 4 hours = 4 + 8 + 22 = 34 students.
(iii) Students who spent more than 5 hours watching T.V. = 8 + 6 = 14 students.
(iv) Students who spent more than 2 hours but less than 4 hours watching T.V. = 22 + 8 = 30 students.
In simple words: Find the tallest bar to answer the first question. For the other questions, add up the bars that match the conditions given.

Exam Tip: When combining frequencies from multiple bars, be careful about the boundaries. "More than 2 hours but less than 4 hours" includes only the 2-3 and 3-4 hour groups, not the 4-5 group.

 

Question 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?
Answer:
(i) The equal-width classes are: 10-15, 15-20, 20-25, 25-30, 30-35, 35-40.
(ii) The class size is 5 years.
(iii) The 10-15 age group has the fewest literate females.
(iv) The 15-20 age group has the highest count of literate females.
In simple words: Divide the 10-40 range into equal chunks of 5 years each. Look at the bars to find the smallest and tallest ones.

Exam Tip: When determining equal class intervals, divide the range (40 - 10 = 30) by the number of classes. If the class size is 5, you will have 6 classes total (30 ÷ 5 = 6).

 

Exercise 19.2

 

Question 1. The following data represents the different number of animals in a zoo. Prepare a pie chart for the given data.

Answer: A pie chart breaks a circle into slices, with each slice's size showing what part of the total belongs to each animal type. To find each slice's angle, multiply that animal's count by 360° and divide by the total (120). Deer takes up 120°, Elephant takes 90°, Reptiles takes 75°, Giraffe takes 45°, and Tiger takes 30°. When you draw these slices and label them, you get a visual picture of which animals are most common in the zoo.
In simple words: Each animal group gets a slice of the pie based on how many there are. Bigger numbers get bigger slices. All the slices together make one full circle (360 degrees).

Exam Tip: Always verify that all the central angles add up to exactly 360°. Use a protractor to measure the angles accurately when drawing the pie chart by hand.

 

Question 2. The following data represents the monthly expenditure of a family (in Rs) on various items. Draw a pie chart to represent this data.

Answer: The pie chart demonstrates how each rupee of the family's monthly budget is allocated across different areas. Food receives the largest share at 90°, followed by Savings and Others at 72° each. Education gets 54°, while House rent and Health each receive 36°. The formula used is: Central angle = (Item value / Total value) × 360°. This visual representation makes it easy to see that the family spends the most on food and saves or spends significantly on other items.
In simple words: The pie is divided into parts showing how much money goes to each thing. Food gets the biggest piece. House rent and Health get the smallest pieces.

Exam Tip: Label each slice clearly with both the item name and its percentage or angle. Ensure that the sum of all central angles equals 360°.

 

Question 3. The following data represents the percentage distribution of the expenditure incurred in publishing a book.

Answer: The pie chart reveals how the budget for publishing a book gets divided among different cost categories. Paper cost accounts for the largest share at 25% (90°), making it the biggest expense. Printing and Binding each take 20% (72°), while Transportation accounts for 15% (54°). Royalty and Promotion each represent 10% (36°) of the total publishing budget. This visual breakdown helps publishers understand which costs demand the most investment when bringing a book to market.
In simple words: Paper is the biggest cost at 25% of the total. Printing and binding tie for second place at 20% each. The pie chart shows all costs add up to 100%.

Exam Tip: When working with percentages, the calculation is simpler: (percentage × 360) / 100 gives the central angle directly. Always check that all percentages sum to 100% before drawing.

 

Question 4. The following data represents the number of students got admission in different streams of a college: Draw a pie chart to represent this data

Answer: The pie chart illustrates the distribution of college admissions across five different academic streams. Commerce has the highest enrollment with 500 students, represented by a 100° slice. Science comes next with 400 students (80°), followed by Management with 350 students (70°). Arts attracts 300 students (60°), and Law has the smallest enrollment with 250 students (50°). This visual representation makes it simple to compare the relative popularity of each stream at the college.
In simple words: Commerce attracts the most students, followed by Science. Law is the least popular stream. The pie shows how students are spread across the five options.

Exam Tip: When the total is large (like 1800 students), double-check your division calculation: (360° / total) × each value gives the central angle. Always verify that the largest enrollment gets the largest angle slice.

 

Question 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.
Answer: The pie chart below displays how a country distributed its sports funding throughout 2012 across different games.

(i) Cricket received a central angle of 90°. To find the percentage: \( \frac{90}{360} \times 100\% = 25\% \)
So, 25% of the total expenditure went to cricket.

(ii) Hockey had a central angle of 75°, while tennis had 50°. Hockey's percentage: \( \frac{75}{360} \times 100\% = 20.83\% \) (or \( \frac{125}{6}\% \)). Tennis's percentage: \( \frac{50}{360} \times 100\% = 13.89\% \) (or \( \frac{125}{9}\% \)). The difference: \( 20.83\% - 13.89\% = 6.94\% \) more was spent on hockey than on tennis.

(iii) Badminton's angle was 60°. With a total of Rs 1,80,00,000: Amount for Badminton = \( \frac{60}{360} \times 1,80,00,000 = Rs. 30,00,000 \)

(iv) Cricket and hockey together had angles: 90° + 75° = 165°. With a total of Rs 2,40,00,000: Amount for cricket and hockey = \( \frac{165}{360} \times 2,40,00,000 = Rs. 1,10,00,000 \)

In simple words: Use the angle of each section divided by 360 to find what part of the total it is. Multiply this fraction by the total amount to get the actual rupees spent on each sport.

Exam Tip: Always use the formula (Angle / 360) × 100% for percentages, and (Angle / 360) × Total for amounts - this is the standard method for pie chart questions.

 

Question 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
Answer: The pie chart shows how students across classes VI through X are distributed in a school.

(i) Class VIII has a central angle of 85°. Number of students = \( \frac{85}{360} \times 1440 = 340 \) students

(ii) Class IX had 75° while class X had 50°. The difference in angle is 75° - 50° = 25°. Number of extra students in class IX = \( \frac{25}{360} \times 1440 = 100 \) students. Therefore, 100 more students are in class IX than in class X.

(iii) Classes VII and VIII combined have angles: 70° + 85° = 155°. Total students = \( \frac{155}{360} \times 1440 = 620 \) students

(iv) Class VI had 80° and class X had 50°. The ratio of angles is 80° : 50° = 8 : 5. Therefore, the ratio of students in class VI to class X is also 8 : 5.

In simple words: Find the angle for each class, divide by 360, then multiply by the total number of students to get the actual count in each class.

Exam Tip: The ratio of central angles is the same as the ratio of quantities they represent - you don't need to calculate the actual student numbers for ratio questions.

 

Exercise 19.3

 

Question 1. List the outcomes you can see in these experiments
Answer:
(i) When spinning the wheel shown, the possible outcomes are: A, A, A, B, C, D. Since A appears three times on the wheel, it is a more likely result than the others.

(ii) When drawing one ball from a bag holding 5 identical balls of different colours, the possible outcomes are: White, Red, Blue, Green, Yellow. Each of these five colours represents a separate possible result.

In simple words: An outcome is any single result that can happen when you do an experiment. Write down all the different results possible.

Exam Tip: List all possible outcomes clearly - if something can occur multiple times (like A on the wheel), list it only once in the outcome set, unless the question specifically asks for frequency.

 

Question 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
Answer: When a standard die is rolled, it can land on any of six faces: 1, 2, 3, 4, 5, or 6.

(i) Even numbers on a die are 2, 4, and 6 - a total of 3 even outcomes. Probability = \( \frac{3}{6} = \frac{1}{2} \)

(ii) Multiples of 3 are 3 and 6 - a total of 2 outcomes. Probability = \( \frac{2}{6} = \frac{1}{3} \)

(iii) Numbers that are not multiples of 3 are 1, 2, 4, and 5 - a total of 4 outcomes. Probability = \( \frac{4}{6} = \frac{2}{3} \)

In simple words: Count how many results match what you want, then divide by the total number of possible results.

Exam Tip: Remember that the total outcomes for a die are always 6 - verify your favourable outcomes add up correctly before finding the probability.

 

Question 3. Two coins are tossed together. Find the probability of getting
(i) two tails
(ii) atleast one tail
(iii) no tail
Answer: When two coins are tossed at the same time, the total possible outcomes are 2 × 2 = 4. These outcomes are: HH, HT, TH, TT.

(i) Two tails means both coins show T. There is only 1 such outcome (TT). Probability = \( \frac{1}{4} \)

(ii) Atleast one tail means the result contains at least one T. These outcomes are: TH, HT, TT - a total of 3 outcomes. Probability = \( \frac{3}{4} \)

(iii) No tail means both coins show H. There is only 1 such outcome (HH). Probability = \( \frac{1}{4} \)

In simple words: Write out all possible results first. Then count which ones fit what you're looking for and divide by 4.

Exam Tip: "Atleast one" means 1 or more - always list these outcomes carefully, as they often include more cases than students initially expect.

 

Question 4. Three coins are tossed together. Find the probability of getting
(i) atleast two heads
(ii) atleast one tail
(iii) atmost one tail
Answer: When three coins are tossed, the total outcomes = 8: HHH, HHT, HTH, THH, HTT, TTH, TTT, THT.

(i) Atleast two heads means 2 or more heads. These outcomes are: HHH, HHT, HTH, THH - a total of 4 outcomes. Probability = \( \frac{4}{8} = \frac{1}{2} \)

(ii) Atleast one tail means 1 or more tails. The only outcome without any tail is HHH, so all other 7 outcomes have atleast one tail: HHT, HTH, HTT, TTT, THH, THT, TTH. Probability = \( \frac{7}{8} \)

(iii) Atmost one tail means 0 or 1 tails. These outcomes are: HHH (0 tails), HHT, HTH, THH (1 tail each) - a total of 4 outcomes. Probability = \( \frac{4}{8} = \frac{1}{2} \)

In simple words: "Atleast" means that amount or more. "Atmost" means that amount or less. Count all outcomes matching each condition carefully.

Exam Tip: For three coins, use a systematic approach - list all 8 outcomes in order to avoid missing any, especially for "atleast" and "atmost" conditions.

 

Question 5. Two dice are rolled simultaneously. Find the probability of getting
(i) the sum as 7
(ii) the sum as 3 or 4
(iii) prime numbers on both the dice
Answer: When two dice are rolled at the same time, the total outcomes = 6 × 6 = 36.

(i) Pairs that add to 7 are: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) - a total of 6 pairs. Probability = \( \frac{6}{36} = \frac{1}{6} \)

(ii) Pairs that sum to 3 are: (1,2), (2,1). Pairs that sum to 4 are: (1,3), (2,2), (3,1). Together: 5 pairs. Probability = \( \frac{5}{36} \)

(iii) Prime numbers on a die are 2, 3, and 5. Pairs with both showing primes are: (2,2), (2,3), (2,5), (3,2), (3,3), (3,5), (5,2), (5,3), (5,5) - a total of 9 pairs. Probability = \( \frac{9}{36} = \frac{1}{4} \)

In simple words: Write out all pairs that match your condition, count them, then divide by 36 (the total number of outcomes with two dice).

Exam Tip: When rolling two dice, always remember the total outcomes are 36 - organize your pairs in a systematic way to avoid missing any.

 

Question 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
Answer: In the box, the number of rusted screws = \( \frac{1}{10} \times 600 = 60 \) screws. The total number of screws is 600.

(i) Rusted screws = 60. Probability of picking a rusted screw = \( \frac{60}{600} = \frac{1}{10} \)

(ii) Non-rusted screws = 600 - 60 = 540. Probability of picking a non-rusted screw = \( \frac{540}{600} = \frac{9}{10} \). Alternatively, using the complement: Probability = 1 - Probability(rusted) = 1 - \( \frac{1}{10} = \frac{9}{10} \)

In simple words: Find how many match what you want, then divide by the total. For the opposite event, subtract from 1.

Exam Tip: When asked for "not" an event, use the complement rule: P(not A) = 1 - P(A) - this is faster and less error-prone than counting.

 

Question 7. A letter is chosen from the word 'TRIANGLE'. What is the probability that it is a vowel?
Answer: The word 'TRIANGLE' has 8 letters: T, R, I, A, N, G, L, E. The vowels among these letters are I, A, and E - a total of 3 vowels. Probability of selecting a vowel = \( \frac{3}{8} \)

In simple words: Count the vowels in the word, then divide by the total number of letters.

Exam Tip: Remember the five vowels (A, E, I, O, U) and identify them carefully in the given word - recount if unsure.

 

Question 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
Answer: The bag contains 5 red + 6 black + 4 white = 15 balls in total.

(i) White balls = 4. Probability = \( \frac{4}{15} \)

(ii) Non-black balls = 5 red + 4 white = 9 balls. Probability = \( \frac{9}{15} = \frac{3}{5} \)

(iii) Red or black balls = 5 red + 6 black = 11 balls. Probability = \( \frac{11}{15} \)

(iv) Balls that are neither red nor black are the white balls = 4. Probability = \( \frac{4}{15} \)

In simple words: Add up the balls that match what you're looking for, then divide by 15.

Exam Tip: For "or" questions, add the two categories. For "neither... nor", you're looking at the remaining category only.

 

Question 9. A box contains 17 cards numbered 1, 2, 3…….., 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
Answer: The box holds 17 cards numbered 1 through 17, giving a total of 17 possible outcomes.

(i) Odd numbers are 1, 3, 5, 7, 9, 11, 13, 15, 17 - a total of 9. Probability = \( \frac{9}{17} \)

(ii) Even numbers are 2, 4, 6, 8, 10, 12, 14, 16 - a total of 8. Probability = \( \frac{8}{17} \)

(iii) Prime numbers are 2, 3, 5, 7, 11, 13, 17 - a total of 7. Probability = \( \frac{7}{17} \)

(iv) Numbers divisible by 3 are 3, 6, 9, 12, 15 - a total of 5. Probability = \( \frac{5}{17} \)

(v) Numbers divisible by both 2 and 3 (i.e., divisible by 6) are 6, 12 - a total of 2. Probability = \( \frac{2}{17} \)

In simple words: Identify which numbers from 1 to 17 fit the condition, count them, and divide by 17.

Exam Tip: For "divisible by 2 and 3 both", find numbers divisible by 6 - this is faster than checking each condition separately.

 

Question 10. A card is drawn from a well-shuffled 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) non-face card of red colour
Answer: A standard deck has 52 cards total.

(i) There are 4 aces (one per suit). Probability = \( \frac{4}{52} = \frac{1}{13} \)

(ii) Red cards include all hearts (13) and all diamonds (13) = 26 red cards. Probability = \( \frac{26}{52} = \frac{1}{2} \)

(iii) Kings and queens = 4 + 4 = 8 cards. Cards that are neither = 52 - 8 = 44. Probability = \( \frac{44}{52} = \frac{11}{13} \)

(iv) Red face cards are jacks, queens, and kings in hearts and diamonds = 6 cards. Red aces = 2 cards (ace of hearts, ace of diamonds). Together = 6 + 2 = 8. Probability = \( \frac{8}{52} = \frac{2}{13} \). However, note that the standard calculation shows 6 red face cards only, giving \( \frac{6}{52} = \frac{3}{26} \), and red face card or an ace gives \( \frac{6 + 2}{52} = \frac{8}{52} = \frac{2}{13} \) or if counting 3 aces in red suits, \( \frac{6 + 3}{52} = \frac{9}{52} \). Using the source's structure of separate parts, red face card (from source) = 6 alone, then combined with ace aspect. From source verification: answer shows \( \frac{6}{52} \) initially. Adding 3 aces (red aces only, 2 total) would give additional count. Standard: \( \frac{8}{52} = \frac{2}{13} \).

(v) Spades total 13 cards. Probability = \( \frac{13}{52} = \frac{1}{4} \)

(vi) Non-face cards are 52 - 12 = 40 cards. Red non-face cards are red number cards: 2-10 in hearts and diamonds = 9 × 2 = 18 plus 2 red aces = 20. Probability = \( \frac{20}{52} = \frac{5}{13} \)

In simple words: Know your deck: 4 suits, 13 cards each, 12 face cards total, and calculate which cards fit each condition.

Exam Tip: Memorize key facts about a standard deck - 26 red, 26 black, 4 aces, 12 face cards - these save calculation time.

 

Question 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.
Answer: The lottery has 5 winning tickets and 995 non-winning tickets, making a total of 5 + 995 = 1000 tickets. The probability of selecting a winning ticket is the number of winning tickets divided by the total number of tickets: Probability = \( \frac{5}{1000} = \frac{1}{200} \). Therefore, a person's chance of winning a prize when buying one ticket is 1 in 200.

In simple words: Divide the number of winning tickets by the total tickets to find the probability of winning.

Exam Tip: Always simplify probability fractions to lowest terms - 1/200 is clearer and more useful than 5/1000.