ML Aggarwal Class 7 Maths Solutions Chapter 17 Data Handling

Exercise 14.1

Question 1. (i) Arrange the following data in ascending order: 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 2, 4, 4, 4, 4, 4, 5, 5, 6, 6
Answer: The data arranged in ascending order is: 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 5, 5, 6, 6
In simple words: Place all the numbers from smallest to biggest, starting with 1 and ending with 6.

Exam Tip: Always check that your count of each number matches the original data before moving to the next step.

Question 1. (ii) Find the range of the data.
Answer: Range of the data = Maximum value - Minimum value = 6 - 1 = 5
In simple words: The range tells you how spread out the numbers are. Subtract the smallest number from the biggest number.

Exam Tip: Range is always the largest number minus the smallest number - this single formula works for all datasets.

Question 1. (iii) Prepare a frequency table for the data.
Answer:

Exam Tip: Group tally marks in sets of five with a diagonal line through them - this makes counting faster and reduces errors.

Question 1. (iv) Find the number of houses which have 4 or more than 4 rooms.
Answer: Number of houses which have 4 or more than 4 rooms = 5 + 3 + 2 = 10 houses
In simple words: Add up the frequencies for all houses with 4 rooms, 5 rooms, and 6 rooms to get the total count.

Exam Tip: When the question asks "4 or more," include the frequency for 4 itself plus all frequencies above it - do not exclude the boundary value.

Question 2. (i) Using the bar graph, find the total number of books sold in 2008 = 15. Also find the number of books sold in 2009 and the number of books sold in 2011.
Answer: From the bar graph: Total number of books sold in 2008 = 15. Number of books sold in 2009 = 8 (approximately, reading from the bar height). Number of books sold in 2011 = 10 (approximately, reading from the bar height).
In simple words: Look at each bar's height on the graph and read the value from the y-axis to find how many books were sold in each year.

Exam Tip: When reading bar graphs, always check the scale on the y-axis first - this tells you what each division represents.

Question 2. (ii) State whether the demand for English books rose faster from 2008 to 2009 and from year 2009 to 2010 it decreased comparatively. Then from 2010 to 2011 it increased.
Answer: Looking at the bar graph, demand for English books went up from 2008 to 2009. From 2009 to 2010, it came down by comparison. From 2010 to 2011, it went up again. This pattern shows periods of growth followed by a decline, and then another rise.
In simple words: The bars show that book sales increased first, then fell, then rose once more over the four-year period.

Exam Tip: To compare changes between years, always look at whether each new bar is taller or shorter than the previous one - this tells you if demand increased or decreased.

Exercise 14.2 - Question 3. Bar Graph on Preferred Colours

Question 3. (i) Blue is the most preferred colour.
Answer: This statement is correct. Looking at the bar graph, the blue bar reaches the highest point compared to all other colours (Red, Green, Yellow, and Orange).
In simple words: The blue bar is taller than all the other bars, which means more people chose blue than any other colour.

Exam Tip: To identify the most preferred item from a bar graph, find the tallest bar - the height shows the highest frequency or preference count.

Question 3. (ii) Green is the least preferred colour.
Answer: This statement is correct. The green bar is shorter than all other bars on the graph, showing that fewer people chose green compared to Red, Blue, Yellow, and Orange.
In simple words: The green bar is the shortest, which means green was picked by the fewest people.

Exam Tip: The least preferred item always has the shortest bar - this represents the smallest frequency or count.

Question 3. (iii) There are five colours in all. They are
(a) Red
(b) Green
(c) Blue
(d) Yellow
(e) Orange
Answer: (a), (b), (c), (d), (e) - All five colours are represented in the graph
In simple words: The bar graph displays five separate colours, and the bars show how many people preferred each one.

Exam Tip: Always count the number of bars in a graph carefully - the number of bars equals the number of categories shown.

Exercise 14.2 (continued) - Question 4. Languages Sold (2008-2011)

Question 4. (i) Find the difference in the sale of two language books least in 2011.
Answer: From the graph, in 2011 the difference between the English books and Hindi books sales is: English books (approximately 650) - Hindi books (approximately 600) = approximately 50 books (or the difference in their bar heights).
In simple words: Look at the two bars for 2011, and see how much taller one bar is compared to the other.

Exam Tip: To find a difference between two years or two categories, read both values from the graph and subtract the smaller from the larger.

Question 4. (ii) The demand for English books rose faster from 2008 to 2009 and from year 2009 to 2010 it decreased comparatively. Then from 2010 to 2011 it increased.
Answer: Examining the graph, English book sales increased between 2008 and 2009. The demand then declined from 2009 to 2010 relative to the previous year's growth. Following this, sales went back up from 2010 to 2011. This pattern reveals that English book sales experienced both upward and downward trends across these consecutive periods.
In simple words: The English book bars show the sales went up, then down, then up again as time went forward.

Exam Tip: When describing trends in a bar graph, focus on whether each new bar is higher or lower than the bar before it.

Exercise 14.2 (continued) - Question 5. Favourite Sports Preference

Question 5. (i) Highest marks obtained by the students = 95; Lowest marks obtained by the students = 39
Answer: Based on the data provided, the highest mark recorded was 95 and the lowest mark recorded was 39. These values represent the maximum and minimum achievement levels among all students in the group.
In simple words: One student got 95 as their best score, and another student got 39 as their lowest score.

Exam Tip: Always identify and state the maximum and minimum values clearly when describing a dataset - these are key statistics for further analysis.

Question 5. (ii) Range of the marks = Maximum mark - Minimum mark
Answer: Range = 95 - 39 = 56
In simple words: The range shows the gap between the highest and lowest marks. Subtract the lowest from the highest to get the spread of the marks.

Exam Tip: Range is calculated the same way every time: take the largest value and subtract the smallest value to show how widely the data is spread.

Question 5. (iii) Mean marks obtained by the students = 85 + 46 + 90 + 85 + 39 + 48 + 56 + 95 + 81 + 75 / 10 = 73
Answer: To find the mean, add all the marks together: 85 + 46 + 90 + 85 + 39 + 48 + 56 + 95 + 81 + 75 = 730. Then divide by the number of students (10): 730 ÷ 10 = 73. The mean mark is 73.
In simple words: Add up all the marks and divide by how many students there are to find the average mark.

Exam Tip: The mean is found by totalling all values and dividing by the count - make sure you count the correct number of data points before dividing.

Question 6. (i) Height of the tallest girl = 151 cm; Height of the shortest girl = 128 cm
Answer: The tallest girl measured 151 cm in height. The shortest girl measured 128 cm in height. These represent the maximum and minimum heights in the group of girls.
In simple words: One girl was 151 cm tall (the tallest) and another was 128 cm tall (the shortest).

Exam Tip: When comparing measurements, always clearly identify which is the maximum and which is the minimum value.

Question 6. (ii) Mean height of the girls = 135 + 150 + 139 + 126 + 151 + 132 + 146 + 149 + 148 + 141 / 10 = 1417
Answer: To calculate mean height, add all heights: 135 + 150 + 139 + 126 + 151 + 132 + 146 + 149 + 148 + 141 = 1417 cm. Divide by the number of girls (10): 1417 ÷ 10 = 141.7 cm. The mean height is approximately 141.7 cm or 141.4 cm (depending on rounding).
In simple words: Add all the heights together, then divide by the total number of girls to find the average height.

Exam Tip: When calculating mean from a list of measurements, double-check your addition before dividing - a single addition error will make your final answer wrong.

Question 6. (iii) Find the range of heights.
Answer: Range = Tallest height - Shortest height = 151 - 128 = 23 cm
In simple words: The range tells you how much taller the tallest girl is than the shortest girl.

Exam Tip: Range for height (or any measurement) is always calculated the same way: maximum value minus minimum value.

Question 6. (iv) How many girls are there?
Answer: There are 10 girls.
In simple words: Count the number of height measurements given - that total is the number of girls in the group.

Exam Tip: Always check the total number of data points in a set before performing calculations - this number is needed for mean and other statistical measures.

Exercise 14.2 (continued) - Questions 7 and 8

Question 7. (i) Arithmetic mean of the data: 8, 4, 6, 2, x + 2, 7 is 5.
Answer: Using the formula for mean: (8 + 4 + 6 + 2 + x + 2 + 7) ÷ 6 = 5

This gives us: (27 + x) ÷ 6 = 5

Multiply both sides by 6: 27 + x = 30

Solve for x: x = 30 - 27 = 3

Therefore, x = 3.
In simple words: Set up the mean formula with the unknown x, multiply to clear the division, and then solve for x by moving numbers around.

Exam Tip: When finding a missing value using mean, always set up the equation correctly with the total sum divided by the count equal to the given mean.

Question 8. Mean from grouped frequency data: Marks (x_i): 2, 3, 4, 7, 10 with No. of students (f_i): 3, 2, 6, 7, 2 and f_i x_i: 6, 6, 24, 49, 20
Answer: To find mean from a frequency table: Mean = (Sum of f_i x_i) ÷ (Sum of f_i) = 105 ÷ 20 = 5.25

Calculation: Sum of f_i x_i = 6 + 6 + 24 + 49 + 20 = 105
Sum of f_i = 3 + 2 + 6 + 7 + 2 = 20
Mean = 105 ÷ 20 = 5.25
In simple words: Multiply each mark by how many students got it, add all those products together, then divide by the total number of students.

Exam Tip: For frequency tables, always remember the formula: Mean = Σ(f × x) ÷ Σf. The numerator is the sum of products, and the denominator is the total frequency.

Exercise 14.3

Question 1. (i) Find the median of the data: 1, 3, 3, 4, 5, 5, 6
Answer: There are 7 values (odd number). The median is the middle value, which is the 4th number when arranged in order. Median = 4
In simple words: Count how many numbers there are. If it is odd, the median is the number right in the middle when they are in order.

Exam Tip: For an odd count of numbers, the median position is (n + 1) ÷ 2. For this dataset, that is (7 + 1) ÷ 2 = position 4, which is 4.

Question 1. (ii) Find the median of the data: 1, 3, 3, 4, 5, 5, 6, 4
Answer: Arranging in order: 1, 3, 3, 4, 4, 5, 5, 6. There are 8 values (even number). The median is the average of the 4th and 5th values: Median = (4 + 5) ÷ 2 = 9 ÷ 2 = 4.5
In simple words: When there is an even count of numbers, the median is found by taking the two middle numbers, adding them, and dividing by 2.

Exam Tip: For an even count, the median formula is: (value at position n/2 + value at position (n/2)+1) ÷ 2.

Question 2. (i) Find the mode of the data: 3, 1, 5, 6, 3, 4, 5, 3
Answer: Arranging the data: 1, 3, 3, 3, 4, 5, 5, 6. The mode is the value that occurs most frequently. The number 3 appears three times, more than any other number. Mode = 3
In simple words: The mode is the number that appears most often in the list.

Exam Tip: Always count how many times each number appears - the one with the highest count is the mode.

Question 2. (ii) In the given frequency distribution, we find that the observation 22 has maximum frequency so the mode is = 22
Answer: From the frequency distribution, the observation with the highest frequency is 22, appearing more times than any other value. Therefore, the mode is 22.
In simple words: Look at the frequency column and find which value has the tallest count - that value is the mode.

Exam Tip: In frequency tables, the mode is always the data value (or class) that has the highest frequency - you do not need to count individual items.

Question 3. Find the median and mode of the data: 12, 12, 13, 13, 14, 14, 14, 16, 19
Answer: This dataset contains 9 values (odd number). The median is the middle value at position 5: Median = 14. For the mode, count each value's frequency: 12 appears 2 times, 13 appears 2 times, 14 appears 3 times, 16 appears 1 time, 19 appears 1 time. The value 14 appears most often. Mode = 14
In simple words: The median is the number in the middle position when arranged in order. The mode is the number that shows up the most times.

Exam Tip: When a dataset has both median and mode to find, identify the middle position first for median, then count frequencies for mode.

Question 4. Find the median and mode of the data: 5, 9, 10, 12, 15, 16, 19, 20, 21, 20, 20, 23, 24, 25, 25
Answer: This dataset has 15 values. Median is at position (15 + 1) ÷ 2 = position 8: Median = 20. For mode, count frequencies: 20 appears 3 times, 25 appears 2 times, and all other values appear 1 time. Mode = 20
In simple words: Find the number exactly in the middle for the median. Count which number appears most to find the mode.

Exam Tip: For larger datasets, use position formula: for odd n, median position = (n + 1) ÷ 2.

Question 5. (i) Find the median and mode of the data: 32, 35, 36, 37, 38, 38, 38, 40, 42, 43, 43, 43, 45, 47, 50
Answer: This dataset has 15 values (odd number). Median position = (15 + 1) ÷ 2 = 8th position: Median = 40. Counting frequencies: 38 appears 3 times, 43 appears 3 times, and others appear fewer times. Mode = 38 and 43 (bimodal - two modes)
In simple words: The median is the 8th number in order. Two numbers (38 and 43) each appear three times, so there are two modes.

Exam Tip: A dataset can have more than one mode - this is called bimodal (two modes) or multimodal (more than two modes).

Question 5. (ii) Is the mode bimodal?
Answer: Yes (38, 43)
In simple words: Bimodal means the data has two different values that appear equally often as the most frequent values.

Exam Tip: When two or more values share the same highest frequency, list all of them as the modes - do not pick just one.

Question 6. Find the median and mode of the data: 6, 8, 10, 10, 15, 15, 15, 50, 80, 100, 120
Answer: This dataset has 11 values (odd number). Median position = (11 + 1) ÷ 2 = 6th position: Median = 15. Counting frequencies: 15 appears 3 times (more than any other), 10 appears 2 times, and the rest appear once each. Mode = 15. Note: All three measures (not the same) - no all three are not the same.
In simple words: Count to find the 6th number for the median. The number 15 appears most often, so it is the mode.

Exam Tip: In a single dataset, median and mode are usually different values - they measure different aspects of central tendency.

Question 7. Mode = 15
Answer: The mode of the dataset is 15, as this is the value that occurs most frequently (3 times) in the given data.
In simple words: The mode is the value that appears most often in your data set.

Exam Tip: Always verify the mode by counting each value's frequency - the highest count wins.

Exercise 14.4

Question 1. (i) Certain to happen
Answer: An event that is certain to happen is one where the outcome will definitely occur, no matter what - the probability is 1 or 100%.
In simple words: A certain event always happens - like the sun rising tomorrow or water boiling at 100°C.

Exam Tip: "Certain" events have a probability of exactly 1.0 - these are also called "sure events."

Question 1. (ii) Impossible to happen
Answer: An event that is impossible to happen is one where the outcome can never occur - the probability is 0 or 0%.
In simple words: An impossible event never occurs - like getting a 7 when you roll a standard six-sided die.

Exam Tip: "Impossible" events have a probability of exactly 0 - nothing can make them happen.

Question 1. (iii) Can happen but not certain
Answer: An event that can happen but is not certain means the outcome might or might not occur - the probability is between 0 and 1.
In simple words: These events may happen or may not - like tossing a coin and getting heads, or rolling a 4 on a die.

Exam Tip: Events with probability between 0 and 1 are called "uncertain" or "probable" events - they have some chance of happening.

Question 1. (iv) Impossible to happen
Answer: This describes an event with zero probability - it cannot occur under any circumstances.
In simple words: Like impossible events above, these never take place.

Exam Tip: When an event is described as "impossible to happen," its probability is 0.

Question 1. (v) Can happen but not certain
Answer: This event has a probability between 0 and 1, meaning it might occur or it might not.
In simple words: There is a real chance of this event happening, but it is not guaranteed.

Exam Tip: Probable events always have 0 < P < 1, never reaching the extremes of 0 or 1.

Question 1. (vi) Can happen but not certain
Answer: An event with probability between 0 and 1 is uncertain - the outcome is not guaranteed but has a real chance of occurring.
In simple words: It might happen, it might not - the result is unknown until it occurs.

Exam Tip: Use the range 0 < P < 1 to identify uncertain events in probability problems.

Question 1. (vii) Can happen but not certain
Answer: This event falls in the uncertain range, with probability strictly between 0 and 1.
In simple words: The event has a real possibility of occurring, but it is not certain.

Exam Tip: Remember the three main probability ranges: P = 0 (impossible), 0 < P < 1 (uncertain), P = 1 (certain).

Question 2. Probability = number of events / Total number of outcomes
Answer: Probability of an event = (Number of favourable outcomes) ÷ (Total number of possible outcomes)

For example, when drawing one card: Number of events = 1, Total outcomes = 2, Probability = 1/2
In simple words: Probability is calculated by counting how many ways the event you want can happen, then dividing by the total number of possible results.

Exam Tip: The probability formula P(E) = (Favourable outcomes) ÷ (Total outcomes) is the foundation for all probability calculations.

Question 3. (i) Event of drawing marble number 5 = 1; Total number of outcomes = 6
Answer: When one marble numbered 5 is drawn from a bag containing 6 marbles numbered 1 to 6: Number of ways to draw marble 5 = 1 (only one marble has this number). Total possible outcomes = 6 (any of the 6 marbles could be drawn). Probability = 1/6
In simple words: Only one marble out of six has the number 5, so the chance of picking it is 1 out of 6.

Exam Tip: For a single draw from a set of items, probability = 1 ÷ (total number of items), if each item is equally likely.

Question 3. (ii) Event of drawing marble number 2 = 1; Total number of outcomes = 6
Answer: To draw the marble numbered 2 from 6 marbles: Number of ways to draw marble 2 = 1. Total possible outcomes = 6. Probability = 1/6
In simple words: Just like marble 5, only one marble is numbered 2, so the probability is also 1 out of 6.

Exam Tip: When all items are equally likely and only one specific item can succeed, the probability is always 1 ÷ n (where n is the total count).

Question 4. (i) Event of getting a number less than 3 = 2; Total number of outcomes = 6
Answer: Numbers less than 3 on a die are: 1 and 2 (2 outcomes). Total possible outcomes when rolling a die = 6. Probability = 2/6 = 1/3
In simple words: Two numbers (1 and 2) out of six are less than 3, so the chance is 2 out of 6, which simplifies to 1 out of 3.

Exam Tip: When multiple outcomes satisfy the condition, add them all up to get the total favourable outcomes.

Question 4. (ii) A prime number; Event of getting a number less prime number = 3; Total number of outcomes = 6
Answer: Prime numbers on a die are: 2, 3, and 5 (3 outcomes). Total possible outcomes = 6. Probability = 3/6 = 1/2
In simple words: Three numbers (2, 3, and 5) out of six are prime, so the chance is 3 out of 6, which reduces to 1 out of 2.

Exam Tip: Always identify all numbers that match your condition - for primes on a die, remember 1 is not prime, but 2, 3, and 5 are.

Question 4. (iii) Event of getting a number greater than 2 = 4; Total number of outcomes = 6
Answer: Numbers greater than 2 on a die are: 3, 4, 5, and 6 (4 outcomes). Total possible outcomes = 6. Probability = 4/6 = 2/3
In simple words: Four numbers (3, 4, 5, and 6) out of six are greater than 2, so the probability is 4 out of 6, or 2 out of 3.

Exam Tip: When finding numbers "greater than" a value, do not include the value itself - start from the next number up.

Question 5. (i) Event of drawing a defective mango = 3; Total number of outcomes = 24
Answer: From a basket of 24 mangoes with 3 defective mangoes: Probability = (Number of defective mangoes) ÷ (Total mangoes) = 3 ÷ 24 = 1/8
In simple words: There are 3 bad mangoes out of 24 total, so the chance of picking a defective one is 3 out of 24, which reduces to 1 out of 8.

Exam Tip: Always reduce fractions to lowest terms unless the question asks for the answer in a specific form.

Question 5. (ii) Event of drawing a good mango = 21; Total number of outcomes = 24
Answer: Good mangoes = Total mangoes - Defective mangoes = 24 - 3 = 21. Probability = 21 ÷ 24 = 7/8
In simple words: If 3 mangoes are bad, then 21 are good. The chance of picking a good mango is 21 out of 24, which simplifies to 7 out of 8.

Exam Tip: Remember that P(good) + P(defective) = 1, so P(good) = 1 - P(defective). Here: 1 - 1/8 = 7/8.

Question 6. (i) Event of drawing a red card = 26; Total number of outcomes = 52
Answer: A standard deck has 26 red cards (13 hearts + 13 diamonds) out of 52 total cards. Probability = 26/52 = 1/2
In simple words: Half of all cards in a deck are red, so the chance of drawing a red card is 1 out of 2, or 50%.

Exam Tip: For playing cards: 26 red (hearts and diamonds) and 26 black (clubs and spades) - always remember this standard fact.

Question 6. (ii) Event of drawing a king = 4; Total number of outcomes = 52
Answer: A standard deck has 4 kings (one of each suit) out of 52 total cards. Probability = 4/52 = 1/13
In simple words: There are 4 kings among 52 cards, so the chance is 4 out of 52, which reduces to 1 out of 13.

Exam Tip: For any specific rank in a deck (king, queen, jack, ace, etc.), there are always exactly 4 - one per suit.

Question 6. (iii) Event of drawing a card of spade = 13; Total number of outcomes = 52
Answer: A standard deck has 13 spades out of 52 total cards. Probability = 13/52 = 1/4
In simple words: One quarter of a deck is spades, so the chance of drawing a spade is 13 out of 52, or 1 out of 4.

Exam Tip: Each suit (hearts, diamonds, clubs, spades) has 13 cards - this means each suit has exactly 1/4 of the deck.