Tales by Dots and LinesClass 8 Mathematics NCERT Solutions

Given: The data is the set of the first 50 natural numbers: $1, 2, 3, \ldots, 50$. Number of values, $n = 50$.

To Find: The mean of the data.

Formula: The sum of the first $n$ natural numbers is given by $S_n = \frac{n(n+1)}{2}$. The mean is $\frac{\text{Sum of values}}{\text{Number of values}}$. Alternatively, for an arithmetic progression, the mean is the average of the first and last term: Mean = $\frac{\text{First term} + \text{Last term}}{2}$.

Solution: Using the arithmetic progression formula: First term = 1, Last term = 50. Mean = $\frac{1 + 50}{2} = \frac{51}{2} = 25.5$.

Observation: The mean is exactly halfway between the first and last number.

Final Answer: The mean is 25.5.

Given: The data is the set of the first 50 odd numbers: $1, 3, 5, \ldots$. The 50th odd number is given by the formula for the nth term of an AP: $a_n = a + (n-1)d$. Here, $a=1, n=50, d=2$. So, $a_{50} = 1 + (50-1)2 = 1 + 49 \times 2 = 1 + 98 = 99$. The data is $1, 3, 5, \ldots, 99$. Number of values, $n = 50$.

To Find: The mean of the data.

Solution: Using the arithmetic progression formula for the mean: First term = 1, Last term = 99. Mean = $\frac{1 + 99}{2} = \frac{100}{2} = 50$.

Observation: The mean of the first $n$ odd numbers is $n$. Here, the mean is 50.

Final Answer: The mean is 50.

Given: The data is the set of the first 50 multiples of 4: $4, 8, 12, \ldots$. The 50th multiple of 4 is $4 \times 50 = 200$. The data is $4, 8, 12, \ldots, 200$. Number of values, $n = 50$.

To Find: The mean of the data.

Solution: This is an arithmetic progression. First term = 4, Last term = 200. Mean = $\frac{4 + 200}{2} = \frac{204}{2} = 102$.

Observation: The mean is also a multiple of 2, but not 4. It is the average of the first and last term.

Final Answer: The mean is 102.

Given: A dot plot with one missing value. The mean of the complete data set is 9. The data points shown are: 2, 4, 5, 5, 8, 11, 13, 14, 16. There are 9 known values. Including the missing value, there will be a total of 10 values. Let the missing value be $x$. Mean = 9. Number of values = 10.

To Find: The value of the missing dot, $x$.

Formula: Mean = $\frac{\text{Sum of all values}}{\text{Number of values}}$

Solution: Sum of all values = Mean $\times$ Number of values Sum of all values = $9 \times 10 = 90$.

Now, let's find the sum of the known values: Sum of known values = $2 + 4 + 5 + 5 + 8 + 11 + 13 + 14 + 16 = 78$.

The sum of all values is the sum of known values plus the missing value $x$. $90 = 78 + x$ $x = 90 - 78$ $x = 12$

Final Answer: The missing value is 12. A dot should be placed at the number 12 on the number line.

To Find: Which of the three albums has a mean song length of 5.57 minutes.

Method: For each album, we will list the song lengths, calculate their sum, and divide by the number of songs to find the mean.

Album 1 Data: The song lengths are: 3, 4, 4, 5, 5, 5, 6, 7, 7, 8. Number of songs = 10. Sum of lengths = $3 + 4 + 4 + 5 + 5 + 5 + 6 + 7 + 7 + 8 = 54$ minutes. Mean = $\frac{54}{10} = 5.4$ minutes.

Album 2 Data: The song lengths are: 3, 4, 4, 5, 6, 6, 7, 7, 8, 9. Number of songs = 10. Sum of lengths = $3 + 4 + 4 + 5 + 6 + 6 + 7 + 7 + 8 + 9 = 59$ minutes. Mean = $\frac{59}{10} = 5.9$ minutes.

Album 3 Data: The song lengths are: 3, 4, 5, 5, 6, 6, 6, 7, 8, 9. Number of songs = 10. Sum of lengths = $3 + 4 + 5 + 5 + 6 + 6 + 6 + 7 + 8 + 9 = 59$ minutes. Mean = $\frac{59}{10} = 5.9$ minutes.

There seems to be a mistake in the question's premise or the provided options, as none of the albums have a mean of 5.57. Let's re-examine the dot plots, assuming the question is correct and there might be a misinterpretation of the dots. Let's assume one of the dot plots has a different number of data points. Let's re-read the dot plots very carefully.

Let's try a different approach from the textbook: the balancing act. Let's check if 5.57 can be the balancing point for any of the datasets. Let's assume there is a typo in the question and one of the means should be 5.57. Let's re-calculate.

Let's assume the question meant one of the dot plots has data points: 3, 4, 5, 5, 6, 6, 7, 7, 8, 9. Wait, this is not in the plots. Let's check the plots again.

Album 1: 3, 4, 4, 5, 5, 5, 6, 7, 7, 8. Sum = 54. Mean = 5.4 Album 2: 3, 4, 4, 5, 6, 6, 7, 7, 8, 9. Sum = 59. Mean = 5.9 Album 3: 3, 4, 5, 5, 6, 6, 6, 7, 8, 9. Sum = 59. Mean = 5.9

It appears there is an error in the question as none of the provided dot plots have a mean of 5.57. However, if we must choose the closest, 5.4 and 5.9 are the calculated means.

Let's assume a typo in the data for Album 1. What if one '5' was a '7'? Data: 3, 4, 4, 5, 5, 6, 7, 7, 7, 8. Sum = 56. Mean = 5.6. Close, but not 5.57. What if a '5' was a '6'? Data: 3, 4, 4, 5, 5, 6, 6, 7, 7, 8. Sum = 55. Mean = 5.5. Closer.

Let's re-examine the question's context. The chapter emphasizes the balancing point. Let's test Album 1 with a mean of 5.57. Distances to the left of 5.57: (5.57-3) + 2(5.57-4) + 3(5.57-5) = 2.57 + 2(1.57) + 3(0.57) = 2.57 + 3.14 + 1.71 = 7.42 Distances to the right of 5.57: (6-5.57) + 2(7-5.57) + (8-5.57) = 0.43 + 2(1.43) + 2.43 = 0.43 + 2.86 + 2.43 = 5.72. They are not balanced.

Let's assume there is a typo in the target mean and it should have been 5.4. In that case, Album 1 would be the answer. If the question is strictly as stated, none of the options are correct. Let's proceed assuming there is a typo in the source material and re-examine the dot plots from the source image to be absolutely sure.

The dot plots are: Album 1: {3, 4, 4, 5, 5, 5, 6, 7, 7, 8} -> Mean = 5.4 Album 2: {3, 4, 4, 5, 6, 6, 7, 7, 8, 9} -> Mean = 5.9 Album 3: {3, 4, 5, 5, 6, 6, 6, 7, 8, 9} -> Mean = 5.9

There is no album with a mean of 5.57. Let's assume the question intended for a different dataset not represented, or there is a significant error in the problem statement.

However, if we are forced to find a dataset that could have this mean, let's construct one. For a dataset of 7 points, say {3, 4, 5, 6, 7, 8, 9}, the sum is 42, mean is 6. To get a mean of 5.57 with 7 points, the sum must be $5.57 \times 7 = 38.99 \approx 39$. The data {3, 4, 5, 6, 7, 8, 6} has sum 39. This is not what is plotted.

Let's assume the question meant one of the plots has 7 points. Let's check the images again. No, they all have 10 points.

Conclusion based on calculation: None of the albums have a mean of 5.57 minutes. There is an error in the question as presented in the textbook. Album 1 has a mean of 5.4, and Albums 2 and 3 both have a mean of 5.9.

Explanation of Method: To determine which album has a specific mean, one must calculate the mean for each album. The mean is found by summing the lengths of all the songs in the album and then dividing by the total number of songs. After performing this calculation for all three albums, none of the resulting means is 5.57.

Given: The data set is already sorted: $8, 10, 19, 23, 26, 34, 40, 41, 41, 48, 51, 55, 70, 84, 91, 92$. Number of values, $n = 16$.

Finding the Median: Since $n$ is even, the median is the average of the two middle values, which are the $(\frac{n}{2})^{th}$ and $(\frac{n}{2}+1)^{th}$ values. These are the 8th and 9th values. 8th value = 41 9th value = 41 Median = $\frac{41 + 41}{2} = 41$.

Final Answer for initial median: The median is 41.

(i) Include one value without affecting the median. If we add one value, the new number of values will be $n' = 17$. Since 17 is odd, the new median will be the middle value, which is the $(\frac{17+1}{2})^{th} = 9^{th}$ value in the new sorted list. To keep the median at 41, the 9th value of the new list must be 41. The original 8th and 9th values are both 41. If we insert the new value, $x$, and we want the 9th term to be 41, the new value $x$ must be placed such that the original 8th term (41) becomes the 9th term, or the original 9th term (41) remains the 9th term. This happens if we insert a value less than or equal to 41. But for the median itself to be 41, the simplest way is to add the value 41 itself. Let's see. Original: ... 40, 41, 41, 48 ... If we add 41: ... 40, 41, 41, 41, 48 ... The 9th value in this new list of 17 is 41. So the median is 41.

Final Answer (i): The value to be included could be 41.

(ii) Include two values without affecting the median. If we add two values, the new number of values will be $n' = 18$. Since 18 is even, the new median will be the average of the 9th and 10th values. The original 9th value is 41 and the 10th is 48. To keep the median at 41, the new 9th and 10th values must average to 41. Let the two new values be $x$ and $y$. We need to insert them such that the new 9th and 10th values in the sorted list average to 41. A simple way is to add two values, one less than or equal to 41 and one greater than or equal to 41, such that the middle two values remain 41 and 41. Let's add 40 and 42. Original list's middle: ..., 40, 41, 41, 48, ... New sorted list: ..., 40, 40, 41, 41, 42, 48, ... The 9th value is 41 and 10th is 41. Their average is 41. Another possibility: add two values whose average is 41, and they are symmetric around the median. For example, add 39 and 43. Original: ..., 40, 41, 41, 48, .... New: ..., 40, 39, 41, 41, 43, 48, .... Sorted: ..., 39, 40, 41, 41, 43, 48,... The 9th and 10th values are still 41 and 41. Their average is 41. A simpler case is to add two values, one smaller than the median and one larger, that don't shift the middle two numbers. For example, add 5 (small) and 100 (large). The middle part of the list remains ..., 40, 41, 41, 48, ... The 9th and 10th values of the new 18-item list are still 41 and 41. Their average is 41. So, one value could be any number $\leq 41$ and the other any number $\geq 41$. For example, 1 and 100.

Final Answer (ii): We could add two values such as 41 and 41, or a pair like 40 and 42, or any pair $x, y$ where $x \leq 41$ and $y \geq 41$.

(iii) Remove one value without affecting the median. If we remove one value, the new number of values will be $n' = 15$. Since 15 is odd, the new median will be the middle value, which is the $(\frac{15+1}{2})^{th} = 8^{th}$ value. To keep the median at 41, the 8th value of the new list must be 41. The original 8th value is 41. If we remove any value other than the 8th value, the 8th value might shift.