OP Malhotra Class 10 Maths Solutions Chapter 18 Arithmetic Mean Median Mode and Quartiles Exercise 18 (C)

 

Question 1. Find the number half-way between 0.2 and 0.02.
Answer: To find the number halfway between two numbers, we calculate their average. So, we add 0.2 and 0.02, then divide by 2. This gives \( \frac{0.2+0.02}{2} = \frac{0.22}{2} = 0.11 \). This middle point helps us find the center of a range.
In simple words: To find the middle point between two numbers, you add them together and then divide by two.

🎯 Exam Tip: Remember that the "half-way number" is always the arithmetic mean of the two numbers. Don't just subtract and divide by two – you must add them first.

 

Question 2. Find the mean of the following sets of numbers:
(a) 4, 5, 7, 8
(b) 3, 5, 0, 2, 8
(c) 2.5, 2.4, 3.5, 2.8, 2.9, 3.3, 3.6
(d) -6, -2, -1, 0, 1, 2, 5, 9
(e) First five prime numbers
(f) First eight even natural numbers
(g) All factors of 30
(h) First five multiples of 8
(i) x, x+1, x+2, x+3, x+4, x+5, x+6
Answer:
To find the mean, we sum all the numbers and divide by how many numbers there are.
(a) For the numbers 4, 5, 7, 8:
There are \( n = 4 \) numbers.
Mean \( = \frac{\text{Sum}}{n} = \frac{4+5+7+8}{4} = \frac{24}{4} = 6 \)
(b) For the numbers 3, 5, 0, 2, 8:
There are \( n = 5 \) numbers.
Mean \( = \frac{3+5+0+2+8}{5} = \frac{18}{5} = 3.6 \)
(c) For the numbers 2.5, 2.4, 3.5, 2.8, 2.9, 3.3, 3.6:
There are \( n = 7 \) numbers.
Mean \( = \frac{2.5+2.4+3.5+2.8+2.9+3.3+3.6}{7} = \frac{21.0}{7} = 3.0 \)
(d) For the numbers -6, -2, -1, 0, 1, 2, 5, 9:
There are \( n = 8 \) numbers.
Mean \( = \frac{-6-2-1+0+1+2+5+9}{8} = \frac{-9+17}{8} = \frac{8}{8} = 1 \)
(e) The first 5 prime numbers are 2, 3, 5, 7, 11.
There are \( n = 5 \) numbers.
Mean \( = \frac{2+3+5+7+11}{5} = \frac{28}{5} = 5.6 \)
(f) The first 8 even natural numbers are 2, 4, 6, 8, 10, 12, 14, 16.
There are \( n = 8 \) numbers.
Mean \( = \frac{2+4+6+8+10+12+14+16}{8} = \frac{72}{8} = 9 \)
(g) All factors of 30 are 1, 2, 3, 5, 6, 10, 15, 30.
There are \( n = 8 \) numbers.
Mean \( = \frac{1+2+3+5+6+10+15+30}{8} = \frac{72}{8} = 9 \)
(h) The first 5 multiples of 8 are 8, 16, 24, 32, 40.
There are \( n = 5 \) numbers.
Mean \( = \frac{8+16+24+32+40}{5} = \frac{120}{5} = 24 \)
(i) For the numbers x, x+1, x+2, x+3, x+4, x+5, x+6:
There are \( n = 7 \) numbers.
Mean \( = \frac{x+(x+1)+(x+2)+(x+3)+(x+4)+(x+5)+(x+6)}{7} \)
\( \implies \) Mean \( = \frac{7x+21}{7} \)
\( \implies \) Mean \( = \frac{7(x+3)}{7} \)
\( \implies \) Mean \( = x+3 \)
In simple words: To find the average (mean) of a group of numbers, first add them all up. Then, count how many numbers you have. Finally, divide the total sum by that count.

🎯 Exam Tip: Always make sure to count the number of data points (n) correctly. For prime numbers, factors, or multiples, list them out first to avoid errors. When working with variables, collect like terms carefully.

 

Question 3. The mean of the numbers 6, y, 7, x, 14 is 8. Express y in terms of x.
Answer: We know that the mean is the sum of numbers divided by the count of numbers. Here, the numbers are 6, y, 7, x, 14, and there are 5 numbers in total. The mean is given as 8. So, the total sum of the numbers is \( 5 \times 8 = 40 \).
We can write this as:
\( 6 + y + 7 + x + 14 = 40 \)
\( \implies y + x + 27 = 40 \)
\( \implies y = 40 - 27 - x \)
\( \implies y = 13 - x \)
This means the value of y changes depending on x.
In simple words: If you know the average and how many numbers there are, you can find their total sum. Then, you can use this total to write one unknown number using another unknown number.

🎯 Exam Tip: Remember the formula for mean: Sum of observations / Number of observations. When asked to express one variable in terms of another, isolate the desired variable on one side of the equation.

 

Question 4. Nisha secured 73, 86, 78 and 75 marks in four tests. What is the least number of points she should secure in her next test if she has to have an average of 80?
Answer: Nisha has taken 4 tests already, and she is about to take her 5th test. For an average of 80 marks across 5 tests, the total marks needed would be \( 5 \times 80 = 400 \). The sum of her marks from the first four tests is \( 73 + 86 + 78 + 75 = 312 \). To find the marks she needs in her fifth test, we subtract her current total from the desired total: \( 400 - 312 = 88 \). She needs 88 marks to reach her target average.
In simple words: First, figure out the total marks she needs for her desired average. Then, add up her current marks. Subtract the current total from the needed total to find out how many more marks she has to get.

🎯 Exam Tip: This type of question often involves calculating the required sum for a target average. Clearly define the total number of items (tests) before calculating the target sum. Don't forget to include the "next test" in the total count.

 

Question 5. A class of 10 students was given a test in Mathematics. The marks, out of 50, secured by the students were as follows : 31, 36, 27, 38, 45, 39, 32, 29, 41, 38 Find the mean score.
Answer: There are \( n = 10 \) students in the class. The marks they scored are 31, 36, 27, 38, 45, 39, 32, 29, 41, 38. To find the mean score, we first add all the marks together. The sum of these marks is \( 31 + 36 + 27 + 38 + 45 + 39 + 32 + 29 + 41 + 38 = 356 \). Now, divide this sum by the number of students: \( \text{Mean} = \frac{356}{10} = 35.6 \). So, the average score for the class is 35.6.
In simple words: Add up all the marks the students got. Then, divide that total by the number of students. The answer is the average mark for the class.

🎯 Exam Tip: Be careful when adding a long list of numbers to avoid calculation errors. A common mistake is miscounting the number of observations, so double-check the value of 'n'.

 

Question 6. Find the mean of the following frequency distributions:
(a)
(b)
(c)
(d)
Answer:
(a) For a frequency distribution, the mean is calculated using the formula \( \text{Mean} = \frac{\sum fx}{\sum f} \). We create a table to find the sum of \( fx \) and the sum of \( f \).

Mean \( = \frac{\sum fx}{\sum f} = \frac{1600}{50} = 32 \) kg.
(b) We follow the same method for the marks distribution.

Mean \( = \frac{\sum fx}{\sum f} = \frac{1190}{40} = 29.75 \)
(c) For the given frequency distribution:

Mean \( = \frac{\sum fx}{\sum f} = \frac{90}{15} = 6 \)
(d) For the given frequency distribution with decimal values:

Mean \( = \frac{\sum fx}{\sum f} = \frac{72}{210} = \frac{12}{35} \approx 0.34 \)
In simple words: For a frequency distribution, you multiply each value by how many times it appears (that's `f x x`). Add all these products together. Then, divide this total by the sum of all the frequencies (the total count of items). This gives you the mean.

🎯 Exam Tip: When calculating the mean for frequency distributions, always create a column for \( f \times x \) in your table. Double-check your multiplication and summation to avoid errors. Ensure you write the correct units if specified, like "kg" or "marks".

 

Question 7. Fill in the blanks : While calculating the mean of the grouped data, we make the assumption that frequency in any class is centred at its _______.
Answer: While calculating the mean of the grouped data, we make the assumption that frequency in any class is centred at its class mark. The class mark is the midpoint of each class interval, which helps estimate the mean for grouped data.
In simple words: When you have data grouped into ranges, you assume all the data in one range is found right in the middle of that range. This middle point is called the class mark.

🎯 Exam Tip: The concept of "class mark" is crucial for calculating the mean of grouped data. Remember that it's the midpoint of a class interval, representing all values within that interval for mean calculations.

 

Question 8. The frequency distribution of marks obtained by 40 students of a class is as under : Calculate the Arithmetic mean.
Answer: To calculate the arithmetic mean for grouped data, we first find the class mark (midpoint) for each class interval. Then, we multiply the class mark (x) by its frequency (f) to get \( fx \). Finally, we use the formula \( \text{Mean} = \frac{\sum fx}{\sum f} \).

Mean \( = \frac{\sum fx}{\sum f} = \frac{936}{40} = 23.4 \) marks.
In simple words: First, find the middle value for each mark range (that's the class mark). Then, multiply each class mark by how many students got marks in that range. Add up all these products. Finally, divide by the total number of students to get the average mark.

🎯 Exam Tip: When calculating class marks, use the formula (Lower Limit + Upper Limit) / 2. Be careful with inclusive vs. exclusive class intervals; for continuous data, the upper limit of one class is the lower limit of the next. Here, the intervals are 0-8, 8-16, etc., indicating continuous classes where the upper limit is not included in the first class but starts the next.

 

Question 9. Find the mean of the following data:
(a)
(b)
Answer:
(a) For the first set of data, we calculate the class mark (midpoint) for each mark range and then use the formula for mean.

Mean \( = \frac{\sum fx}{\sum f} = \frac{645}{30} = 21.5 \) marks.
(b) For the second set of data, which has continuous class intervals, we calculate the class mark and then the mean.

Mean \( = \frac{\sum fx}{\sum f} = \frac{578.5}{20} = 28.925 \approx 28.93 \) marks.
In simple words: For each part, first find the middle number for each group (class mark). Multiply this middle number by how many times it appears. Add up all these products, then divide by the total count of all items to get the average.

🎯 Exam Tip: Pay attention to the type of class intervals. For inclusive intervals (like 10-14, 15-19), the class mark calculation involves finding the true limits or simply taking the midpoint of the given limits. For continuous intervals (like 0-10, 11-20), ensure you adjust the class limits for true continuity if needed, or simply take the midpoint as given.

 

Question 10. In a class of 60 boys the marks obtained in a monthly test were as under : Find the mean marks of the class.
Answer: To find the mean marks, we first determine the class mark (midpoint) for each mark range. Then, we multiply the class mark (x) by the frequency (f), which is the number of students, to get \( fx \). Finally, we sum all \( fx \) values and divide by the total number of students.

Mean \( = \frac{\sum fx}{\sum f} = \frac{1830}{60} = 30.5 \)
In simple words: First, find the middle score for each mark range. Multiply this middle score by how many students achieved it. Add up all these multiplied scores. Finally, divide by the total number of students to get the average mark for the class.

🎯 Exam Tip: When dealing with class intervals like 10-20, 20-30, remember these are continuous. The class mark is found by (lower limit + upper limit) / 2. Ensure your arithmetic is precise, especially when summing up products.

 

Question 11. Calculate the mean of the following frequency table by:
(i) direct-method and
(ii) short-cut method
Answer:
We will create a combined table to calculate the mean using both the direct and short-cut methods. First, find the class marks (x). For the short-cut method, assume a mean \( A \), usually the class mark of the middle class, and calculate deviations \( d = x - A \) and \( fd \).

(i) Direct method:
Mean \( = \frac{\sum fx}{\sum f} = \frac{1100}{50} = 22 \)
(ii) Short-cut method:
Let the assumed mean \( A = 27.5 \)
Mean \( = A + \frac{\sum fd}{\sum f} = 27.5 + \frac{-275}{50} \)
\( \implies \) Mean \( = 27.5 - 5.5 = 22 \)
The mean is the same using both methods, showing the consistency of statistical calculations.
In simple words: The direct method means you multiply each middle value by its count, sum them up, and divide by the total count. The short-cut method makes calculations easier by choosing an estimated average first, finding how much each value differs from it, and then adjusting the estimate. Both ways should give you the same average.

🎯 Exam Tip: Both the direct and short-cut methods are valid for calculating the mean. The short-cut method (also known as the assumed mean method) is often quicker when dealing with large numbers, as it simplifies the calculations. Always clearly state your assumed mean (A) if using the short-cut method. Make sure to check that both methods yield the same result.

 

Question 12. The following table gives the classification of 100 cows of a dairy farm, according to the amount of milk given by each in a dairy. Calculate the mean correct to first place of decimal.
Answer: To find the mean amount of milk, we first determine the class mark (midpoint) for each milk range. Then, we multiply the class mark (x) by the frequency (f), which is the number of cows, to get \( fx \). Finally, we sum all \( fx \) values and divide by the total number of cows (100).

Mean \( = \frac{\sum fx}{\sum f} = \frac{980}{110} \approx 8.9 \) kg.
In simple words: Find the middle value of each milk range. Multiply this by the number of cows in that range. Add up all these products and then divide by the total number of cows to get the average milk produced per cow. Round your final answer to one decimal place.

🎯 Exam Tip: When working with grouped frequency distributions, correctly calculating the class mark (midpoint of each class interval) is the first critical step. Ensure your calculations for \( \sum fx \) and \( \sum f \) are accurate, and remember to round the final answer as specified in the question.

 

Question 13. The weights (in grams) of 50 apples picked out at random from a consignment are given below: 82, 118, 80, 110, 104, 84, 106, 107, 76, 82, 109, 107, 115, 93, 187, 95, 123, 125, 111, 92, 86, 70, 126, 78, 130, 129, 139, 119, 115, 128, 100, 186, 84, 99, 113, 204, 111, 141, 136, 123, 90, 115, 98, 110, 78, 90, 107, 81, 131, 75.
(i) What is the range of the weights ?
(ii) Form a frequency distribution with class intervals 70-89, 90-109, and so on.
(iii) Use your frequency distribution to calculate the mean.
Answer:
(i) To find the range, we identify the maximum and minimum weights from the given data.
Maximum weight \( = 204 \) g
Minimum weight \( = 70 \) g
Range \( = \text{Maximum weight} - \text{Minimum weight} = 204 - 70 = 134 \) g.
(ii) and (iii) We form a frequency distribution table with the specified class intervals (70-89, 90-109, etc.). Then, we calculate the class mark (x) for each interval, assume a mean \( A \), find deviations \( d = x - A \), and compute \( fd \) to find the mean using the short-cut method. Here, \( A = 139.5 \).

Using the assumed mean method:
Mean \( = A + \frac{\sum fd}{\sum f} = 139.5 + \frac{-1460}{50} \)
\( \implies \) Mean \( = 139.5 - 29.2 = 110.3 \) g.
In simple words: (i) The range is the biggest number minus the smallest number. (ii) To make a frequency table, you put the data into groups (like 70-89, 90-109) and count how many items fall into each group. (iii) To find the average from this table, you can pick an estimated average, see how much each group's middle value differs, and then adjust your estimate based on these differences.

🎯 Exam Tip: When forming frequency distributions, ensure class intervals are consistent (same width) and cover the entire range of data. For calculating the mean of grouped data, the assumed mean method is efficient but requires careful calculation of deviations and \( fd \) products. Double-check all counts and sums.

 

Question 14. The following table gives the weekly wages of workers in a factory. Calculate:
(i) the mean,
(ii) the number of workers getting weekly wages below Rs. 80, and
(iii) the number of workers getting Rs. 65 and more but less than Rs. 85 as weekly wages.
Answer:
First, we prepare a frequency distribution table to calculate the mean. We determine the class mark (x) for each wage range, choose an assumed mean \( A = 67.5 \), calculate deviations \( d = x - A \), and find \( fd \).

(i) Mean:
Using the assumed mean method:
Mean \( = A + \frac{\sum fd}{\sum f} = 67.5 + \frac{120}{80} \)
\( \implies \) Mean \( = 67.5 + 1.5 = 69 \) Rs.
(ii) Number of workers getting weekly wages below Rs. 80:
This includes workers in the wage groups 50-55, 55-60, 60-65, 65-70, 70-75, and 75-80.
Number of workers \( = 5 + 20 + 10 + 10 + 9 + 6 = 60 \)
(iii) Number of workers getting Rs. 65 and more but less than Rs. 85 as weekly wages:
This includes workers in the wage groups 65-70, 70-75, 75-80, and 80-85.
Number of workers \( = 10 + 9 + 6 + 12 = 37 \)
In simple words: (i) The average wage is found by picking an assumed average, finding differences for each wage group, and then adjusting the assumed average. (ii) To find workers earning below a certain amount, add up the frequencies of all wage groups below that amount. (iii) To find workers earning within a specific range, add up the frequencies of the groups that fall within that range.

🎯 Exam Tip: Pay close attention to inclusive and exclusive limits in wage ranges when counting workers (e.g., "below Rs. 80" means up to Rs. 79.99, so the 80-85 group is excluded). Ensure your assumed mean (A) is a class mark for simpler deviation calculations.

 

Question 15. Find the mean of the following data:

Answer: First, we convert the cumulative frequency table into a simple frequency distribution table. To do this, we find the number of students for each mark range. Then, we calculate the class mark (midpoint) for each interval and multiply it by its frequency (f x x). Finally, we sum these products and divide by the total number of students to find the mean. The class interval for "Less than 10" can be 0-10, for "Less than 20" it can be 10-20, and so on.

Mean \( = \frac{\sum f x}{\sum f} = \frac{1250}{50} = 25 \)
In simple words: First, turn the "less than" numbers into normal groups. Then, find the middle point of each group and multiply it by how many students are in that group. Add all these up and divide by the total number of students to get the average mark.

🎯 Exam Tip: When converting a "less than" cumulative frequency table, remember to subtract the previous cumulative frequency from the current one to find the actual frequency for each class interval.

 

Question 16. Recast the following cumulative table into the form of an ordinary frequency distribution and determine the value of the mean:

Answer: We need to convert the given cumulative frequency table into a regular frequency distribution. For each mark range, we find the number of recipients in that range. Then, we calculate the class mark (midpoint) for each interval. We'll use the short-cut method for calculating the mean, which involves choosing an assumed mean (A), calculating deviations (d), and then finding \( \sum fd \) and \( \sum f \). The mean is then \( A + \frac{\sum fd}{\sum f} \). The actual range for "less than 5" means 0-5, and so on. Marks are usually positive, so 0 is a sensible lower bound for the first class.

Mean \( = A + \frac{\sum f d}{\sum f} = 27.5 + \frac{-800}{150} \)
\( = 27.5 - 5.3 = 22.2 \)
In simple words: First, create a standard list of mark ranges and how many people are in each range. Then, use a shortcut method by picking an average mark, figuring out how far each group's middle mark is from it, and using those differences to find the overall average. This method simplifies calculations for larger datasets.

🎯 Exam Tip: When using the shortcut method, choose an assumed mean (A) that is close to the middle of the class marks. This helps keep the deviation values smaller and calculations simpler.

 

Question 17. Find the missing frequency in the following data if arithmetic mean is 19.92.

Answer: Let the missing frequency be 'p'. We first create a table to calculate the sum of \( fx \) and the sum of frequencies \( f \). We use the class mark (midpoint) for 'x'. Then, we set up an equation using the formula for the arithmetic mean: Mean \( = \frac{\sum fx}{\sum f} \). We plug in the given mean (19.92) and solve for 'p'. This process helps determine unknown values in data sets when the average is known.

Mean is given \( = 19.92 \)
\( \implies \) \( \frac{\sum f x}{\sum f} = 19.92 \)
\( \implies \) \( \frac{1772+22 p}{90+p} = 19.92 \)
\( \implies \) \( 1772 + 22p = 19.92(90+p) \)
\( \implies \) \( 1772 + 22p = 1792.8 + 19.92p \)
\( \implies \) \( 22p - 19.92p = 1792.8 - 1772 \)
\( \implies \) \( 2.08p = 20.8 \)
\( \implies \) \( p = \frac{20.8}{2.08} \)
\( \implies \) \( p = \frac{208}{208} \times \frac{10}{1} \)
\( \implies \) \( p = 10 \)
Therefore, the missing frequency is 10.
In simple words: We list all the numbers and how often they appear, leaving 'p' for the missing one. We add up all the numbers (with 'p' included) and divide by the total count (also with 'p'). Since we know the average, we set up an equation and solve it to find out what 'p' must be.

🎯 Exam Tip: When solving for a missing frequency, be very careful with algebraic manipulations, especially when distributing the mean value across the sum of frequencies. One small error can lead to a wrong answer.

 

Question 18. Calculate the Arithmetic mean, correct to one decimal place, for the following frequency distribution of marks obtained in an Arithmetic test.

Answer: To find the arithmetic mean, we first need to determine the class mark (midpoint) for each range of marks. Then, we multiply each class mark by its corresponding number of students (frequency) to get \( f \times x \). We sum all the \( f \times x \) values and divide by the total number of students to get the mean. This method is common for grouped data to find the central tendency.

Mean \( = \frac{\sum f x}{\sum f} = \frac{1180}{42} = 28.095 \approx 28.1 \) (approx)
In simple words: Find the middle point for each score range, then multiply it by how many students got scores in that range. Add all these results together and divide by the total number of students. This gives you the average score for the test.

🎯 Exam Tip: Always make sure to calculate the class mark correctly for each interval (midpoint of the lower and upper limits). Errors here will affect the entire mean calculation.

 

Question 19. The following table gives the marks scored by students in the examination:

Calculate the mean mark, correct to two decimal places.
Answer: We will calculate the mean using the short-cut method to handle the grouped frequency data. First, we find the class mark (x) for each interval. Then, we choose an assumed mean (A), which is typically a class mark near the center of the data. We calculate the deviation (d) for each class mark from A. Next, we multiply each deviation by its frequency (f x d) and find the sum of \( fd \). The arithmetic mean is calculated as \( A + \frac{\sum fd}{\sum f} \). This method simplifies calculations especially when numbers are large.

Mean \( = A + \frac{\sum f d}{\sum f} = 17.5 + \frac{85}{80} \)
\( = 17.5 + 1.0625 \)
\( = 18.5625 \approx 18.56 \)
In simple words: To find the average mark, we use a quick way: pick a middle mark as a guess, then see how much other marks are different from it. We multiply these differences by how many students got those marks, add them up, and then use that sum with our guessed average to find the true average.

🎯 Exam Tip: Always double-check your assumed mean selection and the corresponding deviations, especially for negative values. Small arithmetic errors in this table propagate and result in an incorrect final mean.