CBSE Class 10 Mathematics Statistics MCQs Set 10

Answer: (c) I, II and III

Explanation:
1. The highest frequency is 23, so the modal class is 12–15. Its lower limit is 12 (I is true) and the preceding frequency is 10 (II is true). The class width is 15 − 12 = 3, so IV is false.
2. Mode = \( 12 + \frac{23-10}{2(23)-10-21} \times 3 = 12 + \frac{39}{15} = 14.6 \), so III is true.
3. Statements I, II and III are correct.

Teacher's Note:
1. Find the modal class first, then read l, f₁, f₀, f₂ and h from the table.
2. Class width is upper limit minus lower limit, here 3 and not 4.

Answer: (a) 53.18

Explanation:
1. The maximum frequency is 20, so the modal class is 50–60: l = 50, h = 10, f₁ = 20, f₀ = 13, f₂ = 5.
2. Mode = \( 50 + \frac{20-13}{40-13-5} \times 10 = 50 + \frac{70}{22} = 53.18 \).
3. The mode is 53.18.

Teacher's Note:
1. Use f₀ for the class before the modal class and f₂ for the class after it.
2. Substitute carefully in the denominator 2f₁ − f₀ − f₂.

Answer: (d) 61%

Explanation:
1. Combined mean = \( \frac{n_1\bar{x}_1 + n_2\bar{x}_2 + n_3\bar{x}_3}{n_1+n_2+n_3} \).
2. = \( \frac{25 \times 70 + 40 \times 65 + 35 \times 50}{100} = \frac{1750 + 2600 + 1750}{100} = \frac{6100}{100} \).
3. The average is 61%.

Teacher's Note:
1. Never average the three percentages directly; weight each by the number of students.
2. Total students = 25 + 40 + 35 = 100.

Answer: (a) 63 kg

Explanation:
1. Total weight of 35 students = 35 × 45 = 1575 kg.
2. New mean = 45 + 0.5 = 45.5 kg for 36 persons, so the total weight = 36 × 45.5 = 1638 kg.
3. Weight of the teacher = 1638 − 1575 = 63 kg.

Teacher's Note:
1. Convert 500 grams to 0.5 kg before using it.
2. Teacher's weight is the difference of the two totals.

Answer: (d) 19

Explanation:
1. Families with income more than Rs. 16,000 = 69 and more than Rs. 19,000 = 50.
2. Families with income from 16000 to 19000 = 69 − 50 = 19.

Teacher's Note:
1. In a more-than table, subtract consecutive entries to get a class frequency.
2. Always subtract the larger boundary's count from the smaller boundary's count.

Answer: (b) 25.2

Explanation:
1. Class marks are 5, 15, 25, 35, 45, so Σfx = 40 + 180 + 250 + 385 + 405 = 1260.
2. Σf = 50, so the mean = 1260 ÷ 50 = 25.2.

Teacher's Note:
1. Use the class mark (midpoint) of each class as x.
2. Mean = Σfx ÷ Σf.

Answer: (c) 196.8

Explanation:
1. Take the assumed mean a = 200 and h = 80. The deviations d = x − 200 are −160, −80, 0, 80, 160.
2. Σfd = −3520 − 2800 + 0 + 2000 + 3840 = −480 and Σf = 150.
3. Mean = \( 200 + \frac{-480}{150} = 200 - 3.2 = 196.8 \).

Teacher's Note:
1. The assumed mean method keeps the numbers small.
2. Subtract the correction because Σfd is negative.

Answer: (d) Any natural number

Explanation:
1. The sum of the first n odd natural numbers is 1 + 3 + 5 + … + (2n − 1) = n².
2. Mean = n² ÷ n = n for every n.
3. So n can be any natural number.

Teacher's Note:
1. Remember the sum of the first n odd numbers is n².
2. An identity holds for every value, not just one.

Answer: (c) 5.6

Explanation:
1. Σfx = 20 × 4 + 40 × 5 + 30 × 6 + 10 × 10 = 80 + 200 + 180 + 100 = 560.
2. Σf = 20 + 40 + 30 + 10 = 100.
3. Mean = 560 ÷ 100 = 5.6.

Teacher's Note:
1. Treat the number of times each value occurs as its frequency.
2. Divide by the total count, not by 4.

Answer: (a) 75

Explanation:
1. Sum of the first five numbers = 410 + 5x, so \( \frac{410+5x}{5} = 82 \), which gives 410 + 5x = 410 and x = 0.
2. The second set has sum 375 + 5x = 375, so its mean = 375 ÷ 5 = 75.

Teacher's Note:
1. First use the given mean to find x.
2. Then substitute x into the second set.

Answer: (c) 5

Explanation:
1. Σf = (x − 2) + (x + 2) + (x − 3) + (x + 3) = 4x.
2. Σfx = 3(x − 2) + 5(x + 2) + 7(x − 3) + 9(x + 3) = 24x + 10.
3. Mean = \( \frac{24x+10}{4x} = 6.5 \), so 24x + 10 = 26x and x = 5.

Teacher's Note:
1. Write Σf and Σfx in terms of x and equate the ratio to 6.5.
2. Check: frequencies 3, 7, 2, 8 add to 20 and Σfx = 130.

Answer: (d) remains the same as that of the original set

Explanation:
1. For 9 observations the median is the \( \left(\frac{9+1}{2}\right)^{th} = 5^{th} \) observation in order.
2. The largest 4 observations are the 6th to 9th, so the 5th observation is unchanged.
3. The median remains the same.

Teacher's Note:
1. Median depends only on the middle value of the ordered data.
2. Changing values above the middle does not change it.

Answer: (c) 8

Explanation:
1. Frequencies: 4 occurs 3 times, 5 occurs 5 times, 6 occurs 6 times, 8 occurs 8 times, 10 occurs 7 times.
2. The maximum frequency is 8, for the value 8.
3. The mode is 8.

Teacher's Note:
1. Mode is the value that occurs most often.
2. Compare the counts, not the values.

Answer: (b) 8.8

Explanation:
1. Sum of 20 observations = 20 × 19.3 = 386.
2. New mean = 19.3 − 0.5 = 18.8 for 21 observations, so the new sum = 21 × 18.8 = 394.8.
3. New observation = 394.8 − 386 = 8.8.

Teacher's Note:
1. Find the total before and after adding the observation.
2. The difference of the totals is the new observation.

Answer: (b) 7

Explanation:
1. Let the lower limit be y and the upper limit be x. Then \( \frac{x+y}{2} = 10 \), so x + y = 20, and x − y = 6.
2. Adding: 2x = 26, so x = 13 and y = 20 − 13 = 7.
3. The class is 7–13, with mid value 10.

Teacher's Note:
1. Mid value is the average of the two limits.
2. Width is the difference of the limits.

Answer: (c) \( 11\frac{1}{3} \)

Explanation:
1. Mean = \( \frac{5x+25}{5} = 9 \), so 5x + 25 = 45 and x = 4.
2. The last three observations are 9, 11 and 14, with sum 34.
3. Their mean = \( \frac{34}{3} = 11\frac{1}{3} \).

Teacher's Note:
1. Find x from the given mean first.
2. Then average only the last three values.

Answer: (c) 9

Explanation:
1. Sum = 2 + 9 + (x + 6) + (2x + 3) + 5 + 10 + 5 = 3x + 40, and the mean is 7, so 3x + 40 = 49 and x = 3.
2. The data becomes 2, 9, 9, 9, 5, 10, 5.
3. The value 9 occurs three times, so the mode is 9.

Teacher's Note:
1. Use the mean to find x before looking for the mode.
2. Count how often each value appears.

Answer: (a) 18

Explanation:
1. Sum of the first three = 3 × 14 = 42 and sum of the last three = 3 × 17 = 51.
2. Adding them counts the third observation twice, so 42 + 51 = (total of all five) + (third) = 75 + third.
3. Third observation = 93 − 75 = 18.

Teacher's Note:
1. The third observation is common to both groups.
2. Total of all five = 5 × 15 = 75.

Answer: (c) the median decreases

Explanation:
1. Let x₁ < x₂ < … < x₁₁. The median of the other ten is \( \frac{x_5+x_6}{2} \), and the old median is x₆.
2. After replacement the middle (6th) value of the new ordered list is \( \frac{x_5+x_6}{2} \), which is less than x₆.
3. The largest value was replaced by a smaller one, so the mean also decreases. Only (c) is correct.

Teacher's Note:
1. Write the numbers in increasing order first.
2. Compare the new middle value with the old one.

Answer: (d) 58

Explanation:
1. Sum of all 15 = 15 × 50 = 750; sum of the first eight = 8 × 48 = 384; sum of the last eight = 8 × 53 = 424.
2. The eighth observation is counted in both groups, so 384 + 424 − 750 = 58.

Teacher's Note:
1. The first eight and last eight overlap in exactly one observation.
2. Subtract the total of all 15 from the sum of both groups.

Answer: (c) 36.82

Explanation:
1. The maximum cases (23) occur in the class 35–45, so this is the modal class: l = 35, h = 10, f₁ = 23, f₀ = 21, f₂ = 14.
2. Mode = \( 35 + \frac{23-21}{46-21-14} \times 10 = 35 + \frac{20}{11} = 36.82 \).
3. The required age is 36.82 years.

Teacher's Note:
1. The average age for maximum cases means the mode.
2. Check f₀ and f₂ are taken from the neighbouring classes.

Answer: (d) 45

Explanation:
1. The highest frequency is 23, in the class 35–45.
2. Its upper limit is 45.

Teacher's Note:
1. The modal class has the highest frequency.
2. Upper limit is the larger number of the class.

Answer: (d) 35.4

Explanation:
1. Class marks are 10, 20, 30, 40, 50, 60, so Σfx = 60 + 220 + 630 + 920 + 700 + 300 = 2830.
2. Σf = 80, so the mean = 2830 ÷ 80 = 35.375 ≈ 35.4.

Teacher's Note:
1. Find the class mark of each class first.
2. Divide Σfx by Σf.

Answer: (a) 41.4

Explanation:
1. The maximum frequency is 42, so the modal class is 35–45: l = 35, h = 10, f₁ = 42, f₀ = 10, f₂ = 24.
2. Mode = \( 35 + \frac{42-10}{84-10-24} \times 10 = 35 + \frac{320}{50} = 41.4 \).

Teacher's Note:
1. Identify the class with the largest frequency.
2. Use f₀ = 10 (before) and f₂ = 24 (after).

Answer: (b) 40.2

Explanation:
1. Total frequency n = 112, so n/2 = 56. Cumulative frequencies are 8, 24, 34, 76, 100, 112.
2. The median class is 35–45 (first cumulative frequency above 56): l = 35, cf = 34, f = 42, h = 10.
3. Median = \( 35 + \frac{56-34}{42} \times 10 = 35 + 5.24 = 40.2 \) (approximately).

Teacher's Note:
1. Median class is where the cumulative frequency first reaches n/2.
2. cf is the cumulative frequency of the class before it.

Answer: (c) 20 units

Explanation:
1. n = 56, so n/2 = 28. Cumulative frequencies: 16, 28, 46, 52, 56, 56.
2. The median class is 10–20: l = 10, cf = 16, f = 12, h = 10.
3. Median = \( 10 + \frac{28-16}{12} \times 10 = 20 \) units.

Teacher's Note:
1. Median class is the first class whose cumulative frequency is at least n/2.
2. Here the cumulative frequency of 10–20 equals exactly 28.

Answer: (a) 19.64 units

Explanation:
1. Class marks: 5, 15, 25, 35, 45, 55. Σfx = 80 + 180 + 450 + 210 + 180 + 0 = 1100.
2. Σf = 56, so the mean = 1100 ÷ 56 = 19.64 units (approximately).

Teacher's Note:
1. Multiply each frequency by its class mark.
2. Divide the total by 56.

Answer: (c) 20–30

Explanation:
1. The highest frequency is 18, which belongs to the class 20–30.
2. The modal class is 20–30.

Teacher's Note:
1. The modal class has the highest frequency.
2. Compare 16, 12, 18, 6, 4 and 0.

Answer: (b) 43.6 units

Explanation:
1. The maximum frequency is 40, so the modal class is 40–50: l = 40, h = 10, f₁ = 40, f₀ = 20, f₂ = 5.
2. Mode = \( 40 + \frac{40-20}{80-20-5} \times 10 = 40 + \frac{200}{55} = 43.6 \) units (approximately).

Teacher's Note:
1. Find the modal class, then apply the mode formula.
2. Here 2f₁ − f₀ − f₂ = 55.

Answer: (c) 38.75 units

Explanation:
1. Class marks 5, 15, 25, 35, 45, 55: Σfx = 0 + 75 + 250 + 700 + 1800 + 275 = 3100.
2. Σf = 80, so the mean = 3100 ÷ 80 = 38.75 units.

Teacher's Note:
1. Use class marks as x.
2. Mean = Σfx ÷ Σf.

Answer: (b) 5.6

Explanation:
1. The first five prime numbers are 2, 3, 5, 7 and 11, with sum 28.
2. Mean = 28 ÷ 5 = 5.6.

Teacher's Note:
1. List the primes carefully; 1 is not prime.
2. Mean = sum ÷ number of values.

Answer: (b) 7.5

Explanation:
1. The factors of 24 are 1, 2, 3, 4, 6, 8, 12 and 24, which are 8 numbers with sum 60.
2. Mean = 60 ÷ 8 = 7.5.

Teacher's Note:
1. List all eight factors, including 1 and 24.
2. Divide by 8.

Answer: (b) 14

Explanation:
1. The first six multiples of 4 are 4, 8, 12, 16, 20 and 24, with sum 84.
2. Mean = 84 ÷ 6 = 14.

Teacher's Note:
1. Write the multiples out: 4 × 1 to 4 × 6.
2. The mean of an equally spaced list is its middle value.

Answer: (b) 12

Explanation:
1. The y terms cancel in pairs (−5y with +5y, −3y with +3y, −y with +y), so the sum is 6x.
2. Mean = 6x ÷ 6 = x = 12.

Teacher's Note:
1. Pair up the terms that cancel.
2. The mean of terms placed symmetrically about x is x itself.

Answer: (a) 45.4

Explanation:
1. n = 100, so n/2 = 50. Cumulative frequencies: 14, 30, 48, 71, 89, 97, 100.
2. The median class is 45–50: l = 45, cf = 48, f = 23, h = 5.
3. Median = \( 45 + \frac{50-48}{23} \times 5 = 45 + 0.43 = 45.4 \).

Teacher's Note:
1. The median class is the first one whose cumulative frequency is at least 50.
2. Do not confuse the median with the frequency 23 or the lower limit 45.

Answer: (d) 13

Explanation:
1. Cumulative frequency of 150–155 is a = 12.
2. Cumulative frequency of 155–160 is a + b = 25, so b = 25 − 12 = 13.

Teacher's Note:
1. Each cumulative frequency is the running total of the frequencies.
2. Subtract the previous cumulative frequency.

Answer: (a) 5

Explanation:
1. Cumulative frequency of 165–170 is d = 43 − 35 = 8, since c = 25 + 10 = 35.
2. Then 43 + e = 48, so e = 5.
3. Check: f = 48 + 2 = 50, the total number of students.

Teacher's Note:
1. Find c first, then d, then e.
2. The last cumulative frequency must equal 50.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. The Reason is the standard formula for the mean, so it is true.
2. Σfx = 28 + 70 + 150 + 260 + 400 + 570 = 1478 and Σf = 107, so the mean = 1478 ÷ 107 = 13.81. The Assertion is true.
3. The Reason is the formula used to get the Assertion: option (a).

Teacher's Note:
1. Check the Assertion by actually computing Σfx and Σf.
2. If both are true and the Reason is the method used, choose (a).

Answer: (d) A is false, but R is true.

Explanation:
1. Arranged in ascending order: 0, 5, 11, 19, 21, 27, 30, 36, 42, 50, 52.
2. n = 11 is odd, so the median is the \( \frac{11+1}{2} = 6^{th} \) value, which is 27, not 30. The Assertion is false.
3. The Reason states the correct rule for odd n, so it is true. Option (d).

Teacher's Note:
1. Always arrange the data in order before picking the middle value.
2. Count the position carefully, the 6th value here is 27.

Answer: (c) A is true, but R is false.

Explanation:
1. The correct relation is 3 median = mode + 2 mean, so median = \( \frac{1}{3} \)(mode + 2 mean). The Reason, with \( \frac{1}{2} \), is false.
2. Median = \( \frac{1}{3}(60 + 2 \times 66) = \frac{192}{3} = 64 \), so the Assertion is true.
3. Assertion true, Reason false: option (c).

Teacher's Note:
1. The empirical relation is Mode = 3 Median − 2 Mean.
2. Check the fraction in the Reason; it is 1/3, not 1/2.