CBSE Class 10 Mathematics Statistics MCQs Set 09

Answer: (c) Mode

Explanation:
1. The company wants the size that customers buy most often.
2. The value that occurs most often is the mode, so the company should stock the most frequently sold size.
3. Mean and median would give a size that may not even be sold.

Teacher's Note:
1. Mode is the most frequent value, so it answers 'which is the most popular?' questions.
2. Mean and median can be values nobody actually buys.

Answer: (d) Rs. 5083.33

Explanation:
1. \( \sum f_ix_i = 48000+48000+50000+48000+42000+32000+27000+10000 = 305000 \)
2. \( \text{Mean} = \dfrac{305000}{60} = 5083.33 \)

Teacher's Note:
1. Mean of ungrouped data with frequencies is sum of f times x divided by sum of f.
2. Add each product carefully; a slip in one row changes the answer.

Answer: (c) 20

Explanation:
1. \( N = 67,\ N/2 = 33.5 \). Cumulative frequencies: 4, 9, 22, 42, 56, 63, 67. Median class: 125-145.
2. Highest frequency is 20, so the modal class is also 125-145.
3. Upper limit of median class \( - \) lower limit of modal class \( = 145-125 = 20 \).

Teacher's Note:
1. Find the median class from the cumulative frequency column.
2. The modal class is the one with the highest frequency.

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

Explanation:
1. With 9 observations in order, the median is the 5th observation.
2. The largest 4 observations are the 6th to 9th. Increasing them does not change the 5th observation.
3. So the median remains 20.5.

Teacher's Note:
1. The median depends only on the middle value.
2. Changing values above the middle does not change it.

Answer: (a) 5

Explanation:
1. 18 students weigh above the median and none equals it, so 18 weigh below it. Hence the total number of students is 36.
2. Known frequencies: \( 6+8+7+10 = 31 \), so the 37-40 class has \( 36-31 = 5 \) students.

Teacher's Note:
1. The median splits the data in half.
2. Total = 2 x (number above the median) when no student equals the median.

Answer: (c) 30-40

Explanation:
1. Class frequencies: 0-10: 3, 10-20: 9, 20-30: 15, 30-40: 30, 40-50: 18, 50-60: 5.
2. The highest frequency is 30, so the modal class is 30-40.

Teacher's Note:
1. Subtract consecutive cumulative values to get the frequencies.
2. The modal class has the highest frequency.

Answer: (b) Nitesh

Explanation:
1. The highest frequency is 19, so the modal class is 145-155.
2. Only Nitesh's reading, 147 mmHg, lies in 145-155.

Teacher's Note:
1. Find the modal class first, then see whose value falls in it.
2. 121 and 160 lie in other classes.

Answer: (d) 7

Explanation:
1. Mode = 3 Median \( - \) 2 Mean, so \( 7 = 3\,\text{Median}-14 \).
2. \( \text{Median} = 21/3 = 7 \).

Teacher's Note:
1. Use Mode = 3 Median - 2 Mean.
2. Equal mean and mode force the median to be the same.

Answer: (d) 150

Explanation:
1. Class frequencies: below 140: 4, 140-145: 7, 145-150: 18, 150-155: 11, 155-160: 6, 160-165: 5.
2. The highest frequency is 18, so the modal class is 145-150 and its upper limit is 150.

Teacher's Note:
1. Subtract neighbouring cumulative entries to get class frequencies.
2. The upper limit is the larger number of the modal class.

Answer: (c) 37.3

Explanation:
1. Class marks: 15, 25, 35, 45, 55. \( \sum f_ix_i = 30+75+35+180+165 = 485 \), \( \sum f_i = 13 \).
2. \( \text{Mean} = \dfrac{485}{13} = 37.3 \) (approx.).

Teacher's Note:
1. Mean = sum of f x divided by sum of f.
2. Use class marks, not class limits.

Answer: (a) 5.71

Explanation:
1. \( N = 16,\ N/2 = 8 \). Cumulative frequencies: 4, 11, 14, 16, so the median class is 4-7.
2. \( \text{Median} = 4+\dfrac{8-4}{7}\times 3 = 4+1.71 = 5.71 \).

Teacher's Note:
1. Median class is the first with cumulative frequency at least N/2.
2. Substitute l = 4, cf = 4, f = 7, h = 3.

Answer: (b) 40

Explanation:
1. Total number of students = 30, an even number, so the median is the mean of the 15th and 16th values.
2. Cumulative frequencies: 5, 12, 16, 22, 25, 27, 30. Both the 15th and 16th observations are 40.
3. Median \( = \dfrac{40+40}{2} = 40 \).

Teacher's Note:
1. For ungrouped data with frequencies, use cumulative frequencies to locate the middle values.
2. If N is even, average the two middle values.

Answer: (a) 14

Explanation:
1. Sum of 8 numbers = \( 8\times 23 = 184 \).
2. Sum of 9 numbers = \( 9\times 22 = 198 \).
3. Number added = \( 198-184 = 14 \). The number is 14.

Teacher's Note:
1. Mean times count gives the total.
2. The difference of totals is the new number.
3. Always check that the new mean is lower, so the new number is below 23.

Answer: (b) 8

Explanation:
1. Class marks: 12, 14, 16, 18, 20, 22, 24. Take assumed mean a = 18, so \( d_i = x_i-18 \): \( -6,-4,-2,0,2,4,6 \).
2. \( \sum f_id_i = -18-24-18+0+2f+20+24 = 2f-16 \); \( \sum f_i = 40+f \).
3. \( \text{Mean} = 18+\dfrac{2f-16}{40+f} = 18 \Rightarrow 2f-16 = 0 \Rightarrow f = 8 \). f = 8.

Teacher's Note:
1. Choose the assumed mean equal to the given mean to simplify.
2. Sum of f times d equals zero.
3. Verify by recomputing the mean with f = 8.

Answer: (c) 635.97

Explanation:
1. Empirical relation: Mode = 3 Median \( - \) 2 Mean.
2. \( \text{Mode} = 3(635.97)-2(635.97) = 635.97 \). Mode = 635.97.

Teacher's Note:
1. Use Mode = 3 Median - 2 Mean.
2. Equal mean and median means a symmetric distribution.
3. In a symmetric distribution all three measures coincide.

Answer: (d) Rs. 211

Explanation:
1. Take \( a = 225,\ h = 50 \), \( u_i = \dfrac{x_i-225}{50} \). Class marks: 125, 175, 225, 275, 325, so \( u_i = -2,-1,0,1,2 \).
2. \( \sum f_iu_i = -8-5+0+2+4 = -7 \), \( \sum f_i = 25 \).
3. \( \text{Mean} = 225+50\times\dfrac{-7}{25} = 225-14 = 211 \). Mean = Rs. 211.

Teacher's Note:
1. Choose a middle class mark as the assumed mean.
2. Divide each deviation by the class width h.
3. Multiply the final fraction by h before adding to a.

Answer: (a) Mean 32, Median 33, Mode 34.39

Explanation:
1. The classes are not continuous, so subtract 0.5 from each lower limit and add 0.5 to each upper limit: 10.5-15.5, 15.5-20.5, ..., 45.5-50.5. \( N = 50 \).
2. Mean: class marks 13, 18, 23, 28, 33, 38, 43, 48. \( \sum f_ix_i = 26+54+138+196+462+456+172+96 = 1600 \). \( \text{Mean} = \dfrac{1600}{50} = 32 \).
3. Median: cumulative frequencies 2, 5, 11, 18, 32, ... \( N/2 = 25 \), median class 30.5-35.5. \( \text{Median} = 30.5+\dfrac{25-18}{14}\times 5 = 33 \).
4. Mode: modal class 30.5-35.5 (f1 = 14, f0 = 7, f2 = 12). \( \text{Mode} = 30.5+\dfrac{14-7}{28-7-12}\times 5 = 30.5+3.89 = 34.39 \).
5. Mean = 32, Median = 33, Mode = 34.39 (approx.).

Teacher's Note:
1. Discontinuous classes need a correction of 0.5 on each side.
2. The mode is found from the three frequencies around the highest one.
3. Mean is calculated from class marks.

Answer: (b) 36.76 years

Explanation:
1. Convert to class frequencies (age 15 onwards): 15-20 (2), 20-25 (4), 25-30 (18), 30-35 (21), 35-40 (33), 40-45 (11), 45-50 (3), 50-55 (6), 55-60 (2).
2. Modal class: 35-40 with \( l=35,\ h=5,\ f_1=33,\ f_0=21,\ f_2=11 \).
3. \( \text{Mode} = 35+\dfrac{33-21}{66-21-11}\times 5 = 35+\dfrac{60}{34} = 36.76 \). Modal age = 36.76 years (approx.).

Teacher's Note:
1. The lower limit of the first class is 15 because policies start from age 15.
2. Mode uses the highest frequency and its two neighbours.
3. Round sensibly: 36.76 years.

Answer: (c) 51.75

Explanation:
1. Frequencies are differences of consecutive 'and above' entries: 0-10 (3), 10-20 (5), 20-30 (7), 30-40 (10), 40-50 (12), 50-60 (15), 60-70 (12), 70-80 (6), 80-90 (2), 90-100 (8). \( N = 80 \).
2. Class marks 5, 15, ..., 95. \( \sum f_ix_i = 15+75+175+350+540+825+780+450+170+760 = 4140 \).
3. \( \text{Mean} = \dfrac{4140}{80} = 51.75 \). Mean marks = 51.75.

Teacher's Note:
1. Subtract each entry from the one above it.
2. Class mark = (lower + upper) / 2.
3. Check that the frequencies add up to the first entry, 80.

Answer: (d) 3 : 2

Explanation:
1. Let the number of boys be \( b \) and of girls be \( g \).
2. \( \dfrac{71b+73g}{b+g} = 71.8 \Rightarrow 71b+73g = 71.8b+71.8g \Rightarrow 1.2g = 0.8b \).
3. \( \dfrac{b}{g} = \dfrac{1.2}{0.8} = \dfrac{3}{2} \). Boys : Girls = 3 : 2.

Teacher's Note:
1. Total marks of the school = marks of boys + marks of girls.
2. The combined mean lies nearer to the group with more students.
3. Here 71.8 is nearer to 71, so boys are more.

Answer: (a) 4 : 3

Explanation:
1. Let \( X = 9k,\ M = 8k \). Mode \( Z = 3M-2X = 24k-18k = 6k \).
2. \( M:Z = 8k:6k = 4:3 \). M : Z = 4 : 3.

Teacher's Note:
1. Assume the mean and median as multiples of a common k.
2. Substitute into Mode = 3 Median - 2 Mean.
3. Cancel k in the ratio.

Answer: (b) 26 pages

Explanation:
1. Class marks: 17, 20, 23, 26, 29. Take \( a = 23,\ h = 3 \); \( u_i = -2,-1,0,1,2 \).
2. \( \sum f_iu_i = -2-3+0+9+26 = 30 \), \( \sum f_i = 30 \).
3. \( \text{Mean} = 23+3\times\dfrac{30}{30} = 26 \). Average = 26 pages per day.

Teacher's Note:
1. Class mark = (lower + upper)/2.
2. Gaps between classes do not affect the class marks.
3. Check: direct sum 780 divided by 30 gives 26.

Answer: (c) 109.92 seats

Explanation:
1. Class marks: 102, 106, 110, 114, 118. Take \( a = 110,\ h = 4 \); \( u_i = -2,-1,0,1,2 \).
2. \( \sum f_iu_i = -30-20+0+18+30 = -2 \), \( \sum f_i = 100 \).
3. \( \text{Mean} = 110+4\times\dfrac{-2}{100} = 109.92 \). Mean = 109.92 seats.

Teacher's Note:
1. The assumed mean is the middle class mark.
2. Negative sum of deviations lowers the mean below a.
3. Answer is a mean, so a decimal is fine.

Answer: (d) 201.7 g

Explanation:
1. Modal class: 201-202 (highest frequency 26): \( l=201,\ h=1,\ f_1=26,\ f_0=12,\ f_2=20 \).
2. \( \text{Mode} = 201+\dfrac{26-12}{52-12-20}\times 1 = 201+\dfrac{14}{20} = 201.7 \). Modal weight = 201.7 g.

Teacher's Note:
1. Modal class has the highest frequency.
2. f0 is before and f2 is after the modal class.
3. The mode must lie inside the modal class.

Answer: (a) 8.1

Explanation:
1. Class marks: 1, 3, 5, 7, 9, 11, 13. \( \sum f_ix_i = 1+6+5+35+54+22+39 = 162 \), \( \sum f_i = 20 \).
2. \( \text{Mean} = \dfrac{162}{20} = 8.1 \). Mean = 8.1 plants per house.

Teacher's Note:
1. Check frequencies sum to 20.
2. Class marks are the mid-points.
3. Mean can be a decimal even though plants are counted in whole numbers.

Answer: (b) 4608.7 runs

Explanation:
1. Modal class: 4000-5000 (frequency 18): \( l=4000,\ h=1000,\ f_1=18,\ f_0=4,\ f_2=9 \).
2. \( \text{Mode} = 4000+\dfrac{18-4}{36-4-9}\times 1000 = 4000+\dfrac{14}{23}\times 1000 = 4608.7 \). Mode = 4608.7 runs (approx.).

Teacher's Note:
1. Frequencies 4 and 9 are those on either side of 18.
2. Class width is 1000.
3. The value of the 9000-10000 row does not affect the modal class.

Answer: (c) 56.67 kg

Explanation:
1. Cumulative frequencies: 2, 5, 13, 19, 25, 28, 30. \( N/2 = 15 \), so the median class is 55-60.
2. \( l=55,\ cf=13,\ f=6,\ h=5 \). \( \text{Median} = 55+\dfrac{15-13}{6}\times 5 = 55+1.67 = 56.67 \). Median weight = 56.67 kg.

Teacher's Note:
1. Write the cumulative frequency row first.
2. Use N/2 = 15.
3. Answer lies within the class 55-60.

Answer: (d) 17

Explanation:
1. Adding 5 to every observation increases the sum by \( 20\times 5 = 100 \).
2. New mean \( = \dfrac{20\times 12+100}{20} = 12+5 = 17 \). New mean = 17.

Teacher's Note:
1. If each value rises by k, the mean rises by k.
2. The same is true for the median and mode.
3. Multiplying all values by k multiplies the mean by k.

Answer: (a) x = 8, y = 7

Explanation:
1. \( 5+x+20+15+y+5 = 60 \Rightarrow x+y = 15 \). ... (1)
2. Median 28.5 lies in 20-30: \( l=20,\ f=20,\ cf=5+x,\ h=10,\ N/2 = 30 \).
3. \( 28.5 = 20+\dfrac{30-(5+x)}{20}\times 10 \Rightarrow 17 = 25-x \Rightarrow x = 8 \).
4. From (1), \( y = 7 \). x = 8, y = 7.

Teacher's Note:
1. N is given as 60.
2. Median class is decided by 28.5.
3. Check: cumulative frequencies 5, 13, 33 put N/2 = 30 in 20-30.

Answer: (b) 3

Explanation:
1. Class marks: 3, 5, 7, 9, 11, 13. \( \sum f_ix_i = 18+40+105+9p+88+52 = 303+9p \). \( \sum f_i = 41+p \).
2. \( \dfrac{303+9p}{41+p} = 7.5 \Rightarrow 303+9p = 307.5+7.5p \Rightarrow 1.5p = 4.5 \Rightarrow p = 3 \). p = 3.

Teacher's Note:
1. Mean = sum of f x divided by sum of f.
2. Cross-multiply and collect p terms.
3. Check by recomputing the mean with p = 3.

Answer: (c) 12

Explanation:
1. Mode 45 lies in 40-60, so this is the modal class: \( l=40,\ h=20,\ f_0=10,\ f_2=6,\ f_1 = f \).
2. \( 45 = 40+\dfrac{f-10}{2f-10-6}\times 20 \Rightarrow \dfrac{1}{4} = \dfrac{f-10}{2f-16} \Rightarrow 2f-16 = 4f-40 \Rightarrow f = 12 \). The missing frequency is 12.

Teacher's Note:
1. The class containing the mode is the modal class.
2. Form an equation in f.
3. Check: 12 is the largest frequency, as a modal class requires.

Answer: (d) 110.67 runs

Explanation:
1. Class marks: 20, 60, 100, 140, 180. \( \sum f_ix_i = 240+1200+3500+4200+4140 = 13280 \). \( \sum f_i = 120 \).
2. \( \text{Mean} = \dfrac{13280}{120} = 110.67 \). Mean = 110.67 runs (approx.).

Teacher's Note:
1. Add the frequencies first: 120.
2. Multiply each class mark by its frequency.
3. Divide the total by 120.

Answer: (a) 2.08 m

Explanation:
1. Cumulative frequencies: 40, 120, 182, 220, 250. \( N/2 = 125 \), so the median class is 2-3: \( l=2,\ cf=120,\ f=62,\ h=1 \).
2. \( \text{Median} = 2+\dfrac{125-120}{62}\times 1 = 2.08 \) m (approx.).
3. Interpretation: half of the students jumped less than about 2.08 m and the other half jumped more than 2.08 m.

Teacher's Note:
1. Median class has cumulative frequency just above 125.
2. Interpretation should mention 'half of the students'.
3. Use the same unit (metres) in the answer.

Answer: (b) 146.975 mm

Explanation:
1. Class marks: 122, 131, 140, 149, 158, 167, 176. \( \sum f_ix_i = 366+655+1260+1788+790+668+352 = 5879 \).
2. \( \text{Mean} = \dfrac{5879}{40} = 146.975 \). Mean length = 146.975 mm (approx. 147 mm).

Teacher's Note:
1. Class mark of 118-126 is (118+126)/2 = 122.
2. Continuity correction does not change class marks.
3. Total of frequencies is 40.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. Modal class is 12-15 (highest frequency 23), so l = 12, h = 3, f1 = 23, f0 = 21, f2 = 10.
2. \( \text{Mode} = 12+\dfrac{23-21}{46-21-10}\times 3 = 12+\dfrac{2}{15}\times 3 = 12.4 \), so A is true.
3. R is also a true statement, but for grouped data the mode is found by the formula, not by picking one value; R does not explain A.

Teacher's Note:
1. For grouped data use the modal-class formula.
2. A true reason is not always the correct explanation of the assertion.

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

Explanation:
1. Empirical relation: Mode = 3 Median \( - \) 2 Mean.
2. \( 50.5 = 3(45.5)-2\,\text{Mean} \Rightarrow \text{Mean} = \dfrac{136.5-50.5}{2}=43 \), so A is true.
3. R states Median = Mode + 2 Mean, which is not the correct relation, so R is false.

Teacher's Note:
1. Remember Mode = 3 Median - 2 Mean.
2. Check the relation before using it in a problem.

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

Explanation:
1. The maximum frequency is 25, belonging to the class 25-35, so A is true.
2. R is the definition of the modal class, so R is true and explains A.

Teacher's Note:
1. The modal class is simply the class with the highest frequency.
2. Do not confuse it with the median class.

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

Explanation:
1. Cumulative frequencies: 2, 10, 22, 42, 50. \( N/2 = 25 \).
2. The first cumulative frequency greater than 25 is 42, so the median class is 150-160. A is true.
3. The median class is the class whose cumulative frequency is greater than (and nearest to) N/2, not lesser, so R is false.

Teacher's Note:
1. Median class: first class whose cumulative frequency is at least N/2.
2. Always write the cumulative frequency column.

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

Explanation:
1. Arranged in ascending order: \( a-4,\ a-2,\ a-\tfrac{3}{2},\ a-\tfrac{1}{2},\ a+2,\ a+6 \). There are 6 values.
2. Median \( = \dfrac{(a-\frac{3}{2})+(a-\frac{1}{2})}{2} = a-1 \), so A is true.
3. R is false: for an even number of values there is no single middle value; the median is the mean of the two middle values.

Teacher's Note:
1. For an even count, average the two middle values.
2. Always arrange the data in order first.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. First 10 primes: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29. n = 10 is even.
2. Median \( = \dfrac{11+13}{2} = 12 \), so A is true.
3. R is a true statement for odd n, but A concerns an even number of observations, so R does not explain A.

Teacher's Note:
1. Check whether n is odd or even before choosing the median rule.
2. A true reason must actually apply to the assertion.