CBSE Class 10 Mathematics Statistics Worksheet Set 01

Case Based MCQs

Case I : Read the following passage and answer the questions

Toll Tax Collection

On a particular day, National Highway Authority of India (NHAI) checked the toll tax collection of a particular toll plaza in Rajasthan.

The following table shows the toll tax paid by drivers and the number of vehicles on that particular day.

Question. If A is taken as assumed mean, then the possible value of A is
(a) 32
(b) 42
(c) 85
(d) 55

Answer: D

Question. If xi’s denote the class marks and di’s denote the deviation of assumed mean (A) from xi’s, then the minimum value of |di| is
(a) –200
(b) –100
(c) 0
(d) 100

Answer: C

Question. The mean of toll tax received by NHAI by assumed mean method is
(a) ₹ 52
(b) ₹ 52.14
(c) ₹ 52.50
(d) ₹ 53.50

Answer: B

Question. The mean of toll tax received by NHAI by direct method is
(a) equal to the mean of toll tax received by NHAI by assumed mean method
(b) greater than the mean of toll tax received by NHAI by assumed mean method
(c) less than the mean of toll tax received by NHAI by assumed mean method
(d) can't say

Answer: A

Question. The average toll tax received by NHAI in a day, from that particular toll plaza, is
(a) ₹ 21000.60
(b) ₹ 21900.80
(c) ₹ 30000.90
(d) ₹ 21898.80

Answer: D

Case II : Read the following passage and answer the questions

Distance Analysis of Public Transport Buses

Transport department of a city wants to buy some Electric buses for the city. For which they want to analyse the distance travelled by existing public transport buses in a day.

The following data shows the distance travelled by 60 existing public transport buses in a day.

Question. The upper limit of a class and lower limit of its succeeding class is differ by
(a) 9
(b) 1
(c) 10
(d) none of the options

Answer: B

Question. The median class is
(a) 229.5-239.5
(b) 230-239
(c) 220-229
(d) 219.5-229.5

Answer: D

Question. The cumulative frequency of the class preceding the median class is
(a) 14
(b) 18
(c) 26
(d) 10

Answer: B

Question. The median of the distance travelled is
(a) 222 km
(b) 225 km
(c) 223 km
(d) none of the options

Answer: D

Question. If the mode of the distance travelled is 223.78 km, then mean of the distance travelled by the bus is
(a) 225 km
(b) 220 km
(c) 230.29 km
(d) 224.29 km

Answer: D

Case III : Read the following passage and answer the questions

Electric Scooter Manufacturing Company

An electric scooter manufacturing company wants to declare the mileage of their electric scooters. For this, they recorded the mileage (km/charge) of 50 scooters of the same model. Details of which are given in the following table.

Question. The average mileage is
(a) 140 km/charge
(b) 150 km/charge
(c) 130 km/charge
(d) 144.8 km/charge

Answer: D

Question. The modal value of the given data is
(a) 150
(b) 150.91
(c) 145.6
(d) 140.9

Answer: B

Question. The median value of the given data is
(a) 140
(b) 146.67
(c) 130
(d) 136.6

Answer: B

Question. Assumed mean method is useful in determining the
(a) Mean
(b) Median
(c) Mode
(d) All of these

Answer: A

Question. The manufacturer can claim that the mileage for his scooter is
(a) 144 km/charge
(b) 155 km/charge
(c) 165 km/charge
(d) 175 km/charge

Answer: A

Short Answer Type Questions

 

Question. The marks obtained by 40 students of class IX in an examination are given below :

18, 8, 12, 6, 8, 16, 12, 5, 23, 2,16, 23, 2, 10, 20, 12, 9, 7, 6, 5, 3, 5, 13, 21, 13, 15, 20, 24,1, 7, 21, 16, 13, 18, 23, 7, 3, 18, 17, 16. 

Present the data in the form of a frequency distribution using the same class size, one such class being 15-20 (where 20 is not included)

Answer: The minimum and maximum marks in the given raw data are 0 and 24 respectively. It is given that 15-20 is one of the class intervals and the class size is same. So, the classes of

equal size are

0-5, 5-10, 10-15, 15-20 and 20-25

Thus, the frequency distribution is as given under :

 

 

Question 1. The following table gives the distribution of total household expenditure (in Rs) of 200 manual workers in a city:

Find mean, mode and median of the above data. Also compare and interpret the three measures of central tendency. (Ans: Rs.266.25, Rs.184.78~, Rs.255.36~)
Answer: Let class interval width \(h = 50\), assumed mean \(A = 275\).
Class midpoints \((x_i)\): 125, 175, 225, 275, 325, 375, 425, 475.
Deviations \(u_i = \frac{x_i - A}{h}\): -3, -2, -1, 0, 1, 2, 3, 4.
Frequencies \((f_i)\): 24, 40, 33, 28, 30, 22, 16, 7 (Total \(N = 200\)).
Product \(f_i u_i\): -72, -80, -33, 0, 30, 44, 48, 28 -> Sum \(\sum f_i u_i = -35\).
Mean \(= A + \left(\frac{\sum f_i u_i}{N}\right) \times h\)
\(= 275 + \left(\frac{-35}{200}\right) \times 50 = 275 - 8.75 = \text{Rs. } 266.25\)
Mode class is 150 - 200 (highest frequency 40).
Mode \(= l + \left(\frac{f_1 - f_0}{2f_1 - f_0 - f_2}\right) \times h\)
\(= 150 + \left(\frac{40 - 24}{2(40) - 24 - 33}\right) \times 50 = 150 + \left(\frac{16}{80 - 57}\right) \times 50 = 150 + \frac{800}{23} = 150 + 34.78 = \text{Rs. } 184.78\)
Cumulative frequencies: 24, 64, 97, 125, 155, 177, 193, 200.
Median class is 250 - 300 (since \(\frac{N}{2} = 100\)).
Median \(= l + \left(\frac{\frac{N}{2} - cf}{f}\right) \times h\)
\(= 250 + \left(\frac{100 - 97}{28}\right) \times 50 = 250 + \frac{150}{28} = 250 + 5.36 = \text{Rs. } 255.36\)
In simple words: The average household expenditure is Rs. 266.25, most workers spend around Rs. 184.78, and the middle expenditure is Rs. 255.36.
Exam Tip: Always construct a clean step-by-step frequency distribution table with columns for class interval, midpoint, frequency, cumulative frequency, and step-deviation to avoid calculation errors.

Question 2. If the mean of the following distribution is 54. Find the value of p.

(Ans: p=11)
Answer: Class midpoints \((x_i)\): 10, 30, 50, 70, 90.
Frequencies \((f_i)\): 7, p, 10, 9, 13.
Sum of frequencies \(\sum f_i = 7 + p + 10 + 9 + 13 = 39 + p\).
Products \(f_i x_i\): \(7(10)=70\), \(30p\), \(10(50)=500\), \(9(70)=630\), \(13(90)=1170\).
Sum \(\sum f_i x_i = 70 + 30p + 500 + 630 + 1170 = 2370 + 30p\).
Mean \(= \frac{\sum f_i x_i}{\sum f_i}\)
\(54 = \frac{2370 + 30p}{39 + p}\)
\(54(39 + p) = 2370 + 30p\)
\(2106 + 54p = 2370 + 30p\)
\(54p - 30p = 2370 - 2106\)
\(24p = 264\)
\(p = \frac{264}{24} = 11\)
In simple words: Using the given mean formula, we form a linear equation in one variable and solve for the unknown frequency p.
Exam Tip: Cross-multiply carefully when solving equations involving unknown frequencies in grouped data means.

Question 3. The mean of the following frequency table is 50. Find the missing frequencies.

(Ans: \(F_1=28, F_2=24\))
Answer: Total frequency \(\sum f_i = 17 + F_1 + 32 + F_2 + 19 = 120\)
\(68 + F_1 + F_2 = 120 \implies F_1 + F_2 = 52\) ---(Equation 1)
Midpoints \((x_i)\): 10, 30, 50, 70, 90.
Products \(f_i x_i\): \(17(10)=170\), \(30F_1\), \(32(50)=1600\), \(70F_2\), \(19(90)=1710\).
Sum \(\sum f_i x_i = 170 + 30F_1 + 1600 + 70F_2 + 1710 = 3480 + 30F_1 + 70F_2\).
Given Mean = 50.
\(\frac{3480 + 30F_1 + 70F_2}{120} = 50\)
\(3480 + 30F_1 + 70F_2 = 6000\)
\(30F_1 + 70F_2 = 2520 \implies 3F_1 + 7F_2 = 252\) ---(Equation 2)
Multiply Equation 1 by 3: \(3F_1 + 3F_2 = 156\) ---(Equation 3)
Subtract Equation 3 from Equation 2:
\((3F_1 + 7F_2) - (3F_1 + 3F_2) = 252 - 156\)
\(4F_2 = 96 \implies F_2 = 24\)
Substitute \(F_2 = 24\) in Equation 1:
\(F_1 + 24 = 52 \implies F_1 = 28\)
In simple words: We set up two simultaneous linear equations using the total frequency and the given mean, then solve for the two missing frequencies.
Exam Tip: Always use the total frequency sum to form your first simple linear equation before applying the mean formula.

Question 4. Compute the median from the following data:

Also, find the mean and mode of the above data. (Ans: 154.22~, 154.68~, 154.4)
Answer: Since class marks are given with a uniform class size \(h = 10\), the class intervals are: 110-120, 120-130, 130-140, 140-150, 150-160, 160-170, 170-180, 180-190, 190-200.
Total frequency \(N = 6 + 25 + 48 + 72 + 116 + 60 + 38 + 22 + 13 = 400\).
Cumulative Frequencies (cf): 6, 31, 79, 151, 267, 327, 365, 387, 400.
\(\frac{N}{2} = 200\). Median class is 150 - 160.
Median \(= l + \left(\frac{\frac{N}{2} - cf}{f}\right) \times h\)
\(= 150 + \left(\frac{200 - 151}{116}\right) \times 10 = 150 + \frac{490}{116} = 150 + 4.2248 = 154.22\)
For Mean: Let assumed mean \(A = 155\), \(h = 10\).
\(u_i\): -4, -3, -2, -1, 0, 1, 2, 3, 4.
\(f_i u_i\): \(6(-4)=-24\), \(25(-3)=-75\), \(48(-2)=-96\), \(72(-1)=-72\), \(116(0)=0\), \(60(1)=60\), \(38(2)=76\), \(22(3)=66\), \(13(4)=52\).
Sum \(\sum f_i u_i = -24 - 75 - 96 - 72 + 60 + 76 + 66 + 52 = -13\).
Mean \(= A + \left(\frac{\sum f_i u_i}{N}\right) \times h = 155 + \left(\frac{-13}{400}\right) \times 10 = 155 - 0.325 = 154.675 = 154.68\)
For Mode: Modal class is 150 - 160 (frequency 116).
Mode \(= l + \left(\frac{f_1 - f_0}{2f_1 - f_0 - f_2}\right) \times h\)
\(= 150 + \left(\frac{116 - 72}{2(116) - 72 - 60}\right) \times 10 = 150 + \left(\frac{44}{232 - 132}\right) \times 10 = 150 + \frac{440}{100} = 154.4\)
In simple words: We first convert discrete class marks into continuous intervals with class size 10, then apply standard statistical formulas for median, mean, and mode.
Exam Tip: When discrete point data is given with constant gaps, treat them as continuous class intervals taking midpoints as class marks.

Question 5. If the median of the following data is 32.5, find the values of x and y:

Also, find the mean and mode of the above data. (Ans: x=3, y=6; Mean=32.5, Mode=33.33~)
Answer: Total frequency \(\sum f_i = x + 5 + 9 + 12 + y + 3 + 2 = 40\)
\(x + y + 31 = 40 \implies x + y = 9\) ---(Equation 1)
Cumulative frequencies: x, \(x+5\), \(x+14\), \(x+26\), \(x+y+26\), \(x+y+29\), \(x+y+31\).
Given median = 32.5, which lies in the class interval 30 - 40.
Here \(l = 30\), \(h = 10\), \(cf = x + 14\), \(f = 12\), and \(\frac{N}{2} = 20\).
Median \(= l + \left(\frac{\frac{N}{2} - cf}{f}\right) \times h\)
\(32.5 = 30 + \left(\frac{20 - (x + 14)}{12}\right) \times 10\)
\(2.5 = \frac{6 - x}{12} \times 10\)
\(30 = 60 - 10x \implies 10x = 30 \implies x = 3\)
Substitute \(x = 3\) in Equation 1:
\(3 + y = 9 \implies y = 6\)
With \(x = 3, y = 6\), the frequencies are: 3, 5, 9, 12, 6, 3, 2.
Midpoints \((x_i)\): 5, 15, 25, 35, 45, 55, 65.
Products \(f_i x_i\): \(3(5)=15\), \(5(15)=75\), \(9(25)=225\), \(12(35)=420\), \(6(45)=270\), \(3(55)=165\), \(2(65)=130\).
Sum \(\sum f_i x_i = 15 + 75 + 225 + 420 + 270 + 165 + 130 = 1300\).
Mean \(= \frac{1300}{40} = 32.5\)
Modal class is 30 - 40 (frequency 12).
Mode \(= l + \left(\frac{f_1 - f_0}{2f_1 - f_0 - f_2}\right) \times h\)
\(= 30 + \left(\frac{12 - 9}{2(12) - 9 - 6}\right) \times 10 = 30 + \left(\frac{3}{24 - 15}\right) \times 10 = 30 + \frac{30}{9} = 30 + 3.33 = 33.33\)
In simple words: We find x using the median formula and y using the total frequency sum, then compute mean and mode normally.
Exam Tip: Identify the median class correctly by checking where the value \(N/2\) falls in the cumulative frequency column.

Question 6. Find the median of the following cumulative frequency distribution graphically and verify it by actual calculation:

(Ans: 46.5)
Answer: First, convert the "less than" cumulative frequency table into standard class intervals and frequencies:
Class intervals: 36-38, 38-40, 40-42, 42-44, 44-46, 46-48, 48-50, 50-52.
Frequencies: 0, 3-0=3, 5-3=2, 9-5=4, 14-9=5, 28-14=14, 32-28=4, 35-32=3. (Total \(N = 35\)).
\(\frac{N}{2} = \frac{35}{2} = 17.5\).
The cumulative frequency just greater than 17.5 is 28, so the median class is 46 - 48.
Here \(l = 46\), \(h = 2\), \(cf = 14\), \(f = 14\).
Median \(= l + \left(\frac{\frac{N}{2} - cf}{f}\right) \times h\)
\(= 46 + \left(\frac{17.5 - 14}{14}\right) \times 2 = 46 + \frac{3.5 \times 2}{14} = 46 + \frac{7}{14} = 46 + 0.5 = 46.5\)
In simple words: We reconstruct the frequency table from the given cumulative frequency data and apply the median formula.
Exam Tip: For 'less than' ogive/cumulative frequency tables, subtract consecutive cumulative frequencies to obtain individual class frequencies.

Question 7. The Mode of the following frequency table is 26. Find the missing frequencies if the total frequency is 50.

(Ans: \(F_1=9, F_2=11\))
Answer: Total frequency \(\sum f_i = 3 + F_1 + 15 + F_2 + 8 + 4 = 50\)
\(30 + F_1 + F_2 = 50 \implies F_1 + F_2 = 20\) ---(Equation 1)
Given Mode = 26, which lies in the class interval 20 - 30. Thus, modal class is 20 - 30.
Here \(l = 20\), \(h = 10\), \(f_1 = 15\), \(f_0 = F_1\), \(f_2 = F_2\).
Mode \(= l + \left(\frac{f_1 - f_0}{2f_1 - f_0 - f_2}\right) \times h\)
\(26 = 20 + \left(\frac{15 - F_1}{2(15) - F_1 - F_2}\right) \times 10\)
\(6 = \frac{15 - F_1}{30 - (F_1 + F_2)} \times 10\)
Substitute \(F_1 + F_2 = 20\):
\(6 = \frac{15 - F_1}{30 - 20} \times 10\)
\(6 = \frac{15 - F_1}{10} \times 10\)
\(6 = 15 - F_1 \implies F_1 = 15 - 6 = 9\)
Substitute \(F_1 = 9\) in Equation 1:
\(9 + F_2 = 20 \implies F_2 = 11\)
In simple words: Since the mode is 26, the modal class is 20-30. Using the mode formula along with the total frequency equation gives the missing frequencies.
Exam Tip: The value of the mode immediately reveals the modal class, which simplifies finding unknown frequencies.

Question 8. Find the Mean, Modal and Median lifetime of the 100 tube lights from the following data:

(Ans: 1454.5 hours, 1479.5 hours, 1460.61 hours~)
Answer: Notice that the given intervals are discontinuous (1000-1099, 1100-1199). Convert them into continuous intervals by subtracting 0.5 from lower limits and adding 0.5 to upper limits:
Continuous intervals: 999.5-1099.5, 1099.5-1199.5, 1199.5-1299.5, 1299.5-1399.5, 1399.5-1499.5, 1499.5-1599.5, 1599.5-1699.5, 1699.5-1799.5, 1799.5-1899.5.
Class size \(h = 100\).
Midpoints \((x_i)\): 1049.5, 1149.5, 1249.5, 1349.5, 1449.5, 1549.5, 1649.5, 1749.5, 1849.5.
Let assumed mean \(A = 1449.5\), \(h = 100\).
\(u_i\): -4, -3, -2, -1, 0, 1, 2, 3, 4.
Frequencies \((f_i)\): 5, 9, 11, 14, 18, 17, 12, 9, 5 (Total \(N = 100\)).
\(f_i u_i\): -20, -27, -22, -14, 0, 17, 24, 27, 20.
Sum \(\sum f_i u_i = -20 - 27 - 22 - 14 + 17 + 24 + 27 + 20 = 5\).
Mean \(= A + \left(\frac{\sum f_i u_i}{N}\right) \times h = 1449.5 + \left(\frac{5}{100}\right) \times 100 = 1449.5 + 5 = 1454.5 \text{ hours}\).
Modal class is 1399.5 - 1499.5 (highest frequency 18).
Mode \(= l + \left(\frac{f_1 - f_0}{2f_1 - f_0 - f_2}\right) \times h\)
\(= 1399.5 + \left(\frac{18 - 14}{2(18) - 14 - 17}\right) \times 100 = 1399.5 + \left(\frac{4}{36 - 31}\right) \times 100 = 1399.5 + \frac{400}{5} = 1399.5 + 80 = 1479.5 \text{ hours}\).
Cumulative frequencies: 5, 14, 25, 39, 57, 74, 86, 95, 100.
\(\frac{N}{2} = 50\). Median class is 1399.5 - 1499.5.
Median \(= l + \left(\frac{\frac{N}{2} - cf}{f}\right) \times h\)
\(= 1399.5 + \left(\frac{50 - 39}{18}\right) \times 100 = 1399.5 + \frac{1100}{18} = 1399.5 + 61.11 = 1460.61 \text{ hours}\).
In simple words: First convert discontinuous class intervals into continuous ones by applying adjustment factors (±0.5), then compute mean, mode, and median.
Exam Tip: Always check if class intervals are inclusive (discontinuous) or exclusive (continuous) before calculating statistics.