Class 10: Measure of Central Tendency – Board Problems – ICSE / ISC / CBSE Mathematics Portal for K12 Students

Question 1: Use a graph paper for this question. The daily pocket expenses of $200$ students in a school are given below;

Pocket expenses (in Rs.)	0-5	5-10	10-15	15-20	20-25	25-30	30-35	35-40
Number of students (frequency)	10	14	28	42	50	30	14	12

Draw a histogram representing the above distribution and estimate the mode from the graph. [2014]

Answer:

$\\$

Question 2: The marks obtained by 100 students in a Mathematics test are given below;

Marks	0-10	10-20	20-30	30-40	40-50	50-60	60-70	70-80	80-90	90-100
No. of Students	3	7	12	17	23	14	9	6	5	4

Draw an ogive for the distribution on a graph sheet. (Use a scale of 2 cm = 10 units on both axis)

Use the ogive to estimate the:

(i) median

(ii) lower quartile

(iii) number of students who obtained more than 85% marks in the test.

(iv) number of students who did not pass in the if the pass percentage was 35 [2014]

Answer:

Marks

No. of Students

Cumulative Frequency (c.f)

0-10

10-20

20-30

30-40

40-50

50-60

60-70

70-80

80-90

90-100

100

On the graph paper, we plot the following points:

$(10, 3), (20, 10), (30, 22), (40,39), (50, 62), (60,76), (70,85), \\ (80, 91), (90, 96),(100, 100)$

$\displaystyle \text{(i) Medium } = (\frac{n}{2})^{th} \text{ term } = \frac{100}{2}=50^{th} \text{ term }$

From the graph $50^{th} \text{ term }=44$

$\displaystyle \text{(ii) Lower quartile } = (\frac{n}{4})^{th} \text{ term } = \frac{100}{4}=25^{th} \text{ term }$

From the graph $25^{th} \text{ term } =32$

(iii) The number of students who obtained more than $85\%$ marks in test $= 100-94 = 6$ students

(iv) The number of students who did not pass in the test if test if the pass percentage was 35 $= 30$ students.

Question 3: The marks obtained by 30 students in a class assessment of 5 marks is given below:

Marks	0	1	2	3	4	5
No. of Students	1	3	6	10	5	5

Calculate the mean, median and mode of the above distribution. [2015]

Answer:

Below table

$x$	$f$	$fx$	$cf$
0	1	0	1
1	3	3	4
2	6	12	10
3	10	30	20
4	5	20	25
5	5	25	30
	$\Sigma f=30\ \ \ \ \ \ \ \Sigma fx=90$

$\displaystyle \text{Mean } = \frac{\Sigma fx}{\Sigma f}=\frac{90}{30} = 3$

$\displaystyle \text{Median } = size \ of \left(\frac{N}{2}\right)th\ {obs}^n$

$\displaystyle = size \ of \left(\frac{30}{2}\right)th\ {obs}^n$

$\displaystyle = Size \ of 15 {}^{th} \ obs{}^{n} = 3$

Mode = 3 marks (as highest frequency is 10)

$\\$

Question 4: Calculate the mean of the following distribution. [2015]

Class Interval	0-10	10-20	20-30	30-40	40-50	50-60
Frequency	8	5	12	35	24	16

Answer:

Class	Mid Value $x$	$f$	$fx$
0-10	5	8	40
10-20	15	5	75
20-30	25	12	300
30-40	35	35	1225
40-50	45	24	1080
50-60	55	16	880
Total		$\Sigma f=100$	$\Sigma fx=3600$

$\displaystyle \text{Mean } = \bar{x} = \frac{ \Sigma fx}{ \Sigma f} = \frac{3600}{100} = 36$

$\\$

Question 5: The weight of 50 workers is given below:

Weight in kg	50-60	60-70	70-80	80-90	90-100	100-110	110-120
No. of workers	4	7	11	14	6	5	3

Draw on give of the given distribution using a graph sheet. Take 2 cm =10 kg on one axis and 2 cm =5 workers along the other axis. Use a graph to estimate the following:

(i) the upper and lower quartiles

(ii) if weight 95 kg and above is considering find the number of workers who are overweight. [2015]

Answer:

Below the table

Weight	$f$	$cf$
50-60	4	4
60-70	7	11
70-80	11	22
80-90	14	36
90-100	6	42
100-110	5	47
110-120	3	50
Total	50

$\displaystyle \text{(i) Lower quartile= } \frac{N^{th}}{4}{ \ obs}^n$

$\displaystyle = \frac{{50}^{th}}{4}{ \ obs}^n = 12.5th { \ obs}^n = 72 \ kg$

$\displaystyle \text{Upper quartile = } \frac{3N}{4}\ th\ {obs}^n = \frac{150}{4}th\ {obs}^n = 37.5th { \ obs}^n = 92 \ kg$

(ii) No. of over weight workers $= 50-39 = 11$

$\\$

Question 6: The number $6,8,10,12,13$ and $x$ are arranged in an ascending order. If the mean of the observations is equal to the median, find the value of $x$ [2014]

Answer:

Arrange numbers in ascending order are $6,8,10,12,13, x.$

$\displaystyle \text{Mean } = \frac{6+8+10+12+13+x}{6}=\frac{49+x}{6}$

No. of terms $(n) = 6 \ (even)$

$\displaystyle \text{Median } = \frac{ (\frac{n}{2})^{th} \ term + (\frac{n}{2}+1)^{th} \ term }{2}$

$\displaystyle \text{Median } = \frac{ (\frac{6}{2})^{th} \ term + (\frac{6}{2}+1)^{th} \ term }{2} = \frac{3^{rd}+4^{th}}{2} = \frac{10+12}{2} = \frac{22}{2} = 11$

According to given condition

$\displaystyle \frac{49+x}{6}=11 \Rightarrow 49+x=66$

or $x=17$

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Question 8: Calculate the mean of the distribution given below using the short cut method; [2014]

Marks	11-20	21-30	31-40	41-50	51-60	61-70	71-80
No. of students	2	6	10	12	9	7	4

Answer:

Table as follows

$Marks \\ (C.I)$

$f$

Mean Value $x$

$A=45.5$ $d=x-A$

$f \times d$

11-20

21-30

31-40

41-50

51-60

61-70

71-80

15-5

25-5

35-5

45-5

55-5

65-5

75-5

-30

-20

-10

-60

-120

-100

140

120

$\Sigma f=50$

$\Sigma fd=70$

$\displaystyle \text{Mean } = A + \frac{ \Sigma fd}{\Sigma f} = 45.5+\frac{70}{50} = 45.5 + 1.4 = 46.9$

$\\$

Question 8: The median of the following observation $11,12,14 (x-2), (x+4), (x+9), 32, 38, 47$ arranged in ascending order is $24$ . Find the value of $x$ and hence find the mean. [2013]

Answer:

Given observation are $11,12,14, (x-2), (x+4), (x+9), 32, 38, 47$ and mediam $=24$

Since $n=9$ which is odd, therefore

$\displaystyle \text{Median } =\ \frac{n+1}{2}th\ term =\frac{9+1}{2}th\ term$

$24=5th\ term$

$x+4=24$

$\Rightarrow x=24-4 \Rightarrow x=20$

Therefore, $11,12,14, (20-2), (20+4), (20+9),32,38,47$

$=11,12,14,18,24,29,32,38,47$

$\displaystyle \text{Now, Mean } \\ \\ =\frac{\sum{x}}{n}= \frac{11+12+14+18+24+29+32+38+47}{9} =\frac{225}{9} =25$

$\\$

Question 9: Draw a histogram from the following frequency distribution and find the made from the graph: [2013]

Class	0-5	5-10	10-15	15-20	20-25	25-30
Frequency	2	5	18	14	8	5

Answer:

$\\$

Question 10: Find the mean of the following distribution by step deviation method: [2013]

Class Internal	20-30	30-40	40-50	50-60	60-70	70-80
Frequency	10	6	8	12	5	9

Answer:

C.I

Frequency

$(f_i)$

Mid-Value

$(x)$

$d_i = \frac{x-a}{h}$

$(f_i. d_i)$

20-30

30-40

40-50

50-60

60-70

70-80

-2

-1

-20

-6

$\Sigma f_i = 50$

$\Sigma f_i.d_i = 23$

Here, $a=45 \text{ and } h=10$

$\displaystyle \text{Mean } = a + \frac{\Sigma f_i.d_i }{\Sigma f_i} = 45 + \frac{23}{50} . 10 = 49.6$

$\\$

Question 11: The mark obtained by 120 students in a test are given below;

Marks	0-10	10-20	20-30	30-40	40-50	50-60	60-70	70-80	80-90	90-100
Students	5	9	16	22	26	18	11	6	4	3

Draw an ogive for the given distribution on a graph sheet; Using suitable scale for ogive to estimate the following:

(i) The mediam

(ii) The number of students who obtained more than 75% marks in the test.

(iii) The number of students who did not pass the test if minimum marks required to pass is 40. [2013]

Answer:

Marks

No. of students

Cumulative Frequency

0-10

10-20

20-30

30-40

40-50

50-60

60-70

70-80

80-90

90-100

107

113

117

120

N=120

On the graph paper we plot the following points:

$(10,5), (20,14), (30,30), (40,52), (50,78), (60,96), (70,107), (80,113), (90,117), (100,120)$

$\displaystyle \text{(i) Median } = (\frac{n}{2})^{th} \ term = \frac{120}{2} =60^{th} \ term$

From the graph 60^th term $=42$

(ii) The number of students who obtained more than $75\%$ marks in test $= 120-110 = 10$

(iii) The number of students who did not pass the test if the minimum pass marks $40=52$

$\\$

Question 12: Marks obtained by $40$ students in a short assessment is given below, where $a \ and \ b$ are two missing data:

Marks	5	6	7	8	9
No. of Students	6	A	16	13	B

If the mean of the distribution is $7.2$ find $a \ and \ b$ . [2012]

Answer:

Given, the total number of students $= 40$

Therefore $6 + a+ 16+ 13+ b = 40$

$\Rightarrow a + b = 5$ … … … (i)

$\displaystyle \text{Given mean } (\overline{x}) = \frac {\Sigma fx}{\Sigma f} = 7.2$

$\displaystyle \text{Therefore } 7.2 = \frac{5 \times 6 + 6 \times a +7 \times 16 +8 \times 13 +9 \times b}{40}$

$246 + 6a + 9 b = 288$

$6a + 9 b = 42$

$2a + 3b = 14$ … … … (ii)

Solving (i) and (ii) we get $a = 1$ and $b = 4$

$\\$

Question 13: The following distribution represents the height of 160 students of a school

Height	140-145	145-150	150-155	155-160	160-165	165-170	170-175	175-180
Students	12	20	30	38	24	16	12	8

Draw a given for the given distribution taking 2 cm =5 cm of height on one axis and 2 cm=20 students on the other axis. Using the graph, determine:

(i) The medium height

(ii) The interquartile range

(iii) The number of students whose height is above 172 cm. [2012]

Answer:

Following table:

Height

c.f.

140-145

145-150

150-155

155-160

160-165

165-170

170-175

175-180

100

124

140

152

160

(i) Mean $= 157.3$

(ii) Interquartile range $= Q_3 - Q_1 = 163.5 - 152 = 11.5$

(iii) No. of students above $172 \ cm = 160-144 = 16$

$\\$

Question 14: Find the mode and medium of the following frequency distribution: [2012]

X	10	11	12	13	14	15
f	1	4	7	5	9	3

Answer:

Mode is the value of the highest frequency.

Therefore Mode $= 14$

For Median, first write the data in ascending order as follows:

$10, 11, 11, 11, 11, 12, 12, 12, 12, 12, 12, 12, \\ 13, 13, 13, 13,13, 14, 14, 14,14, 14, 14, 14, 14, 14,15, 15, 15$ .

Since Median is the middle most value.

$\displaystyle \text{Median } = (\frac{N+1}{2})^{th} observation = (\frac{29+1}{2})^{th} observation = 15^{th} \ observation = 13$

$\\$

Question 15: A mathematics aptitude test of 50 students was recorded as follows:

Marks	50-60	60-70	70-80	80-90	90-100
No. of students	4	8	14	19	5

Draw a histogram for the above data using a graph paper and locate the mode. [2011]

Answer:

$\\$

Question 16: Using step division method, calculate the mean marks of the following distribution: State the modal class: [2011]

Class Interval	50-55	55-60	60-65	65-70	70-75	75-80	80-85	85-90
Frequency	5	20	10	10	9	6	12	8

Answer:

C.I	$f$	$x$	$d=x-67.5$	$u$	$f.u.$
50-55	5	52.5	-15	-3	-15
55-60	20	57.5	-10	-2	-40
60-65	10	62.5	-5	-1	-10
65-70	10	67.5	0	0	0
70-75	9	72.5	5	1	9
75-80	6	77.5	10	2	12
80-85	12	82.5	15	3	36
85-90	8	87.5	20	4	32
	$\Sigma f = 80$				$\Sigma fu = 24$

$A.M = 67.5$

$\displaystyle \text{(i) } \bar{x} = A.M. + \frac{\Sigma fu}{\Sigma f} \times i = 67.5 + \frac{24}{80} \times 5 = 69$

(ii) Modal class is $55-60$ (class with highest freq.)

$\\$

Question 17: Marks obtained by 200 students in an examination are given below:

Marks	0-10	10-20	20-30	30-40	40-50	50-60	60-70	70-80	80-90	90-100
No. of students	5	11	10	20	28	37	40	29	14	6

Draw an ogive for the given distribution taking $2 \ cm=10$ marks on one axis and $2 \ cm=20$ students on the other axis. Using the graph, determine.

(i) The median marks

(ii) The number of students who failed if minimum marks required to pass is $40$ .

(iii) If scoring $85$ and more marks is considered as grade one, find the number of students who secured grade on in the examination; [2011]

Answer:

Class Interval	Frequency	Cumulative Frequency
0-10	5	5
10-20	11	16
20-30	10	20
30-40	20	46
40-50	28	74
50-60	37	111
60-70	40	151
70-80	29	180
80-90	14	194
90-100	6	200

(i) $n = 200$ (even)

$\displaystyle \text{ Median } = (\frac{n}{2})^{th} \text{ observation } = (\frac{200}{2})^{th} \text{ observation } = 100^{th} \text{ observation } = 57$

(ii) Number of student who failed $= 46$

(iii) Number of students who secured grade one $200 - 1888 = 12$

$\\$

Question 18: The distribution given below shows the marks obtained by $25$ students in an aptitude test. Find the mean, median and mode of the distribution. [2010]

Marks obtained	5	6	7	8	9	10
No. of students	3	9	6	4	2	1

Answer:

$x_i$	$f_i$	$x_if_i$	$cf$
5	3	15	3
6	9	54	12
7	6	42	18
8	4	32	22
9	2	18	24
10	1	10	25
	$\Sigma = 25$	$\Sigma x_if_i = 171$

$\displaystyle \text{Mean } = \frac{\Sigma x_if_i}{N} = \frac{171}{25} = 6.84$

Since $N = 25$ is odd,

$\displaystyle \text{Median } = (\frac{n+1}{2})^2 \ term = 13^{th} \ term = 7$

Mode $= 6$ (maximum frequency)

$\\$

Question 19: The mean of the following distribution is 52 and the frequency of class interval 30-40 is $f$ . Find $f$ . [2010]

Class Interval	10-20	20-30	30-40	40-50	50-60	60-70	70-80
Frequency	5	3	$f$	7	2	6	13

Answer:

Interval	Frequency $(f_i)$	$x-i$	$d_i = x_i - A$	$f_id_i$
10-20	5	15	-30	-150
20-30	3	25	-20	-60
30-40	F	35	-10	-10f
40-50	7	45 (A)	0	0
50-60	2	55	10	20
60-70	6	65	20	120
70-80	13	75	30	390
	$36+f$			$\Sigma f_id_i= 320 - 10f$

$\displaystyle \text{Mean } = A + \frac{\Sigma f_id_i}{N}$

$\displaystyle \Rightarrow 52 = 45 + \frac{320-10f}{36+f}$

$\displaystyle \Rightarrow 7 = \frac{320-10f}{36+f}$

$\displaystyle \Rightarrow 252 + 7f = 320 - 10f$

$\displaystyle \Rightarrow 17f = 68$

$\displaystyle \Rightarrow f = 4$

$\\$

Question 20: The monthly income of a group of 320 employees in a company is given below:

Monthly Income	No. of Employees
6000-7000	20
7000-8000	45
8000-9000	65
9000-10000	95
10000-11000	60
11000-12000	30
12000-13000	5

Draw an ogive of the given. distribution on a graph sheet taking $2 \ cm = Rs. \ 1000$ on one axis and $2 \ cm = 50 \ employees$ on the other axis. From the graph determine:

(i) the median wage

(ii) the number of employees whose income is below $Rs. \ 8500$

(iii) If the salary of a senior employee is above $Rs. \ 11500$ find the number of senior employees in the company.

(iv) The upper quartile [2010]

Answer:

Monthly Income	No. of Employees	Cumulative Frequency
6000-7000	20	20
7000-8000	45	65
8000-9000	65	130
9000-10000	95	225
10000-11000	60	285
11000-12000	30	315
12000-13000	5	320