ML Aggarwal Solution Class 10 Chapter 21 Measures of Central Tendency Test

Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 Q11 Q12 Q13 Q14 Q15 Q16 Q17 Q18 Q19 Q20 Q21 Q22

 Test

Question 1

Arun scored 36 marks in English, 44 marks in Civics, 75 marks in Mathematics and x marks in Science. If he has scored an average of 50 marks, find x.

Sol :

Marks in English = 36

Marks in Civics = 44

Marks in Mathematics = 75

Marks in Science = x

Total marks in 4 subjects = 36 + 44 + 75 + x = 155 + x

average marks $=\frac{155+x}{4}$

But average marks = 50 (given)

$\frac{155+x}{4}=50$

⇒ 155 + x = 200

⇒ x = 200 – 155 = 45

Question 2

The mean of 20 numbers is 18. If 3 is added to each of the first ten numbers, find the mean of new set of 20 numbers.

Sol :

Mean of 20 numbers =18

Total number = 18 × 20 = 360

By adding 3 to first 10 numbers,

The new sum will be = 360 + 3 × 10 = 360 + 30 = 390

New Mean $=\frac{390}{20}=19.5$

Question 3

The average height of 30 students is 150 cm. It was detected later that one value of 165 cm was wrongly copied as 135 cm for computation of mean. Find the correct mean.

Sol :

In first case,

Average height of 30 students = 150 cm

Total height = 150 × 30 = 4500 cm

Difference in copying the number = 165 – 135 = 30 cm

Correct sum = 4500 + 30 = 4530 cm

Correct mean $=\frac{4530}{30}$

=151 cm

Question 4

There are 50 students in a class of which 40 are boys and the rest girls. The average weight of the students in the class is 44 kg and average weight of the girls is 40 kg. Find the average weight of boys.

Sol :

Total students of a class = 50

No. of boys = 40

No. of girls = 50 – 40 = 10

Average weight of 50 students = 44 kg

Total weight = 44 × 50 = 2200 kg

Average weight of 10 girls = 40 kg

.’. Total weight of girls = 40 × 10 = 400 kg

Then the total weight of 40 boys = 2200 – 400 = 1800kg

Average weight of boys $=\frac{1800}{40}$

=45 kg

Question 5

The contents of 50 boxes of matches were counted giving the following results

No. of matches	41	42	43	44	45	46
No. of boxes	5	8	13	12	7	3

Calculate the mean number of matches per box.

Sol :

No. of matches (x)	No. of boxes (f)	f.x
41	5	205
42	8	336
43	13	559
44	12	528
45	7	315
46	5	230
Total	50	2173

Mean $=\frac{\sum f x}{\sum f}=\frac{2173}{50}$

=43.46

Question 6

The heights of 50 children were measured (correct to the nearest cm) giving the following results :

Height (in cm)	65	66	67	68	69	70	71	72	73
No. of children	1	4	5	7	11	10	6	4	2

Sol :

Calculate the mean height for this distribution correct to one place of decimal.

No. of matches (x)	No. of boxes (f)	f.x
65	1	65
66	4	264
67	5	335
68	7	476
69	11	759
70	10	700
71	6	426
72	4	288
72	2	146
Total	50	3459

Mean $=\frac{\sum f x}{\sum f}=\frac{3459}{50}$

=69.18=69.2

Question 7

Find the value of p for the following distribution whose mean is 20.6 :

Variate (xi)	10	15	20	25	35
Frequency(fi)	3	10	p	7	5

Sol :

Variate (xi)	Frequency(fi)	fixi
10	3	30
15	10	150
20	p	20p
25	7	175
35	5	175
Total	25+p	530+20p

Mean$=\frac{\Sigma f_{i} x_{i}}{\Sigma f_{i}}=\frac{530+20 p}{25+p}$

But Mean=20.6 (given)

$\therefore \frac{530+20 p}{25+p}=20 \cdot 6 $

$\Rightarrow 530+20 p=515+20 \cdot 6 p$

$ \Rightarrow 20 \cdot 6 p-20 p=530-515$

$\Rightarrow 0-6 p=15 $

$ \Rightarrow  \frac{6}{10} p=15 $

$ \Rightarrow  p=\frac{15 \times 10}{6}=25$

Question 8

Find the value of p if the mean of the following distribution is 18.

Variate (x)	13	15	17	19	20+p	23
Frequency(f)	8	2	3	4	5p	6

Sol :

Variate (x)	Frequency (f)	f.x
13	8	104
15	2	30
17	3	51
19	4	76
20+p	5p	100p+5p2
23	6	138
Total	23+5p	399+100p+5p2

Mean$=\frac{\sum f x}{\sum f}=\frac{399+100 p+5 p^{2}}{23+5 p}$

But mean=18(given)

$\therefore \frac{399+100 p+5 p^{2}}{23+5 p}=\frac{18}{1}$

⇒399+100p+5p²=414+90p

$\Rightarrow \quad 5 p^{2}+100 p+399-90 p-414=0$

$ \Rightarrow 5 p^{2}+10 p-15=0$

$\Rightarrow \quad p^{2}+2 p-3=0 $

$ \Rightarrow \quad p^{2}+3 p-p-3=0$

$\Rightarrow \quad p(p+3)-1(p+3)=0 $

$ \Rightarrow \quad(p+3)(p-1)=0$

Either p+3=0, then p=-3 but it is not possible as it is negative

or p-1=0 , then p=1

Question 9

Find the mean age in years from the frequency distribution given below:

Age (in years)	25-29	30-34	35-39	40-44	45-49	50-54	55-59
No. of persons	4	14	22	16	6	5	3

Arranging the classes in proper form

Sol :

Class age (in years)	Mid value (xi)	No. of persons (fi)	fi xi
24.5-29.5	27	4	108
29.5-34.5	32	14	448
34.5-39.5	37	22	814
39.5-44.5	42	16	672
44.5-49.5	47	6	282
49.5-49.5	52	5	260
54.5-59.5	57	3	171
Total		70	2755

Mean$=\frac{\sum f_{i} x_{i}}{\sum f_{i}}=\frac{2755}{70}$

=39.357

=39.36 years

Question 10

Calculate the Arithmetic mean, correct to one decimal place, for the following frequency distribution :

Marks	10-20	20-30	30-40	40-50	50-60	60-70	70-80	80-90	90-100
Students	2	4	5	16	20	10	6	8	4

Sol :

Calculate the mean height for this distribution correct to one place of decimal.

Marks	Students (fi)	Class Marks (xi)	fi.xi
10-20	2	15	30
20-30	4	25	100
30-40	5	35	175
40-50	16	45	720
50-60	20	55	1100
60-70	10	65	650
70-80	6	76	450
80-90	8	85	680
90-100	4	95	380
Total	75		4285

Mean$=\frac{\sum f_{i} x_{i}}{\sum f_{i}}=\frac{4285}{75}$

=57.133

=57.1

Question 11

The mean of the following frequency distribution is 62.8. Find the value of p.

Class	0-20	20-40	40-60	60-80	80-100	100-120
Frequency	5	8	p	12	7	8

Sol :

Mean = 62.8

Class	Frequency (fi)	Class marks (x)	fi xi
0-20	5	10	50
20-40	8	30	240
40-60	p	50	50p
60-80	12	70	840
80-100	7	90	630
100-120	8	110	880
Total	40+p		2640+50p

Mean $=\frac{\sum f_{1} x_{i}}{\sum f_{i}}$

$\Rightarrow 62.8=\frac{2640+50 p}{40+p}$

$ \Rightarrow(40+p) \times 62.8=2640+5$

2512.0+62.8p=2640+50p

$\Rightarrow \quad 62.8 p-50 p=2640-2512$

$ \Rightarrow 12.8 p=128$

$p=\frac{128}{12.8}=\frac{128 \times 10}{128}=10$

Hence p=10

Question 12

The daily expenditure of 100 families are given below. Calculate f1, and f2, if the mean daily expenditure is Rs 188.

Expenditure (in Rs)	140-160	160-180	180-200	200-220	220-240
No. of families	5	25	f1	f2	5

Sol 

Mean = 188,

No. of families = 100

Expenditure	Mid value (xi)	No. of persons (fi)	fi xi
140-160	150	5	750
160-180	170	25	4250
180-200	190	f1	190f1
200-220	210	f2	210f2
220-240	230	5	1150
Total		35+f1+f2=100	6150+190f1+210f2

35+f₁+f₂=100

$\Rightarrow f_{1}+f_{2}=100-35=65$..(i)

$\Rightarrow f_{1}=65-f_{2}$

and $\frac{6150+190 f_{1}+210 f_{2}}{100}=188$

$190 f_{1}+210 f_{2}=18800-6150=12650$

$190\left(65-f_{2}\right)+200 f_{2}=12650$

$12350-190 f_{2}+210 f_{2}=12650$

$20 f_{2}=12650-12350$

$20 f_{2}=300$

$f_{2}=\frac{300}{2}=15$

$\therefore f_{1}=65-15=50$

$f_{1}=50, f_{2}=15$

Question 13

The measures of the diameter of the heads of 150 screw is given in the following table. If the mean diameter of the heads of the screws is 51.2 mm, find the values of p and q

Diameter (in mm)	32-36	37-41	42-46	47-51	52-56	57-61	62-66
No. of screws	15	17	p	25	q	20	30

Sol :

Mean = 51.2

No. of screws = 150

Diameter (in mm)	No. of screws (fi)	Class Marks (xi)	fi xi
32-36	15	34	510
37-41	17	39	663
42-46	p	44	44p
47-51	25	49	1125
52-56	q	54	54q
57-61	20	59	1180
62-66	30	64	1920
Total	107+p+q		5498+44p+54q

107+p+q=150

p+q=150-107=43...(i)

Mean $=\frac{\sum f_{1} x_{i}}{\Sigma f_{i}} \Rightarrow \frac{5498+44 p+54 q}{150}$

=51.2

$\Rightarrow 5498+44 p+54 q=7680$

4p+54q=7680-5498

44p+54q=2182

22p+27q=1091...(ii)

Multiplying (i) by 27 and (ii) and by 1

27p+27q=1161...(iii)

22p+27q=1091...(iv)

Subtracting (iv) from (iii) we get

5p=70

$p=\frac{70}{5}=14$

But p+q=43

$\therefore q=43-p=43-14=29$

Hence p=14, q=29

Question 14

The median of the following numbers, arranged in ascending order is 25. Find x, 11, 13, 15, 19, x + 2, x + 4, 30, 35, 39, 46

Sol :

Here, n = 10, which is even

∴Median $=\frac{1}{2}\left[\frac{n}{2} \mathrm{th}+\left(\frac{n}{2}+1\right) \mathrm{th}\right]$ term

$=\frac{1}{2}\left[\frac{10}{2} \mathrm{th}+\left(\frac{10}{2}+1\right) \mathrm{th}\right]$ term

$=\frac{1}{2}$ (5th +6th term)

$=\frac{1}{2}(x+2+x+4)=\frac{2 x+6}{2}$

=x+3

But median is given=25

$\therefore x+3=25$

$ \Rightarrow x=25-3=22$

Question 15

If the median of 5, 9, 11, 3, 4, x, 8 is 6, find the value of x.

Sol :

Arranging in ascending order, 3, 4, 5, x, 8, 9, 11,

Here n = 7 which is odd.

∴ Median $=\frac{n+1}{2}$ th term

$=\frac{7+1}{2}$= 4th term=x

∴ but median = 6

∴ x = 6

Question 16

Find the median of: 17, 26, 60, 45, 33, 32, 29, 34, 56 If 26 is replaced by 62, find the new median.

Sol :

Arranging the given data in ascending order

17, 26, 29, 32, 33, 34, 45, 56, 60

Here n = 9 which is odd

∴Median $=\frac{n+1}{2}$ th term

$=\frac{9+1}{2}=\frac{10}{2}$=5th term=33

(ii) If 26 is replaced by 62, their the order will be

17, 29, 32, 33, 34, 45, 56, 60, 62

Here 5th term is 34

∴ Median = 34

Question 17

The marks scored by 16 students in a class test are : 3, 6, 8, 13, 15, 5, 21, 23, 17, 10, 9, 1, 20, 21, 18, 12

Find

(i) the median

(ii) lower quartile

(iii) upper quartile

Sol :

Arranging the given data in ascending order:

1, 3, 5, 6, 8, 9, 10, 12, 13, 15, 17, 18, 20, 21, 21, 23

Here n = 16 which is even.

(i) $\therefore$ Median $=\frac{1}{2}\left[\frac{16}{2}\right.$ th $\left.+\left(\frac{16}{2}+1\right) \mathrm{th}\right]$ term

$=\frac{1}{2}$(8th+9th) term

$=\frac{12+13}{2}=\frac{25}{2}$

=12.5

(ii) Lower quartile

$=\frac{1}{4} n=\frac{16}{4}$

=4th term

=6

(iii) Upper quartile

$=\frac{3}{4} n=\frac{3}{4} \times 16$

=12th term

=18

Question 18

Find the median and mode for the set of numbers : 2, 2, 3, 5, 5, 5, 6, 8, 9

Sol :

Here n = 9 which is odd.

∴Median $=\frac{n+1}{2}$th term

$=\frac{9+1}{2}=\frac{10}{2}$

=5th term=5

Here 5 occur maximum times

∴Mode = 5

Question 19

Calculate the mean, the median and the mode of the following distribution :

Age (in years)	12	13	14	15	16	17	18
No. of students	2	3	5	6	4	3	2

Sol :

Mean = 51.2

No. of screws = 150

Age (in years) (xi)	No. of screws (fi)	Class Marks (fi)	fi xi
12	2	2	24
13	3	5	39
14	5	10	70
15	6	16	90
16	4	20	64
17	3	23	51
18	2	25	36
Total	25		374

(i) Mean $=\frac{\sum f_{i} x_{i}}{\sum f_{i}}=\frac{374}{25}$

=14.96

(ii) Here n=25 which is odd

∴$=\frac{n+1}{2}$ th term $=\frac{25+1}{2}=\frac{26}{2}$

=13 th term

=15

(iii) Here 15 occurs most i.e. in 6 times

∴Mode=15

Question 20

The daily wages of 30 employees in an establishment are distributed as follows :

Daily wages(in ₹)	0-10	10-20	20-30	30-40	40-50	50-60
No. of employees	1	8	10	5	4	2

Sol :

Estimate the modal daily wages for this distribution by a graphical method.

Daily wages (in ₹)	No. of employees
0-10	1
10-20	8
20-30	10
30-40	5
40-50	4
50-60	2

Taking daily wages on x-axis and No. of employees on the y-axis

and draw a histogram as shown. Join AB and CD intersecting each other at M.

From M draw ML perpendicular to x-axis, L is the mode

∴ Mode = Rs 23

Question 21

Using the data given below, construct the cumulative frequency table and draw the ogive. From the ogive, estimate ;

(i) the median

(ii) the inter quartile range.

Marks	0-10	10-20	20-30	30-40	40-50	50-60	60-70	70-80
Frequency	3	8	12	14	10	6	5	2

Also state the median class

Sol :

Marks	Frequency	c.f
0-10	3	3
10-20	8	11
20-30	12	23
30-40	14	37
40-50	10	47
50-60	6	53
60-70	5	58
70-80	2	60

Now plot the points (10,3),(20,11),(30,23)(40,37),(50,47),(60,53),(70,58),(80,60) on the graph and join them in free hand to form an ogive as shown.

Here n=60 which is an even number

(i) Median $=\frac{1}{2}\left[\frac{n}{2}\right.$ th $\left.+\left(\frac{n}{2}+1\right) \mathrm{th}\right]$

$=\frac{1}{2}\left[\frac{60}{2} t h+\left(\frac{60}{2}+1\right) t h\right]$ term

$=\frac{1}{2}(30+31)$

=30.5 th

Now take a point A(30-5) on y-axis. From A draw a line parallel to x-axis meeting the curve at P and from P, draw a perpendicular to x-axis meeting it in Q. Q is the median which is 35 and median class is 30-40

(ii) Lower quartile

$=\frac{n}{4}=\frac{60}{4}$

=15

Upper quartile

$=\frac{3}{4} n=\frac{3}{4} \times 60=45$

Now take points B(15) and C(45) on y-axis and from B and C draw lines parallel to x-axis meeting the curve at L and M respectively. From L and M, draw lines perpendicular to x-axis meeting it at E and F respectively. E and F are lower and upper quartile which are 22.3 and 47

∴Interquartile range$=Q_{3}-Q_{1}$

=47.0-22.3=24.7

Question 22

Draw a cumulative frequency curve for the following data :

Marks obtained	0-10	10-20	20-30	30-40	40-50
No. of students	8	10	22	40	20

Hence determine:

(i) the median

(ii) the pass marks if 85% of the students pass.

(iii) the marks which 45% of the students exceed.

Sol :

Marks obtained	No. of students	c.f
0-10	8	8
10-20	10	18
20-30	22	40
30-40	40	80
40-50	20	100

Now plot points (10,8),(20,18),(30,40),(40,80) and (50,100) on the graph and join them in free hand to form an ogive.

Here n=100 which is even

(i) $\therefore$ Median $=\frac{1}{2}\left[\frac{n}{2}\right.$ th $+\left(\frac{n}{2}+1\right)$ th $]$ term

$=\frac{1}{2}\left[\frac{100}{2} \mathrm{th}+\left(\frac{100}{2}+1\right) \mathrm{th}\right]$ term

$=\frac{1}{2}(50 t h+51 t h)=\frac{1}{2} \times 101=50 \cdot 5$

Now take a point A(50.5) on y-axis and from A, draw a line parallel to x-axis meeting the curve at P and from P , draw a perpendicular to x-axis meeting it at Q. Q is the median which is 32.5

(i) If 85% students pass, the pass marks will be 18

(ii) Marks which 45% of the students exceeds=34 marks 

ML Aggarwal Solution Class 10 Chapter 21 Measures of Central Tendency MCQs

Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 Q11 Q12 Q13 Q14 Q15 Q16

 MCQs

Question 1

If the classes of a frequency distribution are 1-10, 11-20, 21-30, …, 51-60, then the size of each class is

(a) 9

(b) 10

(c) 11

(d) 5.5

Sol :

In the classes 1-10, 11-20, 21-30, …, 51-60,

the size of each class is 10. (b)

Question 2

If the classes of a frequency distribution are 1-10, 11-20, 21-30,…, 61-70, then the upper limit of the class 11-20 is

(a) 20

(b) 21

(c) 19.5

(d) 20.5

Sol :

In the classes of distribution, 1-10, 11-20, 21-30, …, 61-70,

upper limit of 11-20 is 20-5 as the classes after adjustment are

0.5-10.5, 10.5-20.5, 20.5-30.5, … (d)

Question 3

If the class marks of a continuous frequency distribution are 22, 30, 38, 46, 54, 62, then the class corresponding to the class mark 46 is

(a) 41.5-49.5

(b) 42-50

(c) 41-49

(d) 41-50

Sol :

The class marks of distribution are 22, 30, 38, 46, 54, 62,

then classes corresponding to these class marks 46 is

46.4 – 4 = 42, 46 + 4 = 50

(Class intervals is 8 as 30 – 22 = 8, 38 – 30 = 8

i.e:, 42 – 50 (b)

Question 4

If the mean of the following distribution is 2.6,

xi	1	2	p	4	5
fi	3	3	1	1	2

then the value of P is

(a) 2

(b) 3

(c) 2.6

(d) 2.8

Sol :

Mean = 2.6

xi	1	2	p	4	5	Total
fi	3	3	1	1	2	10
fixi	3	6	p	4	10	23+p

Mean$=\frac{\sum f_{i} x_{i}}{\sum f}=\frac{23+p}{10}$

=2.6

$23+p=2 \cdot 6 \times 10=26 $

$\Rightarrow p=26-23=3$

Ans (b)

Question 5

The measure of central tendency of statistical data which takes into account all the data is

(a) mean

(b) median

(c) mode

(d) range

Sol :

A measure of central tendency of statistical data is mean. (a)

Question 6

In a grouped frequency distribution, the mid-values of the classes are used to measure which of the following central tendency?

(a) median

(b) mode

(c) mean

(d) all of these

Sol :

In a grouped frequency distribution,

the mid-values of the classes are used to measure Mean (c)

Question 7

In the formula: $\bar{x}=a+\frac{\sum f_{i} d_{i}}{\sum f_{i}}$ for finding the mean of the grouped data, d’is are deviations from a (assumed mean) of

(a) lower limits of the classes

(b) upper limits of the classes

(c) mid-points of the classes

(d) frequencies of the classes

Sol :

The formula $\bar{x}=a+\frac{\sum f_{i} d_{i}}{\sum f_{i}}$ is the finding of mean of the grouped data, d’is are mid-points of the classes

Question 8

In the formula: $\bar{x}=a+c\left(\frac{\sum f_{i} u_{i}}{\sum f_{i}}\right)$, for finding the mean of grouped frequency distribution $\text{u}_{i}$

(a) $\frac{y_{i}+a}{c}$

(b) $c\left(y_{i}-a\right)$

(c) $\frac{y_{i}-a}{c}$

(d) $\frac{a-y_{i}}{c}$

Sol :

In $\bar{x}=a+c\left(\frac{\sum f_{i} u_{i}}{\sum f_{i}}\right)$

for finding the mean of grouped frequency, $\mathrm{u}_{\mathrm{i}}$ is $\frac{y_{i}-a}$

Ans (c)

Question 9

While computing mean of grouped data, we assumed that the frequencies are

(a) evenly distributed over all the classes

(b) centred at the class marks of the classes

(c) centred at the upper limits of the classes

(d) centred at the lower limits of the classes

Sol :

For computing mean of grouped data,

we assumed that frequencies are centred at class marks of the classes.

Ans (b)

Question 10

Construction of a cumulative frequency distribution table is useful in determining the

(a) mean

(b) median

(c) mode

(d) all the three measures

Sol :

Construction of a cumulative frequency distribution table

is used for determining the median,

Ans (b)

Question 11

The times, in seconds, taken by 150 athletes to run a 110 m hurdle race are tabulated below:

Class	13.8-14	14-14.2	14.2-14-4	14.4-14.6	14.6-14.8	14.8-15
Frequency	2	4	5	71	48	20

The number of athletes who completed the race in less than 14.6 seconds is

(a) 11

(b) 71

(c) 82

(d) 130

Sol :

Time taken in seconds by 150 athletes to run a 110 m hurdle race as given in the sum,

the number of athletes who completed the race in less then 14.6 second is

2 + 4 + 5 + 71 = 82 athletes. 

Ans (c)

Question 12

Consider the following frequency distribution:

Class	0-5	6-11	12-17	18-23	24-29
Frequency	13	10	15	8	11

The upper limit of the median class is

(a) 17

(b) 17.5

(c) 18

(d) 18.5

Sol :

From the given frequency upper limit of median class is 17.5

as total frequencies 13 + 10 + 15 + 8 + 11 = 57

$\frac{57+1}{2}=\frac{58}{2}=29$

and 13 + 10 + 15 = 28 where class is 12-17

But actual class will be 11.5-17.5

Upper limit is 17.5

Ans (b)

Question 13

Daily wages of a factory workers are recorded as:

Daily wages(in ₹)	131-136	137-142	143-148	149-154	155-160
No. of workers	5	27	20	18	12

The lower limit of the modal class is

(a) Rs 137

(b) Rs 143

(c) Rs 136.5

(d) Rs 142.5

Sol :

In the daily wages of workers of a factory are 131-136, 137-142, 142-148, …

which are not a proper class

So, proper class will be 130.5-136.5, 136.5-142.5, 142.5-148.5, …

Lower limit of a model class is 136.5 as 136.5-142.5 is the modal class. (c)

Question 14

For the following distribution:

Class	0-5	5-10	10-15	15-20	20-25
Frequency	10	15	12	20	9

The sum of lower limits of the median class and modal class is

(a) 15

(b) 25

(c) 30

(d) 35

Sol :

From the given distribution

Sum of frequencies = 10 + 15 + 12 + 20 + 9 = 66

and median is $\frac{66}{2}=33$

Median class will be 10-15 and modal class is 15-20

Sum of lower limits = 10 + 15 = 25 (b)

Question 15

Consider the following data:

Class	65-85	85-105	105-125	125-145	145-165	14
Frequency	4	5	13	20	14	14

The difference of the upper limit of the median class and the lower limit of the modal class is

(a) 0

(b) 19

(c) 20

(d) 38

Sol :

From the given data

Total frequencies = 4 + 5 + 13 + 20 + 14 + 7 + 4 = 67

Median class $\frac{67+1}{2}=34$

which is (4 + 5 + 13 + 20) 125-145 and modal class is 125-145

Difference of upper limit of median class and the lower limit of the modal class

= 145 – 125 = 20 (c)

Question 16

An ogive curve is used to determine

(a) range

(b) mean

(c) mode

(d) median

Sol :

An ogive curve is used to find median. 

Ans (d)

ML Aggarwal Solution Class 10 Chapter 21 Measures of Central Tendency Exercise 21.6

Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 Q11 Q12 Q13 Q14

 Exercise 21.6

Question 1

The following table shows the distribution of the heights of a group of a factory workers.

Height (in cm)	150-155	155-160	160-165	165-170	170-175	175-180	180-185
No. of workers	6	12	18	20	13	8	6

(i) Determine the cumulative frequencies.

(ii) Draw the cumulative frequency curve on a graph paper.

Use 2 cm = 5 cm height on one axis and 2 cm = 10 workers on the other.

(iii) From your graph, write down the median height in cm.

Sol :

Representing the distribution in cumulative frequency distribution :

Height(in cm)	No. of workers (f)	c.f
150-155	6	6
155-160	12	18
160-165	18	36
165-170	20	56
170-175	13	69
175-180	8	77
180-185	6	83

Here, n = 83 which is even.

Now taking points (155, 6), (160, 18), (165, 36), (170, 56),

(175, 69), (180, 77) and (185, 83) on the graph.

Now join them with free hand to form the ogive

or cumulative frequency curve as shown.

Here n = 83 which is odd

Median$=\frac{n+1}{2}$ th observation

$=\frac{83+1}{2}=42$ th observation

Take a point A (42) on y-axis and from A,

draw a horizontal line parallel to x-axis meeting the curve at P.

From P draw a line perpendicular on the x-axis which meets it at Q.

∴Q is the median which is 166.5 cm. Ans.

Question 2

Using the data given below construct the cumulative frequency table and draw the-Ogive. From the ogive determine the median.

Marks	0-10	10-20	20-30	30-40	40-50	50-60	60-70	70-80
No. of students	3	8	12	14	10	6	5	2

Sol :

Representing the given data in cumulative frequency distributions :

Marks	No. of workers (f)	c.f
0-10	3	3
10-20	8	11
20-30	12	23
30-40	14	37
40-50	10	47
50-60	6	53
60-70	5	58
70-80	2	60

Taking points (10, 3), (20, 11), (30, 23), (40, 37),

(50,47), (60,53), (70, 58) and (80, 60) on the graph.

Now join them in a free hand to form an ogive as shown.

Here n = 60 which is even

Median $=\frac{1}{2}\left[\frac{60}{2} t h \right.\text{ term }+\left(\frac{60}{2}+1\right) t h\text{ term ]}$

$=\frac{1}{2}\left[\frac{60}{2}+\left(\frac{60}{2}+1\right)\right.\text{ th term ]}$

$=\frac{1}{2}$[30th term+31th term]

=30.5 observations

Now take a point A (30.5) on y-axis and from A,

draw a line parallel to x-axis meeting the curve at P

and from P, draw a perpendicular to x-axis meeting is at Q.

∴ Q is the median which is 35.

Question 3

Use graph paper for this question.

The following table shows the weights in gm of a sample of 100 potatoes taken from a large consignment:

Weight(gm)	50-60	60-70	70-80	80-90	90-100	100-110	110-120	120-130
Frequency	3	10	12	16	18	14	12	10

(i) Calculate the cumulative frequencies.

(ii) Draw the cumulative frequency curve and from it determine the median weight of the potatoes. (1996)

Sol :

Representing the given data in cumulative frequency table :

Weight (gm)	Frequency (f)	c.f
50-60	8	8
60-70	10	18
70-80	12	30
80-90	16	46
90-100	18	64
100-110	14	78
110-120	12	90
120-130	10	100

Now plot the points (60, 8), (70, 18), (80, 30), (90, 46), (100, 64),

(110, 78), (120, 90), (130, 100) on the graph and join them

in a free hand to form an ogive as shown

Here n =100 which is even.

Median $=\frac{1}{2}\left[\frac{n}{2} t h \text{ term }+\left(\frac{n}{2}+1\right) t h\text{ term ]}$

$=\frac{1}{2}\left[\frac{100}{2}+\left(\frac{100}{2}+1\right) t h\right.\text{ term ]}$

$=\frac{1}{2}$[50th term+51th term]

=50.5

Now take a point A (50.5) on they-axis and from A

draw a line parallel to x-axis meeting the curve at R

From P, draw a perpendicular on x-axis meeting it at Q.

Q is the median which is = 93 gm.

Question 4

Attempt this question on graph paper.

Age (years)	5-15	15-25	25-35	35-45	45-55	55-65	65-75
No. of casualties due to accidents	6	10	15	13	24	8	7

(i) Construct the ‘less than’ cumulative frequency curve for the above data, using 2 cm = 10 years, on one axis and 2 cm = 10 casualties on the other.

(ii) From your graph determine (1) the median and (2) the upper quartile

Sol :

Representing the given data in less than cumulative frequency.

Age	No. of Casualties	Cumulative Frequency
Less than 15	6	6
Less than 25	12	16
Less than 35	15	31
Less than 45	13	44
Less than 55	24	68
Less than 55	8	76
Less than 75	7	83

Now plot the points (15, 6), (25, 16), (35, 31), (45, 44), (55, 68), (65,76)

and (75, 83) on the graph and join these points in free hand

to form a cumulative frequency curve (ogive) as shown.

Here n = 83, which is odd.

(i) Median = $=\frac{n+1}{2}=\frac{83+1}{2}+\frac{84}{2}=42$

Now we take point A (42) on y-axis and from A,

draw a line parallel to x-axis meeting the curve at P

and from P, draw a perpendicular to x-axis meeting it at Q.

Q is the median which is = 43

(ii) Upper quartile $=\frac{3(n+1)}{4}=\frac{3 \times(83+1)}{4}=\frac{252}{4}=63$

Take a point B 63 on y-axis and from B,

draw a parallel line to x-axis meeting the curve at L.

From L, draw a perpendicular to x-axis meeting it at M which is 52.

∴ Upper quartile = 52 years

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 an ogive 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 weighing 95 kg and above is considered overweight find the number of workers who are overweight. (2015)

Sol :

The cumulative frequency table of the given distribution table is as follows:

Height(in cm)	No. of workers (f)	c.f
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

The ogive is as follows:

Plot the points (50, 0), (60, 4), (70, 11), (80, 22), (90, 36), (100, 42),

(110, 47), (120, 50) Join these points by using freehand drawing.

The required ogive is drawn on the graph paper.

Here n = number of workers = 50

(i) To find upper quartile:

Let A be the point on y-axis representing a frequency

$\frac{3 n}{4}=\frac{3 \times 50}{4}=37.5$

Through A, draw a horizontal line to meet the ogive at B.

Through B draw a vertical line to meet the x-axis at C.

The abscissa of the point C represents 92.5 kg.

The upper quartile = 92.5 kg To find the lower quartile:

Let D be the point on y-axis representing frequency$=\frac{n}{4}=\frac{50}{4}=12.5$

Through D, draw a horizontal line to meet the ogive at E.

Through E draw a vertical line to meet the x-axis at F.

The abscissa of the point F represents 72 kg.

∴ The lower quartile = 72 kg

(ii) On the graph point, G represents 95 kg.

Through G draw a vertical line to meet the ogive at H.

Through H, draw a horizontal line to meet y-axis at 1.

The ordinate of point 1 represents 40 workers on the y-axis .

∴The number of workers who are 95 kg and above

= Total number of workers – number of workers of weight less than 95 kg

= 50 – 40 = 10

Question 6

The table shows the distribution of scores obtained by 160 shooters in a shooting competition. Use a graph sheet and draw an ogive for the distribution.

(Take 2 cm = 10 scores on the x-axis and 2 cm = 20 shooters on the y-axis)

Scores	0-10	10-20	20-30	30-40	40-50	50-60	60-70	70-80	80-90	90-100
No. of shooters	9	13	20	26	30	22	15	10	8	7

Use your graph to estimate the following:

(i) The median.

(ii) The interquartile range.

(iii) The number of shooters who obtained a score of more than 85%.

Sol :

Scores	No. of Shooter	c.f
0-10	9	9
10-20	13	22
20-30	20	42
30-40	26	68
40-50	30	98
50-60	22	120
60-70	15	135
70-80	10	145
80-90	8	153
90-100	7	160

Plot the points (10, 9), (20, 22), (30, 42), (40, 68), (50, 98),

(60, 120), (70, 135), (80, 145), (90, 153), (100, 160)

on the graph and join them with free hand to get an ogive as shown:

Here n = 160

$\frac{n}{2}=\frac{160}{2}=80$

Median : Take a point 80 on 7-axis and through it,

draw a line parallel to x-axis-which meets the curve at A.

Through A, draw a perpendicular on x-axis which meet it at B.

B Is median which is 44.

(ii) Interquartile range (Q1)

$\frac{n}{4}=\frac{160}{4}=40$

From a point 40 ony-axis, draw a line parallel to x-axis

which meet the curve at C and from C draw a line perpendicular to it

which meet it at D. which is 31.

The interquartile range is 31.

(iii) Number of shooter who get move than 85%.

Scores : From 85 on x-axis, draw a perpendicular to it meeting the curve at P.

From P, draw a line parallel to x-axis meeting y-axis at Q.

Q is the required point which is 89.

Number of shooter getting more than 85% scores = 160 – 149 = 11.

Question 7

The daily wages of 80 workers in a project are given below

Wages (in Rs)	400-450	450-500	500-550	550-600	600-650	650-700	700-750
No. of workers	2	6	12	18	24	13	5

Use a graph paper to draw an ogive for the above distribution. (Use a scale of 2 cm = Rs 50 on x- axis and 2 cm = 10 workers on y-axis). Use your ogive to estimate:

(i) the median wage of the workers.

(ii) the lower quartile wage of the workers.

(iii) the number of workers who earn more than Rs 625 daily. (2017)

Sol :

Wages (in Rs)	No. of workers	c.f
400-450	2	2
450-500	6	8
500-550	12	20
550-600	18	38
600-650	24	62
650-700	13	75
700-750	5	80

Number of workers = 80

(i) Median$=\left(\frac{n}{2}\right)$th term

=40th term

Through mark 40 on the y-axis, draw a horizontal line

which meets the curve at point A.

Through point A, on the curve draw a vertical line which meets the x-axis at point B.

The value of point B on the x-axis is the median, which is 604.

(ii) Lower Quartile (Q1) $=\left(\frac{80}{4}\right)^{t h}$ term

=20th term=550

(iii) Through mark 625 on x-axis, draw a vertical line which meets the graph at point C.

Then through point C, draw a horizontal line which meets the y-axis at the mark of 50.

Thus, number of workers that earn more Rs 625 daily = 80 – 50 = 30

Question 8

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 shooters	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 one in the examination.

Sol :

Marks	f	c.f
0-10	5	5
10-20	11	16
20-30	10	26
30-40	20	46
40-50	28	74
50-60	37	111
60-70	40	2151
70-80	29	180
80-90	14	194
90-100	6	200
	N=200

(i) Median is 57.

(ii) 44 students failed.

(iii) No. of students who secured grade one = 200 – 188 = 12.

Question 9

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
9000-1000	95
1000-11000	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.

Sol :

Monthly Income	No. of Employees (f)	c.f.
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

Now plot the points (7000,20), (8000,65), (9000,130),

(10000,225), (11000,285), (12000,315) and(13000, 320)

on the graph and join them in order with a free hand

to get an ogive as shown in the figure

(i) Total number of employees = 320

$\frac{N}{2}=\frac{320}{2}=160$

From 160 on the y-axis, draw a line parallel to x-axis meeting the curve at P.

From P. draw a perpendicular on x-axis meeting it at M, M is the median which is 9300

(ii) From 8500 on the x-axis, draw a perpendicular which meets the curve at Q.

From Q, draw a line parallel to x-axis meeting y-axis at N. Which is 98

(iii) From 11500 on the x-axis, draw a line perpendicular to x-axis meeting the curve at R.

From R, draw a line parallel to x-axis meeting y-axis at L. Which is 300

No. of employees getting more than Rs. 11500 = 320

(iv) Upper quartile (Q1)

$\frac{3 N}{4}=\frac{320 \times 3}{4}=240$

From 240 on y-axis, draw a line perpendicular on the x-axis which meets the curve at S.

From S, draw a perpendicular on x-axis meeting it at T. Which is 10250.

Hence Q3 = 10250

Question 10

Using a graph paper, draw an ogive for the following distribution which shows a record of the weight in kilograms of 200 students

Weight	40-45	45-50	50-55	55-60	60-65	65-70	70-75	75-80
Frequency	5	17	22	45	51	31	20	9

Use your ogive to estimate the following:

(i) The percentage of students weighing 55 kg or more.

(ii) The weight above which the heaviest 30% of the students fall,

(iii) The number of students who are :

1. under-weight and

2. over-weight, if 55.70 kg is considered as standard weight.

Sol :

Weight	40-45	45-50	50-55	55-60	60-65	65-70	70-75	75-80
Frequency	5	17	22	45	51	31	20	9
c.f	5	22	44	39	140	171	191	200

Plot the points (45, 5), (50, 22), (55, 44), (60, 89), (65, 140),

(70, 171), (75, 191) and (80, 200) on the graph

and join them in free hand to get an ogive as shown From the graph,

number of students weighing 55 kg or more = 200 – 44 = 156

Percentage $=\frac{156}{200} \times 100=78 \%$

(ii) 30% of 200 $=\frac{200 \times 30}{100}=60$

∴ Heaviest 60 students in weight = 9 + 20 + 31 = 60

(From the graph, the required weight is 65 kg or more but less than 80 kg)

(iii) Total number of students who are

1. under weight = 47 and

2. over weight = 152

(∴ Standard weight is 55.70 kg)

Question 11

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 shooters	3	7	12	17	23	14	9	6	5	4

Draw an ogive on a graph sheet and from it determine 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 test if the pass percentage was 35. We represent the given data in cumulative frequency table as given below :

Sol :

We represent the given data in the

cumulative frequency table as given below:

N=100

Median $=\frac{100}{2}=50^{\text {th }}$ term=45

∴Median=45

(ii) Lower quartile : $(Q_1)$

$\mathrm{N}=100 \Rightarrow \frac{100}{4}=25^{\mathrm{th}}$ term=32

$\therefore Q_{1}=32$

(iii) Number of students with 85%

or less=70

∴More than 85% marks

=100-70=30

(iv) Number of students who did not pass=38

Marks	f	c.f
0-10	3	3
10-20	7	10
20-30	12	22
30-40	17	39
40-50	23	62
50-60	14	76
60-70	9	85
70-80	6	91
80-90	5	96
90-100	4	100

Question 12

The marks obtained by 120 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 shooters	5	9	16	22	26	18	11	6	4	3

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

(i) the median

(ii) the lower quartile

{iii) the number of students who obtained more than 75% marks in the test.

(iv) the number of students who did not pass in the test if the pass percentage was 40. (2002)

Sol :

We represent the given data in cumulative frequency table as given below :

Marks	f	c.f
0-10	5	5
10-20	9	14
20-30	16	30
30-40	22	52
40-50	26	78
50-60	18	96
60-70	11	107
70-80	6	113
80-90	4	117
90-100	3	120

Now we plot the points (10, 5), (20, 14), (30, 30), (40, 52), (50, 78), (60, 96), (70, 107),

(80, 113), (90, 117) and (100, 120) on the graph

and join the points in a free hand to form an ogive as shown.

Here n = 120 which is an even number

(i) Median $=\frac{1}{2}\left[\frac{120}{2}+\left(\frac{120}{2}+1\right)\right]$

$=\frac{1}{2}(60+61)=60.5$

Now take a point A (60.5) on y-axis and from A

draw a line parallel to x- axis meeting the curve in P

and from P, draw a perpendicular to x-axis meeting it at Q.

∴ Q is the median which is 43.00 (approx.)

(ii) Lower quartile $=\frac{n}{4}=\frac{120}{4}=30$

Now take a point B (30) on y-axis and from B,

draw a line parallel to x-axis meeting the curve in L

and from L draw a perpendicular to x-axis meeting it at M.

M is the lower quartile which is 30.

(iii) Take a point C (75) on the x-axis

and from C draw a line perpendicular to it meeting the curve at R.

From R, draw a line parallel to x-axis meeting y-axis at S.

∴S shows 110 students getting below 75%

and 120 – 110 = 10 students getting more than 75% marks.

(iv) Pass percentage is 40%

Now take a point D (40) on x-axis and from D

draw a line perpendicular to x-axis meeting the curve at E

and from E, draw a line parallel to x-axis meeting the y-axis at F.

∴ F shows 52

∴ No of students who could not get 40% and failed in the examination are 52.

Question 13

The following distribution represents the height of 160 students of a school.

Height (in cm)	140-145	145-150	150-155	155-160	160-165	165-170	170-175	175-180
No. of Students	12	20	30	38	24	16	12	8

Draw an ogive 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 median height.

(ii)The inter quartile range.

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

Sol :

The cumulative frequency table may be prepared as follows:

Height (in cm)	No. of Students (f)	c.f.
140-145	12	12
145-150	20	32
150-155	30	62
155-160	38	100
160-165	24	124
165-170	16	140
170-175	12	152
175-180	8	160

Now, we take height along x-axis and number of students along the y-axis.

Now, plot the point (140, 0), (145, 12), (150, 32), (155, 62), (160, 100), (165, 124),

(170, 140), (175, 152) and (180, 160). Join these points by a free hand curve to get the ogive.

(i) Here N = 160

$\frac{N}{2}=80$

On the graph paper take a point A on the y- axis representing 80.

A draw horizontal line meeting the ogive at B.

From B, draw BC ⊥ x-axis, meeting the x-axis at C. The abscissa of C is 157.5

So, median = 157.5 cm

(ii) Proceeding in the same way as we have done in above,

we have, Q1 = 152 and Q3 = 164 So, interquartile range = Q3 – Q1 = 164 – 152 = 12 cm

(iii) From the ogive, we see that the number of students whose height is less than 172 is 145.

No. of students whose height is above 172 cm = 160 – 145 = 15

Question 14

100 pupils in a school have heights as tabulated below :

Height in cm	121-130	131-140	141-150	151-160	161-170	171-180
No. of pupils	12	16	30	20	14	8

Draw the ogive for the above data and from it determine the median (use graph paper).

Sol :

Representing the given data in cumulative frequency table (in continuous distribution):

Height (in cm)	No. of pupils	c.f.
120.5-130.5	12	12
130.5-140.5	16	28
140.5-150.5	30	58
150.5-160.5	20	78
160.5-170.5	14	92
170.5-180.5	8	100

∴ Here n = 100 which is an even number

∴ Median = 

$\frac{1}{2}\left[\frac{n}{2}+\left(\frac{n}{2}+1\right)\right]$

$=\frac{1}{2}\left[\frac{100}{2}+\left(\frac{100}{2}+1\right)\right]$

$=\frac{1}{2}(50+51)=\frac{101}{2}$

=50.5

Now plot points (130.5, 12), (140.5, 28), (150.5, 58), (160.5, 78), (170.5, 92)

and (180.5, 100) on the graph and join them in free hand to form an ogive as shown.

Now take a point A (50-5) on y-axis and from A

draw a line parallel to x-axis meeting the curve at P

and from P, draw a line perpendicular to x-axis meeting it at Q

∴ Q (147.5) is the median.