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

 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
0-101
10-208
20-3010
30-405
40-504
50-602
















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.

Marks0-1010-2020-3030-4040-5050-6060-7070-80
Frequency38121410652
Also state the median class

Sol :

MarksFrequencyc.f
0-1033
10-20811
20-301223
30-401437
40-501047
50-60653
60-70558
70-80260


















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 obtained0-1010-2020-3030-4040-50
No. of students 810224020

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 obtainedNo. of studentsc.f
0-1088
10-201018
20-302240
30-404080
40-5020100

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 

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