Chapter-Test Measures of Central Tendency ML Aggarwal ICSE Class-10 Solutions

Chapter-Test Measures of Central Tendency ML Aggarwal ICSE Class-10 Solutions Chapter-21 . We Provide Step by Step Answer of Exe-21.1,  Exe-21.2,  Exe-21.3,  Exe-21.4,  Exe-21.5,  with MCQs and Chapter-Test of Measures of Central Tendency Questions  / Problems related  for ICSE Class-10 APC Understanding Mathematics  . Visit official Website CISCE  for detail information about ICSE Board Class-10.

Board ICSE
Publications Avichal Publishing Company (APC)
Subject Maths
Class 10th
Chapter-21 Measures of Central Tendency
Writer ML Aggarwal
Book Name Understanding
Topics Solution of  Exe – 21.1  (Mean),   Exe – 21.2 (Median) , Exe – 21.3 (Mode) ,  Exe – 21.4 (Histogram),  Exe – 21.5 (Ogive),  Exe –21.6,  with MCQ Questions  and Chapter-Test
Academic Session 2021-2022

How to Solve Measures of Central Tendency Problems/Questions / Exercise of ICSE Class-10 Mathematics

Before viewing Answer of Chapter-21 Measures of Central Tendency of ML Aggarwal Solutions. Read the Chapter Carefully with formula and then solve all example of Exe-21.1,  Exe-21.2,  Exe-21.3,  Exe-21.4,  Exe-21.5, Exe-21.6  given in  your text book .

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For more practice on Measures of Central Tendency related problems /Questions / Exercise try to solve Measures of Central Tendency exercise of other famous publications also such as Goyal Brothers Prakshan (RS Aggarwal ICSE) Graphical Representations of Statistical Data,   Measure of Central Tendency (Mean) and also  Concise Selina Publications ICSE Mathematics  Graphical Representation (Histograms and Ogives),    Measures of Central Tendency (Mean , Median, Quartiles and Mode). Get the Formula of Measures of Central Tendency   for ICSE Class 10 Maths  to understand the topic more clearly in effective way.


Chapter- Test , Measures of Central Tendency ML Aggarwal Solutions

Chapter Test MCQ Measures of Central Tendency ML Aggarwal ICSE Class-10 Solutions

Page 521-522

Question 1 :  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.
Answer 1 :
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 = (390/20) = 19.5
Question 2 :  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.
Answer 2 : 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 = (4530/30) = 151 cm

Question 3

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.

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Answer 3

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 = 188/40 = 45kg


Page 522

Question 4

The heights of 50 children were measured (correct to the nearest cm) giving the following results :
ML aggarwal class-10 chapter 21 Measure of central tendency img 56

Answer 4

Calculate the mean height for this distribution correct to one place of decimal.
Measures of Central Tendency ML Aggarwal Solutions chap 21 img 60

Mean = Ʃfx/Ʃf

= 3459/50

= 69.18 = 69.2

Question 5  Chapter-Test Measures of Central Tendency ML 

Find the value of p if the mean of the following distribution is 18.
Measures of Central Tendency ML Aggarwal Solutions chap 21 img 61

Answer 5

ML aggarwal class-10 chapter 21 Measure of central tendency img 57

Mean = Ʃfi xi / Ʃfi

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18 = (399+5p2+100p)/( 23+5p) [Given mean = 18]

18(23+5p) = 399+5p2+100p

414 + 90p = 399+5p2+100p

5p2+100p-90p+399-414 = 0

5p2+10p-15 = 0

Dividing by 5, we get

p2+2p-3 = 0

(p-1)(p+3) = 0

p-1 = 0 or p+3 = 0

p = 1 or p = -3

p cannot be negative.

So p = 1

Hence the value of p is 1.

Question 6

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

Measures of Central Tendency ML Aggarwal Solutions chap 21 img 63

Answer 6

Arranging the classes in proper form
Measures of Central Tendency ML Aggarwal Solutions chap 21 img 64

Question 7

The mean of the following frequency distribution is 62.8. Find the value of p.
ML aggarwal class-10 chapter 21 Measure of central tendency img 58

Answer 7

Mean = 62.8
Measures of Central Tendency ML Aggarwal Solutions chap 21 img 65
Hence p = 10

Question 8

The daily expenditure of 100 families are given below. Calculate f1, and f2, if the mean daily expenditure is Rs 188.
ML aggarwal class-10 chapter 21 Measure of central tendency img 59

Answer 8

Mean = 188,
No. of families = 100
Measures of Central Tendency ML Aggarwal Solutions chap 21 img 66

Given no. of families = 100

So 35+f1+f= 100

f1+f= 100-35 = 65

f= 65-f2 ..(i)

Mean = Ʃfi xi / Ʃfi

188 = (6150+190f1+210f2)/100 [Given mean = 188]

188(100) = 6150+190f1+210f2

18800 = 6150+190f1+210f2

18800-6150 = 190f1+210f2

12650 = 190f1+210f2 ..(ii)

Substitute (i) in (ii)

12650 = 190(65-f2)+210f2

12650 = 12350-190f2+210f2

12650-12350 = -190f2+210f2

300 = 20f2

f2 = 300/20 = 15

Put f2 in (i)

f= 65-15

f= 50

Hence the value of f1 and f2 is 50 and 15 respectively.

Question 9

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

Answer 9

Here, n = 10, which is even
Measures of Central Tendency ML Aggarwal Solutions chap 21 img 68

But median is given = 25

So, x + 3 = 25

x = 25 – 3

= 22

Question 10

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

Answer 10

Arranging in ascending order, 3, 4, 5, x, 8, 9, 11,
Here n = 7 which is odd.
∴ Median = (n + 1)/2 th term = (7 + 1)/2 = 4th term = x
but median = 6
Hence ∴ x = 6

Question 11

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

Answer 11

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) So median = ½ ( n/2 th term + ((n/2)+1)th term)

= ½ (16/2 th term + ((16/2)+1)th term)

= ½ (8 th term + (8+1)th term)

= ½ (8 th term + 9th term)

= ½ (12+13)

= ½ ×25

= 12.5

(ii) Lower quartile, Q1 = (n/4) th term

= (16)/4

= 4 th term

= 6

(iii)Upper quartile, Q3 = (3n/4) th term

= (3×16/4) th term

= (3×4)th term

12 th term

= 18

Question 12

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

Answer 12

Measures of Central Tendency ML Aggarwal Solutions chap 21 img 71

Question 13

The daily wages of 30 employees in an establishment are distributed as follows :
Measures of Central Tendency ML Aggarwal Solutions chap 21 img 72
Estimate the modal daily wages for this distribution by a graphical method.

Answer 13

Measures of Central Tendency ML Aggarwal Solutions chap 21 img 73
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 14  Chapter-Test Measures of Central Tendency ML 

Draw a cumulative frequency curve for the following data :
Measures of Central Tendency ML Aggarwal Solutions chap 21 img 74
Hence determine:
(i) the median
(ii) the pass marks if 85% of the students pass.
(iii) the marks which 45% of the students exceed.

Answer 14

Measures of Central Tendency ML Aggarwal Solutions chap 21 img 75

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