# RD Sharma Solutions Class 9 Measures Of Central Tendency Exercise 24.1

## RD Sharma Solutions Class 9 Chapter 24 Exercise 24.1

Q.1: If the heights of 5 persons are 140 cm, 150 cm, 152 cm, 158 cm and 161 cm respectively. Find the mean height.

SOLUTION :

Given : the heights of 5 persons are 140 cm , 150 cm , 152 cm , 158 cm and 161 cm

$∴$ Mean Weight = $\frac{sum \;of\; heights}{total \;no.\; of\; persons}$

= $\frac{140+150+152+158+161}{5}$

= $\frac{761}{5}$ =152.2

Q 2 . Find the mean of 994 , 996 , 998 , 1000 , 1002.

SOLUTION :

Given :

Numbers are 994 , 996 , 998 , 1000 , 1002.

$∴$ Mean   = $\frac{sum\;of\;numbers}{total\;numbers}$

= $\frac{994+996+998+1000+1002}{5}$

= $\frac{4990}{5}$ = 998

Mean = 998

Q 3 . Find the mean of first five natural numbers.

SOLUTION :

The first five odd numbers are 1 , 2 , 3 , 4 , 5.

$∴$ Mean   = $\frac{sum\;of\;numbers}{total\;numbers}$

= $\frac{1+2+3+4+5}{5}$

= $\frac{15}{5}$ = 3

Mean = 3

Q 4 . Find the mean of all factors of 10.

SOLUTION :

All factors of 6 are 1 , 2 , 5 , 10.

$∴$ Mean   = $\frac{sum\;of\;factors}{total\;factors}$

= $\frac{1+2+5+10}{4}$ = 4.5

Mean = 4.5

Q 5 . Find the mean of first ten even natural numbers.

SOLUTION :

The first five even natural numbers are 2 , 4 , 6 , 8 , 10 , 12 , 14 , 16 , 18 , 20

$∴$ Mean    = $\frac{sum\;of\;numbers}{total\;numbers}$

= $\frac{2+4+6+8+10+12+14+16+18+20}{10}$ =11

Mean = 11

Q 6 . Find the mean of x , x + 2 , x + 4 , x + 6 , x + 8.

SOLUTION :

Numbers are x , x + 2 , x + 4 , x + 6 , x + 8.

$∴$ Mean    = $\frac{sum\;of\;numbers}{total\;numbers}$

= $\frac{x+x+2+x+4+x+6+x+8}{5}$

= $\frac{5x+20}{5}$

= $5\left ( \frac{x+4}{5} \right )$

= x + 4

Q 7 . Find the mean of first five multiples of 3.

SOLUTION :

First five multiples of 3 are 3 , 6 , 9 , 12 , 15.

$∴$ Mean    = $\frac{sum\;of\;numbers}{total\;numbers}$

= $\frac{3 + 6 + 9 + 12 + 15 }{5}$

=9

Mean = 9

Q 8 . Following are the weights of 10 new born babies in a hospital on a particular day  : 3.4 , 3 .6 , 4.2 , 4.5 , 3.9 , 4.1 , 3.8 , 4.5 , 4.4 , 3.6 (in kg). Find the mean.

SOLUTION :

The weights (in kg) of 10 new born  babies are : 3.4 , 3 .6 , 4.2 , 4.5 , 3.9 , 4.1 , 3.8 , 4.5 , 4.4 , 3.6

$∴$ Mean Weight = $\frac{sum \;of\; weights}{total \;no.\; of\; babies}$

=$\frac{3.4 +3 .6 + 4.2 + 4.5 + 3.9 + 4.1 + 3.8 + 4.5 + 4.4 + 3.6 }{10}$

=4 kg

Q 9 . The percentage marks obtained by students of a class in mathematics are as follows: 64 , 36 , 47 , 23 , 0 , 19 , 81 , 93 , 72 , 35 , 3 , 1 .Find their mean.

SOLUTION :

The percentage marks obtained by students are 64 , 36 , 47 , 23 , 0 , 19 , 81 , 93 , 72 , 35 , 3 , 1

$∴$ Mean marks    = $\frac{sum\;of\;marks}{total\;numbers\;of\;marks}$

= $\frac{64 + 36 + 47 + 23 + 0 + 19 + 81 + 93 + 72 + 35 + 3 + 1}{5}$ =39.5

Mean Marks = 39.5

Q 10. The numbers of children in 10 families of a locality  are 2 , 4 , 3 , 4 , 2 , 3 , 5 , 1 , 1 , 5 . Find the number of children per family.

SOLUTION :

The numbers of children  in 10 families  are : 2 , 4 , 3 , 4 , 2 , 3 , 5 , 1 , 1 , 5

$∴$ Mean = $\frac{total\;no.\;children}{total\;families}$

= $\frac{2 + 4 + 3 + 4 + 2 + 3 + 5 + 1 + 1 + 5}{10}$ = 3

Q 11 . If M is the mean of $x_{1},x_{2},x_{3},x_{4},x_{5}\;and\;x_{6}$ , Prove that $\left ( x_{1}-M \right )+\left ( x_{2}-M \right )+\left ( x_{3}-M \right )+\left ( x_{4}-M \right )+\left ( x_{5}-M \right )+\left ( x_{6}-M \right )=0$.

SOLUTION :

Let M be the mean of $x_{1},x_{2},x_{3},x_{4},x_{5}\;and\;x_{6}$

Then M= $\frac{x_{1}+x_{2}+x_{3}+x_{4}+x_{5}+x_{6}}{6}$

= $x_{1}+x_{2}+x_{3}+x_{4}+x_{5}+x_{6}$ = 6M

To Prove :- $\left ( x_{1}-M \right )+\left ( x_{2}-M \right )+\left ( x_{3}-M \right )+\left ( x_{4}-M \right )+\left ( x_{5}-M \right )+\left ( x_{6}-M \right )=0$.

Proof :- L . H . S

= $\left ( x_{1}-M \right )+\left ( x_{2}-M \right )+\left ( x_{3}-M \right )+\left ( x_{4}-M \right )+\left ( x_{5}-M \right )+\left ( x_{6}-M \right )$

= $\left ( x_{1}+x_{2}+x_{3}+x_{4}+x_{5}+x_{6} \right )-\left ( M+M+M+M+M+M \right )$

= 6M – 6M

= 0

= R . H . S

Q 12 . Duration of sunshine(in hours) in  Amritsar  for first  10 days of August  1997 as reported by the Meterological Department  are given as  follows : 9.6 , 5.2 , 3.5 , 1.5 , 1.6 , 2.4 , 2.6 , 8.4 , 10.3 , 10.9

1. Find the mean $\bar{X}$

2.Verify that $\sum_{i=1}^{10}\left ( _{x} i-\bar{X}\right )=0$

SOLUTION :

Duration of sunshine (in hours ) for 10 days are =9.6 , 5.2 , 3.5 , 1.5 , 1.6 , 2.4 , 2.6 , 8.4 , 10.3 , 10.9

(i) Mean  $\bar{X} \;= \;\frac{sum\;of\;numbers}{total\;numbers}$

= $\frac{9.6 + 5.2 + 3.5 + 1.5 + 1.6 + 2.4 + 2.6 + 8.4 + 10.3 + 10.9}{10}$

= $\frac{56}{10}$ = 5.6

(ii) L.H.S  =  $\sum_{i=1}^{10}\left ( _{x} i-\bar{X}\right )$

= $(x_{1}-\bar{x})+( x_{2}-\bar{x})+( x_{3}-\bar{x})$+…….. +$( x_{10}-\bar{x})$

= $\left ( 9.6-5.6 \right )+\left ( 5.2-5.6 \right )+\left ( 3.5-5.6 \right )+\left ( 1.5-5.6 \right )+\left ( 1.6-5.6 \right )+\left ( 2.4-5.6 \right )+\left ( 2.6-5.6 \right )+\left ( 8.4-5.6 \right )+\left ( 10.3-5.6 \right )+\left ( 10.9-5.6 \right )$

= 4 – 0.4 – 2.1 – 4.1 – 4 – 3.2 – 3 + 2.8 + 4.7 + 5.3

= 16.8 – 16.8 = 0

= R.H.S

Q 13. Explain,  by taking a suitable example, how the arithmetic mean alters by (i) adding a constant k to each term, (ii) Subtracting a constant k from each term, (iii) multiplying each term by a constant k and (iv) dividing each term by non-zero constant k.

SOLUTION :

Let say numbers are 3 , 4 , 5

$∴$ Mean    = $\frac{sum\;of\;numbers}{total\;numbers}$

= $\frac{3 +4 + 5}{3}$ = 4

(i). Adding constant term k = 2 in each term.

New numbers are = 5 , 6 , 7

$∴$ Mean    = $\frac{sum\;of\;numbers}{total\;numbers}$

= $\frac{ 5 + 6 + 7}{3}$

= 6 = 4 + 2

$∴$ new  mean will be 2 more than the original mean.

(ii). Subtracting constant term k = 2 in each term.

New numbers are = 1 , 2 , 3

$∴$ Mean    = $\frac{sum\;of\;numbers}{total\;numbers}$

= $\frac{ 1 + 2 + 3}{3}$

= 2 = 4 – 2

$∴$ new  mean will be 2 less than the original mean.

(iii) . Multiplying by constant term k = 2 in each term.

New numbers are = 6 , 8 , 10

$∴$ Mean    = $\frac{sum\;of\;numbers}{total\;numbers}$

= $\frac{ 6 + 8 + 10}{3}$

= 8 = $4\times 2$

$∴$ new  mean will be 2 times of  the original mean.

(iv) . Divide the constant term k =2 in each term.

New numbers are = 1.5 , 2 , 2.5.

$∴$ Mean    = $\frac{sum\;of\;numbers}{total\;numbers}$

= $\frac{ 1.5 + 2 + 2.5}{3}$

= 2 = $\frac{4}{2}$

$∴$ new  mean will be half  of  the original mean.

Q 14. The mean of marks scored by 100 students was found to be 40. Later on, it was discovered that a score of 53 was misread as 83. Find the correct mean.

SOLUTION :

Mean marks of 100 students = 40

Sum of marks of 100 students = $100\times 40$

= 4000

Correct value = 53

Incorrect value = 83

Correct sum = 4000 – 83 + 53 = 3970

$∴$ correct  mean = $\frac{3970}{100}$ = 39.7

Q 15 . The traffic police recorded the speed (in km/hr) of 10 motorists as 47 , 53 , 49 , 60 , 39 , 42 , 55 , 57 , 52 , 48 . Later on, an error in recording instrument was found. Find the correct average speed of the motorists if the instrument is recorded 5 km/hr less in each case.

SOLUTION :

The speed of 10 motorists are 47 , 53 , 49 , 60 , 39 , 42 , 55 , 57 , 52 , 48 .

Later on it was discovered that the instrument recorded 5 km/hr less than in each case

$∴$ correct values are = 52 , 58 , 54 , 65 , 44 , 47 , 60 , 62 , 57 , 53.

$∴$ correct  mean = $\frac{52 + 58 + 54 + 65 + 44 + 47 + 60 + 62 + 57 + 53}{10}$

= $\frac{552}{10}$ =55.2 km/hr

Q 16. The mean of five numbers is 27. If one number is excluded, their mean is 25. Find the excluded number.

SOLUTION :

The mean of five numbers is 27

The sum of five numbers = $5\times 27$ = 135

If one number is excluded , the new mean is 25

$∴$Sum of 4 numbers = $4\times 25$ = 100

$∴$ Excluded number = 135 – 100 = 35

Q 17. The mean weight per student in a group of 7 students is 55 kg. The individual weights of 6 of them (in kg) are 52, 54, 55, 53, 56 and 54. Find the weight of the seventh student.

SOLUTION :

The mean weight per student in a group of 7 students = 55 kg

Weight of 6 students (in kg) =  52 , 54 , 55 , 53 , 56 and 54

Let the weight of seventh student = x kg

$∴$ Mean Weight = $\frac{sum \;of\; weights}{total \;no.\; of\;students}$

$\Rightarrow 55$  =$\frac{52 + 54 + 55 + 53 + 56 + 54 + x}{7}$

$\Rightarrow 385=324+x$

$\Rightarrow x=385 -324$

$\Rightarrow x=61$ kg

$∴$ weight of seventh student = 61 kg.

Q 18. The mean weight of 8 numbers is 15. If each number is multiplied by 2 , what will be the new mean?

SOLUTION :

We have ,

The mean weight of 8 numbers is 15

Then , the sum of 8 numbers = $8\times 15$ = 120

If each number is multiplied by 2

Then , new mean = $120\times 2$ = 240

$∴$ new  mean = $\frac{240}{8}$ = 30.

Q 19. The mean of 5 numbers is 18. If one number is excluded, their mean is 16. Find the excluded number.

SOLUTION :

The mean of 5 numbers is 18

Then , the sum of 5 numbers = $5\times 18$ = 90

If one number is excluded

Then , the mean of 4 numbers = 16

$∴$ sum of 4 numbers = $4\times 16$ = 64

Excluded number = 90 – 64 = 26.

Q 20. The mean of 200 items was 50. Later on, it was on discovered that the two items were misread as 92 and 8 instead of 192 and 88. Find the correct mean.

SOLUTION :

The mean of 200 items = 50

Then the sum of 200 items = $200\times 50$ = 10,000

Correct values = 192 and 88.

Incorrect values = 92 and 8.

$∴$ correct sum = 10000 – 92 – 8 + 192 + 88 = 10180

$∴$ correct mean = $\frac{10180}{200}$ = $\frac{101.8}{2}$ = 50.9 .

Q 21 . Find the values of n and $\bar{X}$in each of the following cases :

(i). $\sum_{i=1}^{n}\left ( x_{i}-12 \right )=-10\; and\;\sum_{i=1}^{n}\left ( x_{i}-3 \right )=62$

(ii). $\sum_{i=1}^{n}\left ( x_{i}-10 \right )=30\; and\;\sum_{i=1}^{n}\left ( x_{i}-6 \right )=150$

SOLUTION :

(i). Given $\sum_{i=1}^{n}\left ( x_{i}-12 \right )$=-10

$( x_{1}-12)+( x_{2}-12)$+ ……… +$\left ( x_{n}-12 \right )=-10$

$\Rightarrow \left ( x_{1} +x_{2} +x_{3}+ x_{4}+ x_{5} +\cdot \cdot \cdot +x_{n}\right )-\left ( 12+12+12+12+\cdot \cdot \cdot \cdot +12 \right )=-10$

$\Rightarrow \sum x-12n=-10\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \left ( 1 \right )$

$And \; \sum_{n}^{i=1}\left ( x_{i}-3 \right )=62$

$\Rightarrow \left ( x_{1}-3 \right )+\left ( x_{2}-3 \right )+\cdot \cdot \cdot \cdot \cdot +\left ( x_{n}-3 \right )=62$

$\Rightarrow \left ( x_{1}+x_{2}+\cdot \cdot \cdot +x_{n} \right )-\left ( 3+3+3+\cdot \cdot \cdot +3 \right )=62$

$\Rightarrow \sum x-3n=62\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \left ( 2 \right )$

By subtracting equation (1) from equation(2) , we get

$\sum x-3n-\sum x+12n=62+10$

$\Rightarrow 9n=72$

$\Rightarrow n=\frac{72}{9}=8$

Put value of n in equation (1)

$\sum x-12\times 8=-10$

$\Rightarrow \sum x-96=-10$

$\Rightarrow \sum x=96-10=86$

$∴$ $\bar{x}=\frac{\sum x}{n}=\frac{86}{8}=10.75$

(ii). Given  $\sum_{i=1}^{n}\left ( x_{i}-10 \right )$=30

$( x_{1}-10)+( x_{2}-10)$+ ………… + $( x_{n}-10)$=30

$\Rightarrow \left ( x_{1} +x_{2} +x_{3}+ x_{4}+ x_{5} +\cdot \cdot \cdot +x_{n}\right )-\left ( 10+10+10+10+\cdot \cdot \cdot \cdot +10 \right )=30$

$\Rightarrow \sum x-10n=30\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \left ( 1 \right )$

$And \; \sum_{n}^{i=1}\left ( x_{i}-6 \right )=150$

$\Rightarrow \left ( x_{1}-6 \right )+\left ( x_{2}-6 \right )+\cdot \cdot \cdot \cdot \cdot +\left ( x_{n}-6 \right )=150$

$\Rightarrow \left ( x_{1}+x_{2}+\cdot \cdot \cdot +x_{n} \right )-\left ( 6+6+6+\cdot \cdot \cdot +6 \right )=150$

$\Rightarrow \sum x-6n=150\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \left ( 2 \right )$

By subtracting equation (1) from equation(2) , we get

$\sum x-6n-\sum x+10n=150-30$

$\Rightarrow 4n=120$

$\Rightarrow n=\frac{120}{4}=30$

Put value of n in equation (1)

$\sum x-10\times 30=30$

$\Rightarrow \sum x-300=30$

$\Rightarrow \sum x=30+300$ = 330

$∴$ $\bar{x}=\frac{\sum x}{n}=\frac{330}{30}=11$.

Q 22 . The sum of the deviations of a set of n values $x_{1},x_{2},x_{3},\cdot \cdot \cdot ,x_{n}$ measured from 15 and -3 are -90 and 54 respectively . Find the value of n and mean .

SOLUTION :

Given :

$\sum_{n}^{i=1}\left ( x_{i}-15 \right )=-90$

$\Rightarrow \left ( x_{1}-15 \right )+\left ( x_{2}-15 \right )+\cdot \cdot \cdot \cdot \cdot+ \left ( x_{n}-15 \right )=-90$

$\Rightarrow \left ( x_{1}+x_{2}+\cdot \cdot \cdot \cdot +_{n} \right )-\left ( 15+15+15+\cdot \cdot \cdot \cdot \cdot \cdot +15 \right )=-90$

$\Rightarrow \sum x-15n=-90\cdot \cdot \cdot \cdot \cdot \left ( 1 \right )$

$And\; \sum_{n}^{i=1}\left ( x_{i}+3 \right )=54$

$\Rightarrow \left ( x_{1}+3 \right )+\left ( x_{2}+3 \right )+\cdot \cdot \cdot \cdot \cdot+ \left ( x_{n}+3 \right )=54$

$\Rightarrow \left ( x_{1}+x_{2}+\cdot \cdot \cdot \cdot +_{n} \right )+\left ( 3+3+3+\cdot \cdot \cdot \cdot \cdot \cdot +3\right )=54$

$\Rightarrow \sum x+3n=54\cdot \cdot \cdot \cdot \cdot \left ( 2 \right )$

By subtracting equation (1) from equation(2) , we get

$\sum x+3n-\sum x+15n = 54+90$

$\Rightarrow 18n=144$

$\Rightarrow n=\frac{144}{18}=8$

Put value of n in equation(1)

$\sum x-15\times 8=-90$

$\sum x-120=-90$

$\sum x=120-90=30$

$∴$ $\bar{x}=\frac{\sum x}{n}=\frac{30}{8}=3.75$.

Q 23 . Find the sum of the deviations of the variate values 3 , 4 , 6 , 7 , 8 , 14 from their mean.

SOLUTION :

Values 3 , 4 , 6 , 7 , 8 , 14

$∴ Mean=\frac{sum\;of\;numbers}{total\;numbers}$

$∴ Mean=\frac{3+4+6+7+8+14}{6}$

$∴ Mean=\frac{42}{6}$

=7

$∴$ Sum of deviation of values from their mean

= $\left ( 3-7 \right )+\left ( 4-7 \right )+\left ( 6-7 \right )+\left (7-7 \right )+\left ( 8-7 \right )+\left (14-7 \right )$

= – 4 – 3 – 1 + 0 + 1 + 7

= – 8 + 8 = 0

Q 24 . If $\bar{X}$ is the mean of the ten natural numbers $x_{1},x_{2},x_{3},\cdot \cdot \cdot ,x_{10}$ show that $\left ( x_{1}-\bar{X} \right )+\left ( x_{2}-\bar{X} \right )+\cdot \cdot \cdot \cdot +\left ( x_{10}-\bar{X} \right )=0$

SOLUTION :

We have , $\bar{x}=\frac{x_{1}+x_{2}+\cdot \cdot \cdot +x_{10}}{10}$

$\Rightarrow x_{1}+x_{2}+\cdot \cdot \cdot +x_{10}=10\bar{x}\cdot \cdot \cdot \cdot \cdot \cdot \cdot \left ( 1 \right )$

Now , $\left ( x_{1}-\bar{X} \right )+\left ( x_{2}-\bar{X} \right )+\cdot \cdot \cdot \cdot +\left ( x_{10}-\bar{X} \right )$

=$\left ( x_{1}+x_{2}+\cdot \cdot \cdot \cdot \cdot +x_{10} \right )-\left ( \bar{x}+\bar{x}+\bar{x}+ upto \;10 \;terms \right )$

=$10\bar{x}-10\bar{x}$                                         $\left [ By \;equation \left ( i \right ) \right ]$

$∴$ $\left ( x_{1}-\bar{X} \right )+\left ( x_{2}-\bar{X} \right )+\cdot \cdot \cdot \cdot +\left ( x_{10}-\bar{X} \right )=0$..

#### Practise This Question

A quadrilateral has two of its angles as 30 and 40 .What type of quadrilateral can it be?