RD Sharma Solutions Ex-24.1, (Part - 2), Measures Of Central Tendency, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics PDF Download

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

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

New numbers are = 5 , 6 , 7

∴ 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

∴ 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

∴ 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.

∴ 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 = 100×40

= 4000

Correct value = 53

Incorrect value = 83

Correct sum = 4000 – 83 + 53 = 3970

∴ correct  mean = 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 

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 = 5×27 = 135

If one number is excluded , the new mean is 25

∴Sum of 4 numbers = 4×25 = 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

∴ 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 = 8×15 = 120

If each number is multiplied by 2

Then , new mean = 120×2 = 240

∴ new  mean = 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 = 5×18 = 90

If one number is excluded

Then , the mean of 4 numbers = 16

∴ sum of 4 numbers = 4×16 = 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 = 200×50 = 10,000

Correct values = 192 and 88.

Incorrect values = 92 and 8.

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

Q 21 . Find the values of n and  in each of the following cases :

SOLUTION :

⇒ (x₁−12)+(x₂−12)+ ……… +(x_n−12) = −10

⇒(x₁+x₂+x₃+x₄+x₅+⋅⋅⋅+x_n)−(12+12+12+12+⋅⋅⋅⋅+12)=−10

⇒ ∑ x−12n =−10⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅(1)

⇒ (x₁−3)+(x₂−3)+⋅⋅⋅⋅⋅+(x_n−3) = 62

⇒ (x₁+x₂+⋅⋅⋅+x_n)−(3+3+3+⋅⋅⋅+3) = 62

⇒ ∑x−3n = 62⋅⋅⋅⋅⋅⋅⋅⋅⋅(2)

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

∑x−3n−∑x+12n  62+10

⇒ 9n = 72

⇒ n = 72/9 = 8

Put value of n in equation (1)

∑x−12×8 = −10

⇒∑x−96 = −10

⇒∑x = 96−10 = 86

(x₁−10)+(x₂−10)+ ………… + (x_n−10) = 30

⇒(x₁+x₂+x₃+x₄+x₅+⋅⋅⋅+x_n)−(10+10+10+10+⋅⋅⋅⋅+10) = 30

⇒ ∑x−10n = 30⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅(1)

⇒(x₁−6)+(x₂−6)+⋅⋅⋅⋅⋅+(x_n−6) = 150

⇒(x₁+x₂+⋅⋅⋅+x_n)−(6+6+6+⋅⋅⋅+6) = 150

⇒∑x−6n = 150⋅⋅⋅⋅⋅⋅⋅⋅⋅(2)

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

∑x−6n−∑x+10n = 150−30

⇒ 4n = 120

⇒ n = 120/4 = 30

Put value of n in equation (1)

∑x−10×30 = 30

⇒ ∑x−300 = 30

⇒ ∑x=30+300 = 330

Q 22 . The sum of the deviations of a set of n values x₁,x₂,x₃,⋅⋅⋅,x_n measured from 15 and -3 are -90 and 54 respectively . Find the value of n and mean .

SOLUTION :

Given :

⇒ (x₁−15)+(x₂−15)+⋅⋅⋅⋅⋅+(x_n−15) = −90

⇒ (x₁+x₂+⋅⋅⋅⋅+_n)−(15+15+15+⋅⋅⋅⋅⋅⋅+15) =−90

⇒ ∑x−15n =−90⋅⋅⋅⋅⋅(1)

⇒ (x₁+3)+(x₂+3)+⋅⋅⋅⋅⋅+(x_n+3) = 54

⇒ (x₁+x₂+⋅⋅⋅⋅+_n)+(3+3+3+⋅⋅⋅⋅⋅⋅+3) = 54

⇒ ∑x+3n = 54⋅⋅⋅⋅⋅(2)

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

∑x+3n−∑x+15n = 54+90

⇒ 18n =144

⇒ n =144/18 = 8

Put value of n in equation(1)

∑x−15×8 = −90

∑x−120 = −90

∑x = 120−90 = 30

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

= 7 

∴ Sum of deviation of values from their mean

= (3−7)+(4−7)+(6−7)+(7−7)+(8−7)+(14−7)

= – 4 – 3 – 1 + 0 + 1 + 7

= – 8 + 8 = 0

Q 24 . If  is the mean of the ten natural numbers x₁,x₂,x₃,⋅⋅⋅,x₁₀ show that 

SOLUTION :

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