RD Sharma Solutions (Part - 2) - Ex-23.1, Data Handling II Central Values, Class 7, Math Class 7 Notes | EduRev

RD Sharma Solutions for Class 7 Mathematics

Created by: Abhishek Kapoor

Class 7 : RD Sharma Solutions (Part - 2) - Ex-23.1, Data Handling II Central Values, Class 7, Math Class 7 Notes | EduRev

The document RD Sharma Solutions (Part - 2) - Ex-23.1, Data Handling II Central Values, Class 7, Math Class 7 Notes | EduRev is a part of the Class 7 Course RD Sharma Solutions for Class 7 Mathematics.
All you need of Class 7 at this link: Class 7

Question 14:

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

Answer 14:

We have :

RD Sharma Solutions (Part - 2) - Ex-23.1, Data Handling II Central Values, Class 7, Math Class 7 Notes | EduRev

RD Sharma Solutions (Part - 2) - Ex-23.1, Data Handling II Central Values, Class 7, Math Class 7 Notes | EduRev

Question 15:

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

Answer 15:

The mean number of children per family =  RD Sharma Solutions (Part - 2) - Ex-23.1, Data Handling II Central Values, Class 7, Math Class 7 Notes | EduRev

RD Sharma Solutions (Part - 2) - Ex-23.1, Data Handling II Central Values, Class 7, Math Class 7 Notes | EduRev

Thus, on an average there are 3 children per family in the locality.

 

Question 16:

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.

Answer 16:

We have:
 

n = The number of observations = 100,  Mean = 40

RD Sharma Solutions (Part - 2) - Ex-23.1, Data Handling II Central Values, Class 7, Math Class 7 Notes | EduRev

RD Sharma Solutions (Part - 2) - Ex-23.1, Data Handling II Central Values, Class 7, Math Class 7 Notes | EduRev

⇒ 40 ×100 = Sum of the observations

Thus, the incorrect sum of the observations = 40 x 100 = 4000.

Now,

The correct sum of the observations = Incorrect sum of the observations - Incorrect observation + Correct observation

⇒ The correct sum of the observations = 4000 - 83 + 53 
⇒ The correct sum of the observations = 4000 - 30 =  3970

 ∴   RD Sharma Solutions (Part - 2) - Ex-23.1, Data Handling II Central Values, Class 7, Math Class 7 Notes | EduRev

 

Question 17:

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

Answer 17:

We have:

RD Sharma Solutions (Part - 2) - Ex-23.1, Data Handling II Central Values, Class 7, Math Class 7 Notes | EduRev

So, sum of the five numbers = 5 ×27 = 135.

Now, 

 The mean of four numbers =    RD Sharma Solutions (Part - 2) - Ex-23.1, Data Handling II Central Values, Class 7, Math Class 7 Notes | EduRev

 So, sum of the four numbers = 4 ×25  = 100.

Therefore, the excluded number  = Sum of the five numbers - sum of the four numbers

⇒ The excluded number = 135 - 100 = 35.

 

Question 18:

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.

Answer 18:

We have:

RD Sharma Solutions (Part - 2) - Ex-23.1, Data Handling II Central Values, Class 7, Math Class 7 Notes | EduRev

Let the weight of the seventh student be x kg.

RD Sharma Solutions (Part - 2) - Ex-23.1, Data Handling II Central Values, Class 7, Math Class 7 Notes | EduRev

RD Sharma Solutions (Part - 2) - Ex-23.1, Data Handling II Central Values, Class 7, Math Class 7 Notes | EduRev

Thus, the weight of the seventh student is 61 kg.

 

Question 19:

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

Answer 19:

Let x1, x2x3...x8 be the eight numbers whose mean is 15 kg. Then,

RD Sharma Solutions (Part - 2) - Ex-23.1, Data Handling II Central Values, Class 7, Math Class 7 Notes | EduRev

x1, x2x3...x8  = 15 x 8

⇒ x1, x2x3...x8 = 120

Let the new numbers be 2x1 , 2x2, 2x3, ...2x8. Let M be the arithmetic mean of the new numbers.
Then,

RD Sharma Solutions (Part - 2) - Ex-23.1, Data Handling II Central Values, Class 7, Math Class 7 Notes | EduRev

Question 20:

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

Answer 20:

Let x1,x2,x3,x4 & x5x1,x2,x3,x4 & x5 be five numbers whose mean is 18. Then,
18 = Sum of five numbers ÷ 5
∴ Sum of five numbers = 18 × 5 = 90.

Now, if one number is excluded, then their mean is 16.
So,
16= Sum of four numbers ÷ 4
∴ Sum of four numbers = 16 × 4 = 64.

 

 The excluded number =  Sum of the five observations - Sum of the four observations

∴ The excluded number = 90 - 64
 ∴ The excluded number = 26.

 

Question 21:

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

Answer 21:

n = Number of observations = 200

RD Sharma Solutions (Part - 2) - Ex-23.1, Data Handling II Central Values, Class 7, Math Class 7 Notes | EduRev

⇒Sum of the observations = 50 × 200 = 10,000

Thus, the incorrect sum of the observations = 50 x 200
Now, 
The correct sum of the observations = Incorrect sum of the observations - Incorrect observations + Correct observations
⇒Correct sum of the observations =  10,000 - (92+ 8) + (192 + 88)

⇒ Correct sum of the observations = 10,000 - 100 + 280
⇒ Correct sum of the observations = 9900 +280
⇒ Correct sum of the observations = 10180.

RD Sharma Solutions (Part - 2) - Ex-23.1, Data Handling II Central Values, Class 7, Math Class 7 Notes | EduRev

 

Question 22:

The mean of 5 numbers is 27. If one more number is included, then the mean is 25. Find the included number.

Answer 22:

We have:
Mean =  Sum of  five numbers ÷ 5
⇒ Sum of the five numbers = 27 x 5 = 135.

Now, New mean = 25
25 = Sum of six numbers ÷÷ 6
⇒ Sum of the six numbers = 25 x 6 = 150.

The included number = Sum of the six numbers - Sum of the five numbers
⇒The included number = 150 - 135
⇒The included number =  15.

 

Question 23:

The mean of 75 numbers is 35. If each number is multiplied by 4, find the new mean.

Answer 23:

Let x1, x2, x3, ... x75 be 75 numbers with their mean equal to 35. Then,

RD Sharma Solutions (Part - 2) - Ex-23.1, Data Handling II Central Values, Class 7, Math Class 7 Notes | EduRev

RD Sharma Solutions (Part - 2) - Ex-23.1, Data Handling II Central Values, Class 7, Math Class 7 Notes | EduRev

The new numbers are 4x1, 4x2, 4x3, ...4x75. Let M be the arithmetic mean of the new numbers. Then,

RD Sharma Solutions (Part - 2) - Ex-23.1, Data Handling II Central Values, Class 7, Math Class 7 Notes | EduRev

RD Sharma Solutions (Part - 2) - Ex-23.1, Data Handling II Central Values, Class 7, Math Class 7 Notes | EduRev

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