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.

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**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.**

We have :

**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.**

The mean number of children per family =

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

**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.**

We have:

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

â‡’ 40 Ã—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

â‡’ The correct sum of the observations = 4000 - 83 + 53

â‡’ The correct sum of the observations = 4000 - 30 = 3970

âˆ´

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

We have:

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

Now,

The mean of four numbers =

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

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

â‡’ The excluded number = 135 - 100 = 35.

**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.**

We have:

Let the weight of the seventh student be *x* kg.

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

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

Let *x*_{1},* **x*_{2}, *x*_{3}...*x*_{8} be the eight numbers whose mean is 15 kg. Then,

*x*_{1},* **x*_{2}, *x*_{3}...*x*_{8} = 15 x 8

*â‡’ x*_{1},* **x*_{2}, *x*_{3}...*x*_{8} = 120

Let the new numbers be 2x_{1} , 2x_{2}, 2x_{3}, ...2x_{8. }Let* M* be the arithmetic mean of the new numbers.

Then,

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

Let x1,x2,x3,x4 & x5x1,x2,x3,x4 & x5 be five numbers whose mean is 18. Then,

18 = Sum of five numbers Ã· 5

âˆ´ Sum of five numbers = 18 Ã— 5 = 90.

Now, if one number is excluded, then their mean is 16.

So,

16= Sum of four numbers Ã· 4

âˆ´ Sum of four numbers = 16 Ã— 4 = 64.

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

âˆ´ The excluded number = 90 - 64

âˆ´ The excluded number = 26.

**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.**

*n* = Number of observations = 200

â‡’Sum of the observations = 50 Ã— 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

â‡’Correct sum of the observations = 10,000 - (92+ 8) + (192 + 88)

â‡’ Correct sum of the observations = 10,000 - 100 + 280

â‡’ Correct sum of the observations = 9900 +280

â‡’ Correct sum of the observations = 10180.

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

We have:

Mean = Sum of five numbers Ã· 5

â‡’ Sum of the five numbers = 27 x 5 = 135.

Now, New mean = 25

25 = Sum of six numbers Ã·Ã· 6

â‡’ Sum of the six numbers = 25 x 6 = 150.

The included number = Sum of the six numbers - Sum of the five numbers

â‡’The included number = 150 - 135

â‡’The included number = 15.

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

Let x_{1}, x_{2}, x_{3}, ... x_{75} be 75 numbers with their mean equal to 35. Then,

The new numbers are 4x_{1}, 4x_{2}, 4x_{3}, ...4x_{75. }Let M be the arithmetic mean of the new numbers. Then,