## RS Aggarwal Class 9 Ex 14C Chapter 14

**Q.1: Find the arithmetic mean of:**

** (i) The first eight natural numbers**

** (ii) The first the odd numbers**

** (iii) The first five prime numbers **

** (iv) The first six even numbers **

** (v) The first seven multiples of 5 **

** (vi) All the factors of 20 **

**Sol:**

**(i)** First eight natural numbers are:

1,2,3,4,5,6,7 and 8

Therefore, mean =

=

= Therefore, Mean =4.5

**(ii)** First ten odd numbers are: 1, 3, 5, 7, 9, 11, 13, 15, 17 and 19

Therefore, mean =

=

Therefore, Mean =10

**(iii)** First five prime numbers are : 2,3,5,7,11

Therefore, mean =

=

Therefore, Mean =5.6

**(iv)** First six even numbers are: 2, 4, 6, 8, 10, 12

Therefore, mean =

=

Therefore, Mean = 7

**(v)** First seven multiples of 5 are: 5, 10,15, 20, 25, 30, 35

Therefore, mean =

=

Therefore, Mean = 20

**(vi)** Factors of 20 are: 1, 2, 4, 5, 10, 20

Therefore, mean =

=

Therefore, Mean = 7

**Q.2: The number of children in 10 families of a taxably are: 2, 4, 3, 4, 2, 0, 3, 5, 1, 6. Find the mean number of children per family. **

**Sol: **

Mean =

=

Therefore, Mean = 3

**Q.3: The following are the number of books issued in a school library during a week: 105, 216, 322, 167, 273, 405 and 346. Find the average number of books issued per day.**

**Sol:**

Mean =

=

Therefore, Mean = 262

**Q.4: The daily minimum temperature recorded (in degree F) at a place during a week was as under. **

Monday |
Tuesday |
Wednesday |
Thursday |
Friday |
Saturday |

35.5 |
30.8 |
27.3 |
32.1 |
23.8 |
29.9 |

**Sol:**

Mean Temperature =

=

Therefore, Mean = 29.9

**Q.5: The percentages of mark obtained by 12 students of a class in mathematics are: 64, 36, 47, 23, 0, 19, 81, 93, 72, 35, 3, 1. Find the mean percentage of marks.**

**Sol:**

Mean =

=

Therefore, Mean =39.5

**Q.6: If the arithmetic mean of 7, 9, 11, 13, x, 21 is 13, find the value of x.**

**Sol:**

Mean of the given number =

And, mean = 13 (given)

Therefore,

Therefore, the value of x = 17

**Q.7: The mean of 24 numbers is 35.If 3 is added to each number, what will be the new mean?**

**Sol:**

Let, the given number be

Therefore,

The new number are

Therefore, Mean of the new numbers

Therefore,

**Q.8: The mean of 20 numbers is 43.If 6 is subtracted from each of the numbers, what will be the new mean?**

**Sol:**

Let, the given numbers be

Then, the mean of these numbers =

Therefore,

**The new numbers are:**

**( using (i) )**

Therefore,

**Q.9: The mean of 15 numbers is 27. If each number is multiplied by 4, what will be the mean of the new numbers?**

**Sol:**

Let the given number be

Then, the mean of these numbers =

The new numbers are

Therefore, mean of the new numbers =

Therefore, the new mean = 108

**Q.10: The mean of 12 numbers is ****40. If each number is divided by 8, what will be the mean of the new numbers?**

**Sol:**

Let the given number be

Then, the mean of these numbers = 40

The new numbers are

Therefore, mean of the new numbers =

Therefore, the new mean = 5

**Q.11: The mean of 20 numbers is 18.If 3 is added to each of the first ten numbers; find the mean of the new set of 20 numbers.**

**Sol:**

Let the given number be

Let the mean be

Therefore,

But it is given that 3 is added to each of the first ten numbers.

Therefore, the first new ten numbers are

Let,

From (i), we know that

Therefore, mean of the new set of 20 numbers =

Therefore, mean of the new set of 20 numbers = 19.5

**Q.12: The mean weight of 6boys in a group is 48kg. The individual weights of five of them are 51kg, 45kg, 49kg, 46kg, and 44kg.Find the weight of the sixth boy.**

**Sol:**

Mean weight of the boys =48 kg

Therefore, Mean weight =

Thus Sum of the weight of 6 boys =

Sum of the weight of 5 boys = (51 + 45 + 49 + 46 + 44) kg = 235 kg

Weight of the sixth boy = (Sum of Weight of 6 boys) – (Sum of the weight of 5 boys) = (288 – 235) = 53 kg

**Q.13: The mean of the marks scored by 50students was found to be 39.Later on, it was discovered that a score of 43 was misread as 23.Find the correct mean.**

**Sol:**

Calculated mean marks of 50 students = 39

Calculated sum of these marks = (39 \times 50) =1950

Corrected sum of these marks = [1950-(wrong number)+(correct number) ]

= (1950-23+43) = 1970

Therefore, the correct mean =

**Q.14: The mean of 100 items was found to be 64. Later on it was discovered that two items were misread as 26 and 9 instead of 36 and 90 respectively. Find the correct mean.**

**Sol:**

Calculated mean of 100 items = 64

Sum of 100 items, as calculated =

Correct sum of these items = [6400 – (wrong items) + (correct items)] = [6400 – (26 + 9) + (36 + 90)] = [6400 – 35 + 126]

Therefore, the Correct mean =

**Q.15: The mean of six numbers is 23. If one of the numbers is excluded, the mean of the remaining numbers is 20. Find the excluded number.**

**Sol:**

Mean of 6 numbers = 23

Sum of 6 numbers =

Again, mean of 5 numbers = 20

Sum of 5 number =

The excluded number = (Sum of 6 number) – (sum of 5 numbers) = (138-100) = 38

Therefore, the excluded number = 38