RS Aggarwal Solutions Class 9 Ex 14C

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 =SumofnumbersTotalnumbers

=1+2+3+4+5+6+7+88=368=4.5

= Therefore,  Mean =4.5

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

Therefore, mean =SumofnumbersTotalnumbers

=1+3+5+7+9+11+13+15+17+1910=10010=10

Therefore, Mean =10

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

Therefore, mean =SumofnumbersTotalnumbers

=2+3+5+7+115=285=5.6

Therefore, Mean =5.6

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

Therefore, mean =SumofnumbersTotalnumbers

= 2+4+6+8+10+126=426=7

Therefore, Mean = 7

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

Therefore, mean = SumofnumbersTotalnumbers

=5+10+15+20+25+30+357=1407=20

Therefore, Mean = 20

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

Therefore, mean =SumofnumbersTotalnumbers

= 1+2+4+5+10+206=426=7

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 = SumofnumbersTotalnumbers

= 2+4+3+4+2+0+3+5+1+610=3010=3

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 =SumofnumbersTotalnumbers

=105+216+322+167+273+405+3467=18347=262

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 = SumofTemperaturesNo.ofdays

=35.5+30.8+27.3+32.1+23.8+29.96=179.46=29.9

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 = SumofmarksTotalnumberofstudents

=64+36+47+23+0+19+81+93+72+35+3+112=47412=39.5

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 = 7+9+11+13+x+216

Undefined control sequence \because

And, mean = 13 (given)

Therefore, 13=61+x6

61+x=78
x=7861=17

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 x1,x2,.x24

Mean=x1+x2+.+x2424

Therefore,x1+x2+.+x2424=35

x1+x2+.+x24=840 . . . . . . . . (i)

The new number are (x1+3),(x2+3),.+(x24+3).

Therefore, Mean of the new numbers

(x1+3)+(x2+3)+.+(x24+3)24=840+7224 (using (i) )

=91224=38

Therefore, Thenewmean=38

 

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 x1,x2,.......x20

Then, the mean of these numbers =

Therefore, x1+x2+.........+x2020=43

x1+x2+........+x20=860 . . . . . . . (i)

The new numbers are:

(x16)+(x26)+........+(x206)20=86012020 . . . . . . ( using (i) )

Therefore, 74020=37

 

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 x1,x2,.,x15

Then, the mean of these numbers = x1+x2+.+x1515=27

x1+x2+.+x15=405

The new numbers are (x1×4),(x2×4),.(x15×4)

Therefore, mean of the new numbers = (x1×4)+(x2×4)+.+(x15×4)15=405×415=162015=108

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 x1,x2,.,x12

Then, the mean of these numbers = 40 x1+x2+.+x1212=40

x1+x2+.+x12=480

The new numbers are (x1÷8),(x2÷8),.(x12÷8)

Therefore, mean of the new numbers = (x1÷8)+(x2÷8)+.+(x12÷8)12=480÷812=6012=5

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 x1,x2,.,x20

Let the mean be X¯

Therefore,X¯=x1+x2+.+x2020=18

x1+x2+.+x20=18×20=360 . . . . (i)

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

Therefore, the first new ten numbers are

(x1+3),(x2+3),.,(x10+3)

Let, X¯ be the mean of new numbers

=(x1+x2+.+x20)+3×1020

From (i), we know that =x1+x2+.+x20=360

Therefore, mean of the new set of 20 numbers = 360+3020=39020=19.5

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 = Sumoftheweightofsixboys6=48

Thus Sum of the weight of 6 boys = (48×6)kg = 288 kg

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 = 197050=39.4

 

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 = (100×64)=6400

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

Therefore, the Correct mean = 6491100=64.91

 

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 = 23×6=138

Again, mean of 5 numbers = 20

Sum of 5 number = 20×5=100

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

Therefore, the excluded number = 38


Practise This Question

What is the basic unit of classification?