# 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$\frac{Sum \; of \; numbers}{Total \; numbers}$

=1+2+3+4+5+6+7+88=368=4.5$\frac{1+2+3+4+5+6+7+8}{8}=\frac{36}{8}=4.5$

= Therefore,  Mean =4.5

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

Therefore, mean =SumofnumbersTotalnumbers$\frac{Sum \; of \; numbers}{Total \; numbers}$

=1+3+5+7+9+11+13+15+17+1910=10010=10$\frac{1+3+5+7+9+11+13+15+17+19}{10}=\frac{100}{10}=10$

Therefore, Mean =10

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

Therefore, mean =SumofnumbersTotalnumbers$\frac{Sum \; of \; numbers}{Total \; numbers}$

=2+3+5+7+115=285=5.6$\frac{2+3+5+7+11}{5}=\frac{28}{5}=5.6$

Therefore, Mean =5.6

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

Therefore, mean =SumofnumbersTotalnumbers$\frac{Sum \; of \; numbers}{Total \; numbers}$

= 2+4+6+8+10+126=426=7$\frac{2+4+6+8+10+12}{6}=\frac{42}{6}=7$

Therefore, Mean = 7

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

Therefore, mean = SumofnumbersTotalnumbers$\frac{Sum \; of \; numbers}{Total \; numbers}$

=5+10+15+20+25+30+357=1407=20$\frac{5+10+15+20+25+30+35}{7}=\frac{140}{7}=20$

Therefore, Mean = 20

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

Therefore, mean =SumofnumbersTotalnumbers$\frac{Sum \; of \; numbers}{Total \; numbers}$

= 1+2+4+5+10+206=426=7$\frac{1+2+4+5+10+20}{6}=\frac{42}{6}=7$

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$\frac{Sum \; of \; numbers}{Total \; numbers}$

= 2+4+3+4+2+0+3+5+1+610=3010=3$\frac{2+4+3+4+2+0+3+5+1+6}{10}=\frac{30}{10}=3$

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$\frac{Sum \; of \; numbers}{Total \; numbers}$

=105+216+322+167+273+405+3467=18347=262$\frac{105+216+322+167+273+405+346}{7}=\frac{1834}{7}=262$

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$\frac{Sum \; of \; Temperatures}{No. \; of \; days}$

=35.5+30.8+27.3+32.1+23.8+29.96=179.46=29.9$\frac{35.5+30.8+27.3+32.1+23.8+29.9}{6}=\frac{179.4}{6}=29.9$

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$\frac{Sum \; of \; marks}{Total \; number \; of \; students}$

=64+36+47+23+0+19+81+93+72+35+3+112=47412=39.5$\frac{64+36+47+23+0+19+81+93+72+35+3+1}{12}=\frac{474}{12}=39.5$

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$\frac{7+9+11+13+x+21}{6}$

Undefined control sequence \because$\left ( \because Mean=\frac{Sum\; of \; observation}{Number \; of \; observation} \right )$

And, mean = 13 (given)

Therefore, 13=61+x6$13 =\frac{61+x}{6}$

61+x=78$\Rightarrow 61+x =78$
x=7861=17$\Rightarrow x =78-61 =17$

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$x_{1},x_{2},…….x_{24}$

Mean=x1+x2+.+x2424$\Rightarrow Mean =\frac{x_{1}+x_{2}+…….+x_{24}}{24}$

Therefore,x1+x2+.+x2424=35$\frac{x_{1}+x_{2}+…….+x_{24}}{24}=35$

x1+x2+.+x24=840$\Rightarrow x_{1}+x_{2}+…….+x_{24}=840$ . . . . . . . . (i)

The new number are (x1+3),(x2+3),.+(x24+3)$(x_{1}+3),(x_{2}+3),…….+(x_{24}+3)$.

Therefore, Mean of the new numbers

(x1+3)+(x2+3)+.+(x24+3)24=840+7224$\frac{(x_{1}+3)+(x_{2}+3)+…….+(x_{24}+3)}{24}=\frac{840+72}{24}$ (using (i) )

=91224=38$=\frac{912}{24} =38$

Therefore, Thenewmean=38$The new mean =38$

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$x_{1},x_{2}, . . . . . . . x_{20}$

Then, the mean of these numbers =

Therefore, x1+x2+.........+x2020=43$\frac{x_{1}+x_{2}+ . . . . . . . . . +x_{20}}{20}=43$

x1+x2+........+x20=860$\Rightarrow x_{1}+x_{2}+ . . . . . . . . +x_{20}=860$ . . . . . . . (i)

The new numbers are:

(x16)+(x26)+........+(x206)20=86012020$\Rightarrow \frac{(x_{1}-6)+(x_{2}-6)+ . . . . . . . . +(x_{20}-6)}{20}=\frac{860-120}{20}$ . . . . . . ( using (i) )

Therefore, 74020=37$\frac{740}{20}=37$

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$x_{1},x_{2},……….,x_{15}$

Then, the mean of these numbers = x1+x2+.+x1515=27$\frac{x_{1}+x_{2}+……….+x_{15}}{15}=27$

x1+x2+.+x15=405$\Rightarrow x_{1}+x_{2}+……….+x_{15}=405$

The new numbers are (x1×4),(x2×4),.(x15×4)$(x_{1}\times 4),(x_{2}\times 4),…….(x_{15}\times 4)$

Therefore, mean of the new numbers = (x1×4)+(x2×4)+.+(x15×4)15=405×415=162015=108$\frac{(x_{1}\times 4)+(x_{2}\times 4)+…….+(x_{15}\times 4)}{15}=\frac{405\times 4}{15}=\frac{1620}{15}=108$

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$x_{1},x_{2},……….,x_{12}$

Then, the mean of these numbers = 40 x1+x2+.+x1212=40$\frac{x_{1}+x_{2}+……….+x_{12}}{12}=40$

x1+x2+.+x12=480$\Rightarrow x_{1}+x_{2}+……….+x_{12}=480$

The new numbers are (x1÷8),(x2÷8),.(x12÷8)$(x_{1}\div 8),(x_{2}\div 8),…….(x_{12}\div 8)$

Therefore, mean of the new numbers = (x1÷8)+(x2÷8)+.+(x12÷8)12=480÷812=6012=5$\frac{(x_{1}\div 8)+(x_{2}\div 8)+…….+(x_{12}\div 8)}{12}=\frac{480 \div 8}{12}=\frac{60}{12}=5$

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$x_{1},x_{2},……….,x_{20}$

Let the mean be X¯$\bar{X}$

Therefore,X¯=x1+x2+.+x2020=18$\bar{X}=\frac{x_{1}+x_{2}+……….+x_{20}}{20}=18$

x1+x2+.+x20=18×20=360$\Rightarrow x_{1}+x_{2}+……….+x_{20}=18\times 20=360$ . . . . (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)$(x_{1}+3),(x_{2}+3),……….,(x_{10}+3)$

Let, X¯$\bar {{X}’}$ be the mean of new numbers

=(x1+x2+.+x20)+3×1020$=\frac{(x_{1}+x_{2}+……….+x_{20})+3\times 10}{20}$

From (i), we know that =x1+x2+.+x20=360$=x_{1}+x_{2}+……….+x_{20}=360$

Therefore, mean of the new set of 20 numbers = 360+3020=39020=19.5$\frac{360+30}{20}=\frac{390}{20}=19.5$

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$\frac{Sum \; of \; the \;weight \; of \; six \; boys}{6}=48$

Thus Sum of the weight of 6 boys = (48×6)kg$(48\times 6)kg$ = 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$\frac{1970}{50} =39.4$

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$(100\times 64)=6400$

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$\frac{6491}{100}=64.91$

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$23\times 6=138$

Again, mean of 5 numbers = 20

Sum of 5 number = 20×5=100$20 \times 5=100$

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

Therefore, the excluded number = 38

#### Practise This Question

When alpha particles are sent through a thin metal foil, only one out of ten thousand rebounded. This observation led to the conclusion that: