S Chand Class 10 CHAPTER 18 Arithmetic Mean, Median, Mode and Quartiles Exercise 18 A

 Exercise 18 A 

Question 1

Find the number half way between $0.2$ and $0.02$,

Ans: Halfway number  (average)$\frac{0.2+0.02}{2}=\frac{0.22}{2}=0.11$

Question 2

Find the mean of the following sets of number

(i) $4,5,7,8$ (ii) $3,5,0,2,8$ (iii) $2.5,2.4,3.5,2.8,2.9,3.5$ and 3.6 (iv) $-6,-2,-1,0,1,2,5,9$

(v) first five prime number  (vi) First eight even natural number (ix) x, x+1, x+2,x+3,x+4,x+5 and x+6

 

Sol: (i) here n= 4

$\therefore$ Mean $=\frac{4+5+7+8}{4}=\frac{24}{4}=6$

(ii) $n=5$

$\therefore \text { Mean }=\frac{3+5+0+2 \pi 8}{5}=\frac{18}{5}=3.6$

(iii) $n=7$

$\therefore$ Mean $=\frac{2.5+2.4+3.5+2.8+2.9+3.3+3.6}{7}$

$=\frac{21.0}{7}=3.0$

(iv) n=8

$\begin{aligned} \therefore \text { Mean } &=\frac{-6+(-2)+(-1)+0+1+2+5+9}{8} \\ &=\frac{-6-2-1+0+1+2+5+9}{8}=\frac{8}{8}=1 \end{aligned}$

(v) First 5 prime number are 2,3,,4,711

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

(vi) First 8 even natural number are 2,3,6,8,10,12,14 and 16

Mean = $\frac{2+4+6+8+10+12+14+16}{8}=\frac{72}{8}=9$

(vii) all factors of 30 are 1,2,3,5,6,10,15,30

Here n- 8

Mean = $\frac{1+2+3+5+6+10+15+30}{8}=\frac{72}{8}=9$

(viii) First 5 multiple of 8[ are 8,16,24,32,40 here n = 5

$\therefore$ Mean $=\frac{8+16+24+32+40}{5}=\frac{120}{5}=24$

(ix) Here $n=7$

$\begin{aligned}\therefore \text { Mean } &=\frac{x+x+1+x+2+x+3+x+4+x+5+x+6}{7} \\&=\frac{7 x+21}{7}=\frac{7(x+3)}{7}=x+3\end{aligned}$

Question 3

The mean of the number 6,y,4 , x ,14 is 8 express y in terms of x 

Sol: Mean $=\frac{6+y+7+x+14}{5}$

$8=\frac{x+y+27}{5}$

$40=x+y+27$

$y+x=40-27 \Rightarrow y+x=13$

$\therefore y=13-x$

Question 4

Nisha secured 73,86 ,78 and 75 marks in four tests what is the least number of points she should secure in her test if she has to have an average of 80.

Sol: Number of total tests = 5 

Average of 5 tests -80 

Suppose that the next test secure is x than 

$\begin{aligned} \text { Mean } &=\frac{73+86+78+75+x}{5} \\ \Rightarrow 80 &=\frac{312+x}{5} \\ \Rightarrow 400 &=312+x \\ \therefore x &=400-312=88 \end{aligned}$

Question 5

A class of 10 students was given a test in math . The marks out of 50 secured by the students were as follows  $31,36,27,38,45,39,32,29,41,38$. find the mean Score

Sol: Here n= 10 and scores are $31,36,27,38,45,39,32,29,41,38$

$\begin{aligned} \therefore \text { Mean } &=\frac{31+36+27+38+45+39+32+29+41+38}{10} \\ &=\frac{356}{10}=35.6 \end{aligned}$

Question 6

Find the mean of the following frequency distribution: 

$\begin{aligned}&\text { (a) }\\&\begin{array}{|l|c|c|c|c|c|}\hline \text { Weight } & 30 & 31 & 32 & 33 & 34 \\\hline \text { Noogslucenk } & 8 & 10 & 15 & 8 & 9 \\\hline\end{array}\end{aligned}$

(b) 

$\begin{array}{|l|c|c|c|c|c|}\hline \text { Marks } & 20 & 25 & 30 & 35 & 40 \\\hline \text { Student } & 5 & 10 & 12 & 8 & 5 \\\hline\end{array}$

(c)

 $\begin{array}{|c|c|c|c|c|}\hline x & 2 & 5 & 7 & 8 \\\hline y & 2 & 4 & 6 & 3 \\\hline\end{array}$

(d)

 $\begin{array}{|c|c|c|c|c|c|c|}\hline x & 0.1 & 0.2 & 0.3 & 0.4 & 0.5 & 0.6 \\\hline y & 30 & 60 & 20 & 40 & 10 & 50 \\\hline\end{array}$

Sol: (a)

$\begin{array}{|c|c|c|}\hline \text { Weight } & \text { No of students } & f \times x \\\hline 30 & 8 & 240 \\31 & 10 & 310 \\32 & 15 & 480 \\33 & 8 & 264 \\34 & 9 & 306 \\\hline \text { Total } & 50 & 1600 \\\hline\end{array}$

$\begin{aligned} \therefore \text { Mean } &=\frac{\sum f x}{\sum f} \\ &=\frac{1600}{50} \\ &=32 \mathrm{~kg} \end{aligned}$

(b)

$\begin{array}{|c|c|c|}\hline \text { Marks }(x) & \text { Students }(t) & f \times x \\\hline 20 & 5 & 100 \\25 & 10 & 250 \\30 & 12 & 360 \\35 & 8 & 280 \\40 & 5 & 200 \\\hline \text { total} & 40 & 1190 \\\hline\end{array}$

$\begin{aligned} \text { Mean } &=\frac{\varepsilon f x}{\varepsilon f} \\ &=\frac{1190}{40} \\ &=\frac{119}{4}=29.75 \end{aligned}$

(c) 

 $\begin{array}{|c|c|c|}\hline x & f & f \times x \\\hline 2 & 2 & 4 \\5 & 4 & 20 \\7 & 6 & 42 \\8 & 3 & 24 \\\hline \text { total } & 15 & 90 \\\hline\end{array}$

$\begin{aligned} \therefore \text { Mean } &=\frac{\varepsilon f x}{\varepsilon f} \\ &=\frac{90}{15} \\ &=6 \end{aligned}$

(d)

$\begin{array}{|c|c|c|}\hline x & f & f \times x \\\hline 0.1 & 30 & 3 \\0.2 & 60 & 12 \\0.3 & 20 & 6 \\04 & 40 & 16 \\0.5 & 10 & 5 \\0.6 & 50 & 30 \\\hline \text { Tolat } & 210 & 72 \\\hline\end{array}$

$\begin{aligned}\therefore \text { Mean } &=\frac{\varepsilon f x}{\varepsilon f} \\&=\frac{72}{210}=\frac{12}{35} \\&=0.34 .\end{aligned}$

Question 7

Fill in the blanks 

While calculating the mean of a the grouped data. We make the assumption that frequency is any class is entered at its.......

Sol: Mean of grouped data in any class is called its class mark 

Question 8

The frequency distribution of marks obtained by 40 students of a class is as under calculate the arithmetic mean 

$\begin{array}{|c|c|c|c|c|c|c|}\hline \text { Marks } & 0-8 & 8-16 & 16-24 & 24-32 & 32-40 & 40-48 \\\hline \text { Students } & 5 & 3 & 10 & 16 & 4 & 2 \\\hline\end{array}$

Sol:  

$\begin{array}{|c|c|c|c|}\hline \text { marks } & 8 & x & f x \\\hline 0-8 & 5 & 4 & 20 \\8-16 & 3 & 12 & 36 \\16-24 & 10 & 20 & 200 \\24-32 & 16 & 28 & 448 \\32-40 & 4 & 36 & 144 \\40-48 & 2 & 42 & 88 \\\hline \text { total } & 40 & & 936 \\\hline\end{array}$

$\begin{aligned} \text { Meam } &=\frac{\varepsilon f x}{\varepsilon t} \\ &=\frac{936}{40} \\ &=23.4 \text { Marks } \end{aligned}$

Question 9

Fill the mean of the following data: 

(a)

$\begin{array}{|c|c|c|c|c|c|}\hline \text { Marks } & 10-14 & 15-19 & 20-24 & 25-29 & 30-34 \\\hline \text { Student } & 4 & 6 & 12 & 5 & 3 \\\hline\end{array}$

(b) 

$\begin{array}{|l|c|c|c|c|c|}\hline \text { Class } & 0-10 & 11-20 & 21-30 & 31-40 & 41-50 \\\hline \text { frequency } & 3 & 4 & 2 & 5 & 6 \\\hline\end{array}$

Sol: (a) 

$\begin{array}{|c|c|c|c|}\hline \text { Marks } & \text { Student }(\text { f) } & x & f \times x \\\hline 10-14 & 4 & 12 & 48 \\15-19 & 6 & 17 & 102 \\20-24 & 12 & 22 & 264 \\25-29 & 5 & 27 & 135 \\30-34 & 3 & 32 & 96 \\\hline \text { Tolel } & 30 & & 645 \\\hline\end{array}$

Mean$=\frac{\varepsilon f x}{\varepsilon f}$= $\frac{578.5}{20}$

$=28.925$

Question 10

In a class of 60 boys the marks obtained in a monthly test were as under : 

$\begin{array}{|c|c|c|c|c|c|}\hline \text { Marks } & 10-20 & 20-30 & 30-40 & 40-50 & 50-60 \\\hline \text { Students } & 10 & 25 & 12 & 00 & 05 \\\hline\end{array}$

Find the mean marks of class 

Sol: $\begin{array}{|l|l|l|l|}\hline \text { Marks } & \text { students (f) } & x & f \times x \\\hline 10-20 & 10 & 15 & 150 \\20-30 & 25 & 25 & 625 \\30-40 & 12 & 35 & 420 \\40-50 & 08 & 360 \\56-60 & 05 & 55 & 275 \\\hline & 60 & 1830 & 183 \\\hline\end{array}$

Mean $=\frac{\sum f x}{\sum f}$= $\frac{1830}{60}=\frac{183}{6}=30.5$ 

Question 11

(i) Direct method and (ii) short cut method 

$\begin{array}{|l|c|c|c|c|c|c|c|c|c|}\hline \text { Class } & 5-10 & 10-15 & 15-20 & 20-25 & 25-30 & 30-35 & 35-40 & 40-45 & 45-50 \\\hline \text { frequency } & 10 & 6 & 4 & 12 & 8 & 4 & 2 & 1 & 3 \\\hline\end{array}$

Sol: (image to be added)

(i) Direct method

$\text { Mean }=\frac{\sum \cdot d}{\sum f}=\frac{1100}{50}=22$

(ii) Short cut method 

Mean = A+ =\frac{\sum \cdot d}{\sum f}

$=27.5+\frac{-275}{50}$

$=27.5-5.5=22$

Question 12

The following table givens the classification of 100 cows of a dairy form . According to the amount of milk given by each in a dairy 

$\begin{array}{|c|c|c|c|c|c|c|c|c|c|}\hline \begin{array}{c}\text { Amount of milk in kg} \\\text {no. of cows} 1 \mathrm{~kg}\end{array} & 0-2 & 2-4 & 4-6 & 6-61 & 8-10 & 10-12 & 12-74 & 14-16 & 16-18 \\\hline \text { No.0fectss } & 4 & 14 & 17 & 20 & 10 & 13 & 12 & 10 & 10 \\\hline\end{array}$

Calculate the mean correct to first place of decimal 

Sol: 

$\begin{array}{|c|c|c|c|}\hline \text { Amount of milk} & \text { cows }(f) & x & f \times x \\\hline 0-2 & 4 & 1 & 4 \\2-4 & 14 & 3 & 42 \\4-6 & 17 & 5 & 85 \\6-8 & 20 & 7 & 140 \\8-10 & 10 & 9 & 90 \\10-12 & 13 & 11 & 143 \\12-14 & 12 & 13 & 156 \\14-16 & 10 & 15 & 150 \\16-18 & 10 & 17 & 170 \\\hline \text { Jotel } & 150 & & 980 \\\hline\end{array}$

$\therefore$ Mean $=\frac{\varepsilon f x}{\varepsilon f}=\frac{980}{110}=8.9=8.9$

Question 13

The weight of 50 apples picked out at random from a are given below:-$82,118,80,110,104,84,106,107,76,82,109,107,115$ $93,187,95,123,125,111,92,86,70,126,70,130,129,139,119,115,128,100,186$

$84,99,113,204,111,141,136,123,90,115,98,110,78,90,107,81,131,75 .$

(i) What the range of weights 

(ii) Form a frequency distribution with class- interval $70-89,90-109$ and So on 

(iii) Use your frequency distribution to calculate the mean 

Sol: (i) Maximum weight = 204g

Minimum weight = 70g 

Range= 204-70=134gm

(ii)Frequency distribution:-

(IMAGE TO BE ADDED)

(iii) $\begin{aligned} \text { Mean } &=A+\frac{\varepsilon f d}{\varepsilon f} \\ &=139.5+\frac{(-1460)}{50} \\ &=139.5-\frac{146}{5} \\ &=139.5-29.2 \end{aligned}$

=110.3gm

Question 14

The following table given the weekly wages of workers in a factory 

$\begin{array}{|c|c|c|c|c|c|c|c|c|}\hline \text { Weekly Wages (Rs) } & 50-55 & 55-60 & 60-65 & 65-70 & 70-75 & 75-80 & 80-85 &85-90 \\\hline \text { No. of worker} & 5 & 20 & 10 & 10 & 9 & 6 & 12 & 8 \\\hline\end{array}$

Calculate (i) The mean(ii) The number of workers getting weekly wages below Rs 80 and (iii) the number of workers getting Rs 65 and more but tea than 85 as weekly wages 

Sol:

$\begin{array}{|l|c|c|c|c|}\hline 50-55 & 5 & 52.5 & -15 & -75 \\55-60 & 20 & 57.5 & -10 & -200 \\60-65 & 10 & 62.5 & -5 & -50 \\65-70 & 10 & 67.5=A & 0 & 0 \\70-75 & 9 & 72.5 & 5 . & 65 \\75-80 & 6 & 77.5 & 10 & 15 \\80-85 & 12 & 82.5 & 15 & 160 \\85-90 & 8 & 97.5 & 20 & 120 \\\hline \text { Total } & 80 & & & 120 \\\hline\end{array}$

(i) $\begin{aligned} \text { Mean }=A+\frac{\varepsilon f d}{\varepsilon t} &=67.5+\frac{120}{80} \\ &=67.5+1.5 \\ &=69.0 \end{aligned}$

(ii) Number of workers getting wages below 80 = 60

(iii) Number of workers getting more than Rs 65 but less then Rs 85 = $10+9+6+12=37$

Question 15

Find the mean of following data

(IMAGE TO BE ADDED)

Sol:  

$\begin{array}{|c|c|c|c|c|}\hline \text { Marks obtained } & \text { Class-mark No } & \text { No of workers } & f & f \times x \\\hline 0-10 & 5 & 7 & 7 & 35 \\10-20 & 15 & 19 & 12 & 100 \\20-30 & 25 & 32 & 13 & 325 \\30-40 & 35 & 42 & 10 & 350 \\40-50 & 45 & 50 & 8 & 360 \\\hline & & & 50 & 1250 \\\hline\end{array}$

Mean $=\frac{\varepsilon f x}{\varepsilon f}=\frac{1250}{50}=\frac{125}{5}=25$

Question 16

The following table into the form of an ordinary frequency distribution and determine the value of mean 

(IMAGE TO BE ADDED)

Sol: 

(IMAGE TO BE ADDED)

$\begin{aligned} \text { Mean } &=A+\frac{\varepsilon f f}{\varepsilon f} . \\ &=27.5+\frac{-800}{150} \\ &=27.5-\frac{80}{15}=27.5-\frac{16}{3} \\ &=27.5-5.3=22.2 \end{aligned}$

Question 17

Find the missing frequency in the following 

$\begin{array}{|l|c|c|c|c|c|c|c|c|c|}\hline \text { class } & 4-8 & 8-12 & 12-16 & 16-20 & 20-24 & 24-28 & 28-32 & 32-36 & 36-40 \\\hline \text { Frequency } & 11 & 13 & 16 & 14 & - & 9 & 17 & 6 & 4 \\\hline\end{array}$

Sol: Suppose that the missing frequency is p

$\begin{array}{|c|c|c|c|}\hline \text { Class } & \text { frequency } f & x & f \times x \\\hline 4-8 & 11 & 6 & 66 \\8-12 & 13 & 10 & 130 \\12-16 & 16 & 14 & 224 \\16-20 & 14 & 18 & 252 \\20+24 & p & 22 & 22 p \\24-28 & 9 & 26 & 234 \\22-32 & 17 & 30 & 510 \\32-36 & 6 & 34 & 204 \\36-40 & 4 & 38 & 152 \\\hline \text { Toral } & 90+p & \varepsilon f x=1772+22 p \\\hline\end{array}$

mean $=\frac{\varepsilon f x}{\varepsilon t}$

$19.92=\frac{1772+22 p}{90+p} \Rightarrow 1772+22 p=1792.8+19.92 p$

$\begin{aligned} \Rightarrow 22 p-19.92 p &=1792.8-1772 \\ 2.08 p &=20.8 \Rightarrow p=\frac{20.8}{2.08}=10 \end{aligned}$

Question 18

Calculate the A.M correct to one decimal place for the following frequency distribution of marks obtained in an arithmetic test

$\begin{array}{|l|c|c|c|c|c|}\hline \text { Marks } & 0-10 & 10-20 & 20-30 & 30-40 & 40-50 \\\hline \begin{array}{c}\text { No.of students) } \\\text { }\end{array} & 2 & 5 & 20 & 8 & 7 \\\hline\end{array}$

Sol: 

$\begin{array}{|c|c|c|c|}\hline \text { Marks } & \text { no of students } & x & f \times x \\\hline 0-10 & 2 & 5 & 10 \\10-20 & 5 & 15 & 75 \\20-30 & 20 & 25 & 500 \\30-40 & 8 & 35 & 280 \\40-50 & 7 & 45 & 315 \\\hline & \varepsilon f=42 &\varepsilon x=1180 \\\hline\end{array}$

Mean $=\frac{\varepsilon f x}{\varepsilon f}=\frac{1180}{42}=28.095=28.1$

Question 19

Following table gives marks scored by student in an examination 

4 $$\begin{array}{|c|c|c|c|c|c|c|c|c|}\hline \text { Marks } & 0-5 & 5-10 & 10-15 & 15-20 & 20-25 & 25-30 & 30-35 & 35-40 \\\hline \text { No.of } & 3 & T & 15 & 24 & 16 & 8 & 5 & 2 \\\hline \text { Student } & & & & & & \\\hline\end{array}$$ $

Ans: 

$\begin{array}{|l|c|l|c|c|c|}\hline \text { Marks } & \text { No of student(f)} & x & A & d=x-A & f \times d \\\hline 0-5 & 3 & 2.5 & & -15 & -45 \\5-10 & 7 & 7.5 & & -10 & -70 \\10-15 & 15 & 12.5 & & -5 & -75 \\15-20 & 24 & 17.5 & 17.5 & 0 & 0 \\20-25 & 16 & 22.5 & & 5 & 80 \\25-30 & 8 & 27.5 & & 10 & 80 \\30-35 & 5 & 32.5 & & 15 & 75 \\35-40 & 2 & 37.5 & & 20 & 40 \\\hline total & \sum f=80 & & & & \varepsilon f d=85 \\\hline\end{array}$

$\begin{aligned} \text { Mean } &=A+\frac{\varepsilon t d}{\varepsilon f} \\ &=17.5+\frac{85}{80} \\ &=17.5+1.06 \\ &=18.56 \\ &=18.56 \end{aligned}$