ML Aggarwal Solution Class 10 Chapter 21 Measures of Central Tendency Exercise 21.1

 Exercise 21.1

Question 1

(a) Calculate the arithmetic mean of 5.7, 6.6, 7.2, 9.3, 6.2.

(b) The weights (in kg) of 8 new born babies are 3, 3.2, 3.4, 3.5, 4, 3.6, 4.1, 3.2. Find the mean weight of the babies.

Solution:

(a) Sum of 5 observations = 5.7 + 6.6 + 7.2 + 9.3 + 6.2 = 35.0

$\therefore$ Mean $=\frac{35.0}{5}=7$

(b) Weights of 8 babies (in kg) are 3, 3.2, 3.4, 3.5, 4, 3.6, 4.1, 3.2

∴ Total weights of 8 babies

= 3 + 3.2 + 3.4 + 3.5 + 4 + 3.6 + 4.1 + 3.2 = 28.0 kg

Mean weight $=\frac{\sum x_{i}}{n}$

$=\frac{28.0}{8}$ (Here n=8)

= 3.5 kg


(ii) By increasing 4 marks in each student then increased marks
=15×4=60

New sum=207+60=267

$\therefore \quad$ New mean $=\frac{267}{15}=17 \cdot 8$


(iii) By deducting 2 marks from each students.then total deduction $=15 \times 2=30$

New sum $=207-30=177$

New mean $=\frac{New mean $=\frac{177}{15}=11 \cdot 8$177}{15}=11 \cdot 8$


(iv) The marks being doubled of each student then the new sum

$=207 \times 2=414$

$\therefore$ New mean $=\frac{414}{15}=27 \cdot 6$


Question 2

The marks obtained by 15 students in a class test are 12, 14, 07, 09, 23, 11, 08, 13, 11, 19, 16, 24, 17, 03, 20 find

(i) the mean of their marks.

(ii) the mean of their marks when the marks of each student are increased by 4.

(iii) the mean of their marks when 2 marks are deducted from the marks of each student.

(iv) the mean of their marks when the marks of each student are doubled.

Sol :

Sum of marks of 15 students.

= 12 + 14 + 07 + 09 + 23 + 11 + 08 + 13 + 11 + 19 + 16 + 24 + 17 + 03 + 20

= 207

(i) Mean $=\frac{207}{15}$

= 13.8


Question 3

(a) The mean of the numbers 6, y, 7, x, 14 is 8. Express y in terms of x.

(b) The mean of 9 variates is 11. If eight of them are 7, 12, 9, 14, 21, 3, 8 and 15 find the 9th variate.

Sol :

(a) Sum of numbers = 6 + y + 7 + x + 14

= 27 + x + y …(i)

But mean of 5 numbers = 8

∴ Sum = 8 × 5 = 40 …(ii)

From (i) and (ii)

27 + x + y = 40

⇒ x + y = 40 – 27 = 13

∴ y = 13 – x

(b) Mean of 9 variates = 11

∴ Total sum =11 × 9 = 99

But sum of 8 of these variates

= 7 + 12 + 9 + 14 + 21 + 3 + 8 + 15 = 89

∴ 9th variate = 99 – 89 = 10


Question 4

(a) The mean age of 33 students of a class is 13 years. If one girl leaves the class, the mean becomes $12\frac{15}{16}$  years. What is the age of the girl ?

(b) In a class test, the mean of marks scored by a class of 40 students was calculated as 18.2. Later on, it was detected that marks of one student was wrongly copied as 21 instead of 29. Find the correct mean.

Sol :

(a) Mean age of 33 students = 13 years

Total age = 13 × 33 = 429 years

After leaving one girl, the mean of 32

students $=12 \frac{15}{16}=\frac{207}{16}$ years.

$\therefore$ Total age of 32 students $=\frac{207}{16} \times 32$

=414 years

Hence , the age of the girls=429-414

=15 years


(b) Mean of marks obtained by 40 students

=18.2

$\therefore$ Total marks obtained by than $=18.2 \times 40$

=728

Difference of marks copited wrongly =29-21

=8

$\therefore$ Actual total marks $=728+8=736$

New mean $=\frac{736}{40}=18.4$


Question 5

Find the mean of 25 given numbers when the mean of 10 of them is 13 and the mean of the remaining numbers is 18.

Sol :

Mean of 10 numbers = 13

Sum = 13 × 10 = 130

and mean of remaining 15 numbers = 18

Sum = 18 × 15 = 270

Total sum of 25 numbers = 130 + 270 = 400

Mean of 25 numbers $=\frac{400}{25}=16$


Question 6

Find the mean of the following distribution:

Number 5 10 15 20 25 30 35
Frequency 1 2 5 6 3 2 1
Sol :
x f fx
5 1 5
10 2 20
15 5 75
20 6 120
25 3 75
30 2 60
35 1 35
Total 20 390

$\therefore  $ Mean $=\frac{\sum f x}{\sum f}=\frac{390}{20}=19 \cdot 5$


Question 7

The contents of 100 match boxes were checked to determine the number of matches they contained

No. of matches 35 36 37 38 39 40 41
No. of boxes 6 10 18 25 21 12 8
(i) Calculate, correct to one decimal place, the mean number of matches per box.
(ii) Determine how many extra matches would have to be added to the total contents of the 100 boxes to; bring the mean upto exactly 39 matches. (1997)
Sol :
No. of match
(x)		 No. of boxes
( f )		 fx
35 6 210
36 10 360
37 18 666
38 25 950
39 21 819
40 12 480
41 8 328
Total 100 3813

(i) Mean $=\frac{\sum f x}{\sum f}=\frac{3813}{100}=38 \cdot 13=38 \cdot 1$

(ii) New Mean=39

Total sum=39×100=3900
New matches to be added
=3900-3813
=87


Question 8

Calculate the mean for the following distribution :

Pocket money (in Rs) 60 70 80 90 100 110 120
No. of students 2 6 13 22 24 10 3

Sol :
Pocket money
(x)		 No. of students
( f )		 fx
60 2 120
70 6 420
80 13 1040
90 22 1980
100 24 2400
110 10 1100
120 3 360
Total 80 7420

Mean $=\frac{\sum f x}{\sum f}=\frac{7420}{80}=\operatorname{Rs} .92 \cdot 75$

Question 9

Six coins were tossed 1000 times, and at each toss the number of heads were counted and the results were recorded as under :

No. of heads 6 5 4 3 2 1 0
No. of tosses 20 25 160 283 338 140 34

Sol :
Pocket money
(x)		 No. of students
( f )		 fx
6 20 120
5 25 125
4 160 640
3 283 849
2 338 676
1 140 140
0 34
 0
Total 1000 2550

Mean $=\frac{\sum f x}{\sum f}=\frac{2550}{1000}=2 \cdot 55$

Question 10

Find the mean for the following distribution
Numbers 60 61 62 63 64 65 66
Cumulative frequency 8 18 33 40 49 55 60
Sol :
Pocket money
(x)		 Cumulative frequency
( c.f )		 frequency
(f)		 fx
60 8 8 480
61 18 10 610
62 33 15 930
63 40 7 441
64 49 9 576
65 55 6 390
66 60
 5 330
Total 60 3757

Mean $=\frac{\sum f x}{\sum f}=\frac{3757}{60}=62 \cdot 616=62 \cdot 62$


Question 11

Category A B C D E F G
Wages (in rupees) per day 50 60 70 80 90 100 110
No. of workers 2 4 8 12 10 6 8
(i) Calculate the mean wage correct to the nearest rupee (1995)
(ii) If the number of workers in each category is doubled, what would be the new mean wage ?
Sol :
Category Wages (in Rs)
x		 No. of workers
(f)		 fx
A 50 2 100
B 60 4 240
C 70 8 560
D 80 12 960
E 90 10 900
F 100 6 600
G 110
 8 880
Total 50 4240

(i) Mean$=\frac{\sum f x}{\sum f}=\frac{4240}{50}=84 \cdot 80=85$

(ii) If the workers are doubled, then
Total number of workers
$=50 \times 2=100$

Total wage will also be doubled
$\therefore$ Total wages $=4240 \times 2=8480$

$\therefore $ New mean $=\frac{8480}{100}=84 \cdot 80=85$

Question 12

If the mean of the following distribution is 7.5, find the missing frequency ” f “.
Variate 5 6 7 8 9 10 11 12
Frequency 20 17 f 10 8 6 7 6
Sol :
Sol :
Variate
(x)		 Frequency
(f)		 fx
5 20 100
6 17 102
7 f 7f
8 10 80
9 8 72
10 6 60
11 7 77
12 6 72
Total ∑f=74+f 563+7f

$\because M=\frac{\sum f x}{\sum f}$

$\therefore 7.5=\frac{563+7 f}{74+f} \Rightarrow 555+7.5 f=563+7 f$

$\Rightarrow 0.5 f=8 \Rightarrow f=\frac{8 \times 10}{5} \Rightarrow f=16$

∴Missing frequency (f)=16


Question 13

Find the value of the missing variate for the following distribution whose mean is 10

Variate (xi) 5 7 9 11  _  15 20
Frequency (fi) 4 4 4 7 3 2 1
Sol :
Variate
(x)		 Frequency
( f )		 fx
5 4 20
7 4 28
9 4 36
11 7 77
x 3 3x
15 2 30
20 1
 20
Total 25 211+3x

Mean $=\frac{\sum f x}{\sum f}=\frac{211+3 x}{25}$

But mean=10 (given)

$\therefore \frac{211+3 x}{25}=10$

$\Rightarrow \quad 211+3 x=250$

$\Rightarrow  3 x=250-211=39$

$\Rightarrow \dot{x}=\frac{39}{3}=13$

∴Missing variate=13


Question 14

Marks obtained by 40 students in a short assessment are given below, where a and b are two missing data.

Marks 5 6 7 8 9
No. of students 6 a 16 13 b

If the mean of the distribution is 7.2, find a and b.

Sol :
Marks
(x)		 No. of students
(f)		 fx
5 6 30
6 a 6a
7 16 112
8 13 104
9 b 9b
Total 35+a+b=40 246+6a+9b

35+a+b=40
$\Rightarrow a+b=40-35$
$\Rightarrow a+b=5 \Rightarrow b=5-a$..(i)

$\Rightarrow$ Also, mean $=\frac{\sum(f x)}{\sum f}$

$\Rightarrow 7.2=\frac{246+6 a+9 b}{40}$

$\Rightarrow 246+6 a+9 b=288$

$\Rightarrow 6 a+9 b=42$

$\Rightarrow 2 a+3 b=14$...(ii)

2a+3(5-a)=14

2a+15-3a=14

a=1

Solving (i) and (ii) , we get a=1, b=4


Question 15

Find the mean of the following distribution

Class Interval 0-10 10-20 20-30 30-40 40-50
Frequency 10 6 8 12 5

If the mean of the distribution is

Sol :

Class interval Class mark
(xi)		 Frequency
fi		 fixi
0-10 5 10 50
10-20 15 6 90
20-30 25 8 200
30-40 35 12 420
40-50 45 5 225
Total Σfi=41 Σfixi=985

Mean $=\frac{\sum f_{i} x_{i}}{\sum f_{i}}=\frac{985}{41}=24.02$ (approx)


Question 16

Calculate the mean of the following distribution:

Class Interval 0-10 10-20 20-30 30-40 40-50 50-60
Frequency 8 5 12 35 24 16

Sol :

Consider the following distribution :

Class interval Class mark
(xi)		 Frequency
fi		 fixi
0-10 8 5 40
10-20 5 15 75
20-30 12 25 300
30-40 35 35 1225
40-50 24 45 1080
40-50 24 45 1080
Total fi=100 fixi=3600 

Mean $=\frac{\Sigma f_{i} x_{i}}{\Sigma f_{i}}=\frac{3600}{100}=36$


Question 17

Calculate the mean of the following distribution using step deviation method:
Marks 0-10 10-20 20-30 30-40 40-50 50-60
No. of students  10 9 25 30 16 10

Sol :

Marks Mid values
(xi)		 No. of students
(fi) 		di=xi-A ti$=\frac{d_i}{c}$ fiti
0-10 5 10 -20 -2 -20
10-20 15 9 -10 -1 -9
20-30 25 25=A 0 0 0
30-40 35 30 10 1 30
40-50 45 16 20 2 32
50-60 55 10 30 3 30
Total Σfi=100 Σfiti=63
Let A=25 and C=10

Mean $=\mathrm{A}+\frac{\Sigma f_{t} t_{i}}{\Sigma f_{i}} \times c$
$=25+\frac{63}{100} \times 10$
=25+6.3=31.3

Question 18

Find the mean of the following frequency distribution:
Class interval 0-50 50-100 100-150 150-200 200-250 250-300
Frequency 4 8 16 13 6 3

Sol :

Class interval Frequency
(xi) 		Class Mark
 fx
0-50 4 25 100
50-100 8 75 600
100-150 16 125 2000
150-200 13 175 2275
200-250 6 225 1350
250-300 3 275 825
Total 50 7150         

Arithmetic mean $(\bar{x})$

$=\frac{\sum f x}{\sum f}=\frac{7150}{50}=143$


Question 19

The following table gives the daily wages of workers in a factory:
Wages in ₹ 45-50 50-55 55-60 60-65 65-70 70-75 75-80
No. of workers 5 8 30 25 14 12 6
Calculate their mean by short cut method.
Sol :
Wages in Rs No. of Workers
(f)		 Mid- mark
(x)		 d=(x-A) f×d
45-50 5 47.5 -15 -75
50-55 8 52.5 -10 -80
55-60 30 57.5 -5 -150
60-65 25 62.5=A 0 0
65-70 14 67.5 5 70
70-75 12 72.5 10 120
75-80 6 77.5 15 90
Total Σf=100 Σfd=-25

Mean $=\mathrm{A}+\frac{\sum f d}{\sum f}=62.5+\frac{-25}{100}=62.50-.25=62.25$


Question 20

Calculate the mean of the distribution given below using the short cut method.
Marks 11-20 21-30 31-40 41-50 51-60 61-70 71-80
No. of students 2 6 10 12 9 7 4
Sol :
Marks Frequency
(f)		 Mid-value
(x)		 di=xi-A
A=45.5		 ui$=\frac{x_i-A}{10}$ f×ui
11-20 2 15.5 -30 -3 -6
21-30 6 25.5 -20 -2 -12
31-40 10 35.5 -10 -1 -10
41-50 12 45.5 0 0 0
61-70 7 65.5 20 2 14
71-80 4 75.5 30 3 12
Total Σf=50 Σfui=7         
A=45.5

Mean $=\mathrm{A}+\frac{\sum f u_{i}}{\sum f} \times 10$
$=45.5+\frac{7}{50} \times 10$
=45.5+1.4=46.9

Question 21

A class teacher has the following absentee record of 40 students of a class for the whole term. Find the mean number of days a students was absent.
No. of days 0-6 6-10 10-14 14-20 20-28 28-38 38-40
No. of students 11 10 7 4 4 3 1
Sol :
No. of days Mid value
(x)		 No. of students
(f)		 f×x
0-6 3 11 33
6-10 8 10 80
10-14 12 7 84
14-20 17 4 68
20-28 24 4 96
28-38 33 3 99
38-4039 1 39
Total
40 499

Mean $=\frac{\sum f x}{\sum f}=\frac{499}{40}$

=12.475


Question 22

The mean of the following distribution is 23.4. Find the value of p.

Class interval0-88-1616-2424-3232-4040-48
Frequency5310P42

Sol :

Class intervalFrequency
(xi)		Class Mark
fx
0-85420
8-1631236
16-241020200
24-32p2828p
32-40436144
40-4824488
Total24+p488+28p

Mean $=\frac{\sum f x}{\sum f}$

$\Rightarrow 23.4-\frac{488+28 p}{24+p}$
⇒23.4(24+p)
⇒488+28p
⇒561.6+23.4p
⇒488+28p
⇒28p-23.4p
⇒561.6-488
⇒4.6p
⇒73.6
⇒$p=\frac{73.6}{4.6}=\frac{736}{46}=16$

Hence p=16

Question 23

The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is Rs. 18. Find the value of f
Daily pocket allowance (in Rs)11-1313-1515-1717-1919-2121-2323-25
No. of children36913f54
Sol :
Mean = Rs. 18
Daily PocketAllowance
(in Rs)		No. of students
(f)		class mark
f×x
11-1331236
13-1561484
15-17916144
17-191318234
19-21f2020f
21-23522110
23-2542496
Total
704+20f

Mean $=\frac{\sum f x}{\sum f}$
$ \Rightarrow 18=\frac{704+20 f}{40+f}$
$\Rightarrow 18(40+f)=704+20 f$
$\Rightarrow 720+18 f=704+20 f $
$\Rightarrow 720-704-20 f-18 f $
$\Rightarrow 2 f=16$
$f=\frac{16}{2}=8$

Hence f=8 


Question 24

The mean of the following distribution is 50 and the sum of all the frequencies is 120. Find the values of p and q.

Class Interval0-2020-4040-6060-8080-100
Frequency17P32q19
Sol :
Mean = 50, Total number of frequency = 120
Class intervalClass mark
(xi)		Frequency
fi		fixi
0-201710170
20-40p3030p
40-6032501600
60-80q7070q
80-10019901710
Total68+p+q=3480+30p+70q   

Mean $=\frac{\sum f x}{\sum f}$ and 

68+p+q=120

⇒p+q=120-68=52
$\Rightarrow 50=\frac{3480+30 p+70 q}{120}$

6000=3480+30p+70q
30p+70q=6000-3480=2520
3p+7q=252...(i)
and p+q=52...(ii)

Multiplying (i) by 1 and (ii) by 7 and subtracting (i) from (ii)

$\begin{aligned}7p+7q=&364\\3p+7q=&252\\-\phantom{3p}-\phantom{7q}&-\phantom{252} \\ \hline 4p=&112\end{aligned}$


$p=\frac{-112}{-4}=28$ 

∴q=52-p=52-28=24


Hence p=28, q=24



Question 25

The mean of the following frequency distribution is 57.6 and the sum of all the frequencies is 50. Find the values of p and q.

Class interval0-2020-4040-6060-8080-100100-120
Frequency7p12q85

Sol :

Mean = 57.6

and sum of all frequencies = 50

Class Frequency
(f)		Class Mark
(x)		fx
0-2071070
20-40p3030p
40-601250600
60-80q7070q
80-100890720
100-1205110550
Total32+p+q=501940+30p+70p

32+p+q=50p+q

=50-32

∵p+q=18...(i)

Mean $=\frac{\sum f \times x}{\sum f}$

$\Rightarrow 57.6=\frac{1940+30 p+70 q}{50} $

$ \Rightarrow 2880=1940+30 p+70 q$

$\Rightarrow 30 p+70 q=2880-1940=940$

$\Rightarrow 3 p+7 q=94$...(ii)

Multiplying (i) by 7 and (ii) by 1 and subtracting (ii) from (i) 

$\begin{aligned}7p+7q=&126\\3p+7q=&94\\-\phantom{3p}-\phantom{7q}&-\phantom{252} \\ \hline 4p=&32\end{aligned}$

$\therefore p=\frac{32}{4}=8$

∴q=18-p=18-8=10

Hence p=8, q=10


Question 26

The following table gives the life time in days of 100 electricity tubes of a certain make :

Life time in daysNo. of tubes
Less than 508
Less than 10023
Less than 15055
Less than 20081
Less than 25093
Less than 300100

Find the mean life time of electricity tubes.

Sol :

Life time (in days)
(class interval)		(c.f)Frequency
(f)		Class marks
(x)		u$=\frac{x-A}{h}$f×u
0-508825-3-24
50-100231575-2-30
100-1505532125-1-32
150-200812617500
200-2509312225112
250-3001007275214
Total100-60                  
Let assumed mean (A)=175 and h=50
Mean $=A+h \times \frac{\sum f u}{\sum f}$

$=175+50 \times \frac{-60}{100}$

=175-30=145 days


Question 27

Using the information given in the adjoining histogram, calculate the mean correct to one decimal place.

Sol :

From the histogram given, we represent the information in the following table :

Figure to be added

Class intervalFrequency
(f)		Class mark
(x)		f x
20-3032575
30-40535175
40-501245540
50-60955495
60-70465260
Total331545

Mean $\frac{\sum f x}{\sum f}=\frac{1545}{33}$

=46.81 or 46.8

Comments

Post a Comment

Popular posts from this blog

ML Aggarwal Solution Class 9 Chapter 9 Logarithms MCQs

 MCQs correct Solution from the given four options (1 to 7): Question 1 If log√3 27 = x, then the value of x is (a) 3 (b) 4 (c) 6 (d) 9 Sol : ⇒$\log _{\sqrt{3}} 27=x$ ⇒$(\sqrt{3})^{x}=27$ ⇒$(3)^{\frac{1}{2} x x}=3^{3}$ ⇒$3^{\frac{x}{2}}=3^{3}$ ⇒$\frac{x}{2}=3$ ⇒x=6...(c) Question 2 If log 5 (0.04) = x, then the value of x is (a) 2 (b) 4 (c) -4 (d) -2 Sol : ⇒$\log _{5}(0.04)=x$ ⇒$5^{x}=0.04=\frac{4}{100}=\frac{1}{25}=5^{-2}$ ∴x=-2...(d) Question 3 If log 0.5  64 = x, then the value of x is (a) -4 (b) -6 (c) 4 (d) 6 Sol : ⇒$\log _{0.5} 64=x $ ⇒$0.5^{x}=64$ ⇒$=\left(\frac{1}{2}\right)^{x}=2^{6}$ ⇒$2^{-x}=2^{6}$ ∴-x=6⇒x=-6...(b) Question 4 If $\log _{10} \sqrt[3]{5} x=-3$, then the value of x is (a) $\frac{1}{5}$ (b) $-\frac{1}{5}$ (c) -1 (d) 5 Sol : $\log _{\sqrt[3]{5}} x=-3,(\sqrt[3]{5})^{-3}=x$ $x=\left(5^{\frac{1}{3}}\right)^{-3}=5^{\frac{1}{3}(-3)}=5^{-1}$ $x=\frac{1}{5}$ (b) Question 5 If log (3x + 1) = 2, then the value of x is (a) $\frac{1}{3}$ (b) 99 (c) 33 (d) $\frac{19}{3}$ Sol
Read more

ML Aggarwal Solution Class 9 Chapter 20 Statistics Exercise 20.2

  Exercise 20.2 Question 1 Sol : (i) discrete  (ii) continuous  (iii) discrete  (iv) continuous  (v) continuous  Question 2 Sol : Given data  13, 6, 10 ,5 ,11 ,14 ,2 ,8 ,15 ,16 ,9 ,13 ,17 ,11, 19 ,5 ,7 ,12 ,20 ,21 ,18 ,1 ,8 ,12 ,18 Classes 0-4 5-9 10-14 15-9 20-24 Frequency 2 7 8 6 2 Question 3 Sol : (i)  Classes 1-10 11-20 21-30 31-40 Frequency 8 7 6 6 (ii) Range = Maximum value - Minimum value  =40-2=38 (iii) Third class of frequency table $=\frac{21+30}{2}=\frac{51}{2}$ =25.5 Question 4 Sol : Variate : A particular value of a variable is called variate Class size :  The difference between the actual upper limit and the actual lower li
Read more

ML Aggarwal Solution Class 9 Chapter 3 Expansions Exercise 3.2

Exercise 3.2 Question 1 Sol : Given : x-y=8..(i) xy=5...(ii) Squaring both sides in equation (i) ∴(x-y) 2 =8 2 ⇒x 2 -2.xy+y 2 =64 ⇒x 2 -2(5)+y 2 =64 ⇒x 2 +y 2 =64+10 ⇒x 2 +y 2 =74 Question 2 Sol : ⇒Given : x+y=10...(i) ⇒xy=21....(ii) Squaring both sides in equation (i) ⇒(x+y) 2 =10 2 ⇒x 2 +2xy+y 2 =100 ⇒x 2 +2(21)+y 2 =100 ⇒x 2 +4 2 +y 2 =100 ⇒x 2 +y 2 =100-42 ⇒x 2 +y 2 =58 2(x 2 +y 2 )=2×58=116 Question 3 Sol : ⇒Given : 2a+3b=7..(i) ⇒ab=2....(ii) Squaring both sides in equal (i) ⇒(2a+3b) 2 =7 2 ⇒(2a) 2 +2.2a+3b+(3b) 2 =49 ⇒4a 2 +12ab+9b 2 =49 ⇒4a 2 +12(2)+9b 2 =49 ⇒4a 2 +24+9b 2 =49 ⇒4a 2 +9b 2 =49-24 ⇒4a 2 +9b 2 =25 Question 4 Sol : ⇒Given : 3x-4y=16...(i) ⇒xy=4..(ii) Squaring on both sides in equation (i) ⇒(3x-4y) 2 =16 2 ⇒(3x) 2 -2.3x.4y+(4y) 2 =256 ⇒9x 2 -24xy+16y 2 =256 ⇒9x 2 -24(4)+16y 2 =256 ⇒9x 2 -96+16y 2 =256 ⇒9x 2 +16y 2 =256+96 ⇒9x 2 +16y 2 =352 Question 5 Sol : ⇒Given : x+y=8...(i) ⇒x-y=2...(ii) Since we know that ⇒2(x) 2 +2(y) 2 =(x+y) 2 +(x-y) 2 From (i) ⇒S
Read more