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1013/FY/Pre_Pap/Maths_Soln 6
Vidyalankar F.Y. Diploma : Sem. I Basic Mathematics
Prelim Question Paper Solution
Given:
4 3 9
3 2 7
1 4 x
= 0
4 [2x 28] 3 [3x 7] + 9 [12 2] = 0 8x 112 9x + 21 + 108 18 = 0 x 1 = 0 x = 1
x =1 2
3 4
, y =4 5
1 3
3x + y = 1 2 4 5
33 4 1 3
= 3 6 4 5
9 12 1 3
=
3 4 6 5
9 1 12 3
3x + y = 7 11
8 9
A 2 = A A = 3 9
1 9
=3.3 (9)( 1) (9)(9) ( 9)( 9)
( 1)(3) ( 9)( 1) ( 1)(9) ( 9)( 9)
= 0 0
6 72
2A = 0 0
6 72 = 0
hence A2 is a singular matrix
We have, A =
3 2
3 2
1 1
0 4
B =
3 2
1 1
3 2
4 2
L.H.S. = A + B
L.H.S. =
3 2 3 2
3 2 1 1
1 1 3 2
0 4 4 2
1. (a)
1. (b)
1. (c)
1. (d) Vidyala
nkar
Prelim Question Paper Solution
1013/FY/Pre_Pap/Maths_Soln 7
L.H.S. =
3 2
3 1 2 1
1 3 1 2
0 4 4 2
L.H.S. =
3 2
2 1
4 1
4 2
… (1)
R.H.S. = B + A
R.H.S. =
3 2 3 2
1 1 3 2
3 2 1 1
4 2 0 4
R.H.S. =
3 2
1 3 1 2
3 1 2 1
4 0 2 4
R.H.S. =
3 2
2 1
4 1
4 2
… (2)
From eq. (1) and eq. (2) we get, L.H.S. = R.H.S. (A + B) = (B + A) is verified.
Let 2
1
x x =
1
x x 1=
A Bx x 1
Multiplying throughout by x (x 1), we get, 1 = A (x 1) + Bx
When x = 1, 1 = 0 + B (1) B = 1
When x = 0, 1 = A (0 1) + 0 A = 1
2
1
x x =
1 1x x 1
We know cos(A + B) = cos A cos B sin A sin B put A = B = , cos 2 = cos cos sin sin
= cos2 sin2 But cos2 = 1 sin2 & sin2 = 1 cos2 cos 2 = (1 sin2 ) sin2 = 1 2sin2 cos 2 = cos2 (1 cos2 ) = 1 2 sin2 cos 2 = cos2 (1 cos2 ) = 2 cos2 1
1. (e)
1. (f) Vidyala
nkar
Vidyalankar : F.Y. Diploma Mathematics
1013/FY/Pre_Pap/Maths_Soln 8
Compound Angle The algebraic sum or difference of two or more angles is called compound angle.
Given : 2 sin 40 cos 10 = sin A + sin B
2 sin 40 cos 10 = 2 sin A B
2
cos A B
2
Equating both sides we get
40 = A B
2
A + B = 80 (1)
and 10 = A B
2
A B = 20 (2)
equation (1) + equation (2) 2A = 100 A = 50 equation (1) equation (2) 2B = 60 B = 30
L.H.S. = cos21 sin21cos21 sin21
Dividing numerator and denominator by cos 21°, we get,
= 1 tan211 tan21
= tan45 tan21
1 tan45 tan21
[ tan 45° = 1]
= tan (45° 21°) tanA tanBtan A B
1 tanA tanB
= tan 24° = R.H.S.
L.H.S. = tan117
+ tan1
113
In this case,
Let x = 17
> 0, y = 1
13> 0 and xy =
1 17 13 =
191
< 1
L.H.S. = tan1x y1 xy
= tan1
1 17 13
1 11
7 13
= tan1
13 791
11
91
= tan120 / 9190 / 91
= tan129
= cot192
1 11
tan cot xx
= R.H.S.
1. (g)
1. (h)
1. (i)
1. (j)
Vidyala
nkar
Prelim Question Paper Solution
1013/FY/Pre_Pap/Maths_Soln 9
Let the lines be
1 : 3x 2y + 4 = 0 Its slope, m1 =32
=32
and
2 : 2x 3y 7 = 0 Its slope, m2 =23
=23
It is the a cute angle between lines, then
tan = 1 2
1 2
m m1 m .m
Putting, m1 = 3/2, m2 = 2/3, we get,
tan =
3 22 33 2
12 3
=
9 462
= 5
12
= tan15
12
Range = largest Quantity smallest Quantity = 31 1 = 30.
Then writing into D, Dx, Dy and Dz as required in Cramer’s Rule, we have
D =
3 3
3 3 1
2 1 2
4 3 2
= 3(2 6) 3(4 8) 1(6 + 4) = 3(8) 3(4) (10) = 24 + 12 10 = 22
Dx =
11 3 1
0 1 2
25 3 2
= 11(2 6) 3(0 50) 1(0 + 25) = 11(8) 3(50) 1(25) = 88 + 150 25 = 37
Dy =
3 11 1
2 0 2
4 25 2
= 3(0 50) 11(4 8) 1(50) = 150 11(4) 1(50) = 150 + 44 50 = 200 + 44 = 156
1. (k)
1. (l)
2. (a)
Vidyala
nkar
Vidyalankar : F.Y. Diploma Mathematics
1013/FY/Pre_Pap/Maths_Soln 10
Dz =
3 3 11
2 1 0
4 3 25
= 3(25 0) 3(50 0) + 11(6 + 4) = 75 150 + 110 = 75 40 = 115
Then by Cramer’s rule,
x = xDD
= 3722
= 3722
y = yD
D =
15622
= 15622
z = zDD
= 11522
= 11522
x = 3722
, y = 7811
, and z = 11522
LHS = (x 4) (5 27) 7
(3x 4) (15 2y) 1
= 5 3 7
7 7 1
Comparing a11 and a12 x + 4 = 5 x = 1 5 + 2y = 3 y = (35) / 2 = 4
We have A = 2 2
2 3
1 5
and B = 2 3
3 1 2
1 0 0
Now L.H.S. = (AB)T … (1)
AB = 2 2 2 3
2 3 3 1 2
1 5 1 0 0
AB =
2 3 3 1 2 0 4 0
3 5 1 0 2 0
AB = 2 3
3 2 4
8 1 2
(AB)T =
3 2
3 8
2 1
4 2
= L.H.S. … (1)
We have,
A = 2 2
2 3
1 5
AT = 2
2 1
3 5
B = 2 3
3 1 2
1 0 0
BT =
2 3
3 1
1 0
2 0
R.H.S. = T TB A
2. (b)
2. (c)
Vidyala
nkar
Prelim Question Paper Solution
1013/FY/Pre_Pap/Maths_Soln 11
R.H.S. = 2 2
3 2
3 12 1
1 03 5
2 0
R.H.S. =
6 3 3 5
2 0 1 0
4 0 2 0
R.H.S. =
3 2
3 8
2 1
4 2
… (2)
From eq. (1) and eq. (2) we get L.H.S. = R.H.S. (AB)T = T TB A is proved.
A = 2 3 1
4 5 0
B = 1 2 4
1 3 0
Now, A + B = 2 1 3 2 1 4
4 1 5 3 0 0
A + B = 1 5 3
5 8 0
(A + B)T =
1 5
5 8
3 0
… (i)
AT =
2 4
3 5
1 0
, BT =
1 1
2 3
4 0
AT + BT =
2 1 4 1
3 2 5 3
1 4 0 0
AT + BT =
1 5
5 8
3 0
… (ii)
From (i) and (ii) (A + B)T = AT + BT
3 2
x 5
x x 6x
= 2
x 5
x x x 6
= x 5
x(x 2)(x 3)
2. (d)
2. (e)
Vidyala
nkar
Vidyalankar : F.Y. Diploma Mathematics
1013/FY/Pre_Pap/Maths_Soln 12
= A B Cx x 2 x 3
A = x 0
x 5(x 2)(x 3)
= 5
( 2)( 3)
=
56
B = x 2
x 5x(x 3)
= 3
(2)(5)
= 3
2 5
= 3
10
C = x 3
x 5x(x 2)
= 8
( 3)( 5)
=
83.5
= 8
15
Let 2
3x 2
x 1 x 1
=
3x 2
x 1 x 1 x 1
=
2
3x 2
x 1 x 1
= 2
A B Cx 1 x 1x 1
Multiplying throughout by (x + 1)2 (x 1), we get, 3x + 2 = A (x + 1) (x 1) + B (x 1) + C (x + 1)2
When x = 1, 3 + 2 = 0 + 0 + C (2)2 4C = 5 C = 5/4
When x = 1, 3 + 2 = 0 + B (1 1) + 0 2B = 1 B = 1/2
When x = 0, B = 1/2, C = 5/4
2 = A (1) (1) + 12
(1) + 54
(1)2
2 = A 1 52 4
A = 5 14 2 2 =
5 2 84
=
54
2
3x 2
x 1 x 1
=
2
5 1 54 x 1 4 x 12 x 1
Let A =
3 3
1 3 2
3 2 5
2 3 6
be the given matrix.
A =
3 3
1 3 2
3 2 5
2 3 6
A = 1[(6) (2) (5) (3)] 3[(3) (6) (2) (5)] + 2[(3) (3) (2)(2)]
A = [12 + 15] 3[18 10] + 2[9 + 4]
A = 3 3(8) + 2(5)
2. (f)
3. (a) Vidyala
nkar
Prelim Question Paper Solution
1013/FY/Pre_Pap/Maths_Soln 13
A = 3 24 10
A = 31
A 0
A1 exists To find A1:
Let A =
3 3
1 3 2
3 2 5
2 3 6
= 1 1 1
2 2 2
3 3 3 3 3
a b c
a b c
a b c
Let A1, B1 and C1 ….. are cofactors of elements a1, b1, c1…. respectively.
A1 = 2 5
3 6
= 12 + 15 = 3
B1 = 3 5
2 6 = (18 10) = 8
C1 = 3 2
2 3
= 9 + 4 = 5
A2 = 3 2
3 6
= (18 (6)) = 24
B2 = 1 2
2 6 = 6 4 = 2
C2 = 1 3
2 3
= (3 6) = 9
A3 = +3 2
2 5 = 15 + 4 = 19
B3 = 1 2
3 5 = (5 6) = 1
C3 = +1 3
3 2 = 2 9 = 11
The matrix of cofactors = 1 1 1
2 2 2
3 3 3
A B C
A B C
A B C
=
3
3 8 5
24 2 9
19 1 11
adj A = Transpose of cofactors matrix
adj A =
3
3 24 19
8 2 1
5 9 11
Vidyala
nkar
Vidyalankar : F.Y. Diploma Mathematics
1013/FY/Pre_Pap/Maths_Soln 14
A1 = 1
adj AA =
3 24 191
8 2 131
5 9 11
… (1)
We have x + 3y + 2z = 6 3x 2y + 5z = 5 2x 3y + 6z = 7 are given equations. Writing in matrix form, we get
1 3 2 x
3 2 5 y
2 3 6 z
=
6
5
7
A =
3 3
1 3 2
3 2 5
2 3 6
X =
3
x
y
z
B =
3
6
5
7
… (2)
We have AX = B X = A1B … (3) From eq. (1), eq. (2) and eq. (3), we get,
3 1
x
y
z
=
3 3 3 1
3 24 19 61
8 2 1 531
5 9 11 7
3 1
x
y
z
=
3 3
18 120 1331
48 10 731
30 45 77
3 1
x
y
z
=
3 1
311
3131
62
3 1
x
y
z
=
3 1
1
1
2
x = 1, y = 1, z = 2 which is required answer.
We have,
2
2
x 23x
x 3 x 1
=
2
A Bx Cx 3 x 1
2
2
x 23x
x 3 x 1
=
2
2
A x 1 Bx C x 3
x 3 x 1
x2 + 23x = A(x2) + A + Bx2 + 3Bx + Cx + 3C
3. (b) Vidyala
nkar
Prelim Question Paper Solution
1013/FY/Pre_Pap/Maths_Soln 15
x2 + 23x = (A + B)x2 + x(3B + C) + A + 3C Equating corresponding coefficient (A + B) = 1 … (1) (3B + C) = 23 … (2) A + 3C = 0 … (3) From eq. (3), we get, A = 3C … (4)
From eq. (1) and eq. (4) we get 3C + B = 1 B = 1 + 3C … (5) From eq. (2) and eq. (5), we get 3(1 + 3C) + C = 23 3 + 9C + C = 23 10C = 20 C = 2 … (6) From eq. (4) and eq. (6), we get A = 6 … (7) From eq. (1) and eq. (7) we get B = 7 … (8)
2
2
x 23x
(x 3)(x 1)
= 2
6 7x 2x 3 (x 1)
2
2
x 23x
(x 3) (x 1)
= 2
7x 2 6x 3(x 1)
which are required partial fraction.
Let tan 1
(tan 2) (tan 3)
=
A Btan 2 tan 3
here A = tan 2
tan 1tan 3
= 1
1
= 1
B = tan 3
tan 1tan 2
= 21
= 2
hence,
tan 1
(tan 2) (tan 3)
=
1 2tan 2 tan 3
We know cos(A + B) = cos A cos B + sin A sin B
put A = 2
, = B
hence cos2
= cos cos sin sin
2 2
= 0 + 1 sin
cos2
= sin
3. (c)
3. (d) Vidyala
nkar
Vidyalankar : F.Y. Diploma Mathematics
1013/FY/Pre_Pap/Maths_Soln 16
L.H.S. = sin A sin 2A sin 3A sin 4A
cos A cos 2A cos3A cos 4A
L.H.S. =
sin A sin 4A sin 2A sin 3A
cos A cos 4A cos2A cos 3A
L.H.S. =
4A A A 4A 3A 2A 2A 3A2sin cos 2sin cos
2 2 2 2A 4A A 4A 2A 3A 2A 3A
2cos cos 2cos cos2 2 2 2
L.H.S. =
5A 3A 5A A2 sin cos sin cos
2 2 2 2
5A 3A 5A A2 cos cos cos cos
2 2 2 2
L.H.S. =
5A 3A Asin cos cos
2 2 2
5A 3A Acos cos cos
2 2 2
L.H.S. = 5A
tan2
L.H.S. = R.H.S. Hence proved.
L.H.S. =1 sec2A
tan2A
=
11
cos2Asin2Acos2A
=cos2A 1
sin2A
= 1 cos2A
sin2A
=22cos A
2sinA.cosA =
cosAsinA
= cotA = R.H.S.
L.H.S. = 1 tan2A tanA1 tan2A tanA
=
sin2A sinA1
cos2A cos Asin2A sinA
1cos2A cos A
=
cos2A cosA sin2A sinAcos2A cos A
cos2A cosA sin2A sinAcos2A cos A
= cos2A cosA sin2A sinAcos2A cosA sin2A sinA
L.H.S. = sin 3A L.H.S. = sin(2A + A) L.H.S. = sin2A cos A cos2A sinA L.H.S. = (2 sin A cos A) cos A + [cos2 A – sin2 A] sin A
3. (e)
3. (f)
4. (a)
4. (b)
Vidyala
nkar
Prelim Question Paper Solution
1013/FY/Pre_Pap/Maths_Soln 17
L.H.S. = 2sin A cos2 A + (cos2 A sin2 A) sin A L.H.S. = 2 sin A (1 – sin2 A) + (1 – sin2 A – sin2 A) sin A L.H.S. = 2 sin A – 2 sin3 A + (1 – 2 sin2 A) sin A L.H.S. = 2sin A 2sin3A + sin A 2sin3A L.H.S. = 3sin A 4 sin3A L.H.S. = R.H.S. Hence proved. L.H.S. = sin 20. sin 40. sin 60. sin 80
= sin 20. sin 40.3
2. sin 80
3sin60
2
= 3
4[2 sin 20. sin 40] .sin 80 [Multiplying and dividing by 2]
= 3
4[cos (20° 40°) cos (20° + 40°)] . sin80°
= 3
4[(cos 20° cos 60°) . sin80°]
= 3
4[cos 20° sin 80° cos 60° sin 80°]
= 3
8[2 cos 20° sin 80° 2.
12
sin 80°]
[Multiplying and dividing by 2 and cos 60° = 1/2]
= 3
8[sin (20° + 80°) sin (20° 80°) sin 80°]
= 3
8[sin 100° + sin 60° sin 80°] [ sin (60°) = sin 60°]
= 3
8[sin 80° + sin 60° sin 80°]
[ sin 100° = sin (2 90° 80°) = sin 80°]
= 3
8(sin 60°)
= 3 3
8 2
= 3
16 = R.H.S.
L.H.S. = cosec A cosec A
+cosec A-1 cosec A+1
= cosecA1 1
cosecA-1 cosecA+1
= cosecA2
cosecA+1+cosecA-1
cosec A 1
=
2
cosecA 2 cosecA
cot A
4. (c)
4. (d) Vidy
alank
ar
Vidyalankar : F.Y. Diploma Mathematics
1013/FY/Pre_Pap/Maths_Soln 18
= 2
2
2 cosec A
cot A
=
2
2 2
2 sin A.
sin A cos A
= 2
2
cos A
= 2sec2A = R.H.S.
L.H.S. = cot coseccot cosec 1
L.H.S. =
cos 11
sin sincos 1
1sin sin
L.H.S. = cos 1 sincos 1 sin
L.H.S. =
2
2
2cos 2sin cos2 2 2
2sin 2sin cos2 2 2
L.H.S. = 2cos cos sin
2 2 2
2sin sin cos2 2 2
L.H.S. = 2cos cos sin
2 2 2
2sin cos sin2 2 2
L.H.S. = cos
2
sin2
L.H.S. = cot2
L.H.S. = R.H.S. Hence proved.
L.H.S. = cos145
sin1
513
Let cos145
= 1 cos 1 =
45
sin21 = 1 cos2 1 = 1 16
25
= 925
sin 1 = 35
4. (e)
4. (f) Vidyala
nkar
Prelim Question Paper Solution
1013/FY/Pre_Pap/Maths_Soln 19
Also, Let sin1 513
= 2 sin 2 = 5
13
cos22 = 1 25
169 =
144169
cos 2 = 1213
Next cos (1 2) = cos 1 cos 2 + sin 1 sin 2
= 4 12 3 55 13 5 13
=48 1565 65
=6365
1 2 = cos1 6365
cos1 45
sin1 513
= cos1 6365
L.H.S. = sin A 1 cosA
1 cos A sinA
L.H.S. =
2 2sin A 1 cos A
sinA 1 cosA
L.H.S. =
2 2sin A sin AsinA 1 cosA
L.H.S. =
22sin AsinA 1 cosA
L.H.S. =
22 1 cos A
sinA 1 cosA
L.H.S. =
2 1 cosA 1 cosA
sinA 1 cosA
L.H.S. = 1 cosA
2sinA
L.H.S. = 2 [cosec A cot A] L.H.S. = R.H.S. Hence proved.
We know sin(A + B) = sin A cos B + cos A sin B A = C/2, B = D/2
C Dsin
2
= C D C D
sin cos cos sin2 2 2 2
... (1)
cos(A B) = cos A cos B + sin A sin B Similarly,
C Dcos
2
= C D C D
cos cos sin sin2 2 2 2 ... (2)
5. (a)
5. (b) Vidyala
nkar
Vidyalankar : F.Y. Diploma Mathematics
1013/FY/Pre_Pap/Maths_Soln 20
Multiplying (1) and (2);
C D C D
sin cos2 2
=
C D C Dsin cos cos sin
2 2 2 2
C D C D
cos cos sin sin2 2 2 2
Multiplying by 2 throughout :
C D C D
2sin cos2 2
= 2 2C C D C D D2 sin cos cos cos cos sin
2 2 2 2 2 2
2 2C D D D C Csin cos sin sin cos sin
2 2 2 2 2 2
= 2 2C C D D2 sin cos cos sin
2 2 2 2
2 2D D C Csin cos cos sin
2 2 2 2
C D C D
2sin cos2 2
= D D C C
2 sin cos sin cos2 2 2 2
We know 2 sin cos = sin 2.
hence C D C D
2sin cos2 2
= [sin D + sin C] = RHS
hence proved.
L.H.S. = cot A cot 2Acot A cot 2A
L.H.S. =
cos A cos 2Asin A sin 2Acos A cos 2Asin A sin 2A
L.H.S. =
cos A sin 2A cos2A sinA
cos A sin2A cos2A sinA
L.H.S. =
sin 2A A
sin 2A A
( sin (A + B) = sin A . cos B + cos A sin B
and sin(A B) = sin A cos B cos A sin B)
L.H.S. = sin A
sin3A
L.H.S. = R.H.S. Hence proved.
We have, L1 : 2x + 3y = 13 … (1) L2 : 2x 5y = 7 … (2) are given equation of straight lines.
Let m1 and m2 be slopes of line given by eq. (1) and eq. (2)
5. (c)
5. (d) Vidy
alank
ar
Prelim Question Paper Solution
1013/FY/Pre_Pap/Maths_Soln 21
We get m1 = 2
3
and m2 = 25
If ‘’ be the acute angle between lines, then
tan = 1 2
1 2
m m1 m m
tan =
2 23 5
2 21
3 5
tan =
10 615
15 415
tan =
16151115
tan = 1611
tan = 1611
tan = 1611
= 1 16tan
11
Consider P (x1, y1) = P (5, 4) and the line is 2x + y + 6 = 0. The perpendicular distance from P(x1, y1) on ax + by + c = 0 is
= 1 12 2
ax by c
a b
where a = 2, b = 1, c = 6 x1 = 5, y1 = 4
= 2 5 1 4 6
4 1
= 10 4 6
5
=
20
5
= 4 5 units. Let P(x, y) divide the join of A (+5, 4) and B (2, 3) in the ratio K: 1 m = K and n = 1 Using external division formula,
Px = 2 1mx nxm n
and Py = 2 1my nym n
5. (e)
5. (f)
Vidyala
nkar
Vidyalankar : F.Y. Diploma Mathematics
1013/FY/Pre_Pap/Maths_Soln 22
Px = K 2 1 5
K 1
Py =
K 3 1 4K 1
Px = 2K 5K 1
Py =3K 4K 1
P (x, y) divides the x-axis, Py = 0 0 = 3K + 4 K = +4/3
Px =
42 5
34
13
=
85
313
= 23
The ratio is 4
:13
and the point is (23, 0).
Let for the line, x-intercept = a y- intercept = b Given : a = 2b
The equation of line in two intercept form is, x ya b = 1
Putting a = 2b, we get, x y
2b b = 1, multiplying throughout by 2b.
x + 2y = 2b (i) This line passes through the point (4, 1) Putting x = 4, y = 1, we get, 4 + 2 (1) = 2b 2b = 6 b = 3 a = 2b = 6
Putting b = 3 in equation (1), x + 2y = 2 (3) x + 2y = 6 This is the required line. Let P(x3, y3) and Q(x4, y4) be the points of trisection for the join A(3, 4) and B(2, 3) Then we observe from fig. that the point P(x3, y3) divides the join AB internally in the ratio 1 : 2. m = k, n = 2k k 0, k R Using internal division formula for x and y we get,
Px = 2 1mx nxm n
and Py = 2 1my nym n
A(3, 4) (x1, y1)
P(x3, y3) Q(x4, y4) B(2, 3) (x2, y2)
6. (a)
6. (b) Vidyala
nkar
Prelim Question Paper Solution
1013/FY/Pre_Pap/Maths_Soln 23
Putting values of m, n, x1, y1, x2, y2 we get
Px = k 2 2k 3
k 2k
Py = k 3 2k 4
k 2k
Px = 2k 6k
3k
Py = 3k 8k
3k
Py = 4k3k
Py = 5k3k
P2 = 43
Py = 53
P(x3, y3) = 4 5
,3 3
Now point Q(x4, y4) is the midpoint of PB.
Qx =
42
32
Qy =
53
32
Qx =
4 632
Qy =
43
2
Qx =
23
2
Qy =
46
Qx = 26
Qy = 23
Qx = 13
Q = (x4, y4) = 1 2,
3 3
M = h N
L c.f.F 2
C.I. F CF 9.5 14.5 4 4 14.5 19.5 6 10 19.5 24.5 10 20 24.5 29.5 5 25 29.5 34.5 7 32 34.5 39.5 3 35 39.5 44.5 9 44 44.5 49.5 6 50 N = 50 Median class = 24.5 29.5 h = 5 F = freq. of median class = 5. C.f. = 20
6. (c)
Vidyala
nkar
Vidyalankar : F.Y. Diploma Mathematics
1013/FY/Pre_Pap/Maths_Soln 24
Substituting
Median = 24.5 + 525 20
5
= 24.5 + 5 = 29.5
Range = L S
= 36 25 = 11
Coefficient of Range = L SL S
= 36 2536 25
= 1161
M D = ix x
n
x = fixifi
= 60 150 350 810 495 195
50
= 200050
= 40
MD = 25 15 5 5 15 25
50
= 9050
= 95
Variance = 21fi xi x
n
Coefficient of variance = 100| x |
, where = 5D = Variance
x = fixi
fi
= (9)(15) (17)(45) (75)(43) (105)(82) (195)(24) (135)(81) (165)(44)
300
= 118.7
Variance = 2i1
fi x xn
= 2 2 219(15 118.7) 17(45 118.7) 43(75 118.7)
300
2 2 282(105 118.7) 81(135 118.7) 24(195 118.7)
= 1492.90
Coefficient of variance = 100x =
Variance100
118.7
= 32.54
6. (e)
6. (d)
6. (f)
Vidyala
nkar
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