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Vidyamandir Classes
VMC | JEE Mains-2019 1 Solutions | Morning Session
SOLUTIONS Joint Entrance Exam | IITJEE-2019
9th APRIL 2019 | Morning Session
Vidyamandir Classes
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Joint Entrance Exam | JEE Mains 2019
PART-A PHYSICS
1.(3) Given
2.(3)
3.(1)
4.(3) Clearly
Hence for 4th harmonic,
5.(4)
COM gives : ...(1)
For elastic collision ...(2)
From (1) and (2)
6.(3)
7.(3) We have
8.(3) Conservation of energy gives
5=m Þ 5 5-= Þ = -
v v uu
\1 1 1 32 cm5 40
+ = - Þ =-
uu u
2 2 6net 0 1B B B 10 904-= + = ´
\ 8 6net 3 10 10 904-= ´ = ´ ´ ´E C B
\ 4 8 610 3 10 10 904 0.9 N.- -= = ´ ´ ´ ´ =F QE
\ rns 0.6N2
= =FF
eq12 46 912 4´
= + = W+
R \ main72 8 amp.9
= =i
\ 104 8 2 amp.
12 4W = ´ =+
i
\ 6CV 10 10 (10 2) 200-= = ´ ´ ´ = µQ C
2 0.157p=
l\ 40ml =
4 80m2l
= ´ =l
002 2 mv
4= ´ +
vv
004
- =vv v
1.2 kg=m
system0
1 ( ) ( ) 2 ( )4 / 2 ( / 2)
é ù´ - ´ -= + +ê úpe + -ë û
Q q Q q qUD d D d d
2
20
1 24
é ù= - -ê úpe ë û
q qdd D
22 2
4æ ö
- =ç ÷è ø!
dD D
r.m.s.V Tµ
\200 273 127 V 100 5 m/sV 273 227
+= Þ =
+
2IAOR
1 I mgh2
w =
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For ring
For Cylinder,
For sphere
9.(3) Net repulsive force
10.(1) Theoretical
11.(3) For no drifting,
Hence direction with flow of rivers is
12.(1) Work-done will be equal to rise in potential energy of suspended section
13.(4) H-Cl is diatomic molecule hence degree of freedom is (translational) + 2 (rotational) + 1 (vibration) = 6
14.(3)
(where Clearly it is parabolic functions. Starting from ground momentum first decreases and then becomes zero at highest position. Therefore it
increases in –ve direction. Hence 3rd option is correct.
15.(4)
16.(2)
17.(4) At 0°C,
22
11 2 mgh2
æ ö´ ´ =ç ÷è øVmRR
Þ2
1 =vhg
2 2 2
2 21 R v 3 v3m mgh h2 2 R / 2 2 g
æ ö æ ö´ = Þ =ç ÷ ç ÷è ø è ø2 2
31 7 mgh2 5 4 / 4
æ ö æ ö× ´ =ç ÷ ç ÷è ø è øR Vm
R\
2
3710
=vhg
\ 1 2 33 7: : 1: : 10 :15 : 72 10
= =h h h
0 1 2 0 1 21 2 2 2 2
µ ´ µ= - = - ´
p pi i a i iF F F aa a
01 24
µ=
pi i
2 1sin4 2
q = = =m
uv
\ 30q = °
180 120a = - q = °
\ 22 2= ´ ´ =mg L L MgLWL n n n
3=f
\2
21 6 .2 2 6
= \ =BB
mvmv K T TK
2 2 2= -V u gh Þ 2 2 2 2 2mgh= -m v m u
Þ 2 = -p A Bh 2 2 , 2mg)= =A m u b
[ ]= +g gV i R R34 10 [50 5000] 20.2 volt-= ´ + =
20 volt.»
2 26
1 2
| | 3.75 10 J2 2
-= D = - = ´Q QW UC C
1000m/s3
w= =vK
\1000 / 3 273336 273
=+T
( )µ!V T \ 277 4 C» = °T K
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18.(4)
19.(3)
20.(1)
21.(3) If is contact angle, then for equilibrium of rises water in tube, we have Hence if is doubled, rises mass will also doubled.
22.(3)
23.(1)
24.(1) According to question shift = width of n fringe pattern
25.(2) ( Buoyancy )
26.(4) For a solenoid, self inductance is given as
27.(3) input resistance =
Voltage gain =
28.(3) Clearly
.
29.(1) Area under curve in process `A’ on P-V diagram is more. Hence
30.(4)
Also
PART-B CHEMISTRY
2 21 1 22 2w = q Þ ´ wI K I 2w q
´ = ´ q×d dKdt dt
\2w q
a = =d kdt I
&4 8
= = = = =AB BC CD DE ECR RR R R R R \
374 8 864
æ ö+ ´ç ÷è ø= =eq
R R R
R RR
2 22
2
2 2
1 12 3 488.88 488.9Å1 16602 4
-l= Þ l = »
-
qc cos 2 Mgq´ p =T rr
2 2 2.2
(3 ) (3 ) 3= + =pGM G M GMEa a a
sin ( ) sint = µ q = qB niA B 5 2.5 1100 3 1100 100 2
´= ´ ´ ´ ´
´0.265N 0.27N= »
Þ ( 1)µ - = ´ ×lDt na
\[ 1]
l=
µ -nDta
2 & ' 2( /16)
= p = p-
l lT Tg g g
!16
=mg
\1 1' 4
151116
= ´ = ´-
T T T
self 0æ ö= ´ ´µ ç ÷è øNL N AL
\ self1
µLL
1 610mVr15 10-
=´
3 20.67 10 6.0 10 0.67 K» ´ W = ´ W = W
Rc 3mA 1KB 300 Voltr 15 A 0.67K
W+ = + ´ = +
µ W
72 2
5 10-p p=
l ´\ 5000Ål =
\ = + fE eV Þ12375 2 0.475 V 0.48 volt5000
= - = »V
>A BW W \ &D > D D = DA B A BQ Q U U
4 33
10 10 kg/m(0.1)
+r = =
3r =Ml
\3 0.1 0.01 3
10 0.1Dr D D
= + = + ´r
M lM l
31 3 31 0.31kg/m100 10 100
= + = =
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1.(2) from given eq:
for option -1 6 gm reacts 28 gm i.e. both reactants utilized completely.
for option -2 6 gm reacts 28 gm
10 gm reacts
required, however amount. of given = 56 gm
i.e. present in excess.
So, utilizes completely & hence limiting reagent.
for option-3 6 gm reacts 28 gm
4 gm reacts
required. But given = 14 gm.
is L. R.
for option - 4 reacts 28 gm
6 gm reacts required but given = 35 gm.
2.(2)
major product should be
3.(3) q & w are path functions.
Option – 3 is correct
2 2 3N (g) 3H (g) 2NH (g)1mole 3mole1 28gm 3 2gm28gm 6gm
+ ¾¾®
´ ´= =
! 2H c 2N
! 2H c 2N
\ 2H 228 10gm N6´
246.67gm N= 2N
2N
2H
! 2H c 2N
\ 2H c 228 4 gmN6´
218.67gmN= 2N
2N\
6gm! 2H c 2N
\ 2H c 228 8 gm N6´
237.33gmN= 2N
\
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4.(4) Aerosol is a colloidal solution of solid in air. Eg. Smoke, dust etc. are considered as aerosol.
Option – 4 is correct
5.(1)
Option – 1 is correct
6.(2)
7.(4) from Raoult’s law: from Dalton’s law:
&
…… (i)
& …… (ii)
eq. (i) ÷ eq. (ii)
8.(1)
One molecule is present as water of crystallization.
9.(4) for EMR
From Planck’s formula,
air,3 2Mg MgO Mg ND¾¾¾® +
0M M MP x P= × M M TP Y P= ×
0N N NP x P= × N N TP Y P= ×
\ 0M M M Tx P Y P× = ×
0N N N Tx P Y P× = ×
0M M M T
0N TN N
x P Y PY Px P
× ×=
××
M M
N N
x Y450x 700 Y
´ =
M M
N N
x Y14x 9 Y
= ´
\ M M
N N
x Yx Y
>
4 2CuSO 5H O×
[ ]2 4 4 2Cu(H O) SO H O×
2H O
C = nl C\n =
l
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So,
10.(4)
11.(4)
2-chloro-1-methyl-4-nitrobenzene
12.(1) Increase in CO2 induces Global warming.
E h= nEh
\n =C E
h\ =l
( )( )
( )( )
max min max minLyman Lyman Lyman
Balmer max min max minBalmer Balmer
E E
E E
n - n -Dn= =
Dn n - n -
1 1 1 113.6 13.61 1 41 1 1 113.6 13.64 4 9
æ ö æ ö- - -ç ÷ ç ÷aè ø è ø=æ ö æ ö- - -ç ÷ ç ÷aè ø è ø
11 0 1 1 941 1 1 4 40 14 4 9
9
- - += = =
- - +
9 : 4\
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13.(4) Sucrose is formed due to bonding between.
Glucose from C1 position and fructose from C2 position.
14.(4)
15.(4)
16.(2)
17.(4)
I.P.1st <<< I.P. 2nd.
18.(3)
i.e.
19.(4) Cryolite =
-a -b
4LiAlH3 3 3 2
O||
CH CH CH C O CH CH CH CH CH OH- = - - - ¾¾¾¾® - = - -
DCl3 3
3 2
Cl|
CH C CH CH C CHD
DICl|
CH C CHD|I
- º ¾¾¾® - =
¯
- -
2 2 6 1K 1s 2s 2p 3s=
2 2 6K 1s 2s 2p
inert gas configuration
+ =
CN Cl Me OMeActivating nature
- - - -¾¾¾¾¾¾¾¾®
D A C B< < <
3 6Na [AlF ]
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20.(2)
21.(1)
No symmetry
22.(3) contains 12 Pentagons of 20 hexagons
23.(3)
Polyethylene
Ethanal
Iron oxide
24.(4)
25.(1)
For 1st order reaction
If
or
Similarly
2 2Cu Zn Zn Cu+ ++ ¾¾® +
n 2=
cellG nf E°D ° = - 2 96500 2= - ´ ´ 386 KJ= -
2 2( ( ) )M AA A B type
60C
2 5V O ——— 2 4H SO
4 3/ ( )TiCl Al Me ———
2PdCl ———
——— 3NH
R Pa xa x x
¾¾®
-
1( )dx k a xdt
+ = -
dx kdta x
=-ò òln( )a x kt c- - = +
0t =lnc a= - \ ln( ) lna x kt a- - = -
ln( ) lna x kt ay m x C- = - +
( )0( )¾¾®
AA
R P
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or
or
or
26.(3) For as complete dissociation takes place.
For
As
27.(3)
For Kr the value of is highest
Thus is also highest
28.
29.(3)
30.(1) For For (filling in BMO)
(filling in ABMO)
( )0[ ] [ ]
Ad R k Rdt
-=
( )d R kdt- =
[ ]d R kdt= -ò ò[ ]R kt c= - +
[ ] = - += +A kt cy mx c
xy2i =2xy CRTp = ´
2BaCli 3=
2BaCl 3 0.01 RT 0.03 RTp = ´ ´ =
2xy BaCl4p = ´p
2
xy
BaCl
2 C RT0.03RT
p ´ ´=
p
2
2
BaCl
BaCl
4 2 C0.03
´p ´=
p
24 0.03C 6 102 100
-´= = ´
´
C8aT27Rb
=
ab
æ öç ÷è ø
CT
2 1 4 3
2 2 2 3NO, N O, NO , N O+ + + +
2 2C , Z 12=
BO 2.5=
\ 2 2C , Z 13- =
–2C
BO 2.5=
–2O
BO 2.5=
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PART-C MATHEMATICS
1.(1) The points and
So, slope of line joining the two pts
Again, slope of tangent to is
2.(2) Since S.D. =
Now,
3.(3)
4.(1)
–NOBO 2.0=
–2F
BO 0.5=
(1, (1)) (1, 2)= -f ( 1, ( 1)) ( 1,0)- - = -f
2 12-
= = -
3 2 2= - -y x x x 23 2 2= - -dy x xdx
Þ 23 2 2 1- - = -x x Þ 23 2 1 0- - =x x Þ2 4 12 1 ,1
6 3± +
= = -x
22å åæ ö- ç ÷è ø
xi xix x
22 2 2 2( 1) 0 (1) ( 1) 0 (1)4 4
- + + + - + + +æ ö= - ç ÷è ø
K k
2 224 16+
= -K K 23 8 16 5 80Þ + = ´ =K
23 84+
=K 2 72 24
3Þ = =K
23 8 54+
=K 2 6Þ =K
32
0
sin ( )sin
=+òx dx I say
x cosxÞ
3/2
0
cossin cos
=+òxI dx
x xp
Þ 2 =I3 3/2
0
sin cossin
++òx x dxx cosx
pÞ
/2 2 20
2 (sin cos sin cos )= + -òI x x x x dxp
Þ/2
0
12 1 sin 22
æ ö= -ç ÷è øò
pI x dx Þ
/2
0
cos224
ù= + úû
xI xp
Þ1 1 12
2 4 4 2 2æ ö æ ö= - - = -ç ÷ ç ÷è ø è ø
p pI Þ14-
=pI
2 2cos 10 cos10 cos 50 cos 50° - ° ° + °
2 21 2cos 10 2cos10 cos 50 2 cos 502é ù= ° - ° ° + °ë û [ ]1 1 cos20 cos40 cos60 1 cos100
2= + ° - ° - ° + + °
( ) ( )1 3 cos 20 cos40 cos 90 102 2é ù= + ° - ° + ° + °ê úë û
1 3 32 sin30 sin 10 sin 102 2 4é ù= + × ° ° - ° =ê úë û
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5.(2)
6.(1) Point on the line
Now for p
7.(2)
8.(3)
9.(3)
For
10.(1)
11.(2) Let equation of the plane be
……(1)
22 cos 3 sin 0+ =q q Þ 22 2 sin 3sin 0- + =q q
Þ 22 sin 3 sin 2 0+ - =q q Þ 2 22 sin 4 sin sin 2 0- + - =q q q
(2sin 1) (sin 2) 0+ - =q q
Þ 1sin 2-=q Þ ,
7 11 5, ,6 6 6 6
- -=
p p p pq Þ7 11 5 12 2
6 6+ - -
= =p p p p p p
1 1 22 3 4- + -
= = = lx y zÞ (2 1, 3 1, 4 2)= + - +l l l
(2 1) 2(3 1) 3(4 2) 15+ + - + + =l l l Þ120 5 152
+ = Þ =l l
Þ1 1 1 92, , 4 4 16 202 4 4 2
æ ö= Þ = + + = + =ç ÷è ø
p OP
2
2 21( ) 1 \ [ 1, 0)
1 1= = - Þ = -
- -xf x A Rx x
Þ ( )( ) , 1 [0, )Î -¥ - È ¥f x
1 1 1 2 1 3 1 1 1 78.......
0 1 0 1 0 1 0 1 0 1-é ù é ù é ù é ù é ù
=ê ú ê ú ê ú ê ú ê úë û ë û ë û ë û ë û
n
Þ ( 1) 78S - =n ( 1) 782-
= =n n
Þ ( 1) 156 12 13- = = ´n n Þ 13=n
Þ11 13 1 13
0 1 0 1
- -é ù é ù=ê ú ê ú
ë û ë û
'( ) ( 1)( 1)= - +f x x x xl Þ 3( )¢ = -l lf x x x Þ4 2
( )4 2
= - +x xf x l l µ
( ) (0)=f x f
Þ4 2
04 2
- =l lx x
Þ2
2( 2) 04
- =lx x Þ 0, 2, 2= -x
22
(24 18) 7 3 6 3 24 824 18
+= Þ = -
-
M M MM
Þ 2 236 72 54= -M M Þ 2 290 725
= Þ ±M M
( 0) ( 1) ( 0) 0- + + + - =a x b y c z
Þ + + = -ax by cz b
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Now
……(2)
and
So, equation of plane
satisfy.
12.
Also, and
So, and
and
13.(3)
and
(0 0) (0 1) (1 0) 0- + + + - =a b c
Þ 0+ =b c
2 2 2
0 1 ( 1) 1cos4 22
´ + ´ + ´ -= =
+ +
pa b c
a b c
Þ 2 22- = +b c a b
2 22 2= +b a b
2 2 24 2= +b a b
Þ 2 22=a b Þ 2= ±a b
2 1± + - = -x y z
( 2, 4)+
!1 3= = +b la l l" " !i j Þ ! ! !2 3 (3 ) ( 3 ) .- + = - + + +! !i j k i x j kl µ l d
! !2 3= - +"
!i j kb µ µ µ
1 2= -b b b!" " "
Þ 3 2- =l µ 3= -d
3 1+ = -l µ
12=l 1
2-=µ
Þ !( )11 32
= +b"
!i j ! !2
1 3 32 2
= - + +b"
!i j k
! !
! !1 2
3 3 9 9 11 022 2 2 4 41 3 32 2
æ ö æ ö æ ö´ = = - - - + +ç ÷ ç ÷ ç ÷è ø è ø è ø
- -
b b
!
" "!
i j k
i j k ! !( )1 3 9 32
= - + +!i j k
1( 7)502 --
= + Þ = -n n nn nSn n A T S S
Þ [ ]50 ( 1) ( 7) ( 1) ( 8)2
= - + + - - - -nAT n n n n n n
[ ]50 7 9 82
= + - + -nAT n n
Þ 50 ( 4)= + -nT A n 5 4 (50 ) (50)= - = + - =d T T A A
Þ 50 50 46= +a A
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14.(4)
15.(2)
So, circle with radius 1.
16.(3)
17.(2)
So, g(x) is not differentiable at
18.() The other root is
The other root is
So,
19.(4)
50( , 9 ) ( , 50 46 )= +d A A
33
2
4 4 4
2 cos 1 1 2 cos 1lim lim lim 2 sin 2cot 1 2cosec 2® ® ®
- - 1æ ö= = = = ´ =ç ÷- è øx x x
x xk xx xp p p
| || | | | 1| |
+ += Þ = Þ =
- +a aa a
i iz z zi i
( )833log6 7
4 32 20 8æ ö= = ´ç ÷
è øxT C x
xÞ 8log 64=xx x
Þ( )8
3log7
320 8 20 9´ ´ = ´xx
xÞ 2
0 8(log8 ) log 64 log= +x x
Þ8log
64=xx
xÞ ( ) ( )8 8
1log 2 log 1 0 64,8
- + = Þ =x x x
( ) ( ( )) 15 | ( ) 10 |= = - -g x f f x f x
15 |15 | 10 | 10 |= - - - -x
Þ ( ) 15 | 5 | 10 |= - - -g x x Þ ( ) 15 |15 | ; 10= - - ³g x x x
15 | 5 | 10- - <x x Þ ( ) 10 ; 5= + <g x x x
20 ; 5 10- £ <x x
; 10 15£ <x x
30 ; 15- £x x
5,10,15=x
, Îp q R 2 3+
. Î Þp q R 2 3+
4 1= - =p q
2 4 16 4 12- = - =p q
2 4 12 0- - =p q
22/3 4/3
4/5 2/5 4/3secsec cos
sin cos (tan )= = =ò ò ò
dx xdxI x ecx xdxx x x
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20.(4)
Now
lies on the curve
So, curve is
21.(4) Mid-point
Now line as: is layout
So, locus:
22.(3) Path
23.(2)
and
Point of intersection
Þ
413
4/3(tan ) (tan )
4tan 13
-
= = +-
òd x xI C 1/33(tan )-= - +x C
3= + -y x ax b 23Þ = +dy x adx
(1, 5)1
-
= -dydx
Þ 3 1+ = -a Þ 4= -a
(1, 5)-
Þ 5 1 4- = - - b
5 3 2= - =b
3 4 2= - -y x x
Þ , ,2 2
æ ö= = ç ÷è ø
h kA B P
Þ 2 , 2= =h x k y
1+ =x yh x
Þ
2 2
0 0 11
1 1
+ -=
+
h k
h k
Þ 2 21 11
+ =h x
2 21 14 4
1+ =
x yÞ 2 2 2 2 2 2 2 24 4 0+ = Þ + - =x y x y x y x y
1 1 11 1 1 1 12 3 4 31æ öæ öæ öæ ö= - - - - -ç ÷ç ÷ç ÷ç ÷
è øè øè øè ø
1 2 3 7 7 251 12 3 4 8 32 32
= - ´ ´ ´ = - =
{ }2( , 4); 2= £ £ +lA x y x
2³y x 2 0- - £y x
Þ 2 22 2 0= + Þ - - =x x x x
1, 2= -x
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24.(2)
Also,
25.(2) Roots of are
26.(3)
I.F. =
So,
So, set
( )( )22 2 3
2
1 1
8 12 2 2 4 22 3 3 2 3
- -
é ù 1æ ö æ ö= + - = + - = + - - - +ê ú ç ÷ ç ÷è ø è øê úë û
òx xA x x dx x
18 32
= - -1 952 2
= - =
( ) ( ) ( ) ( )+ = × Þ = xf x y f x f y f x a
(1) 2 ( ) 2= Þ = xf f x
( )1 10102 10
1
2 1( ) 2 2 2 ... 2
2 1
+
=
-é ù+ = + + + =ë û -å
aa
k
zf a k
( ) ( )1 10 102 2 1 16 2 1+ - = -a
Þ 1 4 3+ = Þ =a a
2 1 0+ + =x x 2,w w
Þ 2, 1 0= = Þ + + =a w b w a b
1 11 1 0 1
1 1 0 1
++ = + = + -
+ + - + -
y yy y y y y
y y y y
a b a b a ba b b a bb a a a a b
1 1 2 3( )® + +C C C C 2 2 1( )® -R R R 3 3 1( )® -R R R
1
0 ( 3) 1
0 1 ( 3)
= - -
- +
a b
b
a
y y i
y i
2
2 3
{( 3) (1 )}
{ 3 (2 1)}
= + - - - +
= + - + =
a b aby y
y y y
2 22 æ ö+ = Þ + =ç ÷è ø
dy dyx y x y xdx dx x
22=ò dx
xe x
42 3 2
4= Þ = +ò
xyx x dx yx c
1 3(1) 1 14 4
= Þ = + Þ =y C C
2
23
4 4= +xy
x
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27.(2) if point is (1, 4)
Again, for focal chord
So, length of focal chord =
28.(4) Let the equation of line be
For,
29.(3)
So, ]
30.(3)
2 16=y x 112
Þ =t 21 1( , 2 )= at t
1 2 1= -t t 2 2Þ = -t
2
1 21( ) 4 2 252
æ ö- = + =ç ÷è ø
a t t
2 cos , 3 sin= + = +x r y rq q
24=r
(2 4 cos ) (3 4 sin ) 7+ + + =q q
Þ 4(cos sin ) 2+ =q q Þ 2 24(cos sin 2sin cos ) 1+ + =q q q q
Þ 2 24 tan 8 tan 4 sec+ + =q q q Þ 3tan 2 8tan 3 0+ + =q q
Þ8 64 36tan
6- ± -
=q
Þ8 2 7 2 7 8tan6 6
- ± -= Þ =q qtan
( )27 12 7 8 7 1tan6 (7 1) 7 1
- -- - -= = =
- +q
1 7tan1 7-
=+
q
( ) ( ~ ) ( )Ù º Ù º!pV p q pV p pVq pVq
~ ( ) (~ ) ( )º Ù !pVq p q
8 5 8 5 8 56 5 7 4 8 3= ´ + ´ + ´ =M C C C C C C n Þ 78= =M n