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Qu est ion No : 1 - T opic 1
W h er e does r ou t in g occu r w it h in t h e DoD T CP/IP r efer en ce m odel?
 
A . a pplica t ion
B. in t er n et
C. n et w or k
D. t r a n spor t
A n sw er : B
Ex pla n a t ion : T h e pict u r e below sh ow s t h e com pa r ison bet w een T CP/IP m odel & OSI m odel. Not ice t h a t t h e
In t er n et La y er of T CP/IP is equ iv a len t t o t h e Net w or k La y er w h ich is r espon sible for r ou t in g decision .
 
 
Qu est ion No : 2 - T opic 1
W h ich of t h e follow in g st a t em en t s descr ibe t h e n et w or k sh ow n in t h e g r a ph ic? (Ch oose
 t w o.)
 
 
A . T h er e a r e t w o br oa dca st dom a in s in t h e n et w or k .
B. T h er e a r e fou r br oa dca st dom a in s in t h e n et w or k .
C. T h er e a r e six br oa dca st dom a in s in t h e n et w or k .
D. T h er e a r e fou r collision dom a in s in t h e n et w or k .
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E. T h er e a r e fiv e collision dom a in s in t h e n et w or k .
F. T h er e a r e sev en collision dom a in s in t h e n et w or k .
A n sw er : A ,F
Ex pla n a t ion : On ly r ou t er ca n br ea k u p br oa dca st dom a in s so in t h e ex h ibit t h er e a r e 2 br oa dca st dom a in s: fr om
e0 in t er fa ce t o t h e left is a br oa dca st dom a in a n d fr om e1 in t er fa ce t o t h e r ig h t is a n ot h er br oa dca st dom a in ->.
Bot h r ou t er a n d sw it ch ca n br ea k u p collision dom a in s so t h er e is on ly 1 collision dom a in on t h e left of t h e r ou t er
(beca u se h u b doesn t br ea k u p collision dom a in ) a n d t h er e a r e 6 collision dom a in s on t h e r ig h t of t h e r ou t er (1
collision dom a in fr om e1 in t er fa ce t o t h e sw it ch + 5 collision dom a in s for 5 PCs in Pr odu ct ion ) ->.
 
 
Qu est ion No : 3 - T opic 1
A n et w or k in t er fa ce por t h a s collision det ect ion a n d ca r r ier sen sin g en a bled on a sh a r ed
 t w ist ed pa ir n et w or k . Fr om t h is st a t em en t , w h a t is k n ow n a bou t t h e n et w or k in t er fa ce por t ?
 
A . T h is is a 1 0 Mb/s sw it ch por t .
B. T h is is a 1 0 0 Mb/s sw it ch por t .
C. T h is is a n Et h er n et por t oper a t in g a t h a lf du plex .
D. T h is is a n Et h er n et por t oper a t in g a t fu ll du plex .
E. T h is is a por t on a n et w or k in t er fa ce ca r d in a PC.
A n sw er : C
Ex pla n a t ion : Moder n Et h er n et n et w or k s bu ilt w it h sw it ch es a n d fu ll-du plex con n ect ion s n o lon g er u t ilize
CSMA /CD. CSMA /CD is on ly u sed in obsolet e sh a r ed m edia Et h er n et (w h ich u ses r epea t er or h u b).
 
 
Qu est ion No : 4 - T opic 1
H ost 1 is t r y in g t o com m u n ica t e w it h H ost 2 . T h e e0 in t er fa ce on Rou t er C is dow n .
 
 
Which of the following are true? (Choose two.)
 
A . Rou t er C w ill u se ICMP t o in for m H ost 1 t h a t H ost 2 ca n n ot be r ea ch ed.
B. Rou t er C w ill u se ICMP t o in for m Rou t er B t h a t H ost 2 ca n n ot be r ea ch ed.
C. Rou t er C w ill u se ICMP t o in for m H ost 1 , Rou t er A , a n d Rou t er B t h a t H ost 2 ca n n ot be r ea ch ed.
D. Rou t er C w ill sen d a Dest in a t ion Un r ea ch a ble m essa g e t y pe.
E. Rou t er C w ill sen d a Rou t er Select ion m essa g e t y pe.
F. Rou t er C w ill sen d a Sou r ce Qu en ch m essa g e t y pe.
A n sw er : A ,D
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Ex pla n a t ion : H ost 1 is t r y in g t o com m u n ica t e w it h H ost 2 . T h e e0 in t er fa ce on Rou t er C is dow n . Rou t er C w ill
sen d ICMP pa ck et s t o in for m H ost 1 t h a t H ost 2 ca n n ot be r ea ch ed.
 
 
Qu est ion No : 5 - T opic 1
Refer t o t h e ex h ibit .
 
 
Based on the information given, which switch will be elected root bridge and why?
 
A . Sw it ch A , beca u se it h a s t h e low est MA C a ddr ess
B. Sw it ch A , beca u se it is t h e m ost cen t r a lly loca t ed sw it ch
C. Sw it ch B, beca u se it h a s t h e h ig h est MA C a ddr ess
D. Sw it ch C, beca u se it is t h e m ost cen t r a lly loca t ed sw it ch
E. Sw it ch C, beca u se it h a s t h e low est pr ior it y
F. Sw it ch D, beca u se it h a s t h e h ig h est pr ior it y
A n sw er : E
Ex pla n a t ion : T o elect t h e r oot br idg e in t h e LA N, fir st ch eck t h e pr ior it y v a lu e. T h e sw it ch h a v in g t h e low est
pr ior it y w ill w in t h e elect ion pr ocess. If Pr ior it y V a lu e is t h e sa m e t h en it ch eck s t h e MA C A ddr ess; t h e sw it ch
h a v in g t h e low est MA C A ddr ess w ill becom e t h e r oot br idg e. In t h is ca se, sw it ch C h a s t h e low est MA C A ddr ess so
it becom es t h e r oot br idg e.
 
 
Qu est ion No : 6 - T opic 1
W h ich la y er in t h e OSI r efer en ce m odel is r espon sible for det er m in in g t h e a v a ila bilit y of t h e
 r eceiv in g pr og r a m a n d ch eck in g t o see if en ou g h r esou r ces ex ist for t h a t com m u n ica t ion ?
 
A . t r a n spor t
B. n et w or k
C. pr esen t a t ion
D. session
E. a pplica t ion
A n sw er : E
Ex pla n a t ion : T h is qu est ion is t o ex a m in e t h e OSI r efer en ce m odel. T h e A pplica t ion la y er is r espon sible for
iden t ify in g a n d est a blish in g t h e a v a ila bilit y of t h e in t en ded com m u n ica t ion pa r t n er a n d det er m in in g w h et h er
su ffic ien t r esou r ces for t h e in t en ded com m u n ica t ion ex ist .
 
 
Qu est ion No : 7 - T opic 1
Refer t o t h e g r a ph ic .
 
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Host A is communicating with the server. What will be the source MAC address of the
 frames received by Host A from the server?
 
A . t h e MA C a ddr ess of r ou t er in t er fa ce e0
B. t h e MA C a ddr ess of r ou t er in t er fa ce e1
C. t h e MA C a ddr ess of t h e ser v er n et w or k in t er fa ce
D. t h e MA C a ddr ess of h ost A
A n sw er : A
Ex pla n a t ion : W h er ea s sw it ch es ca n on ly ex a m in e a n d for w a r d pa ck et s ba sed on t h e con t en t s of t h e MA C h ea der ,
r ou t er s ca n look fu r t h er in t o t h e pa ck et t o discov er t h e n et w or k for w h ich a pa ck et is dest in ed. Rou t er s m a k e
for w a r din g decision s ba sed on t h e pa ck et 's n et w or k - la y er h ea der (su ch a s a n IPX h ea der or IP h ea der ). T h ese
n et w or k -la y er h ea der s con t a in sou r ce a n d dest in a t ion n et w or k a ddr esses. Loca l dev ices a ddr ess pa ck et s t o t h e
r ou t er 's MA C a ddr ess in t h e MA C h ea der . A ft er r eceiv in g t h e pa ck et s, t h e r ou t er m u st per for m t h e follow in g
st eps: 1 . Ch eck t h e in com in g pa ck et for cor r u pt ion , a n d r em ov e t h e MA C h ea der . T h e r ou t er ch eck s t h e pa ck et
for MA C-la y er er r or s. T h e r ou t er t h en st r ips off t h e MA C h ea der a n d ex a m in es t h e n et w or k -la yer h ea der t o
det er m in e w h a t t o do w it h t h e pa ck et . 2 . Ex a m in e t h e a g e of t h e pa ck et . T h e r ou t er m u st en su r e t h a t t h e pa ck et
h a s n ot com e t oo fa r t o be for w a r ded. For ex a m ple, IPX h ea der s con t a in a h op cou n t . By defa u lt , 1 5 h ops is t h e
m a x im u m n u m ber of h ops (or r ou t er s) t h a t a pa ck et ca n cr oss. If a pa ck et h a s a h op cou n t of 1 5 , t h e r ou t er
disca r ds t h e pa ck et . IP h ea der s con t a in a T im e t o Liv e (T T L) v a lu e. Un lik e t h e IPX h op cou n t , w h ich in cr em en t s
a s t h e pa ck et is for w a r ded t h r ou g h ea ch r ou t er , t h e IP T T L v a lu e decr em en t s a s t h e IP pa ck et is for w a r ded
t h r ou g h ea ch r ou t er . If a n IP pa ck et h a s a T T L v a lu e of 1 , t h e r ou t er disca r ds t h e pa ck et . A r ou t er ca n n ot
decr em en t t h e T T L v a lu e t o 1 a n d t h en for w a r d t h e pa ck et . 3 . Det er m in e t h e r ou t e t o t h e dest in a t ion . Rou t er s
m a in t a in a r ou t in g t a ble t h a t list s a v a ila ble n et w or k s, t h e dir ect ion t o t h e desir ed n et w or k (t h e ou t g oin g
in t er fa ce n u m ber ), a n d t h e dist a n ce t o t h ose n et w or k s. A ft er det er m in in g w h ich dir ect ion t o for w a r d t h e
pa ck et , t h e r ou t er m u st bu ild a n ew h ea der . (If y ou w a n t t o r ea d t h e IP r ou t in g t a bles on a W in dow s 9 5 /9 8
w or k st a t ion , t y pe ROUT E PRINT in t h e DOS box .) 4 . Bu ild t h e n ew MA C h ea der a n d for w a r d t h e pa ck et . Fin a lly ,
t h e r ou t er bu ilds a n ew MA C h ea der for t h e pa ck et . T h e MA
 
 
Qu est ion No : 8 - T opic 1
A r ou t er h a s t w o Fa st Et h er n et in t er fa ces a n d n eeds t o con n ect t o fou r V LA Ns in t h e loca l
 n et w or k . H ow ca n y ou a ccom plish t h is t a sk , u sin g t h e few est ph y sica l in t er fa ces a n d
 w it h ou t decr ea sin g n et w or k per for m a n ce?
 
A . Use a h u b t o con n ect t h e fou r V LA NS w it h a Fa st Et h er n et in t er fa ce on t h e r ou t er .
B. A dd a secon d r ou t er t o h a n dle t h e V LA N t r a ffic .
C. A dd t w o m or e Fa st Et h er n et in t er fa ces.
D. Im plem en t a r ou t er -on -a -st ick con fig u r a t ion .
A n sw er : D
Ex pla n a t ion : A r ou t er on a st ick a llow s y ou t o u se su b-in t er fa ces t o cr ea t e m u lt iple log ica l n et w or k s on a sin g le
ph y sica l in t er fa ce.
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Qu est ion No : 9 - T opic 1
Ma t ch t h e t er m s on t h e left w it h t h e a ppr opr ia t e OSI la y er on t h e r ig h t . (Not a ll opt ion s a r e
 u sed.)
 
 
A n sw er :
 
 
Qu est ion No : 1 0 - T opic 1
A n et w or k a dm in ist r a t or is v er ify in g t h e con fig u r a t ion of a n ew ly in st a lled h ost by
 est a blish in g a n FT P con n ect ion t o a r em ot e ser v er . W h a t is t h e h ig h est la y er of t h e
 pr ot ocol st a ck t h a t t h e n et w or k a dm in ist r a t or is u sin g for t h is oper a t ion ?
 
A . a pplica t ion
B. pr esen t a t ion
C. session
D. t r a n spor t
E. in t er n et
F. da t a lin k
A n sw er : A
Ex pla n a t ion : FT P belon g s t o A pplica t ion la y er a n d it is a lso t h e h ig h est la y er of t h e OSI m odel.
 
 
Qu est ion No : 1 1 - T opic 1
Refer t o t h e ex h ibit . Com plet e t h is n et w or k dia g r a m by dr a g g in g t h e cor r ect dev ice n a m e or
 descr ipt ion t o t h e cor r ect loca t ion . Not a ll t h e n a m es or descr ipt ion s w ill be u sed.
 
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A n sw er :
 
 
Qu est ion No : 1 2 - T opic 1
A r eceiv in g h ost com pu t es t h e ch eck su m on a fr a m e a n d det er m in es t h a t t h e fr a m e is
 da m a g ed. T h e fr a m e is t h en disca r ded. A t w h ich OSI la y er did t h is h a ppen ?
 
A . session
B. t r a n spor t
C. n et w or k
D. da t a lin k
E. ph y sica l
A n sw er : D
Ex pla n a t ion : T h e Da t a Lin k la y er pr ov ides t h e ph y sica l t r a n sm ission of t h e da t a a n d h a n dles er r or n ot ifica t ion ,
n et w or k t opolog y , a n d flow con t r ol. T h e Da t a Lin k la y er for m a t s t h e m essa g e in t o pieces, ea ch ca lled a da t a
fr a m e, a n d a dds a cu st om ized h ea der con t a in in g t h e h a r dw a r e dest in a t ion a n d sou r ce a ddr ess. Pr ot ocols Da t a
Un it (PDU) on Da t a lin k la y er is ca lled fr a m e. A ccor din g t o t h is qu est ion t h e fr a m e is da m a g ed a n d disca r ded
w h ich w ill h a ppen a t t h e Da t a Lin k la y er .
 
 
Qu est ion No : 1 3 - T opic 1
W h ich of t h e follow in g cor r ect ly descr ibe st eps in t h e OSI da t a en ca psu la t ion pr ocess?
 (Ch oose t w o.)
 
A . T h e t r a n spor t la y er div ides a da t a st r ea m in t o seg m en t s a n d m a y a dd r elia bilit y a n d flow con t r ol
in for m a t ion .
B. T h e da t a lin k la y er a dds ph y sica l sou r ce a n d dest in a t ion a ddr esses a n d a n FCS t o t h e seg m en t .
C. Pa ck et s a r e cr ea t ed w h en t h e n et w or k la y er en ca psu la t es a fr a m e w it h sou r ce a n d dest in a t ion h ost a ddr esses
a n d pr ot ocol-r ela t ed con t r ol in for m a t ion .
D. Pa ck et s a r e cr ea t ed w h en t h e n et w or k la y er a dds La y er 3 a ddr esses a n d con t r ol in for m a t ion t o a seg m en t .
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E. T h e pr esen t a t ion la y er t r a n sla t es bit s in t o v olt a g es for t r a n sm ission a cr oss t h e ph y sica l l in k .
A n sw er : A ,D
Ex pla n a t ion : T h e t r a n spor t la y er seg m en t s da t a in t o sm a ller pieces for t r a n spor t . Ea ch seg m en t is a ssig n ed a
sequ en ce n u m ber , so t h a t t h e r eceiv in g dev ice ca n r ea ssem ble t h e da t a on a r r iv a l. T h e t r a n spor t la y er a lso u se
flow con t r ol t o m a x im ize t h e t r a n sfer r a t e w h ile m in im izin g t h e r equ ir em en t s t o r et r a n sm it . For ex a m ple, in
T CP, ba sic flow con t r ol is im plem en t ed by a ck n ow ledg m en t by t h e r eceiv er of t h e r eceipt of da t a ; t h e sen der
w a it s for t h is a ck n ow ledg m en t befor e sen din g t h e n ex t pa r t . T h e Net w or k la y er (La y er 3 ) h a s t w o k ey
r espon sibilit ies. Fir st , t h is la y er con t r ols t h e log ica l a ddr essin g of dev ices. Secon d, t h e n et w or k la y er det er m in es
t h e best pa t h t o a pa r t icu la r dest in a t ion n et w or k , a n d r ou t es t h e da t a a ppr opr ia t ely .
 
 
Qu est ion No : 1 4 - T opic 1
Refer t o ex h ibit :
 
 
Which two destination addresses will be used by Host A to send data to Host C? (Choose
 two.)
 
A . t h e IP a ddr ess of Sw it ch 1
B. t h e MA C a ddr ess of Sw it ch 1
C. t h e IP a ddr ess of H ost C
D. t h e MA C a ddr ess of H ost C
E. t h e IP a ddr ess of t h e r ou t er 's E0 in t er fa ce
F. t h e MA C a ddr ess of t h e r ou ter 's E0 in t er fa ce
A n sw er : C,F
Ex pla n a t ion : W h ile t r a n sfer r in g da t a t h r ou g h m a n y differ en t n et w or k s, t h e sou r ce a n d dest in a t ion IP a ddr esses
a r e n ot ch a n g ed. On ly t h e sou r ce a n d dest in a t ion MA C a ddr esses a r e ch a n g ed. So in t h is ca se H ost A w ill u se t h e
IP a ddr ess of H ost C a n d t h e MA C a ddr ess of E0 in t er fa ce t o sen d da t a . W h en t h e r ou t er r eceiv es t h is da t a , it
r epla ces t h e sou r ce MA C a ddr ess w it h it s ow n E1 in t er fa ces MA C a ddr ess a n d r epla ces t h e dest in a t ion MA C
a ddr ess w it h H ost Cs MA C a ddr ess befor e sen din g t o H ost C.
 
 
Qu est ion No : 1 5 - T opic 1
W h a t is t h e differ en ce bet w een a CSU/DSU a n d a m odem ?
 
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A . A CSU/DSU con v er t s a n a log sig n a ls fr om a r ou t er t o a lea sed lin e; a m odem con v er t s a n a log sig n a ls fr om a
r ou t er t o a lea sed lin e.
B. A CSU/DSU con v er t s a n a log sig n a ls fr om a r ou t er t o a ph on e lin e; a m odem con v er t s dig it a l sig n a ls fr om a
r ou t er t o a lea sed lin e.
C. A CSU/DSU con v er t s dig it a l sig n a ls fr om a r ou t er t o a ph on e lin e; a m odem con v er t s a n a log sig n a ls fr om a
r ou t er t o a ph on e lin e.
D. A CSU/DSU con v er t s dig it a l sig n a ls fr om a r ou t er t o a lea sed lin e; a m odem con v er t s dig it a l sig n a ls fr om a
r ou t er t o a ph on e lin e.
A n sw er : D
Ex pla n a t ion : CSU/DSU is u sed t o con v er t dig it a l sig n a ls fr om a r ou t er t o a n et w or k c ir cu it su ch a s a T 1 , w h ile a
m odem is u sed t o con v er t dig it a l sig n a ls ov er a r eg u la r POT S lin e.
 
 
Qu est ion No : 1 6 - T opic 1
W h ich of t h e follow in g descr ibes t h e r oles of dev ices in a W A N? (Ch oose t h r ee.)
 
A . A CSU/DSU t er m in a t es a dig it a l loca l loop.
B. A m odem t er m in a t es a dig it a l loca l loop.
C. A CSU/DSU t er m in a t es a n a n a log loca l loop.
D. A m odem t er m in a t es a n a n a log loca l loop.
E. A r ou t er is com m on ly con sider ed a DT E dev ice.
F. A r ou t er is com m on ly con sider ed a DCE dev ice.
A n sw er : A ,D,E
Ex pla n a t ion : T h e idea beh in d a W A N is t o be a ble t o con n ect t w o DT E n et w or k s t og et h er t h r ou g h a DCE n et w or k .
T h e n et w or k s DCE dev ice (in clu des CSU/DSU) pr ov ides c lock in g t o t h e DT E- con n ect ed in t er fa ce (t h e r ou t er s
ser ia l in t er fa ce).
A modem modulates outgoing digital signals from a computer or other digital device to
analog signals for a conventional copper twisted pair telephone line and demodulates the
incoming analog signal and converts it to a digital signal for the digital device. A CSU/DSU
is used between two digital lines - For more explanation of answer D, in telephony the local
loop (also referred to as a subscriber line) is the physical link or circuit that connects from
the demarcation point of the customer premises to the edge of the carrier or
telecommunications service providers network. Therefore a modem terminates an analog
local loop is correct.
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Qu est ion No : 1 7 - T opic 1
Refer t o t h e ex h ibit . PC_1 is sen din g pa ck et s t o t h e FT P ser v er . Con sider t h e pa ck et s a s
 t h ey lea v e Rou t er A in t er fa ce Fa 0 /0 t ow a r ds Rou t er B. Dr a g t h e cor r ect fr a m e a n d pa ck et
 a ddr ess t o t h eir pla ce in t h e t a ble.
 
 
A n sw er :
 
 
Qu est ion No : 1 8 - T opic 1
Refer t o t h e ex h ibit .
 
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After HostA pings HostB, which entry will be in the ARP cache of HostA to support this
 transmission?
 
 
A . Ex h ibit A
B. Ex h ibit B
C. Ex h ibit C
D. Ex h ibit D
E. Ex h ibit E
F. Ex h ibit F
A n sw er : A
Ex pla n a t ion : H ost A k n ow s h ost B is in a n ot h er n et w or k so it w ill sen d t h e pin g s t o it s defa u lt g a t ew a y
1 9 2 .1 6 8 .6 .1 . H ost A sen ds a br oa dca st fr a m e a sk in g t h e MA C a ddr ess of 1 9 2 .1 6 8 .6 .1 . T h is in for m a t ion (IP a n d
MA C a ddr ess of t h e defa u lt g a t ew a y ) is sa v ed in it s A RP ca ch e for la t er u se.
 
 
Qu est ion No : 1 9 - T opic 1
Refer t o t h e ex h ibit .
 
 
Which three statements correctly describe Network Device A? (Choose three.)
 
A . W it h a n et w or k w ide m a sk of 2 5 5 .2 5 5 .2 5 5 .1 2 8 , ea ch in t er fa ce does n ot r equ ir e a n IP a ddr ess.
B. W it h a n et w or k w ide m a sk of 2 5 5 .2 5 5 .2 5 5 .1 2 8 , ea ch in t er fa ce does r equ ir e a n IP a ddr ess on a u n iqu e IP
su bn et .
C. W it h a n et w or k w ide m a sk of 2 5 5 .2 5 5 .2 5 5 .0 , m u st be a La y er 2 dev ice for t h e PCs t o com m u n ica t e w it h
ea ch ot h er .
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D. W it h a n et w or k w ide m a sk of 2 5 5 .2 5 5 .2 5 5 .0 , m u st be a La y er 3 dev ice for t h e PCs t o com m u n ica t e w it h
ea ch ot h er .
E. W it h a n et w or k w ide m a sk of 2 5 5 .2 5 5 .2 5 4 .0 , ea ch in t er fa ce does n ot r equ ir e a n IP a ddr ess.
A n sw er : B,D,E
Ex pla n a t ion : If Su bn et Ma sk is 2 5 5 .2 5 5 .2 5 5 .1 2 8 t h e h ost s v a r y fr om x .x .x .0 - x .x .x .1 2 7 & x .x .x .1 2 8 -
x .x .x .2 5 5 , so t h e IP A ddr esses of 2 h ost s fa ll in differ en t su bn et s so ea ch in t er fa ce n eeds a n IP a n a ddr ess so t h a t
t h ey ca n com m u n ica t e ea ch ot h er . If Su bn et Ma sk is 2 5 5 .2 5 5 .2 5 5 .0 t h e 2 specified h ost s fa ll in differ en t
su bn et s so t h ey n eed a La y er 3 dev ice t o com m u n ica t e. If Su bn et Ma sk is 2 5 5 .2 5 5 .2 5 4 .0 t h e 2 specified h ost s
a r e in sa m e su bn et so a r e in n et w or k a ddr ess a n d ca n be a ccom m oda t ed in sa m e La y er 2 dom a in a n d ca n
com m u n ica t e w it h ea ch ot h er dir ect ly u sin g t h e La y er 2 a ddr ess.
 
 
Qu est ion No : 2 0 - T opic 1
For w h a t t w o pu r poses does t h e Et h er n et pr ot ocol u se ph y sica l a ddr esses? (Ch oose t w o.)
 
A . t o u n iqu ely iden t ify dev ices a t La y er 2
B. t o a llow com m u n ica t ion w it h dev ices on a differ en t n et w or k
C. t o differ en t ia t e a La y er 2 fr a m e fr om a La y er 3 pa ck et
D. t o est a blish a pr ior it y sy st em t o det er m in e w h ich dev ice g et s t o t r a n sm it fir st
E. t o a llow com m u n ica t ion bet w een differ en t dev ices on t h e sa m e n et w or k
F. t o a llow det ect ion of a r em ot e dev ice w h en it s ph y sica l a ddr ess is u n k n ow n
A n sw er : A ,E
Ex pla n a t ion : Ph y sica l a ddr esses or MA C a ddr esses a r e u sed t o iden t ify dev ices a t la y er 2 . MA C a ddr esses a r e
on ly u sed t o com m u n ica t e on t h e sa m e n et w or k . T o com m u n ica t e on differ en t n et w or k w e h a v e t o u se La y er 3
a ddr esses (IP a ddr esses) ->B is n ot cor r ect . La y er 2 fr a m e a n d La y er 3 pa ck et ca n be r ecog n ized v ia h ea der s.
La y er 3 pa ck et a lso con t a in s ph y sica l a ddr ess ->. On Et h er n et , ea ch fr a m e h a s t h e sa m e pr ior it y t o t r a n sm it by
defa u lt ->. A ll dev ices n eed a ph y sica l a ddr ess t o iden t ify it self. If n ot , t h ey ca n n ot com m u n ica t e ->.
 
 
Qu est ion No : 2 1 - T opic 1
In a n Et h er n et n et w or k , u n der w h a t t w o scen a r ios ca n dev ices t r a n sm it ? (Ch oose t w o.)
 
A . w h en t h ey r eceiv e a specia l t ok en
B. w h en t h er e is a ca r r ier
C. w h en t h ey det ect n o ot h er dev ices a r e sen din g
D. w h en t h e m ediu m is idle
E. w h en t h e ser v er g r a n t s a ccess
A n sw er : C,D
Ex pla n a t ion : Et h er n et n et w or k is a sh a r ed en v ir on m en t so a ll dev ices h a v e t h e r ig h t t o a ccess t o t h e m ediu m . If
m or e t h a n on e dev ice t r a n sm it s sim u lt a n eou sly , t h e sig n a ls collide a n d ca n n ot r ea ch t h e dest in a t ion . If a dev ice
det ect s a n ot h er dev ice is sen din g , it w ill w a it for a specified a m ou n t of t im e befor e a t t em pt in g t o t r a n sm it .
W h en t h er e is n o t r a ffic det ect ed, a dev ice w ill t r a n sm it it s m essa g e. W h ile t h is t r a n sm ission is occu r r in g , t h e
dev ice con t in u es t o list en for t r a ffic or collision s on t h e LA N. A ft er t h e m essa g e is sen t , t h e dev ice r et u r n s t o it s
defa u lt l ist en in g m ode.
 
 
Qu est ion No : 2 2 - T opic 1
Refer t o t h e ex h ibit :
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What will Router1 do when it receives the data frame shown? (Choose three.)
 
A . Rou t er 1 w ill st r ip off t h e sou r ce MA C a ddr ess a n d r epla ce it w it h t h e MA C a ddr ess 0 0 0 0 .0 c3 6 .6 9 6 5 .
B. Rou t er 1 w ill st r ip off t h e sou r ce IP a ddr ess a n d r epla ce it w it h t h e IP a ddr ess 1 9 2 .1 6 8 .4 0 .1 .
C. Rou t er 1 w ill st r ip off t h e dest in a t ion MA C a ddr ess a n d r epla ce it w it h t h e MA C a ddr ess 0 0 0 0 .0 c0 7 .4 3 2 0 .
D. Rou t er 1 w ill st r ip off t h e dest in a t ion IP a ddr ess a n d r epla ce it w it h t h e IP a ddr ess of 1 9 2 .1 6 8 .4 0 .1 .
E. Rou t er 1 w ill for w a r d t h e da t a pa ck et ou t in t er fa ce Fa st Et h er n et 0 /1 .
F. Rou t er 1 w ill for w a r d t h e da t a pa ck et ou t in t er fa ce Fa st Et h er n et 0 /2 .
A n sw er : A ,C,F
Ex pla n a t ion : Rem em ber , t h e sou r ce a n d dest in a t ion MA C ch a n g es a s ea ch r ou t er h op a lon g w it h t h e T T L bein g
decr em en t ed bu t t h e sou r ce a n d dest in a t ion IP a ddr ess r em a in t h e sa m e fr om sou r ce t o dest in a t ion .
 
 
Qu est ion No : 2 3 - T opic 1
W h ich t h r ee st a t em en t s a ccu r a t ely descr ibe La y er 2 Et h er n et sw it ch es? (Ch oose t h r ee.)
 
A . Spa n n in g T r ee Pr ot ocol a llow s sw it ch es t o a u t om a t ica lly sh a r e V LA N in for m a t ion .
B. Est a blish in g V LA Ns in cr ea ses t h e n u m ber of br oa dca st dom a in s.
C. Sw it ch es t h a t a r e con fig u r ed w it h V LA Ns m a k e for w a r din g decision s ba sed on bot h La y er 2 a n d La y er 3
a ddr ess in for m a t ion .
D. Micr oseg m en t a t ion decr ea ses t h e n u m ber of collision s on t h e n et w or k .
E. In a pr oper ly fu n ct ion in g n et w or k w it h r edu n da n t sw it ch ed pa t h s, ea ch sw it ch ed seg m en t w ill con t a in on e
r oot br idg e w it h a ll it s por t s in t h e for w a r din g st a t e. A ll ot h er sw it ch es in t h a t br oa dca st dom a in w ill h a v e on ly
on e r oot por t .
F. If a sw it ch r eceiv es a fr a m e for a n u n k n ow n dest in a t ion , it u ses A RP t o r esolv e t h e a ddr ess.
A n sw er : B,D,E
Ex pla n a t ion : Micr oseg m en t a t ion is a n et w or k desig n (fu n ct ion a lit y ) w h er e ea ch w or k st a t ion or dev ice on a
n et w or k g et s it s ow n dedica t ed seg m en t (collision dom a in ) t o t h e sw it ch . Ea ch n et w or k dev ice g et s t h e fu ll
ba n dw idt h of t h e seg m en t a n d does n ot h a v e t o sh a r e t h e seg m en t w it h ot h er dev ices. Micr oseg m en t a t ion
r edu ces a n d ca n ev en elim in a t e collision s beca u se ea ch seg m en t is it s ow n collision dom a in ->. Not e:
Micr oseg m en t a t ion decr ea ses t h e n u m ber of collision s bu t it in cr ea ses t h e n u m ber of collision dom a in s.
 
 
Qu est ion No : 2 4 - T opic 1
Refer t o t h e ex h ibit .
 
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Host A pings interface S0/0 on router 3. What is the TTL value for that ping?
 
A . 2 5 2
B. 2 5 3
C. 2 5 4
D. 2 5 5
A n sw er : B
Ex pla n a t ion : Fr om t h e CCNA ICND2 Ex a m book : Rou t er s decr em en t t h e T T L by 1 ev er y t im e t h ey for w a r d a
pa ck et ; if a r ou t er decr em en t s t h e T T L t o 0 , it t h r ow s a w a y t h e pa ck et . T h is pr ev en t s pa ck et s fr om r ot a t in g
for ev er . I w a n t t o m a k e it c lea r t h a t befor e t h e r ou t er for w a r ds a pa ck et , t h e T T L is st il l r em a in t h e sa m e. For
ex a m ple in t h e t opolog y a bov e, pin g s t o S0 /1 a n d S0 /0 of Rou t er 2 h a v e t h e sa m e T T L. T h e pict u r e below sh ow s
T T L v a lu es for ea ch in t er fa ce of ea ch r ou t er a n d for H ost B. Not ice t h a t H ost A in it ia lizes ICMP pa ck et w it h a T T L
of 2 5 5 :
 
 
Qu est ion No : 2 5 - T opic 1
Refer t o t h e ex h ibit .
 
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What two results would occur if the hub were to be replaced with a switch that is configured
 with one Ethernet VLAN? (Choose two.)
 
A . T h e n u m ber of collision dom a in s w ou ld r em a in t h e sa m e.
B. T h e n u m ber of collision dom a in s w ou ld decr ea se.
C. T h e n u m ber of collision dom a in s w ou ld in cr ea se.
D. T h e n u m ber of br oa dca st dom a in s w ou ld r em a in t h e sa m e.
E. T h e n u m ber of br oa dca st dom a in s w ou ld decr ea se.
F. T h e n u m ber of br oa dca st dom a in s w ou ld in cr ea se.
A n sw er : C,D
Ex pla n a t ion : Ba sica lly , a collision dom a in is a n et w or k seg m en t t h a t a llow s n or m a l n et w or k t r a ffic t o flow ba ck
a n d for t h . In t h e old da y s of h u bs, t h is m ea n t y ou h a d a lot of collision s, a n d t h e old CSMA /CD w ou ld be w or k in g
ov er t im e t o t r y t o g et t h ose pa ck et s r e-sen t ev er y t im e t h er e w a s a collision on t h e w ir e (sin ce Et h er n et a llow s
on ly on e h ost t o be t r a n sm it t in g a t on ce w it h ou t t h er e bein g a t r a ffic ja m ). W it h sw it ch es, y ou br ea k u p
collision dom a in s by sw it ch in g pa ck et s bou n d for ot h er collision dom a in s. T h ese da y s, sin ce w e m ost ly u se
sw it ch es t o con n ect com pu t er s t o t h e n et w or k , y ou g en er a lly h a v e on e collision dom a in t o a PC. Br oa dca st
dom a in s a r e ex a ct ly w h a t t h ey im ply : th ey a r e n et w or k seg m en t s t h a t a llow br oa dca st s t o be sen t a cr oss t h em .
Sin ce sw it ch es a n d br idg es a llow for br oa dca st t r a ffic t o g o u n sw it ch ed, br oa dca st s ca n t r a v er se collision
dom a in s fr eely . Rou t er s, h ow ev er , don 't a llow br oa dca st s t h r ou g h by defa u lt , so w h en a br oa dca st h it s a r ou t er
(or t h e per im et er of a V LA N), it doesn 't g et for w a r ded. T h e sim ple w a y t o look a t it is t h is w a y : sw it ch es br ea k
u p collision dom a in s, w h ile r ou t er s (a n d V LA Ns) br ea k u p collision dom a in s a n d br oa dca st dom a in s. A lso, a
br oa dca st dom a in ca n con t a in m u lt iple collision dom a in s, bu t a collision dom a in ca n n ev er h a v e m or e t h a n on e
br oa dca st dom a in a ssocia t ed w it h it . Collision Dom a in : A g r ou p of Et h er n et or Fa st Et h er n et dev ices in a
CSMA /CD LA N t h a t a r e con n ect ed by r epea t er s a n d com pet e for a ccess on t h e n et w or k . On ly on e dev ice in t h e
collision dom a in m a y t r a n sm it a t a n y on e t im e, a n d t h e ot h er dev ices in t h e dom a in list en t o t h e n et w or k in
or der t o a v oid da t a collision s. A collision dom a in is som et im es r efer r ed t o a s a n Et h er n et seg m en t . Br oa dca st
Dom a in : Br oa dca st in g sen ds a m essa g e t o ev er y on e on t h e loca l n et w or k (su bn et ). A n ex a m ple for Br oa dca st in g
w ou ld be DH CP Requ est fr om a Clien t PC. T h e Clien t is a sk in g for a IP A ddr ess, bu t t h e c lien t does n ot k n ow h ow
t o r ea ch t h e DH CP Ser v er . So t h e c lien t se
 
 
Qu est ion No : 2 6 - T opic 1
Refer t o t h e ex h ibit . PC_1 is ex ch a n g in g pa ck et s w it h t h e FT P ser v er . Con sider t h e
 pa ck et s a s t h ey lea v e Rou t er B in t er fa ce Fa 0 /1 t ow a r ds Rou t er A . Dr a g t h e cor r ect fr a m e
 a n d pa ck et a ddr esses t o t h eir pla ce in t h e t a ble.
 
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A n sw er :
Ex pla n a t ion : Sou r ce Ma c A ddr essDest in a t ion Ma c A ddr essSou r ce IP a ddr essDest in a t ion MA C a ddr ess MA C
0 0 0 0 .0 c8 9 .3 3 3 3 MA C 0 0 0 0 .0 c8 9 .9 9 9 9 IP 1 7 2 .1 6 .3 4 .2 5 0 IP 1 7 2 .1 6 .2 1 .7
 
 
Qu est ion No : 2 7 - T opic 1
Dr a g t h e ca ble t y pe on t h e left t o t h e pu r pose for w h ich it is best su it ed on t h e r ig h t . (Not a ll
 opt ion s a r e u sed.)
 
 
A n sw er :
Ex pla n a t ion :
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To remember which type of cable you should use, follow these tips: - To connect two serial
interfaces of 2 routers we use serial cable To specify when we use crossover cable or
straight-through cable, we should remember: Group 1: Router, Host, Server Group 2: Hub,
Switch One device in group 1 + One device in group 2: use straight-through cable Two
devices in the same group: use crossover cable For example: we use straight-through cable
to connect switch to router, switch to host, hub to host, hub to server and we use crossover
cable to connect switch to switch, switch to hub, router to router, host to host) Topic 2, LAN
Switching Technologies
 
 
Qu est ion No : 2 8 - T opic 2
Refer t o t h e ex h ibit .
 
 
A technician has installed SwitchB and needs to configure it for remote access from the
 
A . Sw it ch B(con fig )# in t er fa ce Fa st Et h er n et 0 /1 Sw it ch B(con fig -if)# ip a ddr ess 1 9 2 .1 6 8 .8 .2 5 2 2 5 5 .2 5 5 .2 5 5 .0
Sw it ch B(con fig -if)# n o sh u t dow n
B. Sw it ch B(con fig )# in t er fa ce v la n 1 Sw it ch B(con fig -if)# ip a ddr ess 1 9 2 .1 6 8 .8 .2 5 2 2 5 5 .2 5 5 .2 5 5 .0
Sw it ch B(con fig -if)# ip defa u lt -g a t ew a y 1 9 2 .1 6 8 .8 .2 5 4 2 5 5 .2 5 5 .2 5 5 .0 Sw it ch B(con fig -if)# n o sh u t dow n
C. Sw it ch B(con fig )# ip defa u lt -g a t ew a y 1 9 2 .1 6 8 .8 .2 5 4 Sw it ch B(con fig )# in t er fa ce v la n 1 Sw it ch B(con fig -if)#
ip a ddr ess 1 9 2 .1 6 8 .8 .2 5 2 2 5 5 .2 5 5 .2 5 5 .0 Sw it ch B(con fig -if)# n o sh u t dow n
D. Sw it ch B(con fig )# ip defa u lt -n et w or k 1 9 2 .1 6 8 .8 .2 5 4 Sw it ch B(con fig )# in t er fa ce v la n 1 Sw it ch B(con fig -if)# ip
a ddr ess 1 9 2 .1 6 8 .8 .2 5 2 2 5 5 .2 5 5 .2 5 5 .0 Sw it ch B(con fig -if)# n o sh u t dow n
E. Sw it ch B(con fig )# ip r ou t e 1 9 2 .1 6 8 .8 .2 5 4 2 5 5 .2 5 5 .2 5 5 .0 Sw it ch B(con fig )# in t er fa ce Fa st Et h er n et 0 /1
Sw it ch B(con fig -if)# ip a ddr ess 1 9 2 .1 6 8 .8 .2 5 2 2 5 5 .2 5 5 .2 5 5 .0 Sw it ch B(con fig -if)# n o sh u t dow n
A n sw er : C
Ex pla n a t ion : T o r em ot e a ccess t o Sw it ch B, it m u st h a v e a m a n a g em en t IP a ddr ess on a V LA N on t h a t sw it ch .
T r a dit ion a lly , w e oft en u se V LA N 1 a s t h e m a n a g em en t V LA N (bu t in fa ct it is n ot secu r e). In t h e ex h ibit , w e
ca n r ecog n ize t h a t t h e Ma n a g em en t W or k st a t ion is in a differ en t su bn et fr om t h e Sw it ch B. For in t er su bn et w or k
com m u n ica t ion t o occu r , y ou m u st con fig u r e a t lea st on e defa u lt g a t ew a y . T h is defa u lt g a t ew a y is u sed t o
for w a r d t r a ffic or ig in a t in g fr om t h e sw it ch on ly , n ot t o for w a r d t r a ffic sen t by dev ices con n ect ed t o t h e sw it ch .
 
 
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Qu est ion No : 2 9 - T opic 2
A sw it ch is con fig u r ed w it h a ll por t s a ssig n ed t o V LA N 2 w it h fu ll du plex Fa st Et h er n et t o
 seg m en t ex ist in g depa r t m en t a l t r a ffic . W h a t is t h e effect of a ddin g sw it ch por t s t o a n ew
 V LA N on t h e sw it ch ?
 
A . Mor e collision dom a in s w ill be cr ea t ed.
B. IP a ddr ess u t iliza t ion w ill be m or e effic ien t .
C. Mor e ba n dw idt h w ill be r equ ir ed t h a n w a s n eeded pr ev iou sly .
D. A n a ddit ion a l br oa dca st dom a in w ill be cr ea t ed.
A n sw er : D
Ex pla n a t ion : Ea ch V LA N cr ea t es it s ow n br oa dca st dom a in . Sin ce t h is is a fu ll du plex sw it ch , ea ch por t is a
sepa r a t e collision dom a in .
 
 
Qu est ion No : 3 0 - T opic 2
Refer t o t h e ex h ibit .
 
 
Switch-1 needs to send data to a host with a MAC address of 00b0.d056.efa4. What will
 Switch-1 do with this data?
 
A . Sw it ch -1 w ill dr op t h e da t a beca u se it does n ot h a v e a n en t r y for t h a t MA C a ddr ess.
B. Sw it ch -1 w ill flood t h e da t a ou t a ll of it s por t s ex cept t h e por t fr om w h ich t h e da t a or ig in a t ed.
C. Sw it ch -1 w ill sen d a n A RP r equ est ou t a ll it s por t s ex cept t h e por t fr om w h ich t h e da t a or ig in a t ed.
D. Sw it ch -1 w ill for w a r d t h e da t a t o it s defa u lt g a t ew a y .
A n sw er : B
Ex pla n a t ion : T h is qu est ion t est s t h e oper a t in g pr in ciples of t h e La y er 2 sw it ch . Ch eck t h e MA C a ddr ess t a ble of
Sw it ch 1 a n d fin d t h a t t h e MA C a ddr ess of t h e h ost does n ot ex ist in t h e t a ble. Sw it ch 1 w ill flood t h e da t a ou t a ll
of it s por t s ex cept t h e por t fr om w h ich t h e da t a or ig in at ed t o det er m in e w h ich por t t h e h ost is loca t ed in .
Sw it ch es w or k a s follow s: ✑ Sw it ch es lea r n t h e MA C a ddr esses of PCs or w or k st a t ion s t h a t a r e con n ect ed t o t h eir
sw it ch por t s by ex a m in in g t h e sou r ce a ddr ess of fr a m es t h a t a r e r eceiv ed on t h a t por t . ✑ Ma ch in es m a y h a v e
been r em ov ed fr om a por t , t u r n ed off, or m ov ed t o a n ot h er por t on t h e sa m e sw it ch or a differ en t sw it ch . ✑ T h is
cou ld ca u se con fu sion in fr a m e for w a r din g . ✑ T h e MA C a ddr ess en t r y is a u t om a t ica lly disca r ded or a g ed ou t
a ft er 3 0 0 secon ds ✑ If t h er e is n ot MA C a ddr ess of dest in a t ion h ost in MA C t a ble, sw it ch sen ds br oa dca st t o a ll
por t s ex cept t h e sou r ce t o fin d ou t t h e dest in a t ion h ost . In ou t pu t t h er e is n o MA C a ddr ess of g iv e h ost so sw it ch
floods t o a ll por t s ex cept t h e sou r ce por t .
 
 
Qu est ion No : 3 1 - T opic 2
W h ich st a t em en t a bou t V LA N oper a t ion on Cisco Ca t a ly st sw it ch es is t r u e?
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A . W h en a pa ck et is r eceiv ed fr om a n 8 0 2 .1 Q t r u n k , t h e V LA N ID ca n be det er m in ed fr om t h e sou r ce MA C
a ddr ess a n d t h e MA C a ddr ess t a ble.
B. Un k n ow n u n ica st fr a m es a r e r et r a n sm it t ed on ly t o t h e por t s t h a t belon g t o t h e sa m e V LA N.
C. Br oa dca st a n d m u lt ica st fr a m es a r e r et r a n sm it t ed t o por t s t h a t a r e con fig u r ed on differ en t V LA N.
D. Por t s bet w een sw it ch es sh ou ld be con fig u r ed in a ccess m ode so t h a t V LA Ns ca n spa n a cr oss t h e por t s.
A n sw er : B
Ex pla n a t ion : Ea ch V LA N r esides in it s ow n br oa dca st dom a in , so in com in g fr a m es w it h u n k n ow n dest in a t ion s
a r e on ly t r a n sm it t ed t o por t s t h a t r eside in t h e sa m e V LA N a s t h e in com in g fr a m e.
 
 
Qu est ion No : 3 2 - T opic 2
W h ich com m a n d en a bles RST P on a sw it ch ?
 
A . spa n n in g -t r ee u plin k fa st
B. spa n n in g -t r ee m ode r a pid-pv st
C. spa n n in g -t r ee ba ck bon efa st
D. spa n n in g -t r ee m ode m st
A n sw er : B
Ex pla n a t ion : Ra pid Spa n n in g T r ee Pr ot ocol (RST P) is a n en h a n cem en t of t h e or ig in a l ST P 8 0 2 .1 D pr ot ocol. T h e
RST P 8 0 2 .1 w pr ot ocol is a n IEEE open im plem en t a t ion . Cisco h a s it s ow n pr opr iet a r y im plem en t a t ion of RST P,
t h a t in clu des t h e ben efit s of it s Per -V LA N spa n n in g t r ee pr ot ocols, ca lled Ra pid-PV ST +. T o a ct iv a t e t h e Ra pid-
PV ST + pr ot ocol: sw it ch (con fig )#spa n n in g -t r ee m ode r a pid-pv st
 
 
Qu est ion No : 3 3 - T opic 2
Refer t o t h e ex h ibit .
 
 
Each of these four switches has been configured with a hostname, as well as being
 configured to run RSTP. No other configuration changes have been made. Which three of
 these show the correct RSTP port roles for the indicated switches and interfaces? (Choose
 three.)
 
A . Sw it ch A , Fa 0 /2 , desig n a t ed
B. Sw it ch A , Fa 0 /1 , r oot
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C. Sw it ch B, Gi0 /2 , r oot
D. Sw it ch B, Gi0 /1 , desig n a t ed
E. Sw it ch C, Fa 0 /2 , r oot
F. Sw it ch D, Gi0 /2 , r oot
A n sw er : A ,B,F
Ex pla n a t ion : T h e qu est ion sa y s n o ot h er con fig u r a t ion ch a n g es h a v e been m a de so w e ca n u n der st a n d t h ese
sw it ch es h a v e t h e sa m e br idg e pr ior it y . Sw it ch C h a s low est MA C a ddr ess so it w ill becom e r oot br idg e a n d 2 of
it s por t s (Fa 0 /1 & Fa 0 /2 ) w ill be desig n a t ed por t s. Beca u se Sw it ch C is t h e r oot br idg e so t h e 2 por t s n ea r est
Sw it ch C on Sw it ch A (Fa 0 /1 ) a n d Sw it ch D (Gi0 /2 ) w ill be r oot por t s. Now w e com e t o t h e m ost difficu lt pa r t of
t h is qu est ion : Sw it ch B m u st h a v e a r oot por t so w h ich por t w ill it ch oose? T o a n sw er t h is qu est ion w e n eed t o
k n ow a bou t ST P cost a n d por t cost . In g en er a l, cost is ca lcu la t ed ba sed on ba n dw idt h of t h e lin k . T h e h ig h er t h e
ba n dw idt h on a lin k , t h e low er t h e v a lu e of it s cost . Below a r e t h e cost v a lu es y ou sh ou ld m em or ize:
SwitchB will choose the interface with lower cost to the root bridge as the root port so we
must calculate the cost on interface Gi0/1 & Gi0/2 of SwitchB to the root bridge. This can
be calculated from the cost to the root bridge of each switch because a switch always
advertises its cost to the root bridge in its BPDU. The receiving switch will add its local port
cost value to the cost in the BPDU. One more thing to notice is that a root bridge always
advertises the cost to the root bridge (itself) with an initial value of 0. Now lets have a look
at the topology again
SwitchC advertises its cost to the root bridge with a value of 0. Switch D adds 4 (the cost
value of 1Gbps link) and advertises this value (4) to SwitchB. SwitchB adds another 4 and
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learns that it can reach SwitchC via Gi0/1 port with a total cost of 8. The same process
happens for SwitchA and SwitchB learns that it can reach SwitchC via Gi0/2 with a total
cost of 23 -> Switch B chooses Gi0/1 as its root port ->. Now our last task is to identify the
port roles of the ports between SwitchA & SwitchB. It is rather easy as the MAC address of
SwitchA is lower than that of Switch
 
 
Qu est ion No : 3 4 - T opic 2
W h ich por t st a t e is in t r odu ced by Ra pid-PV ST ?
 
A . lea r n in g
B. list en in g
C. disca r din g
D. for w a r din g
A n sw er : C
Ex pla n a t ion : PV ST + is ba sed on IEEE8 0 2 .1 D Spa n n in g T r ee Pr ot ocol (ST P). Bu t PV ST + h a s on ly 3 por t st a t es
(disca r din g , lea r n in g a n d for w a r din g ) w h ile ST P h a s 5 por t st a t es (block in g , l ist en in g , lea r n in g , for w a r din g
a n d disa bled). So disca r din g is a n ew por t st a t e in PV ST +.
 
 
Qu est ion No : 3 5 - T opic 2
Refer t o t h e ex h ibit .
 
 
A frame on VLAN 1 on switch S1 is sent to switch S2 where the frame is received on VLAN
 2. What causes this behavior?
 
A . t r u n k m ode m ism a t ch es
B. a llow in g on ly V LA N 2 on t h e dest in a t ion
C. n a t iv e V LA N m ism a t ch es
D. V LA Ns t h a t do n ot cor r espon d t o a u n iqu e IP su bn et
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A n sw er : C
Ex pla n a t ion : Un t a g g ed fr a m es a r e en ca psu la t ed w it h t h e n a t iv e V LA N. In t h is ca se, t h e n a t iv e V LA Ns a r e
differ en t so a lt h ou g h S1 w ill t a g it a s V LA N 1 it w ill be r eceiv ed by S2 .
 
 
Qu est ion No : 3 6 - T opic 2
Refer t o t h e ex h ibit .
 
 
Which switch provides the spanning-tree designated port role for the network segment that
 services the printers?
 
A . Sw it ch 1
B. Sw it ch 2
C. Sw it ch 3
D. Sw it ch 4
A n sw er : C
Ex pla n at ion : Pr in t er s a r e con n ect ed by h u bs. Decide t h e sw it ch t h a t pr ov ides t h e spa n n in g -t r ee desig n a t ed por t
r ole bet w een Sw it ch 3 a n d Sw it ch 4 . T h ey h a v e t h e sa m e pr ior it y 3 2 7 6 8 . Com pa r e t h eir MA C a ddr esses.
Sw it ch 3 w it h a sm a ller MA C a ddr ess w ill pr ov ide a desig n a t ed por t for pr in t er s.
 
 
Qu est ion No : 3 7 - T opic 2
W h ich sw it ch w ou ld ST P ch oose t o becom e t h e r oot br idg e in t h e select ion pr ocess?
 
A . 3 2 7 6 8 : 1 1 -2 2 -3 3 -4 4 -5 5 -6 6
B. 3 2 7 6 8 : 2 2 -3 3 -4 4 -5 5 -6 6 -7 7
C. 3 2 7 6 9 : 1 1 -2 2 -3 3 -4 4 -5 5 -6 5
D. 3 2 7 6 9 : 2 2 -3 3 -4 4 -5 5 -6 6 -7 8
A n sw er : A
Ex pla n a t ion : T h e r oot br idg e of t h e spa n n in g t r ee is t h e br idg e w it h t h e sm a llest (low est ) br idg e ID. Ea ch br idg e
h a s a con fig u r a ble pr ior it y n u m ber a n d a MA C A ddr ess; t h e br idg e ID con t a in s bot h n u m ber s com bin ed
t og et h er - Br idg e pr ior it y + MA C (3 2 7 6 8 .0 2 0 0 .0 0 0 0 .1 1 1 1 ). T h e Br idg e pr ior it y defa u lt is 3 2 7 6 8 a n d ca n on ly
be con fig u r ed in m u lt iples of 4 0 9 6 (Spa n n in g t r ee u ses t h e 1 2 bit s ex t en ded sy st em ID). T o com pa r e t w o br idg e
IDs, t h e pr ior it y is com pa r ed fir st , a s if look in g a t a r ea l n u m ber a n y t h in g less t h a n 3 2 7 6 8 . . .w ill becom e t h e
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t a r g et of bein g t h e r oot . If t w o br idg es h a v e equ a l pr ior it y t h en t h e MA C a ddr esses a r e com pa r ed; for ex a m ple, if
sw it ch es A (MA C=0 2 0 0 .0 0 0 0 .1 1 1 1 ) a n d B (MA C=0 2 0 0 .0 0 0 0 .2 2 2 2 ) bot h h a v e a pr ior it y of 3 2 7 6 8 t h en sw it ch
A w ill be select ed a s t h e r oot br idg e. In t h is ca se, 3 2 7 6 8 : 1 1 -2 2 -3 3 -4 4 -5 5 -6 6 w ou ld be t h e br idg e beca u se it h a s
a low er pr ior it y a n d MA C a ddr ess.
 
 
Qu est ion No : 3 8 - T opic 2
Refer t o t h e ex h ibit .
 
 
Why has this switch not been elected the root bridge for VLAN1?
 
A . It h a s m or e t h a n on e in t er fa ce t h a t is con n ect ed t o t h e r oot n et w or k seg m en t .
B. It is r u n n in g RST P w h ile t h e elect ed r oot br idg e is r u n n in g 8 0 2 .1 d spa n n in g t r ee.
C. It h a s a h ig h er MA C a ddr ess t h a n t h e elect ed r oot br idg e.
D. It h a s a h ig h er br idg e ID t h a n t h e elect ed r oot br idg e.
A n sw er : D
Ex pla n a t ion : T h e r oot br idg e is det er m in ed by t h e low est br idg e ID, a n d t h is sw it ch h a s a br idg e ID pr ior it y of
3 2 7 6 8 , w h ich is h ig h er t h a n t h e r oot s pr ior it y of 2 0 4 8 1 .
 
 
Qu est ion No : 3 9 - T opic 2
Cisco Ca t a ly st sw it ch es CA T 1 a n d CA T 2 h a v e a con n ect ion bet w een t h em u sin g por t s
 FA 0 /1 3 . A n 8 0 2 .1 Q t r u n k is con fig u r ed bet w een t h e t w o sw it ch es. On CA T 1 , V LA N 1 0 is
 ch osen a s n a t iv e, bu t on CA T 2 t h e n a t iv e V LA N is n ot specified.
 W h a t w ill h a ppen in t h is scen a r io?
 
A . 8 0 2 .1 Q g ia n t s fr a m es cou ld sa t u r a t e t h e lin k .
B. V LA N 1 0 on CA T 1 a n d V LA N 1 on CA T 2 w ill sen d u n t a g g ed fr a m es.
C. A n a t iv e V LA N m ism a t ch er r or m essa g e w ill a ppea r .
D. V LA N 1 0 on CA T 1 a n d V LA N 1 on CA T 2 w ill sen d t a g g ed fr a m es.
A n sw er : C
Ex pla n a t ion : A n a t iv e V LA N m ism a t ch er r or w ill a ppea r by CDP if t h er e is a n a t iv e V LA N m ism a t ch on a n
8 0 2 .1 Q lin k . V LA N m ism a t ch ca n ca u se t r a ffic fr om on e v la n t o lea k in t o a n ot h er v la n .
 
 
Qu est ion No : 4 0 - T opic 2
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W h a t a r e t h r ee a dv a n t a g es of V LA Ns? (Ch oose t h r ee.)
 
A . V LA Ns est a blish br oa dca st dom a in s in sw it ch ed n et w or k s.
B. V LA Ns u t ilize pa ck et filt er in g t o en h a n ce n et w or k secu r it y .
C. V LA Ns pr ov ide a m et h od of con ser v in g IP a ddr esses in la r g e n et w or k s.
D. V LA Ns pr ov ide a low -la t en cy in t er n et w or k in g a lt er n a t iv e t o r ou t ed n et w or k s.
E. V LA Ns a llow a ccess t o n et w or k ser v ices ba sed on depa r t m en t , n ot ph y sica l loca t ion .
F. V LA Ns ca n g r ea t ly sim plify a ddin g , m ov in g , or ch a n g in g h ost s on t h e n et w or k .
A n sw er : A ,E,F
Ex pla n a t ion : V LA N t ech n olog y is oft en u sed in pr a ct ice, beca u se it ca n bet t er con t r ol la y er 2 br oa dca st t o
im pr ov e n et w or k secu r it y . T h is m a k es n et w or k m or e flex ible a n d sca la ble. Pa ck et filt er in g is a fu n ct ion of
fir ew a ll in st ea d of V LA N.
 
 
Qu est ion No : 4 1 - T opic 2
Refer t o t h e ex h ibit .
 
 
How should the FastEthernet0/1 ports on the switches that are shown in the exhibit be
 configured to allow connectivity between all devices?
 
A . T h e por t s on ly n eed t o be con n ect ed by a cr ossov er ca ble.
B. Sw it ch X (con fig )# in t er fa ce fa st et h er n et 0 /1 Sw it ch X (con fig -if)# sw it ch por t m ode t r u n k
C. Sw it ch X (con fig )# in t er fa ce fa st et h er n et 0 /1 Sw it ch X (con fig -if)# sw it ch por t m ode a ccess Sw it ch X (con fig -if)#
sw it ch por t a ccess v la n 1
D. Sw it ch X (con fig )# in t er fa ce fa st et h er n et 0 /1 Sw it ch X (con fig -if)# sw it ch por t m ode t r u n k Sw it ch X (con fig -if)#
sw it ch por t t r u n k v la n 1 Sw it ch X (con fig -if)# sw it ch por t t r u n k v la n 1 0 Sw it ch X (con fig -if)# sw it ch por t t r u n k
v la n 2 0
A n sw er : B
Ex pla n a t ion : IN or der for m u lt iple V LA Ns t o cr oss sw it ch es, t h e con n ect ion bet w een t h e sw it ch es m u st be a
t r u n k . T h e sw it ch por t m ode t r u n k com m a n d is a ll t h a t is n eeded, t h e in div idu a l V LA Ns sh ou ld n ot be list ed
ov er t h a t t r u n k in t er fa ce.
 
 
Qu est ion No : 4 2 - T opic 2
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W h y w ill a sw it ch n ev er lea r n a br oa dca st a ddr ess?
 
A . Br oa dca st s on ly u se n et w or k la y er a ddr essin g .
B. A br oa dca st fr a m e is n ev er for w a r ded by a sw it ch .
C. A br oa dca st a ddr ess w ill n ev er be t h e sou r ce a ddr ess of a fr a m e.
D. Br oa dca st a ddr esses u se a n in cor r ect for m a t for t h e sw it ch in g t a ble.
E. Br oa dca st fr a m es a r e n ev er sen t t o sw it ch es.
A n sw er : C
Ex pla n a t ion : Sw it ch es dy n a m ica lly lea r n MA C a ddr esses ba sed on t h e sou r ce MA C a ddr esses t h a t it sees, a n d
sin ce a br oa dca st is n ev er t h e sou r ce, it w ill n ev er lea r n t h e br oa dca st a ddr ess.
 
 
Qu est ion No : 4 3 - T opic 2
W h a t pa r a m et er ca n be differ en t on por t s w it h in a n Et h er Ch a n n el?
 
A . speed
B. DT P n eg ot ia t ion set t in g s
C. t r u n k en ca psu la t ion
D. du plex
A n sw er : B
Ex pla n a t ion : For a n et her ch a n n el t o com e u p, t h e speed, du plex a n d t h e t r u n k en ca psu la t ion m u st be t h e sa m e
on ea ch en d.
 
 
Qu est ion No : 4 4 - T opic 2
Refer t o t h e ex h ibit .
 
 
All switch ports are assigned to the correct VLANs, but none of the hosts connected to
 SwitchA can communicate with hosts in the same VLAN connected to SwitchB. Based on
 the output shown, what is the most likely problem?
 
A . T h e a ccess lin k n eeds t o be con fig u r ed in m u lt iple V LA Ns.
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B. T h e lin k bet w een t h e sw it ch es is con fig u r ed in t h e w r on g V LA N.
C. T h e lin k bet w een t h e sw it ch es n eeds t o be con fig u r ed a s a t r u n k .
D. V T P is n ot con fig u r ed t o ca r r y V LA N in for m a t ion bet w een t h e sw it ch es.
E. Sw it ch IP a ddr esses m u st be con fig u r ed in or der for t r a ffic t o be for w a r ded bet w een t h e sw it ch es.
A n sw er : C
Ex pla n a t ion : In or der t o pa ss t r a ffic fr om V LA Ns on differ en t sw it ch es, t h e con n ect ion s bet w een t h e sw it ch es
m u st be con fig u r ed a s t r u n k por t s.
 
 
Qu est ion No : 4 5 - T opic 2
Refer t o t h e ex h ibit .
 
 
A technician is troubleshooting host connectivity issues on the switches. The hosts in
 VLANs 10 and 15 on Sw11 are unable to communicate with hosts in the same VLANs on
 Sw12. Hosts in the Admin VLAN are able to communicate. The port-to-VLAN assignments
 are identical on the two switches. What could be the problem?
 
A . T h e Fa 0 /1 por t is n ot oper a t ion a l on on e of t h e sw it ch es.
B. T h e lin k con n ect in g t h e sw it ch es h a s n ot been con fig u r ed a s a t r u n k .
C. A t lea st on e por t n eeds t o be con fig u r ed in V LA N 1 for V LA Ns 1 0 a n d 1 5 t o be a ble t o com m u n ica t e.
D. Por t Fa st Et h er n et 0 /1 n eeds t o be con fig u r ed a s a n a ccess lin k on bot h sw it ch es.
E. A r ou t er is r equ ir ed for h ost s on SW 1 1 in V LA Ns 1 0 a n d 1 5 t o com m u n ica t e w it h h ost s in t h e sa m e V LA N on
Sw 1 2 .
A n sw er : B
Ex pla n a t ion : In or der for h ost s in t h e sa m e V LA N t o com m u n ica t e w it h ea ch ot h er ov er m u lt iple sw it ch es, t h ose
sw it ch es n eed t o be con fig u r ed a s t r u n k s on t h eir con n ect ed in t er fa ces so t h a t t h ey ca n pa ss t r a ffic fr om
m u lt iple V LA Ns.
 
 
Qu est ion No : 4 6 - T opic 2
W h a t v a lu e is pr im a r ily u sed t o det er m in e w h ich por t becom es t h e r oot por t on ea ch
 n on r oot sw it ch in a spa n n in g -t r ee t opolog y ?
 
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A . pa t h cost
B. low est por t MA C a ddr ess
C. V T P r ev ision n u m ber
D. h ig h est por t pr ior it y n u m ber
E. por t pr ior it y n u m ber a n d MA C a ddr ess
A n sw er : A
Ex pla n a t ion : T h e pa t h cost t o t h e r oot br idg e is t h e m ost im por t a n t v a lu e t o det er m in e w h ich por t w ill becom e
t h e r oot por t on ea ch n on -r oot sw it ch . In pa r t icu la r , t h e por t w it h low est cost t o t h e r oot br idg e w ill becom e r oot
por t (on n on -r oot sw it ch ).
 
 
Qu est ion No : 4 7 - T opic 2
Refer t o t h e ex h ibit .
 
 
The output that is shown is generated at a switch. Which three statements are true?
 (Choose three.)
 
A . A ll por t s w ill be in a st a t e of disca r din g , lea r n in g , or for w a r din g .
B. T h ir t y V LA Ns h a v e been con fig u r ed on t h is sw it ch .
C. T h e br idg e pr ior it y is low er t h a n t h e defa u lt v a lu e for spa n n in g t r ee.
D. A ll in t er fa ces t h a t a r e sh ow n a r e on sh a r ed m edia .
E. A ll desig n a t ed por t s a r e in a for w a r din g st a t e.
F. T h is sw it ch m u st be t h e r oot br idg e for a ll V LA Ns on t h is sw it ch .
A n sw er : A ,C,E
Ex pla n a t ion : Fr om t h e ou t pu t , w e see t h a t a ll por t s a r e in desig n a t ed r ole (for w a r din g st a t e). T h e com m a n d
sh ow spa n n in g -t r ee v la n 3 0 on ly sh ow s u s in for m a t ion a bou t V LA N 3 0 . W e don t k n ow h ow m a n y V LA N ex ist s
in t h is sw it ch ->. T h e br idg e pr ior it y of t h is sw it ch is 2 4 6 0 6 w h ich is low er t h a n t h e defa u lt v a lu e br idg e
pr ior it y 3 2 7 6 8 . A ll t h r ee in t er fa ces on t h is sw it ch h a v e t h e con n ect ion t y pe p2 p, w h ich m ea n s Poin t -t o- poin t
en v ir on m en t n ot a sh a r ed m edia . T h e on ly t h in g w e ca n specify is t h is sw it ch is t h e r oot br idg e for V LA N 3 o bu t
w e ca n n ot g u a r a n t ee it is a lso t h e r oot br idg e for ot h er V LA Ns.
 
 
Qu est ion No : 4 8 - T opic 2
W h ich t w o lin k pr ot ocols a r e u sed t o ca r r y m u lt iple V LA Ns ov er a sin g le lin k ? (Ch oose t w o.)
 
A . V T P
B. 8 0 2 .1 q
C. IGP
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D. ISL
E. 8 0 2 .3 u
A n sw er : B,D
Ex pla n a t ion : Cisco sw it ch es ca n u se t w o differ en t en ca psu la t ion t y pes for t r u n k s, t h e in du st r y st a n da r d 8 0 2 .1 q
or t h e Cisco pr opr iet a r y ISL. Gen er a lly , m ost n et w or k en g in eer s pr efer t o u se 8 0 2 .1 q sin ce it is st a n da r ds ba sed
a n d w ill in t er oper a t e w it h ot h er v en dor s.
 
 
Qu est ion No : 4 9 - T opic 2
W h ich t w o of t h ese a r e ch a r a ct er ist ics of t h e 8 0 2 .1 Q pr ot ocol? (Ch oose t w o.)
 
A . It is u sed ex clu siv ely for t a g g in g V LA N fr a m es a n d does n ot a ddr ess n et w or k r econ v er g en ce follow in g
sw it ch ed n et w or k t opolog y ch a n g es.
B. It m odifies t h e 8 0 2 .3 fr a m e h ea der , a n d t h u s r equ ir es t h a t t h e FCS be r ecom pu t ed.
C. It is a La y er 2 m essa g in g pr ot ocol w h ich m a in t a in s V LA N con fig u r a t ion s a cr oss n et w or k s.
D. It in clu des a n 8 -bit field w h ich specifies t h e pr ior it y of a fr a m e.
E. It is a t r u n k in g pr ot ocol ca pa ble of ca r r y in g u n t a g g ed fr a m es.
A n sw er : B,E
Ex pla n a t ion : 8 0 2 .1 Q pr ot ocol, or V ir t u a l Br idg ed Loca l A r ea Net w or k s pr ot ocol, m a in ly st ipu la t es t h e
r ea liza t ion of t h e V LA N. 8 0 2 .1 Q is a st a n da r dized r ela y m et h od t h a t in ser t s 4 by t es field in t o t h e or ig in a l
Et h er n et fr a m e a n d r e-ca lcu la t e t h e FCS. 8 0 2 .1 Q fr a m e r ela y su ppor t s t w o t y pes of fr a m e: m a r k ed a n d n on -
m a r k ed. Non -m a r k ed fr a m e ca r r ies n o V LA N iden t ifica t ion in for m a t ion .
 
 
Qu est ion No : 5 0 - T opic 2
W h a t a r e t h e possible t r u n k in g m odes for a sw it ch por t ? (Ch oose t h r ee.)
 
A . t r a n spa r en t
B. a u t o
C. on
D. desir a ble
E. c lien t
F. for w a r din g
A n sw er : B,C,D
Ex pla n a t ion : T h ese a r e t h e differ en t t y pes of t r u n k m odes: ✑ ON: T h is m ode pu t s t h e por t in t o per m a n en t t r u n k
m ode a n d n eg ot ia t es t o con v er t t h e lin k in t o a t r u n k lin k . T h epor t becom es a t r u n k por t ev en if t h e a dja cen t
por t does n ot a g r ee t o t h e ch a n g e. ✑ OFF: T h is m ode pu t s t h e por t in t o per m a n en t n on -t r u n k m ode a n d
n eg ot ia t es t o con v er t t h e lin k in t o a n on -t r u n k lin k . T h e por t becom es a n on -t r u n k por t ev en if t h e a dja cen t por t
does n ot a g r ee t o t h e ch a n g e. ✑ Desir a ble: T h is m ode ca u ses t h e por t t o a ct iv ely a t t em pt t o con v er t t h e lin k in t o
a t r u n k lin k . T h e por t becom es a t r u n k por t if t h e a dja cen t por t is set t o on , desir a ble, or a u t o m ode. ✑ A u t o:
T h is m ode en a bles t h e por t t o con v er t t h e lin k in t o a t r u n k lin k . T h e por t becom es a t r u n k por t if t h e a dja cen t
por t is set t o on or desir a ble m ode. T h is is t h e defa u lt m ode for Fa st a n d Gig a bit Et h er n et por t s. ✑ Non eg ot ia t e:
T h is m ode pu t s t h e por t in t o per m a n en t t r u n k m ode, bu t does n ot a llow t h e por t t o g en er a t e Dy n a m ic T r u n k in g
Pr ot ocol (DT P) fr a m es. T h e a dja cen t por t m u st be con fig u r ed m a n u a lly a s a t r u n k por t t o est a blish a t r u n k lin k .
 
 
Qu est ion No : 5 1 - T opic 2
Refer t o t h e ex h ibit .
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Which two statements are true about interVLAN routing in the topology that is shown in the
 exhibit? (Choose two.)
 
A . H ost E a n d h ost F u se t h e sa m e IP g a t ew a y a ddr ess.
B. Rou t er 1 a n d Sw it ch 2 sh ou ld be con n ect ed v ia a cr ossov er ca ble.
C. Rou t er 1 w ill n ot pla y a r ole in com m u n ica t ion s bet w een h ost A a n d h ost D.
D. T h e Fa st Et h er n et 0 /0 in t er fa ce on Rou t er 1 m u st be con fig u r ed w it h su bin t er fa ces.
E. Rou t er 1 n eeds m or e LA N in t er fa ces t o a ccom m oda t e t h e V LA Ns t h a t a r e sh ow n in t h e ex h ibit .
F. T h e Fa st Et h er n et 0 /0 in t er fa ce on Rou t er 1 a n d t h e Fa st Et h er n et 0 /1 in t er fa ce on Sw it ch 2 t r u n k por t s m u st
be con fig u r ed u sin g t h e sa m e en ca psu la t ion t y pe.
A n sw er : D,F
Ex pla n a t ion : In or der for m u lt iple V LA Ns t o con n ect t o a sin g le ph y sica l in t er fa ce on a Cisco r ou t er ,
su bin t er fa ces m u st be u sed, on e for ea ch V LA N. T h is is k n ow n a s t h e r ou t er on a st ick con fig u r a t ion . A lso, for
a n y t r u n k t o be for m ed, bot h en ds of t h e t r u n k m u st a g r ee on t h e en ca psu la t ion t y pe, so ea ch on e m u st be
con fig u r ed for 8 0 2 .1 q or ISL.
 
 
Qu est ion No : 5 2 - T opic 2
Refer t o t h e ex h ibit .
 
 
At the end of an RSTP election process, which access layer switch port will assume the
 discarding role?
 
A . Sw it ch 3 , por t fa 0 /1
B. Sw it ch 3 , por t fa 0 /1 2
C. Sw it ch 4 , por t fa 0 /1 1
D. Sw it ch 4 , por t fa 0 /2
E. Sw it ch 3 , por t Gi0 /1
F. Sw it ch 3 , por t Gi0 /2
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A n sw er : C
Ex pla n a t ion : In t h is qu est ion , w e on ly ca r e a bou t t h e A ccess La y er sw it ch es (Sw it ch 3 & 4 ). Sw it ch 3 h a s a low er
br idg e ID t h a n Sw it ch 4 (beca u se t h e MA C of Sw it ch 3 is sm a ller t h a n t h a t of Sw it ch 4 ) so bot h por t s of Sw it ch 3
w ill be in for w a r din g st a t e. T h e a lt er n a t iv e por t w ill su r ely belon g t o Sw it ch 4 . Sw it ch 4 w ill n eed t o block on e of
it s por t s t o a v oid a br idg in g loop bet w een t h e t w o sw it ch es. Bu t h ow does Sw it ch 4 select it s block ed por t ? W ell,
t h e a n sw er is ba sed on t h e BPDUs it r eceiv es fr om Sw it ch 3 . A BPDU is su per ior t o a n ot h er if it h a s: 1 . A low er
Root Br idg e ID 2 . A low er pa t h cost t o t h e Root 3 . A low er Sen din g Br idg e ID 4 . A low er Sen din g Por t ID T h ese fou r
pa r a m et er s a r e ex a m in ed in or der . In t h is specific ca se, a ll t h e BPDUs sen t by Sw it ch 3 h a v e t h e sa m e Root
Br idg e ID, t h e sa m e pa t h cost t o t h e Root a n d t h e sa m e Sen din g Br idg e ID. T h e on ly pa r a m et er left t o select t h e
best on e is t h e Sen din g Por t ID (Por t ID = por t pr ior it y + por t in dex ). In t h is ca se t h e por t pr ior it ies a r e equ a l
beca u se t h ey u se t h e defa u lt v a lu e, so Sw it ch 4 w ill com pa r e por t in dex v a lu es, w h ich a r e u n iqu e t o ea ch por t on
t h e sw it ch , a n d beca u se Fa 0 /1 2 is in fer ior t o Fa 0 /1 , Sw it ch 4 w ill select t h e por t con n ect ed w it h Fa 0 /1 (of
Sw it ch 3 ) a s it s r oot por t a n d block t h e ot h er por t -> Por t fa 0 /1 1 of Sw it ch 4 w ill be block ed (disca r din g r ole).
 
 
Qu est ion No : 5 3 - T opic 2
Refer t o t h e ex h ibit .
 
 
Which WAN protocol is being used?
 
A . A T M
B. H DLC
C. Fr a m e Rela y
D. PPP
A n sw er : C
Ex pla n a t ion : T h is qu est ion is t o ex a m in e t h e sh ow in t com m a n d. A ccor din g t o t h e in for m a t ion pr ov ided in t h e
ex h ibit , w e ca n k n ow t h a t t h e da t a lin k pr ot ocol u sed in t h is n et w or k is t h e Fr a m e Rela y pr ot ocol. LMI en q sen t
 
 
Qu est ion No : 5 4 - T opic 2
Refer t o t h e ex h ibit .
 
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Switch port FastEthernet 0/24 on ALSwitch1 will be used to create an IEEE 802.1Q-
 compliant trunk to another switch. Based on the output shown, what is the reason the trunk
 does not form, even though the proper cabling has been attached?
 
A . V LA Ns h a v e n ot been cr ea t ed y et .
B. A n IP a ddr ess m u st be con fig u r ed for t h e por t .
C. T h e por t is cu r r en t ly con fig u r ed for a ccess m ode.
D. T h e cor r ect en ca psu la t ion t y pe h a s n ot been con fig u r ed.
E. T h e “ n o sh u t dow n ” com m a n d h a s n ot been en t er ed for t h e por t .
A n sw er : C
Ex pla n a t ion : A ccor din g t o t h e ou t pu t sh ow n t h e sw it ch por t (la y er 2 Sw it ch in g ) is en a bled a n d t h e por t is in
a ccess m ode. T o m a k e a t r u n k lin k t h e por t sh ou ld con fig u r ed a s a t r u n k por t , n ot a n a ccess por t , by u sin g t h e
follow in g com m a n d: (Con fig -if)#sw it ch por t m ode t r u n k .
 
 
Qu est ion No : 5 5 - T opic 2
W h ich t er m descr ibes a spa n n in g -t r ee n et w or k t h a t h a s a ll sw it ch por t s in eit h er t h e
 block in g or for w a r din g st a t e?
 
A . con v er g ed
B. r edu n da n t
C. pr ov ision ed
D. spa n n ed
A n sw er : A
Ex pla n a t ion : Spa n n in g T r ee Pr ot ocol con v er g en ce (La y er 2 con v er g en ce) h a ppen s w h en br idg es a n d sw it ch es
h a v e t r a n sit ion ed t o eit h er t h e for w a r din g or block in g st a t e. W h en la y er 2 is con v er g ed, r oot br idg e is elect ed
a n d a ll por t r oles (Root , Desig n a t ed a n d Non -Desig n a t ed) in a ll sw it ch es a r e select ed.
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Qu est ion No : 5 6 - T opic 2
W h a t is on e ben efit of PV ST +?
 
A . PV ST + su ppor t s La y er 3 loa d ba la n cin g w it h ou t loops.
B. PV ST + r edu ces t h e CPU cy cles for a ll t h e sw it ch es in t h e n et w or k .
C. PV ST + a llow s t h e r oot sw it ch loca t ion t o be opt im ized per V LA N.
D. PV ST + a u t om a t ica lly select s t h e r oot br idg e loca t ion , t o pr ov ide opt im ized ba n dw idt h u sa g e.
A n sw er : C
Ex pla n a t ion : T h e PV ST + pr ov ides La y er 2 loa d-ba la n cin g for t h e V LA N on w h ich it r u n s. Y ou ca n cr ea t e
differ en t log ica l t opolog ies by u sin g t h e V LA Ns on y ou r n et w or k t o en su r e t h a t a ll of y ou r lin k s a r e u sed bu t t h a t
n o on e lin k is ov er su bscr ibed. Ea ch in st a n ce of PV ST + on a V LA N h a s a sin g le r oot sw it ch . T h is r oot sw it ch
pr opa g a t es t h e spa n n in g -t r ee in for m a t ion a ssocia t ed w it h t h a t V LA N t o a ll ot h er sw it ch es in t h e n et w or k .
Beca u se ea ch sw it ch h a s t h e sa m e in for m a t ion a bou t t h e n et w or k , t h is pr ocess en su r es t h a t t h e n et w or k
t opolog y is m a in t a in ed a n d opt im ized per V LA N. Refer en ce:
h t t p://w w w .cisco.com /en /US/docs/sw it ch es/la n /ca t a ly st 3 7 5 0 x _3 5 6 0 x /soft w a r e/r elea se/1 2 .2
_5 5 _se/con fig u r a t ion /g u ide/sw st p.h t m l
 
 
Qu est ion No : 5 7 - T opic 2
W h ich t w o ben efit s a r e pr ov ided by cr ea t in g V LA Ns? (Ch oose t w o.)
 
A . a dded secu r it y
B. dedica t ed ba n dw idt h
C. pr ov ides seg m en t a t ion
D. a llow s sw it ch es t o r ou t e t r a ffic bet w een su bin t er fa ces
E. con t a in s collision s
A n sw er : A ,C
Ex pla n a t ion : A V LA N is a sw it ch ed n et w or k t h a t is log ica lly seg m en t ed on a n or g a n iza t ion a l ba sis, by fu n ct ion s,
pr oject t ea m s, or a pplica t ion s r a t h er t h a n on a ph y sica l or g eog r a ph ica l ba sis. Secu r it y : V LA Ns a lso im pr ov e
secu r it y by isola t in g g r ou ps. H ig h -secu r it y u ser s ca n be g r ou ped in t o a V LA N, possible on t h e sa m e ph y sica l
seg m en t , a n d n o u ser s ou t side t h a t V LA N ca n com m u n ica t e w it h t h em . LA N Seg m en t a t ion V LA Ns a llow log ica l
n et w or k t opolog ies t o ov er la y t h e ph y sica l sw it ch ed in fr a st r u ct u r e su ch t h a t a n y a r bit r a r y collect ion of LA N
por t s ca n be com bin ed in t o a n a u t on om ou s u ser g r ou p or com m u n it y of in t er est . T h e t ech n olog y log ica lly
seg m en t s t h e n et w or k in t o sepa r a t e La y er 2 br oa dca st dom a in s w h er eby pa ck et s a r e sw it ch ed bet w een por t s
desig n a t ed t o be w it h in t h e sa m e V LA N. By con t a in in g t r a ffic or ig in a t in g on a pa r t icu la r LA N on ly t o ot h er
LA Ns in t h e sa m e V LA N, sw it ch ed v ir t u a l n et w or k s a v oid w a st in g ba n dw idt h .
 
 
Qu est ion No : 5 8 - T opic 2
W h a t does a La y er 2 sw it ch u se t o decide w h er e t o for w a r d a r eceiv ed fr a m e?
 
A . sou r ce MA C a ddr ess
B. sou r ce IP a ddr ess
C. sou r ce sw it ch por t
D. dest in a t ion IP a ddr ess
E. dest in a t ion por t a ddr ess
F. dest in a t ion MA C a ddr ess
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A n sw er : F
Ex pla n a t ion : W h en a fr a m e is r eceiv ed, t h e sw it ch look s a t t h e dest in a t ion h a r dw a r e a ddr ess a n d fin ds t h e
in t er fa ce if it is in it s MA C a ddr ess t a ble. If t h e a ddr ess is u n k n ow n , t h e fr a m e is br oa dca st on a ll in t er fa ces
ex cept t h e on e it w a s r eceiv ed on .
 
 
Qu est ion No : 5 9 - T opic 2
Refer t o t h e ex h ibit .
 
 
Which two statements are true of the interfaces on Switch1? (Choose two.)
 
A . Mu lt iple dev ices a r e con n ect ed dir ect ly t o Fa st Et h er n et 0 /1 .
B. A h u b is con n ect ed dir ect ly t o Fa st Et h er n et 0 /5 .
C. Fa st Et h er n et 0 /1 is con n ect ed t o a h ost w it h m u lt iple n et w or k in t er fa ce ca r ds.
D. Fa st Et h er n et 0 /5 h a s st a t ica lly a ssig n ed MA C a ddr esses.
E. Fa st Et h er n et 0 /1 is con fig u r ed a s a t r u n k lin k .
F. In t er fa ce Fa st Et h er n et 0 /2 h a s been disa bled.
A n sw er : B,E
Ex pla n a t ion : Ca r efu lly obser v e t h e in for m a t ion g iv en a ft er com m a n d sh ow . Fa 0 /1 is con n ect ed t o Sw it ch 2 ,
sev en MA C a ddr esses cor r espon d t o Fa 0 /1 , a n d t h ese MA C a r e in differ en t V LA N. Fr om t h is w e k n ow t h a t Fa 0 /1
is t h e t r u n k in t er fa ce. Fr om t h e in for m a t ion g iv en by sh ow cdp n eig h bor s w e fin d t h a t t h er e is n o Fa 0 /5 in CDP
n eig h bor . H ow ev er , F0 /5 cor r espon ds t o t w o MA C a ddr esses in t h e sa m e V LA N. T h u s w e k n ow t h a t Fa 0 /5 is
con n ect ed t o a H u b. Ba sed on t h e ou t pu t sh ow n , t h er e a r e m u lt iple MA C a ddr esses fr om differ en t V LA Ns
a t t a ch ed t o t h e Fa st Et h er n et 0 /1 in t er fa ce. On ly t r u n k s a r e a ble t o pa ss in for m a t ion fr om dev ices in m u lt iple
V LA Ns.
 
 
Qu est ion No : 6 0 - T opic 2
Ba sed on t h e n et w or k sh ow n in t h e g r a ph ic
 
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Which option contains both the potential networking problem and the protocol or setting
 that should be used to prevent the problem?
 
A . r ou t in g loops, h old dow n t im er s
B. sw it ch in g loops, split h or izon
C. r ou t in g loops, split h or izon
D. sw it ch in g loops, V T P
E. r ou t in g loops, ST P
F. sw it ch in g loops, ST P
A n sw er : F
Ex pla n a t ion : T h e Spa n n in g -T r ee Pr ot ocol (ST P) pr ev en t s loops fr om bein g for m ed w h en sw it ch es or br idg es a r e
in t er con n ect ed v ia m u lt iple pa t h s. Spa n n in g -T r ee Pr ot ocol im plem en t s t h e 8 0 2 .1 D IEEE a lg or it h m by
ex ch a n g in g BPDU m essa g es w it h ot h er sw it ch es t o det ect loops, a n d t h en r em ov es t h e loop by sh u t t in g dow n
select ed br idg e in t er fa ces. T h is a lg or it h m g u a r a n t ees t h a t t h er e is on e a n d on ly on e a ct iv e pa t h bet w een t w o
n et w or k dev ices.
 
 
Qu est ion No : 6 1 - T opic 2
A t w h ich la y er of t h e OSI m odel is RST P u sed t o pr ev en t loops?
 
A . ph y sica l
B. da t a lin k
C. n et w or k
D. t r a n spor t
A n sw er : B
Ex pla n a t ion : RST P a n d ST P oper a t e on sw it ch es a n d a r e ba sed on t h e ex ch a n g e of Br idg e Pr ot ocol Da t a Un it s
(BPDUs) bet w een sw it ch es. On e of t h e m ost im por t a n t fields in BPDUs is t h e Br idg e Pr ior it y in w h ich t h e MA C
a ddr ess is u sed t o elect t h e Root Br idg e -> RST P oper a t es a t La y er 2 Da t a Lin k la y er ->.
 
 
Qu est ion No : 6 2 - T opic 2
W h ich t w o com m a n ds ca n be u sed t o v er ify a t r u n k lin k con fig u r a t ion st a t u s on a g iv en
 Cisco sw it ch in t er fa ce? (Ch oose t w o.)
 
A . sh ow in t er fa ce t r u n k
B. sh ow