CBSE Class 10 Question Paper 2025 Solution Manipuri

Marking Scheme
Strictly Confidential
(For Internal and Restricted use only)
Secondary School Examination, 2025
SUBJECT NAME MANIPURI (Q.P. CODE 16)
General Instructions: -

1 You are aware that evaluation is the most important process in the actual and correct
assessment of the candidates. A small mistake in evaluation may lead to serious
problems which may affect the future of the candidates, education system and teaching
profession. To avoid mistakes, it is requested that before starting evaluation, you must
read and understand the spot evaluation guidelines carefully.

2 “Evaluation policy is a confidential policy as it is related to the confidentiality of
the examinations conducted, Evaluation done and several other aspects. Its’
leakage to public in any manner could lead to derailment of the examination
system and affect the life and future of millions of candidates. Sharing this
policy/document to anyone, publishing in any magazine and printing in News
Paper/Website etc may invite action under various rules of the Board and IPC.”
3 Evaluation is to be done as per instructions provided in the Marking Scheme. It should
not be done according to one’s own interpretation or any other consideration. Marking
Scheme should be strictly adhered to and religiously followed. However, while
evaluating, answers which are based on latest information or knowledge and/or are
innovative, they may be assessed for their correctness otherwise and due marks
be awarded to them. In class-X, while evaluating two competency-based questions,
please try to understand given answer and even if reply is not from marking
scheme but correct competency is enumerated by the candidate, due marks
should be awarded.

4 The Marking scheme carries only suggested value points for the answers
These are in the nature of Guidelines only and do not constitute the complete answer.
The students can have their own expression and if the expression is correct, the due
marks should be awarded accordingly.
5 The Head-Examiner must go through the first five answer books evaluated by each
evaluator on the first day, to ensure that evaluation has been carried out as per the
instructions given in the Marking Scheme. If there is any variation, the same should be
zero after delibration and discussion. The remaining answer books meant for evaluation
shall be given only after ensuring that there is no significant variation in the marking of
individual evaluators.

6 Evaluators will mark ( √ ) wherever answer is correct. For wrong answer CROSS ‘X” be
marked. Evaluators will not put right (✓) while evaluating which gives an impression that
answer is correct and no marks are awarded. This is most common mistake which
evaluators are committing.
7 If a question has parts, please award marks on the right-hand side for each part. Marks
awarded for different parts of the question should then be totaled up and written in the
left-hand margin and encircled. This may be followed strictly.

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8 If a question does not have any parts, marks must be awarded in the left-hand margin
and encircled. This may also be followed strictly.

9 If a student has attempted an extra question, answer of the question deserving more
marks should be retained and the other answer scored out with a note “Extra Question”.

10 No marks to be deducted for the cumulative effect of an error. It should be penalized only
once.
11 A full scale of marks __________(example 0 to 80/70/60/50/40/30 marks as given in
Question Paper) has to be used. Please do not hesitate to award full marks if the answer
deserves it.
12 Every examiner has to necessarily do evaluation work for full working hours i.e., 8 hours
every day and evaluate 20 answer books per day in main subjects and 25 answer books
per day in other subjects (Details are given in Spot Guidelines).This is in view of the
reduced syllabus and number of questions in question paper.
13 Ensure that you do not make the following common types of errors committed by the
Examiner in the past:-
● Leaving answer or part thereof unassessed in an answer book.
● Giving more marks for an answer than assigned to it.
● Wrong totaling of marks awarded on an answer.
● Wrong transfer of marks from the inside pages of the answer book to the title page.
● Wrong question wise totaling on the title page.
● Wrong totaling of marks of the two columns on the title page.
● Wrong grand total.
● Marks in words and figures not tallying/not same.
● Wrong transfer of marks from the answer book to online award list.
● Answers marked as correct, but marks not awarded. (Ensure that the right tick mark
is correctly and clearly indicated. It should merely be a line. Same is with the X for
incorrect answer.)
● Half or a part of answer marked correct and the rest as wrong, but no marks awarded.
14 While evaluating the answer books if the answer is found to be totally incorrect, it should
be marked as cross (X) and awarded zero (0)Marks.
15 Any un assessed portion, non-carrying over of marks to the title page, or totaling error
detected by the candidate shall damage the prestige of all the personnel engaged in the
evaluation work as also of the Board. Hence, in order to uphold the prestige of all
concerned, it is again reiterated that the instructions be followed meticulously and
judiciously.
16 The Examiners should acquaint themselves with the guidelines given in the “Guidelines
for spot Evaluation” before starting the actual evaluation.
17 Every Examiner shall also ensure that all the answers are evaluated, marks carried over
to the title page, correctly totaled and written in figures and words.
18 The candidates are entitled to obtain photocopy of the Answer Book on request on
payment of the prescribed processing fee. All Examiners/Additional Head
Examiners/Head Examiners are once again reminded that they must ensure that
evaluation is carried out strictly as per value points for each answer as given in the
Marking Scheme.

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Marking Scheme (For Examiners/Evaluators)
All India Secondary School Certificate Examination, 2024-25
Class-X
Subject:- Manipuri ( Code 011)
vQkxQD oKshVxVAsKsYS
I. fPxaYS saVYF usYk lPpgUFsYSyY uEowEAl rPc nQdVrVU lEoKHsYShfA
lYxY. fgEo usY yPobdPoKsYS ( dFvYS-dFhPAsYS) uEok sYvYKkkcyY
ufsUS fsYk flUSiPc lPpgUF uEobcsU rPo.ftWxEosYSyY fsPyY dEKbP
ufsUS fiESbEA dWxAl hPxcbY, fbUyY ujUF-jUFb ufsUS fiEAbEA ubU
rQScYxy jUkc fPxA lYcYc rPo.

II. oaPogUFkxYc sUlVxVYF aExGHaY rPwT fhUSoKk, aQKbYbQHsYSk
lVxVQsaBVxVPoc hRxc iY wYbEAdy lPpgUF-jQyY iEhEaElY dRc rPcyY urPc
dWxQ. oAvPfYkx/ tQb oAvPfYkx lUfkfAk fPxaYS saVYFb lYxYcbUyUFk
lPpgUF gUbYSfAaY uEok , nQdVrVU lEoKHsYSyY fhUSoKk lPpgUF-jQ rQScYkc
ufUAa tKk kYSsYSvxY.
III. lPpgUF-J jQ rQSc fhFb axYyUFc lPpgUF ubU flUSiPk dPKc hPxcbY, jUFbc
lPpgUF ubUb (X) uEok fPxA hRcYbUk ‘0’ fPxA lYcYkc,oAvPfYkx / tQb
oAvPfYkx lUFkfAh gTtKvxY.

3

Marking Scheme ( For Examiners/Evaluators)
All India Secondary School Certificate Examination, 2024-25
Class-X
Subject:- Manipuri ( Code 011)

aPSdUl--- a ( SECTION-A)
lPc--(READING COMPREHENSION)
Instructions for ePtT jQb lYxYc ePxQS fhQAsYS usY lPxy tTdYc
Q.No. 1,2&3
ePtTsYS usYyY ujUFc gDdy lPpgUF lYc uEoybckY.

Q.Nos Expected Answer/Value Points Mark
Distribution
Q.1
i. A) fhYA jPk kWkxc fhFb. 1
ii. D) uqPSyY. 1
iii. C) uqPSyY lUAjQDy. 1
iv. A) wcAaY ftWk. 1
v. B) iTiF wEAo. 1
Q.2
I. D) fY fYbyY. 1
ii. A) gKwc qFcyY. 1
iii. D) ePxAaY hrPDyUFckY. 1
iv. B) yrVPK,gKwc qFcyY sAhY dWxcsYS lUKsYKybckY. 1
v. C) wAsY-gPsY rPpbVxVc fYy. 1
Q.3
i. A) ‘lVxVvP fsPk ulPFc fYtUH gKbUk dWcPA qPAjc’ tPockY. 1
ii. C) kYShFckY. 1
Iii. B) dUjYSc ufk ePrQKc dFcU. 1
iv. D) utUFfA jUFfY. 1
v. A) kYShFc iTkcyYkY. 1
SECTION-B ( WRITING)
aPSdUl- s ( oc)
Instruct ulPFc ufgA gDdy lPpgUF lYybckY.ePtRbEAh fPxA ukY(2).,flUS
ions for uEoc sxUAaYbfAhbY ftWxEoyY gUHo,fiESbEA ufbY sYvYKkxYc
Q.No.4
ePtW-ePhP ubU rQSdy fPxA 3 (utUF) . ePxEosYKb fPxA 1(uf)
lYcYybckY.

4

Q.Nos. Expected Answer / Value Points Distribution
4. ufgA gDdy (ePxQS / usQ) lPpgUF lYybckY. 2+3+1=6
Instruct jYwY ocyY arPHsYS rQSdy fPxA lYcYykY.jYwY oxYc fYyY dWiF,
ions for hPxYA, jYwY iTybRxYc fYbUb pHl oaPogUFkc,jYwYyY ePxED jYwY
Q.No.5
oxYc fYbUy iTybRxYcy dWkc fxY,jYwY oxYcyY stY,jYwY
iTybRxYcyY ffYS,dWiFkjYSc fiF jPk wFybckY. jYwY oxYcyY
dWiF, hPxYA ufbY jYwY iTybRxYc fYbUb pHl oaPogUFkc fiF jPk
ocb fPxA 2(ukY) lYcYxE ubUy jYwY ubUyY fiESbEA rQSdy fPxA
3(utUF) lYcYxE.
Q.Nos. Expected Answer / Value Points Distribution
5. jYwY oc. 2+3=5
Instruct ulPFc ukYgA gDdy lPpgUF lYybckY. lPpxRyY kUSyY ePtKwEA
ions for lYcb fPxA 1(uf) sKbEAk hPAlb fPxA ukY(2) lYcYykY.
Q.No.6
Q.Nos. Expected Answer / Value Points Distribution
6. a) iFc jDdy tYLl jDdY. 1+2=3
kUSyY uxw; gUbES jPc uf iTdcbY fwT fwT gUbESjPc
iTdAo tPockY.
sKBbEAk hPAl rQScYrU.
b) ktPk fPoc sPc fT wDdY. 1+2=3
kUSyY uxw;- tWiPHhck hRc wcA sEock kFfY tPockY.
sKbEAk hPAl rQScYrU.
c) wKbc jiUk fgED wEAo. 1+2=3
kUSyY uxw;- tWiPHhck kPlD hRk gx tWvcbU pHwEAo.
sKbEAk hPAl rQScYrU.
Instucti ulPFc uf gDdy lPpgUF lYybckY.
ons for
Q.No.7
a) tPov-jQxED(Application) ocyYbfA fwAaY dWiF jUFcxVP rQcYc,
gUHoyY fiESbEA, ePxEosYD rQScYxy fPxA 1(uf) lYcYykY.
tPovxYc fYyY ffYSkjYSc jUFk ocxVP rQScYxy fPxA 2(ukY)
lYcYykY.
Q.Nos. Expected Answer / Value Points Distribution
7 a) tPov-jQxED oc. 1+2=3
Instruct b) gTtK jQxED / gTtKvc / kEhYs oc usYyYbfA ePtRbEA, hPxYA,
ions for fiESbEA, oxYc fYyY fiF ufbY dWiF usYkjYSc rQScYxy fPxA
Q.No.7

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1(uf) lYcYykY.adVc aY ffYS fwAh fgPb lUKk oxFdcsU fPxA
2(ukY) lYcYykY.
Q.Nos. Expected Answer / Value Points Distribution
7 b) gTtKvc / kEhYs oc. 1+2=3
Instruct ePtT 8 byY 12 iPpc lPpgUF A,B,C,D lYxYcsYbyY ujUFc gDdy
ions oybckY.
forQ.N
o.8
to12
8. C) dPSbk rPFk oTk dWc fuES. 1
9. A) pkcb tQAh gHkc rQSwYkc ukY. 1
10. D) iYSc. 1
11. B) aPKkbVxVc fxY fhP. 1
12. D) uEobck uuEoc sPc. 1
aPSdUl- d ( dEKfYHdED)
SECTION- C ( GRAMMAR)
Instruct ePtT 13 byY 22 iPpc lPpgUF A,B,C,D lYxYcsYbyY ujUFc gDdy
ions oybckY.
forQ.N
o.13
to22
13. D) utUFfA jUFfY. 1
14. B) 2 dW. 1
15. C) fkYlUxY dEDb sYdQcD fgD hxUA gPobEAo. 1
16. A) ‘pH’ kY. 1
17. A) uf rPpo. 1
18. D) fqP kY. 1
19. B) i ~ n kY. 1
20. C) scvQA uf ufsUS nxc uf rPpybckY. 1
21. D) -sP,-cU,-gEo tPLhVxVcsU rPo. 1
22. B) 3 dW. 1
23. fkYlUxY dEKb nPeD iEkYFsYSk bYiwES sQFfY. 1+1=2
gUbF;- uEo.
24. aQs fgD 8(kYlPD) dW. 1+1=2
gUbF;- uW fPcU gTqY.
25. a) uW jPpccUbY gTqY ubUy fkYcUbY gTbQ. 1+1=2
hYKkc ePtWlxQS.
b) wEocY saVUD jHakY.

6

ujFc ePtW lxQS.
26. fkYS hFc adVEv uf rPpc,fgP lEKc aVdVEv uf kHhVxVy ufbyY 1+1=2
tQKk ePpc ePtW lxQScU rPKkc ePtW lxQS aRo.
gUbF:- kTk pksY tPoc fiFb uW dPAaQ.

27. gQKkW. 1+1=2
geVPobyY lYAl ePtKwEA dWc ePtW fjQH kHhxy ePtWsYS ubUU
fExiYFkY ubUy ePtW ufyY fkUSb fxU uEoc ePtKwEA hPAl
fExiYFcUk xUHkY.
SECTION- D ( PROSE)
aPSdUl-f ( ePxQS)
Instruct ulPFc ufhT gDdy lPpgUF lYybckY. fxY fgES hPAl ufsUS
ions for uocyY ffYS ocb fPxA 1(uf) lYcYykY. rPpiF ufsUS
Q.Nos.
uxw(ePtKwEA) hPAlyY fiESbEA ocb fPxA 1(uf) ufsUS sKbEAk
28
hPAlb fPxA 3(utUF) lYcYc rPo.

Q.Nos. Expected Answer / Value Point Distribution
28. a) ePxQSyY ffYS ---- fYk axY tPogYybyQ. 1+1+3=5
uocyY ffYS ---- geVPoxPAiF jPpc sYSt.
uxw(ePtKwEA) ---- fYuEoc tPocsY fPk -fPk hRc qFc fhYA
fyUK sAhY uffFbY osExk sQFcYxAl fhFb rPpxAo
tPoc pHdY.
sKbEAk hPAdAlb ftWxEosYSk ocyY fiESbEAh fPxA
lYcYykY.
b) ePxQSyY ffYS ------ oFiPD hUxQDyY ohPfjP. 1+1+3=5
uocYyY ffYS ---- ftPxPv aUfPxY cYkEbYkY.

uxw(ePtKwEA) ---- ePiF usY ashUxYk fRlVeVP fUKkPcU aKgHl
uEok tPogYc ePiFkY.
sKbEAk hPAdAlb ftWxEosYSk ocyY fiESbEAh fPxA
lYcYykY.
Instruct ukYgA gDdy lPpgUF lYybckY.ePtW lxQS ukY kHhVxBVy utUFb
ions for lPpgUF lYc uEoybckY.ftWxEosYSyY fiESbEA rQSdy fPxA
Q.Nos.
lYcYykY.
29

Q.Nos. Expected Answer / Value Point Distribution

7

29 a) ePiF usY hYaQKbxVvYHk tPogYckY. 2
uTyVxVQv axfjPxY sUcYbPx gQdQKbVxVk fYhW fjP uf uEok
fYk oKvYKc dRc fxFk tUo tPok jWxYckY.
b) fPolPAlYk hYaQKbVxVvYHlU rQAkcyY fgUHhyY 2
aKkcyYbfAh wPbtKgYckY.

c) ‘gESkPS tEyPocY’ yY ePxYb rPpxYc fxUuEoc ePiFbY ePsA 2
qPAhc fYb osExk jWxPA lY tPoc usYkY. fYyY dEKjH
ufsUS dPoyY dEKjH ufsUS dPoyY dEKjH ukY usYyY
jPSrQSkcb fYyY dEKjHk tQKk rPbQ tPockY.
Instruct ufgA gDdy lPpgUF lYybckY.ePtW lxQS uf kHhVxBVy ukYb
ions for lPpgUF lYc uEoybckY.ftWxEosYSyY fiESbEA rQSdy fPxA
Q.Nos.
lYcYykY.
30
Q.Nos Expected Answer / Value Point Distribution
30 a) jxPoxESck uxEoc dPKbPgYcbU hUsUA gUDbkY. 1

b) kUSsQD jPocY lPFtWcyY ffPkY kHhxVy kUSsQD jPocY 1
jxPoxEScyY kUlYkY.
SECTION- D ( POETRY)
aPSdUl- f ( sWxQS )
Instruct ulPFc ufhT gDdy lPpgUF lYybckY. fxY fgES hPAl ufsUS
ions for uocyY ffYS ocb fPxA 1(uf) lYcYykY. rPpiF ufsUS
Q.Nos.
uxw(ePtKwEA) hPAlyY fiESbEA ocb fPxA 1(uf) ufsUS sKbEAk
31
hPAlb fPxA 3(utUF) lYcYc rPo.

Q.Nos Expected Answer / Value Point Distribution
31. a) sWxQSyY ffYS ---- uWyY gUKvP. 1+1+3=5
acYyY ffYS ---- dPosVxVF sFxQKbVxV sYSt.

uxw(ePtKwEA) ---- acYk kUfYH kYkYbyY tUFkY sUxEo tPoc
gTo.fsYk gUKvPsYS ubUyY ojF jFvc uic ePgD gDdY
tPoc pHdY.
sKbEAk hPAdAlb ftWxEosYSk ocyY fiESbEAh fPxA
lYcYykY.
b) sWxQSyY --- bPcY. 1+1+3=5
acYyY ffYS --- kESwEFcF sVxVYcYxQK .

8

uxw(ePtKwEA) ---- acYk fhFsYyY akPk wRkPic kYlP wEAl
fYsAkE ubUy akPk fYjT-sTy dWxy fhYA fwRkP dWc fsP
fxEFbEF dQLdyk lPSbc fYsAkE tPoc pHdY.
sKbEAk hPAdAlb ftWxEosYSk ocyY fiESbEAh fPxA
lYcYykY.
Instruct ufgA gDdy lPpgUF lYybckY.ePtW lxQS utUF kHhVxVy fxYb
ions for lPpgUF lYc uEoybckY. ftWxEosYSyY fiESbEA rQSdy fPxA
Q.Nos.
lYcYykY.
32.
Q.Nos. Expected Answer / Value Point Distribution
32. a) uWgEo tPoxYcsY tRvYAaY fhFyY ftW fsYS fhYA jPk 3
tWxc,fYjT sTybY dWvbVxVc dPoxc fYjF lxVvPsYScUkY.
jYxEDyY aEkUSyY sPKklEJH uEoxYckY.
b) lEH ufyY uuEoc fsA gTkcyYbfAh lEH ubUyY fkUS flPK 3
ukYfA hUk rQSc hPo.fxFbY flPKwESyY sAhF ufbY kUSyY
sAhF ukY usY rPFk gQKkW. gUbF uEok qPSgVxVc dWxPSyY
fkUS lsEH uEoxc ufbY ufUcyY(aEaYD) fgEDk rPFk kUSsYc
usYkjYSckY.
Instruct ePtW lxQS uf kHhVxVy ukYb lPpgUF lYc uEoybckY.
ions for
Q.Nos.
33
Q.Nos. Expected Answer / Value Point Distribution
33. a) dPK tRxFbk kQFwgYcyY fxFbY dPKfY uEoybc wRkP iPc 1
ktPxEDsYS dEokfA kUfYH wEAdAlb sYSyPxW dWxPSk
aQKcyUF aQKbUk fPSgYcbyYkY.
b) osExyY kUS-lPK kPoc uxQLl sAhFcU dPokYS utD fxULk 1
gTgY tPok acYk tPo.

9