Rajasthan Board 10th Model Paper 2025 Maths

RAJASTHAN BOARD

MODEL
PAPER
2025

PRACTICE PAPERS

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iz’u&i= dh ;kstuk2024&2025
d{kk &10th
fo"k; & xf.kr
vof/k & 3?k.Vs 15 feuV iw.kkZd
a & 80
1- mn~ns'; gsrqvd
a Hkkj&
Ø-l-a mn~n's ; vadHkkj izfr'kr
1- Kku 24 30
2- vocks/k 24 30

3- Kkuksi;ksx 16 20
4- dkS'ky 8 10
5- fo’ys"k.k 8 10
;ksx 80 100

2- iz'uksa ds izdkjokjvadHkkj&
Ø-la- iz'uksa dk izdkj iz'uksa dh vad dqyvad izfr'kr izfr'kr laHkkfor
la[;k izfriz'u ¼vadks dk½ ¼iz'uksa le;
dk½
1- cgqfodYikRed 18 1 18 22.5 33.96 36
2- fjDrLFkku 6 1 6 7.5 11.32 15
3- vfry?kqÙkjkRed 12 1 12 15.0 22.64 42
4- y?kqÙkjkRed 10 2 20 25.0 18.87 40
5- nh?kZmÙkjh; 4 3 12 15.0 7-55 32
6- fuca/kkRed 3 4 12 15.0 5-66 30
;ksx 53 80 100 100 195
feuV
fodYi ;kstuk % [k.M ^l* ,oa ^n* esa gSa
3- fo"k; oLrq dk vadHkkj&
Ø-l-a fo"k; oLrq vadHkj izfr'kr
1 okLrfod la[;k, 4 5
2 cgqin 4 5
3 nks pj okys jSf[kd lehdj.k ;qXe 4 5
4 f}?kkr lehdj.k 4 5
5 lekUrj Jsf.k;ka 6 7.50
6 f=Hkqt 4 5
7 funsZ'kkad T;kfefr 7 8.75
8 f=dks.kfefrdk ifjp; 8 10.00
9 f=dks.kfefr ds dqN vuqiz;ksx 5 6.25
10 o`r 6 7.50
11 o`Ùkks ls lacf/kr {ks=Qy 5 6.25
12 i`"Bh; {ks=Qy vkSj vk;ru 6 7.50
13 lakf[;dh 13 16.25
14 izkfFkdrk 4 5
;ksx 80 100

iz'u&i= CY;wfizUV 2024&2025
d{kk &10th fo"k; %&xf.kr le; %& 3?k.Vs 15 feuV iw.kkZad& 80
Ø mÌs'; bdkbZ@mibdkbZ Kku vocks/k Kkuksi;ksx dkS'ky fo’ys"k.k ;ksx
-

vfry?kqÙkjkRed

vfry?kqÙkjkRed

vfry?kqÙkjkRed

vfry?kqÙkjkRed

vfry?kqÙkjkRed
cgqfodYikRed

cgqfodYikRed

cgqfodYikRed

cgqfodYikRed

cgqfodYikRed
nh?kZmÙkjkRed

nh?kZmÙkjkRed

nh?kZmÙkjkRed

nh?kZmÙkjkRed

nh?kZmÙkjkRed
la-

y?kqÙkjkRed

y?kqÙkjkRed

y?kqÙkjkRed

y?kqÙkjkRed

y?kqÙkjkRed
fucU/kkRed

fucU/kkRed

fucU/kkRed

fucU/kkRed

fucU/kkRed
fjDrLFkku

fjDrLFkku

fjDrLFkku

fjDrLFkku

fjDrLFkku
1 okLrfod la[;k, 1(1) 1(1) 2(1)
4(3)

2 cgqin 1(1) 1(1) 2(1) 4(3)

3 nks pj okys jSf[kd lehdj.k 1(1) 1(1) 2(1) 4(3)
;qXe
4 f}?kkr lehdj.k 4(1)* 4(1)

5 lekUrj Jsf.k;ka 1(1) 3(1)* 2(1) 6(3)

6 f=Hkqt 1(1) 1(1) 2(1) 4(3)

7 funsZ'kkad T;kfefr 1(1) 1(1) 2(1) 3(1)
*
7(4)

8 f=dks.kfefrdk ifjp; 1(1) 4(1)* 1(1) 2(1) 8(4)

9 f=dks.kfefr ds dqN vuqiz;ksx 1(1) 1(1) 1(1) 2(1) 5(4)

10 o`r 3(1)* 1(1) 2(1) 6(3)

11 o`Ùkks ls lacf/kr {ks=Qy 1(1) 1(1) 1(1) 2(1) 5(4)

12 i`"Bh; {ks=Qy vkSj vk;ru 1(1) 1(2) 1(2) 1(1) 6(6)

13 lakf[;dh 1(1) 1(2) 1(1) 1(2) 3(1)* 4(1)* 13(8)

14 izkfFkdrk 1(1) 1(1) 1(2) 4(4)

;ksx 9(9) 2(2) 6(6) 3(1) 4(1) 3(3) 3(3) 6(6) 2(1) 6(2) 4(1) 2(2) 1(1) 6(3) 3(1) 4(1) 2(2) 6(3) 2(2) 6(3)

loZ;ksx 24(19) 24(16) 16(8) 8(5) 8(5) 80(53)

ekè;fed f'k{kk cksMZ jktLFkku] vtesj
e‚My ç'u i= ekè;fed ijh{kk 2025

fo"k;% xf.kr ¼Maths½
d{kk& 10
le;% 3 ?k.Vs 15 feuV iw.kkZd% 80
[k.M & v
SECTION- A

1- cgqfodYih ç'u ¼i ls xvi½% fuEu ç'uksa ds mÙkj dk lgh fodYi p;u dj mÙkj iqfLrdk esa
fyf[k,A
Choose the correct answer from multiple choice question- (i to xvi) and write in given
answers book.

¼i½ la[;kvksa 6 vkSj 20 dk vHkkT; xq.ku[k.M fof/k ls LCM gS & 1
¼v½ 2 ¼c½ 120 ¼l½ 60 ¼n½ 6
LCM of 6 and 20 by the prime factorization method is -

¼A½ 2 ¼B½ 120 ¼C½ 60 ¼D½ 6
¼ii½ cgqin x  7 x  12 ds 'kwU;dksa dk ;ksx gsS &
2
1
¼v½ &7 ¼c½ 7 ¼l½ 12 ¼n½ -12
Sum of the zeros of polynomial x  7 x  12 is -
2

(A) -7 (B) 7 (C) 12 (D) -12
¼iii½ lehdj.k 3x  2 y  7 esa ;fn x dk eku 5 gks rks y dk eku gS & 1
¼v½ 4 ¼c½ &8 ¼l½ &4 ¼n½ 0
In the equation 3x  2 y  7 , if the value of x is 5, then the value of y is -

(A) 4 (B) -8 (C) -4 (D) 0
¼iv½ nh xbZ vkd`fr esa DE II BC, AD = 1.5 cm, DB = 3 cm rFkk AE = 2cmgS rks EC dk eku gS 1

¼v½ 6lseh ¼c½ 4 lseh ¼l½ 4-5 lseh ¼n½ 3-5 lseh
In the given figure DE II BC, AD = 1.5 cm, DB = 3 cm and AE = 2 cm, then the value of EC is -

(A) 6 cm (B) 4 cm (C) 4.5 cm (D) 3.5 cm
¼v½ fcUnqvksa ¼2]3½ vkSj ¼5]6½ ds chp dh nwjh gS& 1
¼v½ 2 3 ¼c½ 3 2 ¼l½ 18 ¼n½ 6
The distance between the points (2,3) and (5,6) is -

(A) 2 3 (B) 3 2 (C) 18 (D) 6
1  cos 2θ
¼vi½ ;fn θ  45o gks rks dk eku gS & 1
sin 2θ

¼v½ 1@2 ¼c½ 2 ¼l½ 0 ¼n½ 1
1  cos 2θ
If θ  45 then the value of
o
is -
sin 2θ
(A) 1/2 (B) 2 (C) 0 (D) 1
¼vii½ ,d m/okZ/kj NM+ dh yEckbZ rFkk bldh Nk;k dh yEckbZ dk vuqikr 1: 3
gks rks lw;Z dk mUu;u dks.k gS& 1
¼v½ 300 ¼c½ 450 ¼l½ 600 ¼n½ 900
If the ratio of length of vertical rod and length of its shadow is 1: 3 then the angle of elevation of sun
is-

(A) 300 (B) 450 (C) 600 (D) 900
¼viii½ ;fn ,d fcUnq P ls O dsUnz okys fdlh o`Ÿk ij PA, PB Li’kZ js[kk,sa ijLij 80o ds dks.k ij
>qdh gks rks POA cjkcj gS& 1
¼v½ 500 ¼c½ 600 ¼l½ 700 ¼n½ 800
o
If tangents PA and PB from a point P to a circle with center O are inclined to each other at angle of 80
then POA is equal to -

(A) 500 (B) 600 (C) 700 (D) 800
¼ix½ 14 lseh Hkqtk okys oxZ esa cus vUr% o`Ÿk dh ifjf/k gksxh& 1
¼v½ 22 lseh ¼c½ 44 lseh ¼l½ 33 lseh ¼n½ 55 lseh
The circumference of incircle of a square with side 14 cm is -

(A) 22 cm (B) 44 cm (C) 33 cm (D) 55 cm

¼x½ dks.k θ okys f=T;[k.M ds laxr pki dh yEckbZ gS& 1
θ θ θ θ
¼v½  2 r ¼c½  2 r ¼l½  2 r ¼n½  2 r
180o 360o 90o 720o
Length of an arc of a sector of angle is -

θ θ θ θ
(A)  2 r (B)  2 r (C)  2 r (D)  2 r
180o 360o 90o 720o

¼xi½ ?ku ds ,d i`"B dk ifjeki 28 lseh gS rks ?ku dk vk;ru gS& 1
¼v½ 343 lseh3 ¼c½ 196 lseh3 ¼l½ 294 lseh3 ¼n½ 4144 lseh3
The perimeter of surface of a cube is 28 cm, then the volume of the cube is -

(A) 343 cm3 (B) 196 cm3 (C) 294 cm3 (D) 4144 cm3
¼xii½ ,d flDds dks nks ckj mNkyk tkrk gSA de ls de ,d fpr vkus dh izkf;drk gS & 1
¼v½ 1@4 ¼c½ 1@2 ¼l½ 3@4 ¼n½ 1
A coin is tossed twice. The probability of getting at least one head is -

(A) 1/4 (B) 1/2 (C) 3/4 (D) 1
¼xiii½ lcls U;wure la[;k ftlls 27 dks xq.kk djus ij ,d izkd`r la[;k izkIr gksrh gS]og gS&1
¼v½ 3 ¼c½ 3 ¼l½ 9 ¼n½ 3 3
The lowest number, when multiplied by 27 gives a natural number is -

(A) 3 (B) 3 (C) 9 (D) 3 3
¼xiv½ cgqin ds x 2  x  6 'kwU;d gS& 1
¼v½ 1] 6 ¼c½ 2] &3 ¼l½ 3]&2 ¼n½ 1]&6
Zeros of polynomial x  x  6 are -
2

(A) 1,6 (B) 2,-3 (C) 3,-2 (D) 1,-6
¼xv½ lehdj.k 3x-5y+K=0 esa ;fn x=2 rFkk y=-2 gks rks K dk eku gS & 1
¼v½ 16 ¼c½ &6 ¼l½ 6 ¼n½ &16
In equation is 3x-5y+K =0, if x=2 and y=-2 then the value of K is -

(A) 16 (B) -6 (C) 6 (D) -16
¼xvi½ AD vkSj PM f=Hkqtksa ABC vkSj PQR dh Øe’k% ekf/;dk,sa gS] tcfd ABCPQR
gS ;fn AB:PQ=3:5 gS rks AD:PM gS & 1
¼v½ 5: 3 ¼c½ 9:25 ¼l½ 3:5 ¼n½ 25:9
AD and PM are medians of triangle ABC and PQR respectively, where ABCPQR, if AB:PQ = 3:5
then AD:PM is -

(A) 5:3 (B) 9:25 (C) 3:5 (D) 25:9
¼xvii½ 10 ehVj ÅWaph ehukj ds f’k[kj ls i`Foh ij ,d fcUnw dk voueu dks.k 30 gSA fcUnq dh 0

ehukj ds vk/kkj ls nwjh gS& 1
10
¼v½ 10 3 ehVj ¼c½ ehVj ¼l½ 10 ehVj ¼n½ 5 3 ehVj
3

From the top of 10m high tower, angle of depression at a point on earth is 30 0, distance of point from
base of tower is -

10
(A) 10 3 m (B) m (C) 10 m (D) 5 3 m
3
¼xviii½,d m/okZ/kj [kEcs dh ijNkbZ][kEcs dh ÅWapkbZ ds cjkcj gS rks lw;Z dk mUu;u dks.k gksxk&1
¼v½ 450 ¼c½ 300 ¼l½ 600 ¼n½ 500

The shadow of a vertical pillar is same the height of pillar, the angle of elevation of sun will be -

(A) 450 (B) 300 (C) 600 (D) 500

2- fjä LFkkuksa dh iwÆr dhft, ¼ i ls vi½
Fill in the blanks ¼ i to vi ½

¼i½ ;fn 3]7]9]x]5 dk lekUrj ek/; 6 gks rks x dk eku --------------- gksxkA 1
The arithmetical mean of distribution 3,7,9,x,5 is 6, then value of x will be …………

¼ii½ fcUnqvksa ¼3]a½ vkSj ¼4]1½ dh chp dh nwjh √10 gks rks a dk eku ------------- gksxkA 1
If distance between points (3,a) and (4,1) is √10 then a will be .................

¼iii½ 2sin 2 60o cos 60o dk eku gS-------------------- 1
The value of 2sin 2 60o cos 60o is ................

¼iv½ ,d csyu ds fljs dk {ks=Qy 154lseh2 rFkk bldh ÅpkbZ 21 lseh gS csyu dk vk;ru --------
-- gksxkA 1
The volume of cylinder whose height is 21cm and area of its one end is 154cm 2, will be
.........

¼v½ caVu 3]5]7]4]2]1]4]3]4 dk cgqyd --------------- gSA 1
The mode of the distribution 3,5,7,4,2,1,4,3,4 is .............

¼vi½ 10 vkSj 250 ds chp esa --------- la[;k,sa] 4 ds xq.kt gSA 1
............... numbers between 10 and 250 are multiples of 4.

3- vfr y?kqRrjkRed iz'u
Very short answer type questions-

¼i½ ml o`Ÿk dh f=T;k Kkr dhft, ftldk {ks=Qy 49 oxZ lseh gSA 1
Find the radius of the circle whose area is 49 cm2.

¼ii½,d ikls dks QSdus ij le vad vkus dh izkf;drk Kkr dhft,A 1
Find the probability of getting an even number on throwing a dice.

¼iii½ 10lseh Hkqtk okys ?ku ds fod.kZ dh yEckbZ Kkr dhft,A 1
Find the length of the diagonal of a cube of side 10cm.

¼iv½ ,d yEco`Ÿkh; 'kadq ds vk/kkj dh f=T;k 5lseh rFkk ÅpkbZ 12 lseh gS rks 'kadq dh fr;Zd
Å¡pkbZ Kkr dhft,A 1

If the radius of base of a right circular cone is 5cm and height is 12cm then find the slant
height of the cone.

¼v½ caVu 1]3]2]5]9]11 dk ek/;d Kkr dhft,A 1
Find the median of the distribution 1,3,2,5,9,11.

¼vi½ caVu 3]7]5]2]6 dk ek/; Kkr dhft,A 1
Find the mean of the distribution 3,7,5,2,6.

¼vii½ ;fn P(A)=0.65 gS] rks ÞA ughÞ dh izkf;drk Kkr dhft,A 1
If P(A)=0.65, then find the probability of "not A".

¼viii½ 'kadq dk vk;ru Kkr djus dk lw= fyf[k,A 1
Write the formula to find the volume of a cone.

¼ix½ oxZ vUrjky ¼10&25½ dk oxZ fpUg Kkr dhft,A 1
Find the class mark of the class interval (10-25).

¼x½ ,d cDls esa 7 uhys rFkk 3 lQsn daps gSA ;fn bl cDls esa ls ,d dapk ;kn`PN;k fudkyk
tkrk gS] rks bldh D;k izkf;drk gS fd ;g dapk ÞlQsnÞ gS\ 1
A box contains 7 blue and 3 white marbles. If a marble drawn at random from the box, what
is the probability that it will be "white"?

¼xi½ ;fn ,d csyu dk vk;ru 448 lseh3 vkSj ÅpkbZ 7 lseh gS rks csyu dh f=T;k Kkr
dhft,A 1
If volume of cylinder is 448 cm3 and height is 7 cm then find radius of cylinder.

¼xii½ izFke 10 fo"ke izkd`r la[;kvksa dk lekUrj ek/; Kkr dhft,A 1
Find the arithmetic mean of the first 10 odd natural numbers.

[k.M& c
SECTION-B

4- fl) dhft, 3  2 5 ,d vifjes; la[;k gSA 2

Prove that 3  2 5 is irrational.

5- cgqin x 2  3 ds 'kwU;d Kkr dhft, vkSj 'kwU;dksa rFkk xq.kkadks ds chp ds lEcU/k dh lR;rk dh tk¡p
dhft,A 2

Find the zeros of the polynomial x 2  3 and verify the relationship between the zeroes and the
coeffcient.

6- nks vadks dh ,d la[;k ,oa mlds vadks dks myVus ij cuh la[;k dk ;ksx 66 gSA ;fn la[;k
ds vadks vUrj 2 gks rks la[;k Kkr dhft,A ,slh la[;k fdruh gSA 2
The sum of a two-digit number obtained by reversing the digit is 66, if the digit of number differ
by 2, find the number, how many such numbers are there.

7- ;fn fdlh A.P. ds izFke 14 inks dk ;ksx 1050 gS rFkk bldk izFke in 10 gS] rks 20 ok¡ in
Kkr dhft,A 2
If the sum of the first 14 terms of an A.P. is 1050 and its first term is 10, find the 20 th term.

8- nh xbZ vkd`fr esa ODCOBA , ∠𝐵𝑂𝐶 = 125 vkSj CDO  70o gS] rks OAB Kkr
dhft,A 2

In given figure ODCOBA , ∠𝐵𝑂𝐶 = 125 and CDO  70o then find OAB .

9- fcUnqvksa A(2,-2) vkSj B(-7,4) dks tksM+us okys js[kk[k.M dks le&f=Hkkftr djus okys fcUnqvksa ds
funsZ'kkad Kkr dhft,A 2
Find the coordinates of the points of trisection of the line segment joining the points A(2,-2) and
B(-7,4).

1 o
10- ;fn tan(A B)  3 vkSj tan(A  B)  ; 0  A+B  90o ; A  B gS rks A vkSj B dk eku
3
Kkr dhft,A 2

1 o
If tan(A B)  3 and tan(A  B)  ; 0  A+B  90o ; A  B Find A and B.
3

11- 1-5 ehVj yack ,d izs{kd ,d fpeuh ls 28-5 ehVj dh nwjh ij gSA mldh vk¡[kksa ls fpeuh
ds f'k[kj dk mUu;u dks.k 450 gSA fpeuh dh m¡pkbZ crkbZ;As 2

A tower stands vertically on the ground. from a point on the ground which is 15m away from the
foot of the tower, the angle of elevation of the top of the tower is found to be 60 0, find the height
of the tower.

12 nks ladUs nzh; o`Ÿkksa dh f=T;k,sa 5 lseh rFkk 3 lseh gSA cMs+ o`Ÿk dh ml thok dh yEckbZ Kkr
dhft, tks NksVs o`Ÿk dks Li'kZ djrh gksA 2
Two concentric circle are of radil 5cm and 3cm. find the length of the chord of the larger circle
which touches the similler circle.

13- f=T;k 12 lseh okys ,d o`Ÿk dh dksbZ thok dsUnz ij 1200 dk dks.k varfjr djrh gSA laxr
o`Ÿk[k.M dk {ks=Qy Kkr dhft,A 2
A chord of a circle of radius 12cm subtends an angle of 120 0 at the centre. find the area of the
corresponding segment of the circle.

[k.M& l
SECTION-C

14- ;fn fdlh A.P. ds izFke 7 inksa dk ;ksx 49 gS vkSj izFke 17 inks dk ;ksx 289 gS] rks blds
izFke n inksa dk ;ksx Kkr dhft,A 3
If the sum of first 7 terms of an A.P. is 49 and that of 17 terms is 289, find the sum of first n
terms.
vFkok @ OR
og A.P. Kkr dhft, ftldk rhljk in 16 gS] vkSj 7 ok¡ in 5 osa in ls 12 vf/kd gSA 3
th th
Determine the A.P. whose third term is 16 and the 7 term exceeds the 5 term by 12.

15- n'kkZb;sa fd fcUnq ¼1] 7½] ¼4] 2½] ¼&1]&1½ vkSj ¼&4] 4½ ,d oxZ ds 'kh"kZ gSA 3
Show that the point (1, 7), (4, 2), (&1,&1) and (&4,4) are the vertices of a square.
vFkok/ OR
;fn A vkSj B Øe'k% ¼&2]&2½ vkSj ¼2]&4½ gks rks fcUnq P ds funsZ'kkad Kkr dhft, rkfd
3
AP  AB gks vkSj P js[kk[k.M AB ij fLFkr gksA 3
7
3
If A and B are (-2,-2) and (2,-4) respectively, find the coordinate of P such that AP  AB and P lies on
7
the line segment AB.
16- fl) dhft, fd fdlh o`Ÿk ds ifjxr lekUrj prqHkqZt leprqHkqZt gksrk gSA 3
Prove that the parallelogram circumscribing a circle is rhombus.
vFkok / OR
fl) dhft, fd o`Ÿk ds ifjxr cus prqHkqZt dh lkeus&lkeus dh Hkqtk,¡ dsUnz ij laiwjd dks.k

varfjr djrh gSA 3
Prove that opposite sides of a quadrilateral circumscribing subtends supplementary angles at the centre
of the circle.

17- ,d d{kk ds Nk=ksa ds Hkkj fuEu lkj.kh esa fn;s x;s gS&
Hkkj ¼fdxzk es½a 20 21 22 23 24 25 26 27 28

Nk=ksa dh 1 2 6 7 4 2 3 2 3
la[;k

budk lekUrj ek/; Kkr dhft,A 3
The weight of students in a class are given in the following table.
Weight 20 21 22 23 24 25 26 27 28
(in
Kgs.)

No. of 1 2 6 7 4 2 3 2 3
Students

vFkok @ OR
;fn fuEu caVu dk ek/; 7-5 gks] rks P dk eku Kkr dhft,A 3

x 3 5 7 9 11 13

f 6 8 15 P 8 4
If mean of the following distribution is 7.5, then find the value of P.
x 3 5 7 9 11 13

f 6 8 15 P 8 4

[k.M& n
SECTION-D
18- ,d ledks.k f=Hkqt dh Å¡pkbZ mlds vk/kkj ls 7 lseh de gSA ;fn d.kZ 13 lseh dk gks] rks
vU; nks Hkqtk,¡ Kkr dhft,A 4
The attitude of a right angled triangle is 7cm less than its base.If the hypotenuse is 13cm, find the
other two sides.
vFkok @ OR
1
lehdj.k 3x2  2 x   0 dk fofoDrdj Kkr dhft, vkSj fQj ewyksa dh izd`fr Kkr dhft,A
3
;fn os okLrfod gS] rks mUgsa Kkr dhft,A 4
Find the dircriminant of the equation 3x -2x= /3=0 and hence find the nature of its roots. Find them, if
2 1

they are equal.

1  cos A
19- fl) dhft, &  cosecA  cotA 4
1  cos A

1  cos A
Prove That  cosecA  cotA
1  cos A
vFkok @ OR
fl) dhft, & 4
sin A  cos A=1  3sin A cos A
6 6 2 2

Prove That
sin 6 A  cos 6 A=1  3sin 2 A cos 2 A
20- fuEu ckjEckjrk caVu ls cgqyd Kkr dhft,& 4
izkIrkad 20&30 30&40 40&50 50&60 60&70

Nk=ksa dh la[;k 4 28 42 20 6

Find the mode of following frequency distribution.
Marks Obtained 20-30 30-40 40-50 50-60 60-70

No. of Students 4 28 42 20 6

vFkok @ OR
100 ifjokjksa esa cPpksa dh la[;k fuEu izdkj gS] budk ek/;d Kkr dhft, & 4
cPpksa dh la[;k 0&2 2&4 4&6 6&8 8&10

ifjokjksa dh la[;k 38 36 18 5 3

The number of children in 100 families are as follows, find their median.
No. of Children 0-2 2-4 4-6 6-8 8-10

No. of Families 38 36 18 5 3

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