i'u i= dk Cy fiVmpbse.nic.in/10th/MATHEMATICS/Question Answer 2.pdf · i'u i= dk Cy fiV d{kk & 10 oh fo"k; & xf.kr le; & 3 ?kV iw.kkd & 100 b- Ø- bdkbZ bdkb ij vkofVr vd oLrfu"B

1

iz'u i= dk Cyw fizaV

d{kk & 10 oha fo"k; & xf.kr

le; & 3 ?kaVs iw.kkZad & 100

b- Ø-

bdkbZ bdkbZ ij vkoafVr vad

oLrqfu"B iz’u

vadokj vU; iz’uksa dh la[;k

04 vad

05 vad

06 vad

dqy la[;k 01 vad

1 nks pj jkf’k;ksa dk jSf[kd lehdj.k 10 2 2 & & 2

2 cgqin ,oa ifjes; O;atd 07 2 & 1 & 1

3 vuqikr ,oa lekuqikr 05 1 1 & & 1

4 oxZ lehdj.k 10 1 1 1 & 3

5 Okkf.kfT;d xf.kr 08 3 & 1 & 1

6 le:i f=Hkqt 08 2 & & 1 1

7 oÙk 10 4 & & 1 1

8 jpuk,W 05 & & 1 & 1

9 f=dksa.kfefr 10 5 & 1 & 1

10 ÅWpkbZ ,oa nwjh 05 1 1 & & 1

11 {ks=fefr 10 2 2 & & 2

12 lkaf[;dh] izkf;drk] dafMdk] iqujkofRr 12 2 1 & 1 2

dqy iz’u 01 08 05 03 17

dqy vad 25 32 25 18 100

funsZ’k %& 1- lHkh iz’u gy djuk gSaA 2- izR;sd iz’u ij fu/kkZfjr vad muds lEeq[k n’kkZ;s gSaA 3- iz’u Ø- 1 esa oLrqfu"B izdkj ds 25 iz’u gksxsA izR;sd iz’u ij 1 vad fu/kkZfjr gSA bu iz’uksa

esa lgh fodYi pquuk] fjDr LFkkuksa dh iwfrZ] rFkk lgh tksM+h bR;kfn izdkj ds iz’uksa dk lekos’k fd;k gSA

4- iz’u Ø- 2 ls 17 rd lHkh iz’uksa esa vkarfjd fodYi fn;k tkuk gSA izR;sd iz’uksa esa fodYi leku bdkbZ ,oa leku Lrj ds gSaA

5- iz’uksa dk dfBukbZ Lrj] ljy 50 % ] lekU; 35 % ,oa dfBu 15 % fn;k x;k gSA

2

vkn’kZ iz’u i= G d{kk 10 oh

fo"k; & xf.kr Mathematics

(Hindi and English Version)

Time : 3 Hours MM : 100

funsZ'k%

1- lHkh iz'u vfuok;Z gSaA

2- iz'u i= esa fn;s x;s funsZ'k lko/kkuhiwoZd i<+dj iz'uksa ds mRrj fyf[k,A

3- iz'u i= esa nks [k.M fn;s x, gSa] [k.M ^v* vkSj [k.M ^c^

4- [k.M&^v* esa iz'u dzekad 1 esa oLrqfu"B izdkj ds iz'u fn, x, gSaA

funsZ'kkuqlkj gy dhft,A

5- [k.M & ^c^ esa iz'u dzekad 2 ls 17 esa vkarfjd fodYi fn;s x, gSaA

6- tgka vko';drk gks] LoPN js[kkfp= cukb,A

7- izR;sd iz'u ds fy, vkoafVr vad mlds lEeq[k vafdr gSA

Instructions:

1- All questions are compulsory.

2- Read the instructions of question paper carefully and answer the

questions.

3- There are two parts; Section-‘A’ and ‘B’ in the question paper.

4- Q. No. 1 is objective type question in Section-A. Do as directed.

5- Internal options are given in Q. Nos. 2 to 17 of Section -B.

6- Draw neat and clean diagrams wherever required.

7- Marks allotted to each question are mentioned against the question.

3

[k.M & ¼v½ Section – (A)

oLrqfu"B iz'u

(Objective Type Questions)

1(A) lgh fodYi pqudj viuh mRrj iqfLrdk esa fyf[k, % (1×5 = 5)

Choose the correct option and write in your Answer Book.

(i) jSf[kd lehdj.k a1x+b1y+c1=0 rFkk a2x+b2y+c2=0 ds vuar gy ds fy, çfrca/k gS&

(a) 2

2

1

1

ba

ba

(b) 2

2

1

1

ba

ba

(c) 2

1

2

1

2

1

cc

bb

aa

(d) 2

1

2

2

1

1

cc

ba

ba

The condition for infinite solution of the system of two simultaneous linear equation

a1x+b1y+c1=0 and a2x+b2y+c2=0 is :

(a) 2

2

1

1

ba

ba

(b) 2

2

1

1

ba

ba

(c) 2

1

2

1

2

1

cc

bb

aa

(d) 2

1

2

2

1

1

cc

ba

ba

(ii) nks vadks dh la[;k esa bdkbZ dk vad x rFkk ngkbZ dk vad y gS rks la[;k gksxh&

(a) 10x+y (b) 10y+x (c) 10x-y (d) 10y-x

In a two digit number, X is at unit place and Y is at tens place, them the number will be:

(a) 10x+y (b) 10y+x (c) 10x-y (d) 10y-x

(iii) x

x 1 dk ;ksT; çfrykse gksxk&

(a) x

x 1

(b)

xx 1 (c) x

x

1 (d) x

x

1

The additive inverse of x

x 1 will be:

(a) x

x 1 (b)

xx 1 (c) x

x

1 (d) x

x

1

4

(iv) 9 vkSj 12 dk rrh;uqikrh gSA

(a) 10 (b) 8 (c) 15 (d) 16

The third proportional of 9 and 12 is

(a) 10 (b) 8 (c) 15 (d) 16

(v) 392

xx dk ljyre :i gksxk&

(a) x-3 (b) x+3 (c) x+9 (d) x-9

The simplest form of 392

xx

will be

(a) x-3 (b) x+3 (c) x+9 (d) x-9

1(B) lgh fodYi pqudj viuh mÙkj&iqfLrdk esa fyf[k, & (1×5 = 5)

Choose the correct option and write in your Answer Book.

(i) /kwi esa [kM+s ,d O;fDr dh Nk;k mldh Å¡pkbZ dh 3 xquh gS rks ml le; lw;Z

dk mUu;u dks.k gksxk&

(a) 300 (b) 450 (c) 600 (d) 750

Find the angle of elevation of the sun (sun's attitude) when the length of the shadow

of a person is equation to 3 times of his height.

(a) 300 (b) 450 (c) 600 (d) 750

(ii) csyu dk oØi"B gS&

(a) r2h (b) 2r(r+h) (c) 2rh (d) 31 r2h

The curved surface of the cylinder is –

(a) r2h (b) 2r(r+h) (c) 2rh (d) 31 r2h

(iii) xksys ds vk;ru dk lw= gS&

(a) r2h (b) 34 r3 (c)

31 r3 (d) 4a2h

Formula for the volume of sphive is

(a) r2h (b) 34 r3 (c)

31 r3 (d) 4a2h

5

(iv) 2,4,2,6,8,10,4,3,4 dk cgqyd gS&

(a) 4 (b) 6 (c) 5 (d) 3

The mode of the given number 2,4,2,6,8,10,4,3,4 is

(a) 4 (b) 6 (c) 5 (d) 3

(v) ,d flDds dks mNkyus ij gsM vkus dh çkf;drk gksxh&

(a) 0 (b) 1 (c) 21 (d)

31

The probability of getting head in a throw of single coin, will be –

(a) 0 (b) 1 (c) 21 (d)

31

1(C) fjDr LFkkuksa dh iwfrZ dhft,& (1×5=5)

(i) ewy/ku ,oa pØof) C;kt ds ;ksx dks ---------------- dgrs gSaA

(ii) okgu ,oa e'khujh dh le; ds lkFk ewY; esa deh gksuk ---------------- dgykrk gSA

(iii) çR;sd foÙkh; o"kZ esa 'kkldh; deZpkfj;ksa dks vius osru ds vuq:i dqN /kujkf'k

'kklu dks dj ds :i esa nsuk gksrh gSA ftls ----------------- dgrs gSA

(iv) ;fn nks f=Hkqtksa dh laxr Hkqtk;sa vkuqikfrd gksa rks os f=Hkqt ---------------- gksrs gSaA

(v) nks le:i f=Hkqtksa ds {ks=Qyksa esa 9%25 dk vuqikr gS rks muds 'kh"kZ yEcksa dk

vuqikr -------------------- gksxkA

Fill up the blanks –

(i) The sum of principle and compound interest is called...............

(ii) The cost of vehicles and machine is reduced during with time is called...............

(iii) Every government servant has to pay certain amount as tax according to salary, to the

government. This tax is called............

(iv) If the ratio of two corresponding side are proportional then triangles are....................

(v) The ratio of areas of two similar triangles is 9:25 then ratio of their altitudes is..............

1(D) lgh tksfM+;ka cukb;s& (1×5=5)

LraHk ^v* LraHk ^c*

(i) Sec2-tan2 (a) 2

(ii) 2tan- cot (b) cot2

(iii) 2sin1 (c) 1

6

(iv) cosec2 -1 (d) cot

(v) tan(900-) (e) cos ,

(f) sec2

(g) cosec2

Match the following

Column A Column B

(i) Sec2-tan2 (a) 2

(ii) 2tan -cot (b) cot2

(iii) 2sin1 (c) 1

(iv) cosec2 -1 (d) cot

(v) tan(900-) (e) cos ,

(f) sec2

(g) cosec2

1(E) fuEufyf[kr esa lR;&vlR; fyf[k,& (1×5=5)

(i) oxZ&lehdj.k ds vusd gy gks ldrs gSaA

(ii) fdlh ckgjh fcUnq ls oÙk ij [khaph xbZ Li'kZ js[kk,sa cjkcj gksrha gSaA

(iii) oÙk dh lcls cM+h thok f=T;k dgykrh gSA

(iv) v)Zo`Ùk eas vUrfjd dks.k ledks.k gksrk gSA

(v) oÙk dh ,d gh vo/kk esa cus dks.k leku gksrs gSaA

Write true of false in the following

(i) The quadratic equation has many solutions.

(ii) The length of two tangents; drawn from an external point to a circle are equal.

(iii) The longest chord of the circle is called radius.

(iv) The angle subtended in a semicircle is right angle.

(v) Angles in the same segment of a circle are equal.

7

[k.M ¼c½ Section (B)

vfr y?kqmÙkjh; ç'u (Very short answer type Questions)

2- foyksiu fof/k ls fuEu lehdj.k fudk; dks gy dhft,& ¼4 vad½

8x+5y=9......................... (i)

3x+2y=4......................... (ii)

Solve the following system of equation by elimination method.

8x+5y=9......................... (i)

3x+2y=4......................... (ii)

vFkok ¼OR½

fuEufyf[kr lehdj.k fudk; dks çfrLFkkfir fof/k ls gy dhft,&

x+y=7 .........................(i)

3x-2y=11 ...................(ii)

Solve the following system of equation by substitution method.

x+y=7 .........................(i)

3x-2y=11 ....................(ii)

3- k ds fdl eku ds fy, lehdj.k fudk; kx+y=5 ,oa 3x+y=1 dk vf}rh; gy

gksxkA ¼4 vad½

For what value of K the system of linear equation kx+y=5 and 3x+y=1 will has unique

solution.

vFkok ¼OR½

2 dqlhZ vkSj 3 estksa dk ewY; 800 :i;s gS vkSj 4 dqlhZ vkSj 3 estksa dk ewY;

1000 gSA 2 dqlhZ vkSj 2 estksa dk ewY; Kkr dhft,A

The cost of 2 chairs and 3 tables is Rs. 800 and the cost of 4 chair and 3 tables

is Rs. 1000. Find the cost of 2 chairs and 2 tables.

4- nks la[;kvksa dk vuqikr 3%4 gS ;fn çR;sd la[;k esa 5 tksM+ fn;k tkos rks

vuqikr 4%5 gks tkrk gS la[;k;sa Kkr dhft,A ¼4 vad½

8

The ratio of two numbers is 3:4. If 5 is added to each number them the ratio

becomes 4:5 find the number.

vFkok ¼OR½

;fn ba

zac

ycb

x

gks rks fl) djks fd& (b-c)x+(c-a)y+(a-b)z=0

If ba

zac

ycb

x

then prove that - (b-c)x+(c-a)y+(a-b)z=0

5- fuEufyf[kr lehdj.k dks lw=&fof/k ls gy dhft, ¼4 vad½

3y2 = y+1

Solve the following equation by formula method.

3y2 = y+1

vFkok ¼OR½

oxZ lehdj.k Kkr dhft, ftlds ewy 73 ,oa 73 gSA

Construct the quadratic equation whose roots are 73 and 73 .

6- ,d Hkou ds ikn ls 25 ehVj dh nwjh ls Hkou ds f'k[kj dk mUu;u dks.k 600

gS rks Hkou dh Å¡pkbZ Kkr dhft,A ¼4 vad½

Form a point 25 meter a way from the foot of the building, the angle of elevation of

the top of the building is 60º. Find the height of the building.

vFkok ¼OR½

50 ehVj Å¡ph igkM+h ds f'k[kj ls fdlh ehukj dh pksVh vkSj vk/kkj ds vou;u

dks.k Øe'k% 300 vkSj 450 gS rks ehukj dh Å¡pkbZ Kkr dhft,A

Form the top of 50 meter high hill, the angle of depression of the top and the

bottom of a tower are 30º and 45º. Find the height of the tower.

7- ml cM+s ls cM+s ckal dh yEckbZ Kkr dhft, tks 9 ehVj yEcs] 6 ehVj pkSM+s

vkSj 2 ehVj Å¡ps dejs esa j[kk tk ldsA ¼4 vad½

What will be length of longest bamboo which can be put in a room, 9 meter long, 6

meter broad and 2 meter high?

9

vFkok ¼OR½

,d lkbfdy ds ifg;s dk O;kl 1-4 ehVj gS ifg;k 1 fefuV ea 100 pDdj yxkrk gS crkb;s 1 ?kaVs esa lkbfdy fdruh nwj tk;sxhA The diameter of cycle wheel is 1.4 meter; the wheel revolves 100 times in one

minute. Find how much distance will be cycle cover in one hour.

8- ,d csyu ds vk/kkj dk O;kl 14 ls-eh- gS Å¡pkbZ 20 ls-eh- gS] csyu dk lEiw.kZ i"B Kkr dhft,A ¼4 vad½ The diameter of base of a cylinder is 14 cm. and its height is 20 cm. Find whale

surface area and volume.

vFkok ¼OR½

24 ls-eh- Å¡pkbZ vkSj 6 ls-eh- f=T;k okys 'kadw dks xksys dk :i fn;k x;k gS] xksys dh f=T;k Kkr dhft,A A cone of height 24cm. and radius 6cm. is recast into sphere. Find the radius of the

sphere.

9- fuEufyf[kr ekuksa dh ekf/;dk Kkr dhft,A ¼4 vad½

15, 35, 18, 26, 29, 27, 20, 19, 25

Find the median of the following values of variate-

15, 35, 18, 26, 29, 27, 20, 19, 25

vFkok ¼OR½

,d ikWls dks mNkyus ij fo"ke vad vkus dh çkf;drk Kkr dhft,A

Find the probability of getting an odd number in a single throw of a dia.

10- ;fn 1623

2

x

xA vkSj 2)4(5

xxB rks A+B dk eku Kkr dhft,A ¼5 vad½

If A = 1623

2

x

xA and 2)4(5

xxB then find the value of (A+B)

vFkok ¼OR½

pØh; xq.ku[k.M Kkr dhft, & xy(x-y)+yz(y-z)+zx(z-x)

Find cyclic factors of; xy(x-y)+yz(y-z)+zx(z-x)

11- ,d la[;k vkSj mlds O;qRØe dk ;ksx 750 gks rks la[;k Kkr dhft,A ¼5 vad½

The sum of a number and its reciprocal is 750

, then find the number.

10

vFkok ¼OR½

;fn vkSj oxZ lehdj.k ax2+bx+c=0 ds ewy gks rks (3+3) dk eku Kkr

dhft,A

If and are the roots of quadratic equation ax2+bx+c=0 then find the value of

(3+3).

12- 8000 :i;s dk 10 çfr'kr izfro"kZ C;kt dh nj ls 1½ o"kZ dk pØof) C;kt

Kkr dhft;s ;fn C;kt dh x.kuk Nekgh dh xbZ gksA ¼5 vad½

Find the compound interest on Rs. 8000 for 1½ years at the rate of 10% per annum.

If the interest; is compound half yearly.

vFkok ¼OR½

,d ?kM+h 960 :i;s uxn ;k 480 :i;s vkaf'kd Hkqxrku dj 245 :i;s dh 2

ekfld fdLrksa ij nh xbZA fdLr ;kstuk dh C;kt dh nj Kkr dhft,A

A watch is sold for Rs. 960 cash or for Rs. 480 cash down payment and two monthly

installments of Rs. 245 each. Find the rate of interest charged under the installment

plan.

13- ml f=Hkqt ds ifjoÙk dh jpuk dhft, ftldh Hkqtk;sa 6-5 ls-eh-] 7 ls-eh- ,oa 7-5 ls-eh- gSA ¼5 vad½ Construct the circum circle of the triangle whose sides are 6.5cm, 7cm. and 7.5 cm.

and measure its radius.

vFkok ¼OR½

,d pØh; prqHkqZt dh jpuk dhft, ftlesa 'kh"kZ B=700, AC=5 lseh- AB=2 lseh- vkSj AD = 3 lseh- gSA Construct a cyclic quadrilateral in which vertical angle B=70º, AC=5cm., AB=2cm.

and AD=3cm.

14- fuEu loZlfedk dks fl) dhft,& ¼5 vad½

sin2

cos1sin

sincos1

Prove the following identity

sin2

cos1sin

sincos1

11

vFkok ¼OR½

fl) dhft, ¼fcuk lkj.kh ds ç;ksx ls½

sin480sec420 + cos480cosec420 = 2

Prove that (without using table)

sin480sec420 + cos480cosec420 = 2

15- nks le:i f=Hkqtksa ds {ks=Qyksa ds vuqikr mudh laxr Hkqtkvksa ds oxksZ ds

vuqikr ds cjkcj gksrk gSA ¼6 vad½

The ratio of the areas of two similar triangle is equal to the ratio of the squares of

corresponding sides.

vFkok ¼OR½

f=Hkqt ABC esa B U;wudks.k gS AD 'kh"kZ yEc gS rks fl) dhft,&

AC2 = AB2 + BC2 - 2BC.BD

In triangle ABC, B is an acute angle. AD is an attitude. Prove that

AC2=AB2+BC2-2BC.BD

16- fl) djks fd pØh; prqHkqZt ds lEeq[k dks.kksa dk ;ksx 1800 gksrk gSA ¼6 vad½

Prove that the sum of opposite angles of a cyclic quadrilateral is 1800.

vFkok ¼OR½

5 ls-eh- v)ZO;kl ds ,d oÙk esa nks thok;sa Øe'k% 8 lseh- vkSj 6 lseh- yEckbZ

dh gSaA nksuksa thok;sa lekUrj vkSj dsUæ ds ,d gh vksj gSaA nksuksa thokvksa ds chp

dh nwjh Kkr dhft,A

The length of two chords in a circle with 5cm. radius; are 8cm. and 6cm. respectively.

Both the chords are parallel and on the same side of the center. Find the distance

between the two chords.

17- 100 fo|kfFkZ;ksa ds fuEufyf[kr çkIrkadks ls lekUrj ek/; Kkr dhft,A ¼6 vad½

çkIrkad 0-10 10-20 20-30 30-40 40-50

fo|kfFkZ;ksa dh la[;k 8 30 40 12 10

12

Find the mean of the following marks obtained by 100 students.

Marks obtained 0-10 10-20 20-30 30-40 40-50

No. of Students 8 30 40 12 10

vFkok ¼OR½

1996 dks vk/kkj o"kZ ekudj ,d e/;e oxZ ifjokj ds ctV ls fuEufyf[kr

tkudkjh ds vk/kkj ij 1999 dk fuokZg [kpZ lwpdkad Kkr dhft,&

oLrq ek=k ¼bdkbZ½ ewY; çfr bdkbZ ¼:i;ksa esa½

1996 1999

A 8 22 25

B 12 35 40

C 5 25 30

D 15 20 25

E 10 15 20

Calculate the cost of living index number for the year 1999 on the basis of 1996 of a

medium family from the following information.

Items Quantity

(Unit)

Price per unit (in Rs.)

1996 1999

A 8 22 25

B 12 35 40

C 5 25 30

D 15 20 25

E 10 15 20

13

vkn'kZ mÙkj xf.kr & 10 oha

[k.M & ^^v**

1(A) lgh fodYi (i) (c) = =

(ii) (b) 10y+x (iii) (c) - x (iv) (d) 16 (v) (b) (x+3)

1(B) lgh fodYi (i) (b) 300 (ii) (c) 2πrh (iii) (b) πr3 (iv) (a) 4 (v) (c)

1(C) lgh mÙkj ¼fjDr LFkku½ (i) pØof) feJ/ku

(ii) ewY; ál ¼?klkjk½

(iii) O;kolkf;d dj

(iv) le:i (v) 3:5

1(D) lgh tksfM+;k¡ [A] [B]

(i) 푠푒푐 휃 − 푡푎푛 휃 (c) 1

(ii) 2푡푎푛휃 푐푎푡휃 (a) 2

(iii) √1 − 푠푖푛 휃 (e) cos휃

(iv) 퐶표푠푒푐 휃 − 1 (b) 푐표푡 휃

(v) tan (90°− 휃) (d) 푐표푡휃

1(E) lgh mÙkj (i) vlR;

(ii) lR;

(iii) vlR;

(iv) lR; (v) lR;

14

[k.M ¼c½ iz'u Øekad & 02 dqy 4 vad

gy%& 8x + 5y = 9 (i) 3x + 2y = 4 (ii)

lehdj.k (i) ls 3 rFkk lehdj.k (ii) esa 8 dk xq.kk djus ij 24x + 15y = 27 (iii) 24x + 16y = 32 (iv) ¼1vad½ ;k] 24x + 15y = 27 24x + 16y = 32 ?kVkus ij (-) (-) (-) -y = -5

y = 5 ¼1vad½ y dk eku lehdj.k ¼i½ esa j[kus ij

⇒ 8푥 + 5푦 = 9 ----------------¼i½ ¼1vad½ ⇒ 8푥 + 5 × 5 = 9

⇒ 8푥 + 25 = 9

⇒ 8푥 = 9 − 25 ⇒ 8푥 = −16

⇒ 푥 = -2 ¼½vad½

mÙkj x = - 2 ,oa y = 5 ¼½vad½ uksV & mijksDrkuqlkj fy[ks tkus ij 1$1$1$½$½ ¾ 4 vad izkIr gksaxsA

iz'u Øekad & 02 vFkok gy & x + y = 7 (i) 3x - 2y = 11 (ii) lehdj.k (i) ls y = 7-x (iii) ¼1vad½ y dk eku lehdj.k (ii) esa j[kus ij => 3x - 2y = 11 => 3x-2(7-x) = 11 ¼1vad½

15

=> 3x - 2 (7-x) = 11 => 3x - 14 + 2x = 11 => 3x+ 2x = 11+14 => 5x = 25 => x = 5 ¼1vad½ x dk eku leh- ¼iii½ esa j[kus ij y = 7 – x

y = 7-5 y = 2 ¼½vad½

mÙkj & x = 5 ,oa y = 2 ¼½vad½

uksV & mijksDrkuqlkj fy[ks tkus ij 1$1$1$½$½ ¾ 4 vad izkIr gksaxsA

iz'u Øekad & 03 dqy 4 vad

gy & kx + y = 5 (i) 3x + y = 1 (ii) lkekU; lehdj.k fudk; ls rqyuk djus ij a1x + b1 y = c1 (iii) a2x + b2y = c2 (iv) ¼1vad½ rqyuk ls a1 = k , b1 = 1 c1 = 5 a2 = 3 , b2=1 c2=1 ¼1vad½ fudk; dk ,d vf}rh; gy gksxk ;fn

≠ ¼1vad½

≠

k 3 ¼½vad½ mÙkj fudk; dk ,d vf}rh; gy gksxk ;fn k 3 ¼½vad½


16

iz'u Øekad & 03 vFkok gy & ekuk fd ,d dqlhZ dk ewY; x :i;s gS rFkk

,d est dk Ø; ewY; y :i;s gS iz'ukuqlkj

2x + 3y = 800 (i) 4x + 3y = 1000 (ii) ¼1vad½

lehdj.k (ii) ls lehdj.k (i) dks ?kVkus ij 2x = 200 x = 100 ¼1vad½ x dk eku leh- (i) esa j[kus ij 2x + 3y = 800 2× 100 + 3y = 800 200 + 3y = 800 3y = 800-200 3y = 600 y = 200 ¼1vad½

2 dqlhZ ,oa 2 estksa dk ewY; ¾ 2x + 2y ¾ 2× 100 + 2× 200 = 200 + 400 = 600 ¼½vad½

mÙkj & 2 dqlhZ ,oa 2 estksa dk ewY; ¾ 600 :i;s gksxkA ¼½vad½


iz'u Øekad & 04 dqy 4 vad gy & ekuk la[;k,sa 3x vkSj 4x gSa izR;sd la[;k esa 5 tksM+us ij

ubZ la[;k,sa 3x + 5 vkSj 4x + 5 gksaxh ¼1vad½ iz'ukuqlkj

= ¼1vad½

=> 5 (3x+5) = 4 (4x+5)

17

=> 15x + 25 = 16x + 20 => 15x - 16x = 20-25 => -x = -5 x = 5 ¼1vad½ la[;k;sa 3x => 3 × 5 = 15

4x => 4 × 5 = 20 ¼½vad½

mRrj & vHkh"V la[;k;sa 15 vkSj 20 gksaxh ¼½vad½


iz'u Øekad & 04 vFkok

gy & fn;k gS& = =

ekuk = = = k ¼1vad½

rc x = k(b+c) y = k(c+a) z = k(a+b) ¼1vad½

L.H.S. = (b-c)x + (c-a)y + (a-b)z x, y, z dk eku j[kus ij

= (b-c)k (b+ c) + (c-a)k(c+a) + (a-b)k (a+b) ¼1vad½ = k[ (b2-c2) + (c2-a2) + (a2-b2)] = k[ b2-c2 + c2-a2 + a2- b2]

= k × o ¼½vad½ = o = R.H.S. ¼½vad½


iz'u Øekad & 05 dqy 4 vad gy & fn;k gS& 3y2 = y+1 ;k 3y2-y-1= 0 mDr lehdj.k dh rqyuk lkekU; lehdj.k ay2+by+c=0 ls djus ij a = 3 , b = -1, c = -1 ¼1vad½

ge tkurs gS fd y = ±√

18

y = ( )± ( ) × ××

¼1vad½

y = ±√

y = ±√ ¼1vad½

(+) fpUg ysus ij y = √

(–) fpUg ysus ij y = –√ ¼½vad½

mÙkj y = √ vkSj y = –√ ¼½vad½



fn;k x;k gS& = 3+√7 rFkk = 3-√7 ¼1vad½

ewyksa dk ;ksxQy + = 3+√7+ 3-√7

+ = 6 ¼1vad½

ewyksa dk xq.kuQy . = (3+√7)(3-√7)

. = (3)2-(√7)2

. = 9-7

. = 2 ¼1vad½

rc oxZ lehdj.k gksxh x2 - (+) x + . = 0 (i) ¼½vad½

mDr lehdj.k ¼i½ esa (+) o . ds eku j[kus ij

x2 - 6x + 2 = 0

mÙkj x2 - 6x + 2 = 0 ¼½vad½


19

iz'u Øekad & 06 dqy 4 vad gy &

¼fp= 1vad½ fn;k x;k gS& AC= 25 eh-

ekuk Hkou dh Å¡pkbZ h eh- gS

voueu dks.k ¾ mUu;udks.k ¼,dkUrj dks.k gSa½ ¼1vad½

ledks.k ABC esa 푡푎푛휃 =

푡푎푛휃 = ¼1vad½

= tan60°

= √

h = 25√3 eh- ¼½vad½

mÙkj & Hkou dh Å¡pkbZ ¾ 25√3 ehVj

;k 25 × 1.732 ¾ 43.3 ehVj gksxhA ¼½vad½


h

25 eh-

D

C

B

A 60º

60º

90º

yac vk/kkj

20


¼fp= 1vad½

igkM+ dh Å¡pkbZ AB = 50 eh- ekuk ehukj dh Å¡pkbZ h eh- gS rFkk AC = FD = 푥 eh-

voueu dks.k ¾ mUu;udks.k ¼,dkUrj dks.k gSa½ BF = (50 − ℎ) eh- ¼1vad½

ledks.k ∆ABC esa = tan45°

= 1

x = 50 ehVj ¼1vad½

iqu ledks.k BDF esa = tan30°

= √

50 − h = √

50 − √

= ℎ

√√

= ℎ

(√ )√

= ℎ ¼½vad½

( . ).

= ℎ ;k 21.13 eh-

mRrj & ehukj dh ÅapkbZ h= 21.13ehVj gksxhA ¼½vad½


E

C

x

x A

50-h

50eh- D

h ehukj

30º

30º 45º

45º 90º

B

F

yac vk/kkj

yac vk/kkj

21

iz'u Øekad & 07 dqy 4 vad fn;k gS %& dejs dh yEckbZ a = 9 eh-

pkSM+kbZ b = 6 eh- Å¡pkbZ c = 2 eh-

Kkr djuk gS & cM+s ls cM+s ck¡l dh yEckbZ vFkkZr ?kukdkj dejs dk fod.kZ ¼1vad½

ck¡l dh yEckbZ ¼fod.kZ½ ¾ √푎 + 푏 + 푐 ¼1vad½

¾ (9) + (6) + (2)

¾ √81 + 36 + 4 ¼1vad½

¾ √121

¾ 11 ¼½vad½

mÙkj & cM+s ls cM+s ck¡l dh yEckbZ 11 ehVj gksxhA ¼½vad½


iz'u Øekad & 07 vFkok fn;k gS & ifg;s dk O;kl = 1.4 eh-

ifg;s dh f=T;k = . = = eh-

,d pDdj esa pyh x;h nwjh ¾ ifg;s dh ifjf/k ¼1vad½ ¾ 2휋푟

¾ 2 × ×

¾ ¼1vad½

100 pDdjksa esa pyh x;h nqjh ¾ × 100 ehVj

1 fefuV esa pyh x;h nwjh ¾ × 100 ehVj ¼iz’ukuqlkj½ ¼1vad½

1 ?k.Vs esa pyh x;h nwjh ¾ × 100 × 60 ehVj ¼½vad½

¾ 26400 ehVj

mRrj & lkbfdy 1 ?kaVs esa 26-40 fdeh- nwj tk;sxh ¼½vad½


22

iz'u Øekad & 08 dqy 4 vad fn;k gS & csyu dk O;kl ¾ 14 lseh-

r = lseh- ¼fp= 1vad½

r = 7 lseh- ¼1vad½ ,oa h = 20 lseh- Kkr djuk gS & csyu dk lEiw.kZ i"B csyu dk lEiw.kZ i"B ¾ 2휋푟(푟 + ℎ)

¾ 2 × × 7 × (20 + 7) ¼1vad½

¾ 44 × 27

¾ 1188 ¼½vad½

mÙkj & csyu dk lEiw.kZ i"B 1188 oxZ lseh- ¼½vad½



fn;k gS &

'kadq dh Å¡pkbZ ℎ = 24 ls-eh-

'kadq dh f=T;k 푟 = 6 ls-eh- ¼1vad½

Kkr djuk gS & xksys dh f=T;k

ekuk fd 'kadw ls cus xksys dh f=R;k R lseh- gSA

'kadq dk vk;ru ¾ 휋푟 ℎ

xksys dk vk;ru ¾ 휋푅 ¼1vad½

h = 20cm

2r = 14cm

h = 24 cm.

r = 6 cm.

23

iz'ukuqlkj &

xksys dk vk;ru ¾ 'kadq dk vk;ru

πR = π × 6 × 6 × 24 ¼1vad½

R3 = × × ×

R3 = 63

R = 6 c.m. ¼½vad½

mÙkj & 'kadw ls cus xksys dh f=T;k R = 6 lseh- gksxhA ¼½vad½


iz'u Øekad & 09 dqy 4 vad fn;s x;s eku ¼vkadM+s½ &

15] 35] 18] 26] 29] 27] 20] 19] 25 gSaA

vkadM+ksa dks vkjksgh Øe esa fy[kus ij

15] 18] 19] 20] 25] 26] 27] 29] 35 ¼1vad½

inksa dh la[;k n = 9 fo"ke la[;k gSA ¼½vad½

rc ekf/;dk ¾ oka in gksxh ¼1vad½

¾ oka in

¾ oka in

¾ 5 oka in ¼½vad½

5 oka in 25 gSA vr% ekf/;dk 25 gksxhA mÙkj ¼1vad½

uksV & mijksDrkuqlkj fy[ks tkus ij 1$½$1$½$1 ¾ 4 vad izkIr gksaxsA

24


ikWls dks ,dokj mNkyus ij

laHkkfor dqy ifj.kke 푆 = {1, 2, 3, 4, 5, 6}

dqy ifj.kkeksa dh la[;k 푛(푠) = 6 ¼1vad½

fo"ke vad vkus gsrq vuqdwy ifj.kke 퐴 = {1, 3, 5, }

vuqdwy ifj.kkeksa dh la[;k 푛(퐴) = 3 ¼1vad½

fo"ke vad vkus dh izkf;drk P(A) ¾ ( )( ) ¼1vad½

¾ = 1

2 ¼½vad½

mRrj & fo"ke vad vkus dh izkf;drk gksxhA ¼½vad½



fn;k gS & A = ( )

rFkk B = ( )

rc A + B = ( )

+ ( )

¼1vad½

= ( )( )+

( ) ¼1vad½

= ( )( ) ( )( )( ) ( )

¼1vad½

= )( ) ( )

¼1vad½

= )( ) ( )

¼1vad½

uksV & mijksDrkuqlkj fy[ks tkus ij 1$1$1$1$1 ¾ 5 vad izkIr gksaxsA

25


= xy (x-y)+yz (y-z) + zx (z-x)

dks"Vdksa dks ljy djus ij

= x2y - xy2 +y2z - yz2 + z2x- zx2 ¼1vad½

x dh ?kkrksa dks ?kVrs Øe esa fy[kus ij

= x2y - zx2+ z2x- xy2 +y2z - yz2 ¼1vad½

nks&nks ds lewg cukus ij

= x2(y - z)+ x(z2-y2) +yz (y-z)

= x2(y-z) - x (y2-z2)+ yz(y-z) ¼1vad½

= x2 (y-z)- x (y+z)(y-z)+yz(y-z)

(y-z) mHk;fu"B ysus ij

= (y-z){x2-x(y+z)+yz} ¼1vad½

lkekU; xq.k[k.M

= (y-z){x2-xy- xz+yz}

y dh ?kkrksa dks ?kVrs Øe esa fy[kus ij

= (y-z){yz - xy+ x2- xz}

= (y-z){y(z-x)+ x( x-z)}

= (y-z) {y(z-x)-x(z-x)}

= (y-z)(z-x)(y-x) ¼½vad½

pØh; Øe esa tekus ij

= -(x-y)(y-z)(z-x) ¼½vad½

uksV & mijksDrkuqlkj fy[ks tkus ij 1$1$1$1$½$½ ¾ 5 vad izkIr gksaxsA

26


ekuk fd la[;k x gS ,oa mldk O;qRØe ¾ gSA ¼1vad½

iz'ukuqlkj x + = ¼1vad½

=

7x2 + 7 = 50x

7x2 - 50x+7 = 0 ¼1vad½

7x2-(49+1) x+7 = 0

7x2 - 49x - x+7 = 0

7x (x-7) -1(x-7) = 0

(x-7) (7x-1) = 0 ¼1vad½

vc ;fn x -7 = 0 ;k x = 7

vkSj ;fn 7x - 1 = 0 ;k x = ퟏퟕ ¼½vad½

mRrj & vHkh"V la[;k 7 ;k ퟏퟕ gksxhA ¼½vad½

uksV & mijksDrkuqlkj fy[ks tkus ij 1$1$1$1$½$½ ¾ 5 vad izkIr gksaxsA

mDr esa oxZ lehdj.k dks lw= fof/k ls Hkh gy fd;k tk ldrk gSA


fn;k gS & vkSj oxZ lehdj.k ax2 + bx + c = 0 ds ewy gSaA rks

+ = rFkk . = ¼1vad½

rc 3+3 = ( + )3– 3 ( + ) ¼1vad½

= − 3 ( ) ¼1vad½

27

= + ¼1vad½

= mÙkj ¼1vad½



fn;k gS & ewy/ku P ¾ 8000 :i;s

nj R ¾ 10% izfro"kZ ;k 5% N%ekgh

le; T ¾ 1 o"kZ ;k 3 ¼N%ekgh½ ¼1vad½

feJ/ku A = P (1+ )T ¼1vad½

A = 8000 1 +

A = 8000

A = 8000 ¼1vad½

A = 8000 × × ×

A = 21× 21 × 21

A = 9261 :i;s ¼1vad½

pØof) C;kt ¾ feJ/ku & ewy/ku

¾ 9261 & 8000

pØof) C;kt ¾ 1261 :i;s mRrj ¼1vad½


28


?kM+h dk uxn ewY; ¾ 960 :i;s fd'r ;kstuk esa nh x;h jkf'k ¾ 480 :i;s 'ks"k jkf'k ¾ 480 :i;s dqy fd'rksa esa fn;k x;k ewY; ¾ 2 × 245 ¾ 490 :i;s ¼1vad½

dqy C;kt ¾ 490 & 480 ¾ 10 :i;s izFke ekg dk ewy/ku ¾ 480 :- nwljs ekg dk ewy/ku ¾ 480 & 245 ¾ 235 :i;s ,d ekg gsrq dqy ewy/ku ¾ igys ekg dk ewy/ku $ nwljs ekg dk ewy/ku ¾ 480 $ 235 ¾ 715 :i;s ¼1vad½

nj ¾ ¼1vad½

¾ ¾ ¼1vad½

¾

¾ 16.7%

mÙkj & fd'r ;kstuk esa C;kt dh nj 16.7% ¼1vad½


Lkk/kkj.k O;kt × 100 ewy/ku × le;

10 × 100 715 × 1@12

10 × 100 ×12 715

29


jpuk ds in%&

¼1½ f=Hkqt ABC dh jpuk dh ftlesa AB=6.5, BC=7.5 rFkk AC = 7 lseh- gSA

¼2½ Hkqtk BC rFkk AC dk yEc lef}Hkktd Øe'k PQ o RS [khapkA

¼3½ ;g yEc lef}Hkktd O fcanq ij izfPNsn djrs gSaA

¼4½ OB feykdj O dks dsUnz ekudj rFkk OB f=T;k ysdj vHkh"V ifjoÙk cuk;kA

uksV & ¼1½ lgh cukus ij 1 vad ¼2½ lgh yEc lef}Hkktd [khpus ij 1 vad

¼3½ lgh ifjoÙk cukus ij 1 vad ¼4½ lgh ukekadu djus ij 1 vad

¼5½ jpuk ds in fy[kus ij 1 vad izkIr gksxkA

B C C

A

P

O

S

6-5 ls-eh-

Q

7-5 ls-eh-

R

7 ls-eh-

30


jpuk ds in%&

(1) loZizFke ,d js[kk [k.M AC=5 lseh- [khaphA

(2) fcUnq A ij AC ds uhps dh vksj dks.k CAE= 70º [khapkA (3) AC dk yEc v)Zd PQ [khapkA (4) fcUnq A ij AE yEc [khpk tks yEc v)Zd dks fcUnq O ij izfrPNsn djrk gSA (5) OA f=T;k ls O dks dsUnz ekudj ,d oÙk [khapk tks oÙk dks fcUnq C ls gksdj

tkrk gSA (6) A dks dsUnz ekudj 2 lseh- f=T;k dk pki [khapk tks oÙk ij fcUnq B izfrPNsn

djrk gS rFkk fcUnq A ls 3 lseh dk pki foijhr fn'kk esa [khapk tks oÙk ds fcUnq D ij dkVrk gSA

(7) AB, AD, BC vkSj DC dks feyk;kA (8) blh izdkj ABCD ,d pØh; prqHkqZt cuk ftldk dks.k A= 70º gSA

uksV & lgh oÙk cukus ij 2 vad] lgh pØh; prqHkqZt cukus ij 2 vad] jpuk ds in fy[kus ij 1 vad izkIr gksxkA

y

C A

B P

O

D

Q

E

70º

70º

5CM

3CM

31


fn;k gS + =

LHS. = 1+cosθsinθ + sinθ

1+cosθ ¼1vad½

= ( )( ) ×

( ) ¼1vad½

= ( )

¼1vad½

= ( )

¼1vad½

=

( )

= ( )

( ) =

= RHS. ¼1vad½



fn;k gS sin48° sec42° + cos48° cosec42° = 2

LHS ¾ sin48° sec42° + cos48° cosec42° ¼1vad½

¾ sin48° sec(90° − 48°) + cos48° cosec(90° − 48°) ¼1vad½

¾ sin48° cosec48° + cos48° sec48° ¼1vad½

¾ ° ×

°+ ° ×

° ¼1vad½

¾ 1 $ 1 ¾ 2 ¾ RHS ¼1vad½


32


¼1vad½

fn;k gS & ∆ABC~∆PQR ¼½vad½

fl)djuk %& ¾ ퟐ = = ¼1vad½

jpuk %& AD BC rFkk BC QR ¼½vad½

miifÙk %& ¾

= ½ × × ½ × ×

= × (i) ¼1vad½

vc ABD o PQS esa ABD = PQS ¼fn;k x;k gS½

ADB = PQS ¼jpuk ls ½

rFkk BAD = QPS ¼ds rhuksas dks.kksa dk ;ksxQy ls ½

vFkkZr~ ∆ABC~∆PQR

QyLo:i = ¼le:i dh laxr Hkqtk;sa½ (ii)

rFkk = ¼le:i dh laxr Hkqtk;sa½ (iii) ¼1vad½

leh- (ii) o (iii) dh rqyuk ls

=

vc dk eku leh- (i) esa j[kus ij

{kS- ¼ABC½ {kS- ¼PQR½


½ × vk/kkj × Å¡pkbZ ½ × vk/kkj × Å¡pkbZ


B D C Q S R

A P

33

¾ × ×

¾ ¼½vad½

blh izdkj vU; Hkqtkvksa ds oxksZa dk vuqikr Kkr fd;k tk ldrk gSA

¾ ퟐ = ¼½vad½

uksV & mijksDrkuqlkj fy[ks tkus ij 1$½$1$½$1$1$½$½ ¾ 6 vad izkIr gksaxsA


¼1vad½

Kkr gS & ∆ABC esa dks.k B U;wudks.k gS rFkk ADBC ¼1vad½

fl) djuk & AC2 = AB2 + BC2- 2BC.BD ¼1vad½

miifÙk & ledks.k ADC esa dks.k D ledks.k gSA

d.kZ2 ¾ yEc2 $ vk/kkj2 ¼ikbZFkkxksjl izes; ls½

AC2 = AD2 + DC2 ¼1vad½

fp= ls [DC = BC - BD]

AC2 = AD2 + (BC-BD)2

AC2 = AD2 + BC2 + BD2- 2BC.BD

AC2 = AD2 + BD2 + BC2- 2BC.BD

AC2 = (AD2 + BD2) + BC2- 2BC.BD ¼1vad½

fdUrq ledks.k ABD esa [AB2 = AD2 + BD2]

AC2 = AB2+ BC2 - 2BC.BD ¼1vad½

bfrfl)e~

uksV & mijksDrkuqlkj fy[ks tkus ij 1$1$1$1$1$1 ¾ 6 vad izkIr gksaxsA

{kS- ¼ABC½

{kS- ¼PQR½

B D C

A

U;wudks.k


34


¼1vad½

fn;k gS & ABCD pØh; prqHkqZt rFkk oÙk dk dsUnz O gSA ¼1vad½

fl) djuk %& B + D = 180º ;k

A +C = 180º ¼1vad½ jpuk %& O dks A o C ls feyk;kA

miifÙk %& oÙk ds fdlh pki }kjk dsUnz ij cuk dks.k mlh pki }kjk oÙk dh ifjf/k ds fdlh fcUnq ij cus dks.k dk nqxquk gksrk gSA

fp= ls 2B = x

B = ∠푥 ¼1½

blh izdkj 2D = y D = ∠푦 ¼2½ ¼1vad½

leh- ¼1½ o ¼2½ dks tksM+us ij

B+D = ∠푥 + ∠푦

B+D = (∠푥 + ∠푦)

B+D = (360°) B+D = 180° ¼1vad½

ge tkurs gSa fd prqHkqZt ABCD esa

A+B + C + D = 360° A+C + D + D = 360°

A+C + 180° = 360° A+C = 360°- 180°

A+C = 180° ¼1vad½


C

B

A

D X Y

O

35


fn;k gS & thok AB = 6 cm. rFkk CD = 8 cm. ¼fp= 1vad½

AF = 3 cm. rFkk CE = 4 cm. f=T;k OA = OC = 5 cm. ¼fn;k 1vad½ Kkr djuk gS &

nksuksa thokvksa AB vkSj AB ds chp dh nwjh EF ¼1vad½ gy%&

ledks.k AFO esa ikbZFkkxksjl izes; ls AO2 = OF2 + AF2

52 = OF2 + 32

25 = OF2 + 9 16 = OF2

OF = √16 OF = 4 cm. ¼1vad½

iqu ledks.k CEO esa ikbZFkkxksjl izes; ls OC2 = EC2+ OE2

52 = 42 + OE2

25 = 16 + OE2 9 = OE2 OE2 = √9 OE = 3 cm. ¼1vad½

nksuksa thovksa ds chp dh nwjh EF = OF OE

EF = 4 3 EF = 1 cm. Ans. ¼1vad½


O

C

A F 3 B

D E

4

5

S

36


izkIrkad fo|kfFkZ;ksa dh la[;k

¼ f ½ e/;eku ¼x½ f x

0-10 8 5 40

10-20 30 15 450

20-30 40 25 1000

30-40 12 35 420

40-50 10 45 450

∑풇 = 100 ∑풇풙 = 2360

¼3 vad½

Vsfoy ls izkIr ∑푓푥 ¾ 2360

∑푓 ¾ 100

Kkr djuk gS & lkekUrj ek/; 푥

lkekUrj ek/; 푥 ¾ ∑∑ ¼1vad½

푥 ¾ ¼1vad½

푥 ¾ 23.60

mRrj & lkekUrj ek/; 23-6 gksxk ¼1vad½

uksV & mijksDrkuqlkj fy[ks tkus ij 3$1$1$1 ¾ 6 vad izkIr gksaxsA

37

iz'u Øekad & 17 vFkok gy &

oLrq ek=k bdkbZ qoi

ewY; izfr bdkbZ ¼:- esa½ poi × qoi pli × qoi

1996 esa Poi

1996 esa Pli

A 8 22 25 176 200

B 12 35 40 420 480

C 5 25 30 125 150

D 15 20 25 300 375

E 10 15 20 150 200

;ksx 1171 1405

¼3 vad½

vk/kkj o"kZ esa dqy [kpkZ ¾ ∑ poi × qoi = 1171

orZeku o"kZ esa dqy [kpkZ ¾ ∑ pli × qoi = 1405 ¼1 vad½

fuokZg [kpZ lwpdkad ¾ × 100 ¼1 vad½

¾ × 100

¾ 120 ¼yxHkx½

mRrj & fuokZg [kpZ lwpdkad yxHkx 120 gksxkA ¼1 vad½

uksV & mijksDrkuqlkj fy[ks tkus ij 3$1$1$1 ¾ 6 vad izkIr gksaxsA

orZeku o"kZ esa dqy [kpkZ

vk/kkj o"kZ esa dqy [kpkZ

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i'u i= dk Cy fiVmpbse.nic.in/10th/MATHEMATICS/Question Answer 2.pdf · i'u i= dk Cy fiV d{kk & 10 oh fo"k; & xf.kr le; & 3 ?kV iw.kkd & 100 b- Ø- bdkbZ bdkb ij vkofVr vd oLrfu"B