gkbZ Ldwy ijh{kk& 2012&13 HIGHER SECONDARY SCHOOL EXAMINATION

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Model Question Paper

xf.kr Time& 3 ÄaVs

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

MATHEMATICS (Hindi and English Versions)

Maximum Marks–100

lÒh ç'u gy djuk vfuok;Z gSA Á'u&i«k esa fn, x, funsZ'k lko/kkuhiwoZd i<+dj Á'u¨a ds mÙkj fYkf[k,A Á'u i«k esa n¨ [k.M fn, x, gSa& [k.M v ,oa [k.M cA [k.M&v esa Á'u Øekad 1 esa oLrqfu"B Ádkj ds Á'u fn, x, gSaA funsZ'kkuqlkj gYk dhft,A [k.M c esa Á'u Ø- 2 ls 17 rd vkarfjd fodYi fn, x, gSaA tgk¡ vko';d g¨] LoPN js[kkfp«k cukb,A ÁR;sd Á'u ds fYk, vkoafVr vad mlds lEeq[k vafdr gSaA Á'u&i«k esa xzkQ dh vko';drk ugÈ gSA

(5) (6) (7) (8) Instructions (1) All questions are compulsory. (2) Read the instruction of question paper carefully and write their answers. (3) There are two Sections– Section A and Section B in the question paper (4) Question No. 1 is objective type question in section– A do as directed (5) Internal options are given in Question No. 2 to 17 in Section–B (6) Draw neat and clean diagrams wherever required. (7) Marks alloted to each question are mentioned against the question. (8) Graph is not required in the paper

[k.M&v oLrqfu"B Á'u (Objective Type Questions) Á'u&1 ¼v½- fjä LFkku¨a dh iwfrZ dhft,& (1 × 5 = 5) (i) fdlh o`Ùk dh f«kT;k 7 lseh gS r¨ mldk {¨«kQYk ---------------- g¨xkA (ii) /kukÒ ds fod.kZ dh YkackbZ dk lw«k = -------------------- m (iii) loge   = ............... n

(iv) ewYk/ku + C;kt = -----------------------(v) fdlh f«kÒqt dk {¨«kQYk Kkr djus dk gSj¨ dk lw«k = ------------------------------(A) Fill in the Blanks—

(1) 10-I

(i)

The radius of a circle is 7 cm. then area of circle is = ....................

(ii) Length of diagonal of a cuboid ..........................  m (iii) loge   = -------------------n

(iv) Principal + Interest = -----------------------(v) Hero's formula for area of triangle is ........................

Á'u&1 ¼c½ lgh fodYi pqudj viuh mÙkj iqfLrdk esa fYkf[k,& (1 × 5 = 5) (i) jSf[kd lehdj.k a2x + b1y + c1 = 0 v©j a2x + b2y + c2 = 0 ds vf}rh; gYk dk Áfrca/k gS& (a)

a1 b1 ≠ a 2 b2

(b)

a1 b1 c1 = = (d) a 2 b 2 c2 (ii) lehdj.k x + 2y = 5 esa ;fn x = 1 g¨ r¨ (a) 3 (b) (d) 2 (d) (iii) 7, 9, 21 dk prqFkZ vuqikrh g¨xk& (a) 22 (b) (c) 25 (d)

(c)

(iv) ifjes; O;atd

a1 b1 = a 2 b2 a1 b1 c1 = ≠ a 2 b 2 c2 y dk eku gS& –2 1

24 27

x2 − 9 dk ljYkre :i g¨xk& x−3 (b) x – 3 (d) x – 9

(a) x + 3 (c) x + 9 1 dk ;¨T; ÁfrYk¨e g¨xk& (v) x + x 1 1 (a) x – (b) –x – x x 1 1 (c) x + (d) –x + x x Choose the correct option and write in your answer book (i) In linear equation a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0, the condition for unique solution will be— a1 b1 a1 b1 ≠ (a) a (b) a = b b2 2 2 2 (c)

a1 b1 c1 = = a 2 b 2 c2

(d)

(2) 10-I

a1 b1 c1 = ≠ a 2 b 2 c2

(ii) FInd the value of y in equation x + 2y = (a) 3 (b) (d) 2 (d) (iii) The fourth proportional to 7, 9, 21 is — (a) 22 (b) (c) 25 (d)

x2 − 9 (iv) Raional expression in the simplest x −3 (a) x + 3 (b) (c) x + 9 (d) 1 will be— (v) Additive inverse of x + x 1 (a) x – (b) x 1 (c) x + (d) x

5, if value of x = 1 is — –2 1 24 27 form will be — x – 3 x – 9

1 x 1 –x + x

–x –

Á'u&1 ¼l½- lgh fodYi pqudj viuh mÙkj&iqfLrdk esa fYkf[k,& (i) oxZ lehdj.k x2 + 5x + 6 = 0 ds fofofädj dk eku gS& (a) 1 (c) 5

(b) (d)

–1 –6

(b) (d)

60° 45°

(b) (d)

6 3

(b)

1

(d)



(1 × 5 = 5)

(ii) ;fn ehukj dh Å¡pkbZ ,oa mldh Nk;k dh YkackbZ leku g¨ r¨ lw;Z ds mé;u

d¨.k dk eku g¨xk

(a) 30° (d) 90° (iii) 2, 4, 6, 8, 10 dk lekUrj ek/; gS& (a) 4 (c) 5 (iv) ,d flôs d¨ mNkYkus ij gSM vkus dh (a) 0 1 (c) 2 (v) fuEu fp«k esa ∠B dk eku gS&

(3) 10-I

Ákf;drk gS& 1 3

(a) 50° (b) 70° (c) 100° (d) 10° Choose the correct option and write in your answer book— (i) The discriminant of quadratic equation x2 + 5x + 6 = 0 — (a) 1 (b) –1 (c) 5 (d) –6 (ii) If the height of a tower and the length of its shadow is equal, then the value of the angle of elevation of the sun is— (a) 30° (b) 60° (d) 90° (d) 45° (iii) The arithmetic mean of 2, 4, 6, 8, 10 is — (a) 4 (b) 6 (c) 5 (d) 3 (iv) One coin is tossed simultaneously the probability of head is— (a) 0 (b) 1 1 1 (c) (d) − 2 3 (v) The value of ∠B in the following figure is—

(a) 50° (c) 100°

(b) (d)

(4) 10-I

70° 10°

Á'u&1 ¼n½- lgh t¨fM+;k¡ cukb,& [k.M&v

(1 × 5 = 5)

[k.M&c

(i)

sin θ cosec θ

(a)

sec2 θ

(ii) (iii)

1 + tan2 θ sin (90° – θ)

(b) (c)

1 tan θ

(iv)

cot (90° – θ)

(d)

cosec2 θ

(e)

cos θ

(i)

Section-A sin θ cosec θ

(a)

Section-B sec2 θ

(ii)

1 + tan2 θ

(b)

1

(iii)

sin (90° – θ)

(c)

tan θ

(iv) (v)

cot (90° – θ) 1 + cot2 θ

(d) (e)

cosec2 θ cos θ

(v) 1 + cot2 θ Match the correct pairs

Á'u&1 ¼b½- fuEufYkf[kr esa lR;@vlR; fYkf[k,& (1 × 5 = 5) (i) o`Ùk dh lcls cM+h thok f«kT;k dgYkkrh gSA (ii) v/kZo`Ùk esa vUrfjr d¨.k led¨.k g¨rk gSA (iii) o`Ùk ds cká fcUnq ls [kÈph xbZ n¨ Li'kZ js[kk,¡ vleku g¨rh gSaA (iv) le:i f«kÒqt¨a ds {¨«kQYk¨a dk vuqikr mudh laxr Òqtkv¨a ds ox¨± ds vuqikr ds cjkcj g¨rk gSA (v) ;fn n¨ f«kÒqt led¨f.kd g¨a r¨ f«kÒqt le:i g¨axsA Write true or false in the following— (i)

The largest chord of a circle is called radius.

(ii) Angle subtended in a semi-circle is a right angle. (iii) the length of two tangents drawn from an external point to a circle are unequal (iv) The ratio of the areas of similar traingles is equal to the ratio of the squares of their corresponding sides. (v) If two traingles are equiangular, the traingles are similar

[k.M&c (Section-B) vfrYk?kq mÙkjh; Á'u (Very Short Answer Type Questions) Á'u&2- ,d ik¡ls d¨ Qsadus ij le vad dh Ákf;drk Kkr dhft,A

4

Find the probability that an even number turns up in a single throw of a die (5) 10-I

vFkok (or) fuEufYkf[kr vk¡dM+¨a dh ef/;dk Kkr dhft,& 15, 35, 18, 26, 19, 25, 29, 20, 27 Find the median of the following value of variate— 15, 35, 18, 26, 19, 25, 29, 20, 27

Á'u&3- fuEufYkf[kr lehdj.k fudkYk; d¨ foYk¨iu fof/k ls gYk dhft,& 4 7x – 2y = 1 3x + 4y = 15 Solve the following sustem of equation by elimination method— 7x – 2y = 1 3x + 4y = 15

vFkok (or) fuEufYkf[kr lehdj.k fudk; d¨ ÁfrLFkkiu fof/k ls gYk dhft,& 8x + 5y = 9 3x + 2y = 4 Solve the following system of equaiton by substitution method— 8x + 5y = 9 3x + 2y = 4

Á'u&4- n¨ la[;kv¨a dk ;¨x 100 gS rFkk igYkh la[;k nwljh ls 2 vf/kd gSA la[;k,¡ Kkr dhft,A 4 If sum of two numbers is 100 and the first is 2 more than the second find the numbers.

vFkok (or) 2 dqlÊ v©j 3 est¨a dk ewY; 800 #i, gS rFkk 4 dqlÊ v©j 3 est¨a dk ewY; 1]000 #i, gSA 3 dqlÊ v©j 3 est¨a dk ewY; Kkr dhft,A The cost of 2 chairs and 3 tables is Rs. 800 and the cost of 4 chairs and 3 tables is Rs. 1000. Find the cost of 3 charis and 3 tables.

Á'u&5- fuEufYkf[kr esa x : y dk eku Kkr dhft,& 5x − 3y 2 = 10x − 7y 5 Find the value of x : y from the following– 5x − 3y 2 = 10x − 7y 5

vFkok (or) a c = ;fn g¨ r¨ fl) dhft, fd& b d (6) 10-I

a 2 + b2 b2 = c2 + d 2 d 2

4

a 2 + b2 b2 a c = , then prove that— 2 = If b d c + d2 d2

Á'u&6- fuEufYkf[kr lehdj.k d¨ xq.ku[kaM fof/k ls gYk dhft,&

4

x2 + 3x – 18 = 0 Solve the following equation by factorizing method— x2 + 3x – 18 = 0

vFkok (or) ,d la[;k v©j mlds O;qRØe dk ;¨x

50 gSA la[;k Kkr dhft,A 7

The sum of a number and its reciprocal is

50 . Find the number 7

Á'u&7- fdlh fcUnq ls 200 eh- dh nwjh ij fLFkr fdlh LraÒ ds 'kh"kZ dk mé;u d¨.k 45° gSA LraÒ dh Å¡pkbZ Kkr dhft,A 4 The angle of elevation of the top of the tower from a point 200 m away from teh tower is 45°. Find the height of the tower.

vFkok (or) 60 eh- Å¡ps Ádk'k LraÒ ls ,d tgkt dk voueu d¨.k 60° gS] r¨ LraÒ ds ikn ls tgkt dh nwjh Kkr dhft,A From the top of 60 m high light-house, the angle of depression of the ship is 60°. Find the distance between the ship and the foot of the light-house

Á'u&8- ,d o`Ùk dh f«kT;k 3-5 lseh gSA blds fdlh pki }kjk dsæa ij 36° d¨ d¨.k curk gSA f«kT;[kaM dk {¨«kQYk Kkr dhft,A 4 The radius of circle is 3.5 cm., the arc of the circle subtends an angle 36° at centre. Find the area of sector

vFkok (or) ,d /kukÒ dh YkackbZ] p©M+kbZ v©j Å¡pkbZ Øe'k% 12 lseh-] 11 lseh-] rFkk 10 lseh gSA /kukÒ dh i`"Bh; {¨«kQYk Kkr dhft,A The length, breadth and height of a cuboid are 12 cm, 11 cm and 10 cm. respectively. Find the surface area of the cuboid.

Á'u&9- 8 lseh- f«kT;k ds Yk¨gs ds x¨Y¨ d¨ xYkkdj 1 lseh f«kT;k ds fdrus N¨Vs x¨Y¨ cuk, tk ldrs gSa\ 4 An iron sphere of radius 8 cm is melted then recasted into small spheres each of radius 1 cm. Find the number of small spheres

(7) 10-I

vFkok (or) ,d 8 lseh O;kl okY¨ /kkrq ds csYku d¨ fi?kYkkdj.k 12 lseh O;kl okY¨ fdrus x¨Y¨ cuk, tk ldrs gSa] tcfd csYku dh Å¡pkbZ 90 lseh gSA How many spheres of diameter 12 cm each can be made from a metallic cylinder of diameter 8 cm after the same? The height of cylinder is 90 cm.

Yk?kq mÙkjh; Á'u (Short Answer Type Questions) Á'u&10- fuEufYkf[kr loZlfedk d¨ fl) dhft,&

5

1 + cos A sin A 2 + = sin A 1 + cos A sin A Prove the following identity—

1 + cos A sin A 2 + = sin A 1 + cos A sin A

vFkok (or) fuEufYkf[kr loZlfedk d¨ fl) dhft,& (1 – cos θ) (1 + cos θ) (1 + cot2 θ) = 1 Prove the following identity— (1 – cos θ) (1 + cos θ) (1 + cot2 θ) = 1

Á'u&11- ab (a – b) + bc (b – c) + ca (c – a) ds xq.ku[kaM Kkr dhft,A

5

Factorise– ab (a – b) + bc (b – c) + ca (c – a)

vFkok (or) x2 + 1 esa ls d©u lk ifjes; O;atd ?kVk;k tk, fd x −1

x−3 ÁkIr g¨\ x +1

x−3 x2 + 1 Which rational expression should be substracted from to get ? x +1 x −1

Á'u&12- n¨ Øekxr ÁkÑr la[;k,¡ Kkr dhft,] ftuds ox¨± dk ;¨x 313 gSA

5

Find two consecutive netural numbers whose squares have the sum 313.

vFkok (or) oxZ lehdj.k 2x2 + Px + 4 = 0 dk ewYk 1 g¨] r¨ nwljk ewYk Kkr dhft, v©j P dk eku Kkr dhft,A If 1 is a root of the quadratic equation 2x2 + Px + 4 = 0, then find the other root and also find the value of P—

Á'u&13- 8]000 #i;s dk 3 o"kZ dk 5% okf"kZd C;kt dh nj ls pØo`f) C;kt Kkr dhft,A 5 Find the compound interest on Rs. 8,000 at the rate of interest 5% per annum for 3 years.

vFkok (or) (8) 10-I

,d flYkkbZ e'khu 1]600 #- udn ;k 1]200 #- udn Òqxrku nsdj '¨"k 6 ekg ckn 460 #- nsdj feYkrh gS r¨ fdLr ;¨tuk ds vk/kkj ij C;kt dh nj dh x.kuk dhft,A A sewing machine is available for Rs. 1,600 cash or for Rs. 1,200 cash down payment and Rs. 460 to be paid after 6 months. Find the rate of interest charged under the installment plan.

Á'u&14- ,d f«kÒqt dh Òqtk,¡ 4 lseh] 6 lseh] 8 lseh gSaA f«kÒqt ds ifjxr o`Ùk dh jpuk dhft,A 5 Construct a triangle whose sides are 4 cm., 6 cm and 8 cm. Draw the circumciscle of the triangle–

vFkok (or) ,d f«kÒqt ABC dh jpuk dhft, ftlesa BC = 6 cm. , ∠A = 45° v©j f«kÒqt dh Å¡pkbZ AD = 5 lsehA Construct a triangle ABC in which BC = 6 cm. , ∠A = 45° and altitude AD = 5 cm.

Á'u&15- 100 fo|kfFkZ;¨a ds fuEufYkf[kr ÁkIrkad¨a ls ef/;dk Kkr dhft,& ÁkIrkad 0&10 10&20 20&30 30&40 40&50 fo|kfFkZ;¨a dh la[;k 8 30 40 12 10

6

Find the median of following marks obtained by 100 students— Marks 0–10 10–20 20–30 30–40 40–50 No. of Students 8 30 40 12 10

vFkok (or) 1996 d¨ vk/kkj o"kZ ekudj ,d e/;e oxZ ifjokj ds ctV ls fuEufYkf[kr tkudkjh ds vk/kkj ij o"kZ 1999 dk fuokZg [kpZ lwpdkad Kkr dhft,& ewY; Áfr bdkbZ ¼#- esa½ oLrq ek«kk ¼bdkbZ½ 1996 esa 1999 esa A B C D E

8 12 5 15 10

22 35 25 20 15

25 40 30 25 20

Calculate the cost of living index number for the year 1999 on the basis of 1996 of a medium family from the follwing information—

(9) 10-I

Item A B C D E

Price (in Rs.) per unit Year 1996 Year 1999 22 25 35 40 25 30 20 25 15 20

Quantity (unit) 8 12 5 15 10

Á'u&16- n¨ le:i f«kÒqt¨a dk ifjeki Øe'k% 30 lseh v©j 20 lseh gSA ;fn ,d f«kÒqt dh ,d Òqtk dh YkackbZ 12 lseh g¨ r¨ nwljs f«kÒqt dh laxr Òqtk dh YkackbZ Kkr dhft,A 6 The perimeters of two similar triangles are 30 cm and 20 cm respectively. If one side of one triangle is 12 cm. Find the corresponding side of other triangle—

vFkok (or) ,d lh<+h bl rjg j[kh xbZ gS fd mldh fupYkk fljk nhokj ls 5 eh- nwjh ij gS v©j mldh Åijh fljk tehu ls 10 eh- Å¡ph gS f[kM+dh rd tkrk gSA lh<+h dh YkackbZ Kkr dhft,A A ladder is placed in such a way that its foot is at a distance of 5 m. from a wall and its top reaches a window 10 m above the ground. Find the length of the ladder.

Á'u&17- fdlh o`Ùk ds cká fcUnq ls [kÈph xbZ n¨ Li'kZ js[kk,¡ rqY; g¨rh gSaA fl) dhft,A Prove that the lenghts of two tangents are drawn from an external point to a circle are equal.

vFkok (or) 10 lseh- v/kZO;kl ds ,d o`Ùk esa 16 lseh YkackbZ dh thok [kÈph xbZ gSA o`Ùk ds dsUæ ls thok dh nwjh Kkr dhft,A The radius of a circle is 10 cm and the chord of length 16 cm. is drawn in it. Find the distance of the chord from the centre.

vkn'kZ mÙkj

mÙkj&1 ¼v½ (i) 154 cm2 (iii) loge m – logen (v) ∆ =

(Model Answer) (ii) a 2 + b 2 + c2 (iv) feJ/ku

S(S − a) (S − b) (S − c)

(10) 10-I

mÙkj&1 ¼c½ a1 b1 ≠ a 2 b2 (iii) (d) 27 1 (v) –x + x

(i)

(ii) (c) 2 (iv) (a) x + 3

mÙkj&1 ¼l½ (i)

(a) 1

(ii) (d) 45° 1 (iv) (c) 2

(iii) (b) 6 (v) (c) 100°

mÙkj&1 ¼n½

(ii) (a) sec2 θ (iv) (c) tan θ

(i) (b) 1 (iii) (e) cos θ (v) (d) cosec2θ

mÙkj&1 ¼b½ (i) vlR; (iii) vlR; (v) lR; mÙkj&2-

(ii) lR; (iv) lR;

[k.M&c ¼vfr Yk?kqmÙkjh; Á'u½ n (S) = 6 n (A) = 3 P (A) =

n(A) 1 = n(S) 2

¼1 vad½ ¼1 vad½ ¼2 vad½

Á'u&2 dk vFkok dk mÙkj& vk¡dMksa dks c
ekfè;dk

N +1 9 + 1 10 10 = = = 5 ok¡ in ¼osa in dk eku½ = 2 2 2 2 = 25

mÙkj&3foYk¨iu fof/k& 7x – 2y = 1 3x + 4y = 15

lehdj.k ¼1½ d¨ 4 ls rFkk lehdj.k ¼2½ d¨ 2 ls xq.kk djus ij& (11) 10-I

¼1 vad½ ¼2 vad½ ¼1 vad½

.........(1) ........(2)

28x – 8y = 4 6x + 8y = 30 34x = 34 34 x = 34 = 1

x dk eku lehdj.k ¼1½ esa 7x 7 × 1 7

Á'u&3 dk vFkok dk mÙkj& ÁfrLFkkiu fof/k lehdj.k ¼2½ ls

¼2 vad½ ¼1 vad½

j[kus ij&

– 2y – 2y – 2y –2y –2y

= = = = =

1 1 1 1 – 7 –6 −6 y = −2 x = 1, y = 1

8x + 5y = 3x + 2y = 3x + 2y = 3x = 3x =

9 4 4 4 – 2y 4 – 2y 4 − 2y x = 3

[y = 3]

¼1 vad½

---------¼1½ ---------¼2½ ¼1 vad½

lehdj.k ¼1½ esa x dk eku j[kus ij

8x + 5y = 9  4 − 2y  8   + 5y = 9 3 

32 − 16y 5y + 3 1 32 − 16y + 15y 3 32 − y 3 32 – y –y –y y

= 9

¼1 vad½

= 9 = 9 = = = =

9 × 3 27 – 32 –5 5

(12) 10-I

¼1 vad½

y dk eku ¼1½ esa j[kus ij 8x + 5y 8x + 5 (5) 8x + 25 8x 8x

= = = = =

9 9 9 9 – 25 –16

−16 8

x =

x = −2   = –2  Ans y = 5   

mÙkj&4 ekuk fd la[;k,¡ x rFkk y gSA igYkh la[;k + nwljh la[;k = 100 x + y = 100 -------¼1½ Á'ukuqlkj x – y = 2 -------¼2½ x + y = 100 x – y = 2 2x = 102 102 = 51 x = 2

¼1 vad½

¼2 vad½

¼1 vad½

x dk eku lehdj.k ¼1½ esa j[kus ij x + y = 100 51 + y = 100

y = 100 – 51 = 49

Á'u&4 dk vFkok dk mÙkj ekuk fd dqlÊ dk ewY; x rFkk est dk ewY; y gSA 2x + 3y = 800 – 4x + 3y = 1000 –2x = –200 200 = 100 x = 2

x = 51  Ans.  y = 49 ¼1 vad½

-------¼1½ -----------------¼2½ ¼2 vad½ ¼1 vad½

lehdj.k ¼1½ esa x dk eku j[kus ij 2x + 3y 2 × 100 + 3y 3y 3y

= = = =

y =

3 dqlÊ dk ewY; 3 est¨a dk ewY;

3x = 200 × 3 = dqYk ewY; = =

800 800 800 – 200 600 600 = 200 3 3 × 100 = 300 #600 #300 + 600 900 #-

(13) 10-I

¼1 vad½

mÙkj&5 Økl xq.kk djus ij&

5x − 3y 2 = 10x − 7y 5

5 × (5x – 3y) 25x – 15y 25x – 20x 5x

Á'u&5 dk vFkok dk mÙkj&

= = = =

2 × (10x – 7y) 20x – 14y –14y + 15y 1y

x 1 = mÙkj y 5

¼1 vad½ a = bk , c = dk R.H.S.

L.H.S. a 2 + b2

=

c2 + d 2

=

b2 k 2 + b2 d2 k 2 + d2

=

b2 (k 2 + 1 d 2 (k 2 + 1)

= ∴

¼1 vad½ ¼1 vad½

a c = = k b d

ekuk

¼2 vad½

=

b2

¼1 vad½

d2

b2 d2

b2

¼2 vad½

d2

L.H.S. = R.H.S.

mÙkj&6& xq.k[k.M fofèk

mÙkj

x2 + 3x – 18 = x2 + (6 – 3) x + 8 = x2 + 6x – 3x – 18 = x (x + 6) – 3 (x + 6) = x + 6 = x = x = (–6, 3)

0 0 0 (x + 6) (x – 0 rFkk 0 – 6 –6

(14) 10-I

3) x – 3 = 0 x = 0 + 3 x = 3

¼2 vad½ ¼1 vad½ ¼1 vad½

Á'u&6 dk vFkok dk mÙkj 1 gSA x x 1 50 + = 1 x 7

ekuk fd la[;k x rFkk O;qRØe Á'ukuqlkj

¼2 vad½

50 x2 + 1 = 7 x 2 7 × (x + 1) = 50 × x

7x2 7x2 –50x 7x2 – (49 + 1) x 7x2 – 49 x – x

+ + + +

7 7 7 7

= = = =

¼1 vad½

50x 0 0 0

7x (x – 7) – (x – 7) = 0 (7x – 1) (x – 7) 7x – 1 7x 7x

0 0 0 + 1 1 1 x = 7

mÙkj&

= = = =

rFkk x – 7 = 0

¼1 vad½

x = 0 + 7 x = 7

 1  7,  7

mÙkj&7 A

VkWoj h

¼1 vad½ B

45° 2 0 0 eh-

C

ekuk AB V‚oj gSA Å¡pkbZ AB = h ] BC = 200 m ∆ACB es& a ∴

∠CB = 45° θ = 45° = c tan θ =

yca vkèkkj

(15) 10-I

¼1 vad½

AB BC h 1 = 200 1 × 200 = h h = 200 m tan 45° =

(tan 45° = 1)

¼1 vad½

mÙkj mÙkj&7 dk vFkok dk mÙkj&

k izdk' k LraH

D

60°

60 eh-

A

¼1 vad½

¼2 vad½ 60°

B

x

C

ekuk AB Ádk'k LrEÒ gSA AB = 60 m , BC = x θ = 60° tan θ =

yca vkèkkj

AB BC 60 3 = x

¼1 vad½

tan 60° =

(tan 60° =

3)

3 × x = 60 60 x = 3 x =

60 3 × 3 3

=

60 3 3

¼1 vad½

= 20 × 3 = 20 × 1.732 = 34.64 ehVj

mÙkj& Ádk'k LrEÒ ds ikn ls tgkt dh nwjh 34.64 ehVj gSA mÙkj&8 fn;k gS] o`Ùk dh f«kT;k r = 3.5 lseh θ = 36° θ f«kT;k [k.M dk {¨«kQYk = × πr2 360 (16) 10-I

¼1 vad½ ¼1 vad½

36 22 3.5 3.5 × × × 360 7 10 10 11 × 7 77 = = 20 20 2 = 3.85 eh =

mÙkj& mÙkj&8 dk vFkok dk mÙkj /kukÒ dh YkackbZ a = 12 lseh /kukÒ dh p©M+kbZ b = 11 lseh /kukÒ dh Å¡pkbZ c = 10 lseh /kukÒ dh i`"Bh; {¨«kQYk

mÙkj&

S = S= S = S = S =

¼1 vad½

¼1 vad½

¼1 vad½

2 (ab + bc + ca) ¼1 vad½ 2 [(12 × 11) + (11 × 10) + (10 × 12)] 2 [132 + 110 + 120] ¼1 vad½ 2 × 362 724 lseh2 ¼1 vad½

mÙkj&9 ,d cM+s x¨Y¨ ls ftrus N¨Vs x¨Y¨ cusaxs] mudk vk;ru cM+s x¨Y¨ ds vk;ru ds cjkcj g¨xk cM+s x¨Y¨ dk vk;ru = = = =

N¨Vs x¨Y¨ dk vk;ru = = = =

N¨Vs x¨Yk¨a dh la[;k =

4 π r3 3 4 × π × 83 3 4 × π × 512 3 4 π × 512 cm3 3 4 3 πr 3 4 × π × (1)3 3 4 × π × 1 3 4 π 3

cMs xky s s dk vk;ru NkVss xky s s dk vk;ru

(17) 10-I

¼1 vad½

¼1 vad½

¼1 vad½

4π × 512 3 = 4 π 3 = 512 mÙkj & N¨Vs x¨Yk¨a dh la[;k 512 g¨xhA

¼1 vad½

mÙkj& 9 dk vFkok dk mÙkj& csYku dk O;kl = 8 lseh csYku dh f«kT;k r = x¨Y¨ dk O;kl = x¨Y¨ dh f«kT;k R = csYku dh Å¡pkbZ h = csYku dk vk;ru v = v = v =

x¨Y¨ dk vk;ru v = = r = v =

x¨Yk¨a dh la[;k =

8 = 4 lseh 2 12 lseh 12 = 6 lseh 2 90 lseh π r2 h π × 4 × 4 × 90 1440 π cm3 4 π r3 3 4 × π × (6)3 3 4 × π × 6 × 6 × 6 3 288 π

cy s u dk vk;ru xky s s dk vk;ru

1440 π 288 π = 5 x¨Y¨

¼1 vad½ ¼1 vad½

¼1 vad½

=

mÙkj& mÙkj&10 fl) djs& a L.H.S.

Yk?kq mÙkjh; Á'u 1 + cos A sin A 2 + = sin A 1 + cos A sin A 1 + cos A sin A + sin A 1 + cos A (1 + cos A) × (1 + cos A) + sin A × sin A = sin A(1 + cos A) (18) 10-I

¼1 vad½

¼1 vad½

(1 + cos A) 2 + sin 2 A = sin A(1 + cos A) 12 + cos2 A + 2cos A + sin 2 A = sin A(1 + cos A)

=

= = = ∴

L.H.S. =

Á'u&10 dk vFkok dk mÙkj& L.H.S.

¼1 vad½

1 + 2 cos A + cos2 A + sin 2 A sin A (1 + cos A) ∴ cos2 θ + sin2 θ = 1 1 + 2 cos A + 1 sin A (1 + cos A) 2 + 2cos A ¼2 vad½ sin A (1 + cos A) 2 (1 + cos A) 2 = (R.H.S) sin A (1 + cos A) sin A R.H.S. ¼1 vad½

= (1 – cos θ) (1 + cos θ) (1 + cot2 θ) = (12 – cos2 θ) (1 + cot2 θ) ¼1 vad½ = sin2 θ × cosec2 θ ¼1 vad½ =

1 × cosec2 θ cos ec 2 θ

 1  sin θ =  cosec θ  

¼2 vad½

mÙkj&11& xq.ku[k.M&

= 1 × 1 = 1 (R.H.S.)

¼1 vad½

ab (a – b) + bc (b – c) + ca (c – a) = a2b – ab2 + b2c – bc2 + c2a – ca2 = a2b – ca2 – ab2 + c 2a + b2c – bc 2

¼1 vad½

= a2 (b–c) – a (b2 – c2) + bc (b – c) = a2 (b – c) – a (b + c) (b – c) + bc (b – c)

¼1 vad½

= (b – c)

[a2

– a (b + c) + bc]

¼1 vad½ = (b – c) [a (a – b – c (a – b)] ¼1 vad½ = (b – c) [a2 – ab – ac + bc]

= (b – c) (a – b) (a – c)

mÙkj

= (a – b) (b – c) (c – a)

(19) 10-I

¼1 vad½

Á'u&11 dk vFkok dk mÙkj& ekuk fd A la[;k ?kVkbZ tk, & x−3 x2 + 1 – A = x +1 x −1

¼1 vad½

x − 3 x2 + 1 − –A = x +1 x −1 (x − 3) × (x − 1) − (x 3 + 1) (x + 1) –A = (x + 1) (x − 1)

¼1 vad½

(x 2 − 4x + 3) − (x 3 + x + x 2 + 1) –A = (x + 1) (x − 1) x 2 − 4x + 3 − (x 3 + x + x 2 + 1) –A = (x + 1) (x − 1)

¼1 vad½

x 2 − 4x + 3 − x 3 − x − x 2 + 1 –A = (x + 1) (x − 1) −5x + 2 − x 3 –A = (x + 1) (x − 1)

¼1 vad½

5x − 2 + x 3 A = (x + 1) (x − 1) x 3 + 5x − 2 A = (x + 1) (x − 1)

mÙkj&12 ekuk fd Øekxr la[;k,¡ x rFkk x + 1 gSA Á'ukuqlkj x2 + (x + 1)2 = 313  x2 + x2 + 2x + 1 = 313 2x2 + 2x + 1 2x2 + + 2x 2x2 + 2x x2 + x – 156 x2 + (13 – 12) x – 156 x2 + 13x – 12x – 156 x (x + 13) – 12 (x + 13) (x + 13) (x – 12) x + 13 x

= = = = = = = = = =

313 313 – 1 312 2 ls Òkx nsus ij 0 0 0 0 0 0 ;k x – 12 = 0 x = 0 + 12 0 – 13 ;k

(20) 10-I

¼1 vad½

¼1 vad½

¼1 vad½

¼1 vad½

ÁFke Øekxr la[;k nwljh la[;k

x = –13 ¼vlaÒo½ x = 12 x + 1 = 12 + 1 = 13 (12, 13)

x = 12

mÙkj& Á'u&12 dk vFkok dk mÙkj& nwljk ewYk = α 1+ α = α = 1 + 2 = 3 =

mÙkj&13-

–P = P =

mÙkj

−P 2 2 −P 2 −P 2 6 –6

¼1 vad½

¼1 vad½ ¼1 vad½ ¼1 vad½

R   A = P 1 +  100 

n

5   A = 8000 1 +  100 

pØo`f) C;kt

¼1 vad½ 3

3

21 × 21 × 21 20 × 20 × 20 A = 21 × 21 × 21 A = 9261 #= feJ/ku & ewYk/ku = 9261 – 8000 = 1261 #A = 8000 ×

mÙkj& Á'u&13 dk vFkok dk mÙkj& vkaf'kd Òqxrku ds ckn '¨"k jkf'k&

1600 – 1200 = 400 #400 #- dk N% ekg dk C;kt = 60 #400 #- dk 1 o"kZ dk C;kt = 120 #(21) 10-I

¼1 vad½

¼2 vad½

ewYk/ku = 8000 #- ] nj = 5% le; = 3 o"kZ ] C.I. = ?

 21  A = 8000    20 

¼1 vad½

¼1 vad½ ¼1 vad½ ¼1 vad½ ¼1 vad½ ¼1 vad½

1 #- dk 1 o"kZ dk C;kt 100 #- dk 1 o"kZ dk C;kt mÙkj&14

mÙkj

120 #400 120 = × 100 400 = 30% =

¼1 vad½ ¼1 vad½ ¼1 vad½

¼3 vad½

jpuk ds in& (1) ∆ABC dh jpuk dh (2) BC rFkk AC dk v)Zd [kÈpk (3) O d¨ dsUæ ekudj OA f«kT;k dk pki Y¨dj o`Ùk cuk;kA Á'u&14 dk vFkok dk mÙkj&

¼1 vad½ ¼1 vad½

¼3 vad½

jpuk ds in& ¼1½ BC = 6 lseh dh js[kk [kÈph ¼2½ ∠CBE = 45° dk cuk;k

¼2 vad½ (22) 10-I

¼3½ BC dk v)Zd [kÈpk ¼2 vad½ ¼4½ BC ij Ykac [kÈpk t¨ O ij feYkrk gSA ¼5½ O d¨ dsUæ ekudj o`Ùk [kÈpk ¼6½ M ls 5 cm dk pki MN [kÈpk ¼7½ N ls BC ds lekarj js[kk [kÈph t¨ A rFkk A' ij feYkrh gSA ¼8½ A rFkk A' d¨ B o C ls feYkk;kA ¼9½ A' BC , ABC ∆ ÁkIr g¨rs gSaA mÙkj&15 lap;h vko`fÙk (C.F.) ÁkIrkad fo|kfFkZ;¨a dh la[;k (F) 0–10 10–20 20–30 30–40 40–50

8 30 40 12 10

8 38 78 90 100

N = 100 N 100 = 2 2

=

50

50ok¡ in 20&30 ds oxZ varjkYk esa fLFkr gSA L1 = 20

Fm = 40

ekf/;dk

qo A B C D E

8 12 05 15 10

1996 P0 22 35 25 20 15

¼1 vad½

C = 38 i = 10 N  c −   = L1 +  2  × i f  m     50 − 38  = 20 +  × 10  40  12 = 20 + × 10 40 = 20 + 3 = 23

mÙkj& Á'u&15 dk vFkok dk mÙkj& oLrq ek«kk ewY; iwfrZ bdkbZ 1999 Pt 25 40 30 25 20

(23) 10-I

¼2 vad½

¼2 vad½ ¼1 vad½

oLrq dk dqYk ewY; 1996 P0 q0 176 420 125 300 150 ΣP0 q0 1171

1999 Pt q0 200 480 150 375 200 ΣPt q0 1405

¼3 vad½

∴ fuokZg [kpZ lwpdkad Σ Pt q o = Σ P q × 100 0 0

=

¼1 vad½

1405 × 100 1171

= 120 #- YkxÒx

mÙkj&16 le:irk dh 'krZ ls iFzke f=Hkqt dk ifjeki iFzke f=Hkqt dh ,d Hkqtk = f}rh; f=Hkqt dk ifjeki f}rh; f=Hkt q dh lx a r 30 12 = 20 x 30 × x = 20 × 12 20 × 12 x = 30 x = 8 lseh

mÙkj& Á'u&16 dk vFkok dk mÙkj&

¼2 vad½

¼2 vad½ ¼2 vad½ ¼1 vad½ ¼1 lseh½

¼1 vad½

ikbFkkx¨jl Áes; ls&

AB = 10 eh BC = 5 eh∠ ABC = 90°

¼1 vad½

¼d.kZ½2 = ¼Ykac½2 + ¼vk/kkj½2 AC2

AB2

¼2 vad½

BC2

= + 2 = (10) + (5)2 AC2 = 100 + 25 AC2 = 125 AC2

AC =

mÙkj&

125

AC = 5

(24) 10-I

5 lseh-

¼1 vad½ ¼1 vad½

mÙkj&17

¼2 vad½

fn;k gS& O o`Ùk dk dsUæ gS P cká fcUnq gS fcUnq P ls o`Ùk ij n¨ Li'kZ js[kk,¡ PA rFkk PB [kÈph xbZ gSA fl) djuk gS PA = PB jpuk& dsUæ O d¨ fcUnqv¨a A, B rFkk P ls feYkk;k ¼1 vad½ miifÙk& fdlh fcanq ij [kÈph xbZ Li'kZ js[kk] Li'kZ fcUnq ls [kÈph xbZ f«kT;k ij Ykac g¨rh gSA ∠AOP rFkk ∠BOP esa

∆AOP rFkk ∠BOP

∠OAP = ∠OBP

= 90°

OA = OB OP = OP ∠OAP = ∠OBP

= 90°

¼1 vad½

AP = BP

n¨u¨a Li'kZ js[kk,¡ vkil esa cjkcj gSaA

∠AOP = BOP PA = PB

¼1 vad½

Á'u&17 dk vFkok dk mÙkj& A C

L

¼1 vad½

B D



O

O o`Ùk dk dsUæ gS AB = 16 lsehO ls AB ij Ykac MkYkk OA d¨ feYkk;k OL ⊥ AB 16 AL = LB = 2

(25) 10-I

¼1 vad½

= 8 lseh-

¼1 vad½

OA = 10 lseh OL = ? ∆OAL esa&

OA2 = AL2 + OL2

¼1 vad½

(10)2 = (8)2 + (OL)2 100 = 64 + (OL)2 100 – 64 = (OL)2 OL2 = 36 OL =

vr% o`Ùk ds dsUæ ls thok dh nwjh

¼2 vad½

36

= 6 lseh

(26) 10-I

¼1 vad½

Á'u&i«k CYkw fÁUV

le;&3 ?k.Vk Ø-  bdkbZ

1-

n¨ pj jkf'k;¨a dk e©f[kd lehdj.k 2- cgqin ,oa ifjes; O;atd 3- vuqikr ,oa lekuqikr 4- oxZ lehdj.k 5- okf.kfT;d xf.kr 6- le:i f«kÒqt 7- o`Ùk 8- jpuk,¡ 9- f«kd¨.kfefr 10- Å¡pkbZ ,oa nwjh 11- {¨«kfefr 12- lkaf[;dh] Ákf;drk] dafMdk] iqujko`fÙk dqYk Á'u dqYk vad

ijh{kk & gkbZ LdwYk d{kk&10 fo"k;&xf.kr bdkbZ ij fu/kkZfjr vad

iw.kk±d&100

oLrqfu"B Á'u vadokj Á'u¨a dh la[;k

dqYk Á'u

10

1 vad 02

4 vad 02

5 vad 6 vad & & 02

07

02

&

01

&

01

05 10 08 08 10 05 10 05 10 12

01 01 03 02 04 & 05 01 02 02

01 01 & & & & & 01 02 01

& 01 01 & & 01 01 & & &

& & & 01 01 & & & & 01

01 02 01 01 01 01 01 01 02 02

01@25 25

08 32

05 25

03 18

16 100

(27) 10-I

Maths-sample-paper 8.pdf

x + 2y = 5 x = 1 ' 1' y. (a) 3 (b) –2. (d) 2 (d) 1. (iii) 7, 9, 21 =12K 21. '" (a) 22 (b) 24. (c) 25 (d) 27. (iv) L5. 2 x 9. x 3. −. −. *1 M '". (a) x + 3 (b) x – 3. (c) x + 9 (d) x – 9.

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