100-R10 Maths Past Paper June 2022 PDF

dqy iz’uksa dh la[;k % 23 dqy i`"Bksa dh la[;k % 15 Total No. of Questions: 23 Total No. of Pages: 15 gkbZ Ldwy ijh{kk] twu&2022...

dqy iz’uksa dh la[;k % 23 dqy i`"Bksa dh la[;k % 15 Total No. of Questions: 23 Total No. of Pages: 15 gkbZ Ldwy ijh{kk] twu&2022 100 fo"k; % xf.kr Subject: MATHEMATICS (Hindi & English Versions) le; % 03 ?k.Vs iw.kk±d % 80 Time: 03 Hours Maximum Marks: 80 funsZ’k %& (i) lHkh iz’u vfuok;Z gSaA (ii) iz’u Øekad 01 ls 05 rd oLrqfu"B izdkj ds iz’u gSAa (iii) iz’u Øekad 06 ls 23 esa vkarfjd fodYi fn;s x;s gSaA Instructions :- (i) All questions are compulsory. (ii) Question Nos. 01 to 05 are objective type questions. (iii) Internal options are given in question number 06 to 23. R-10 Page 1 of 15  iz-1 lgh fodYi pqudj fyf[k, & ¼6×1¾6 1¾6½ (i) 96 vkSj 404 dk HCF gksxk & (a) 120 (b) 4 (c) 10 (d) 3 (ii) cgqin x2 – 3 ds 'kwU;d gksaxs & (a) ±√3 (b) ±3 (c) 3 (d) 9 (iii) fdlh iw.kkZd a q ds fy, izR;sd fo"ke iw.kkZd a dk :Ik gksxk & (a) q (b) q+1 (c) 2q (d) 2q + 1 (iv) lehdj.k fudk; a1x + b1y + c1 = 0 rFkk a2x + b2y + c2 = 0 dk ,d vf}rh; gy gksxk] ;fn &   (a) ≠      (b)  = ≠      (c)  = =   buesa ls dksbZ ugha (d) (v) lekarj Js.kh 10] 7] 4 ----------------------- dk 10oka in gS & (a) 14 (b) 17 (c) &14 (d) &17 (vi) fdlh Hkh lekarj Js.kh dk lkoZ varj gksrk gS & (a) /kukRed (b) _.kkRed (c) 'kwU; (d) mijksDr lHkh R-10 Page 2 of 15   Choose the correct option and write it - (i) HCF of 96 and 404 - (a) 120 (b) 4 (c) 10 (d) 3 (ii) The zeroes of the polynomial x2 – 3 will be - (a) ±√3 (b) ±3 (c) 3 (d) 9 (iii) For some integer q every odd integer is of the form - (a) q (b) q+1 (c) 2q (d) 2q + 1 (iv) In system of equation a1x + b1y + c1 = 0, a2x + b2y + c2 = 0 will be an unique solution, if -   (a)  ≠     (b) = ≠       (c)  = =   (d) None of these (v) The 10th term of the series 10, 7, 4 ………… will be - (a) 14 (b) 17 (c) –14 (d) –17 (vi) Common difference of any arithmetic progression will be – (a) Positive (b) Negative (c) Zero (d) All of these R-10 Page 3 of 15  iz-2 fjDr LFkkuksa dh iwfrZ dhft, & ¼7×1¾7 1¾7½ (i) nks cgqinksa dk ;ksx ------------------------------------------------------------------------------------------------- gksrk gSA (ii) lHkh oxZ ----------------------------------------------------------------------------------------------------------------------- gksrs gSAa (iii) Li’kZ fcanq ls tkus okyh f=T;k Li’kZ js[kk ij ------------------------------------------------ gksrh gSA (iv) v/kZxksys dk vk;ru ------------------------------------------------------------------------------------------------- gksrk gSA (v) ?ku dk dqy i`"Bh; {ks=Qy ---------------------------------------------------------------------------------------------------- (vi) laHko ?kVukvksa dh izkf;drk lnSo ----------------------------------------------------------------------- gksrh gSA (vii) 1 – P (E) = ---------------------------------------------------------------------------------------- Fill in the blanks - (i) Sum of two polynomials is a ………………………………………………....... (ii) All square are ……………………………………………………..…………….. (iii) The radius passing through the point of contact is ………………………… to the tangent. (iv) Volume of hemisphere is ………………………………………..……………... (v) Whole surface area of cube is ………………………………….……………… (vi) The probability is sure even is always ………………………..……………… (vii) 1 – P (E) = ………………………………………………………….……………. R-10 Page 4 of 15  iz-3 lgh tksfM+;ka feykb, & ¼6×1¾6 1¾6½ ‘A’ ‘B’ (i) cos60° (a) tanθ (ii) √sec  θ − 1 (b) cosθ  (iii) secθ (c)   (iv) 1 + tan2θ (d)  (v) sin90° (e) sec2θ (vi) √1 − sin θ (f) cotθ (g) 1 Match the correct column - ‘A’ ‘B’ (i) cos60° (a) tanθ (ii) √sec  θ − 1 (b) cosθ  (iii) secθ (c)   (iv) 1 + tan2θ (d)  (v) sin90° (e) sec2θ (vi) √1 − sin θ (f) cotθ (g) 1 R-10 Page 5 of 15  iz-4 izR;sd dk ,d okD;@'kCn esa mÙkj nhft, & ¼7×1¾7 1¾7½ (i) 4 ,oa 5 dk egÙke lekiorZd (HCF) D;k gksxk\ (ii) oxZ lehdj.k (x – 4) (x – 3) = 0 ds ewy D;k gksx a \s (iii) FksYl ize; s dk dFku fy[kksA (iv) fcanq ¼3] 4½ dh ewy fcanq ls nwjh D;k gS\ (v) o`Ùk ds pki dh yackbZ dk lw= fy[kksA (vi) ekf/;dk dk lw= fyf[k,A (vii) 1] 2] 3] 4] 5 dk lekarj ek/; D;k gksxk\ Write the answer in one word/sentence of each - (i) What is the HCF of 4 and 5? (ii) What are the roots of quadratic equation (x – 4) (x – 3) = 0? (iii) Write the statement of Thales theorem. (iv) What is the distance of a point (3, 4) from origin? (v) Write the formula of circumference of a circle. (vi) Write the formula of Median. (vii) What is the Mean of 1, 2, 3, 4, 5? R-10 Page 6 of 15  iz-5 fuEufyf[kr esa lR;@vlR; fyf[k, & ¼6×1¾6 1¾6½ (i) √x + 2 ,d cgqin gSA (ii) lehdj.k x + 2y = 5 esa ;fn x = 1 gS] rks y = 2 gksxkA (iii) x  − 4x + 4 = 0 ds ewy cjkcj gSA (iv) fcanq ¼&8] 6½ f}rh; prqFkkZ’a k esa fLFkr gSA (v) fdlh cká fcanq ls o`Ùk ij [khaph xbZ nksuksa Li’kZ js[kkvksa dh yackbZ leku gksrh gSA (vi) le:i o`Ùkksa ds {ks=Qy lnSo cjkcj gksrs gSaA Write true/false in the following - (i) √x + 2 is a polynomial. (ii) In equation x + 2y = 5 if x = 1, then y = 2. (iii) x  − 4x + 4 = 0 has equal roots. (iv) Point (–8, 6) lies in second quadrant. (v) The tangent drawn from an external point to a circle are equal. (vi) Area of similar circles are always equal. iz-6 vHkkT; xq.ku[kaM fof/k }kjk 12, 15, 21 dk LCM Kkr djksA ¼2 ½ Find the LCM of 12, 15 and 21 using the prime factorisation method. vFkok@ vFkok@OR HCF (306, 657) = 9 fn;k gS] LCM (306, 657) Kkr djksA Find LCM (306, 657), if HCF (306, 657) = 9 given. R-10 Page 7 of 15  iz-7 f}?kkr O;atd x2 + 7x + 10 ds 'kwU;d Kkr djksA ¼2 ½ Find the zeroes of the polynomial x2 + 7x + 10. vFkok@ vFkok@OR fdlh cgqin p(x) ds fy,] y = p(x) dk xzkQ uhps vkÑfr esa fn;k gSA p(x) ds 'kwU;dksa dh la[;k Kkr djksA y x' x o y' The graph of y = p(x) is given below. Find the number of zeroes of p(x). y x' x o y' iz-8 lekarj Js.kh 7, 13, 19…………….205 esa fdrus in gS\ ¼2 ½ Find the number of terms in A.P. 7, 13, 19…………….205. vFkok@ vFkok@OR lekarj Js.kh 3, 1, –1, –3 ……………………. ds fy, izFke in rFkk lkoZ varj fyf[k,A Write the first term and the common difference in the following A.P. 3, 1, –1, –3 ……………………. R-10 Page 8 of 15  iz-9 ikbFkkxksjl ize; s dk dFku fy[kksA ¼2 ½ Write the statement of Pythagoras theorem. vFkok@ vFkok@OR ∆ABC ,d lef}ckgq f=Hkqt gS] ftldk dks.k C ledks.k gSA fl) dhft, AB2 = 2AC2 ∆ABC is an isosceles triangle, right angled at C. Prove that AB2 = 2AC2 iz-10 k dk eku Kkr dhft, ;fn fcanq A (8, 1), B (k, –4) vkSj C (2, –5) laj[ s kh gSaA ¼2 ½ Find the value of k which the points A (8, 1), B (k, –4) and C (2, –5) are collinear. vFkok@ vFkok@OR fcanqvksa (–5, 7) o (–1, 3) ds chp dh nwjh Kkr djksA Find the distance between the points (–5, 7) and (–1, 3). iz-11 x&v{k ij og fcanq Kkr dhft,] tks (2, –5) vkSj (–2, 9) ls lenwjLFk gSA ¼2 ½ Find the point on the x-axis, which is equidistant from (2, –5) and (–2, 9). vFkok@ vFkok@OR fcanq A ds funsZ’kkad Kkr dhft,] tgka AB ,d o`Ùk dk O;kl gS ftldk dsUnz (2, –3) gS rFkk B ds funsZ’kkad (1, 4) gSA Find the coordinates of point A, where AB is the diameter of a circle whose centre is (2, –3) and B is (1, 4). % iz-12 ;fn tanA = & gks] rks sinA dk eku Kkr djksA ¼2 ½ % If tanA = & , then find the value of sinA. R-10 Page 9 of 15   vFkok@ vFkok@OR eku Kkr djks & sin60°. cos30° + sin30°. cos60° Find the value – sin60°. cos30° + sin30°. cos60° iz-13 6 lseh- f=T;k okys ,d o`Ùk ds ,d f=T;[kaM dk {ks=Qy Kkr dhft,] ftldk dks.k 60° gSA ¼2 ½ Find the area of sector of a circle with radius 6 cm, if angle of the sector is 60°. vFkok@ vFkok@OR nks o`Ùkksa dh f=T;k,a Øe’k% 8 lseh- vkSj 6 lseh- gSA ml o`Ùk dh f=T;k Kkr djks ftldk {ks=Qy bu nksuksa o`Ùkksa ds {ks=Qy ds ;ksx ds cjkcj gSA The radii of two circles are 8 cm and 6 cm respectively. Find the radius of the circle having area equal to the sum of the areas of the two circles. iz-14 nks f[kykM+h] laxhrk vkSj js’kek Vsful dk ,d eSp [ksyrh gSA laxhrk }kjk eSp thrus dh izkf;drk 0.62 gSA js’kek ds thrus dh D;k izkf;drk gS\ ¼2 ½ Two players, Sangeeta and Reshma, play a tennis match. It is known that the probability of Sangeeta winning the match is 0.62. What is the probability of Reshma winning the match? vFkok@ vFkok@OR gjizhr nks fHkUu&fHkUu flDdksa dks ,d lkFk mNkyrh gS] bldh D;k izkf;drk gS fd og de ls de ,d fpr izkIr djsxh\ Harpreet tosses two different coins simultaneously, what is the probability that she gets at least one head? R-10 Page 10 of 15  iz-15 ,d cSx esa 3 yky] vkSj 5 dkyh xsna s gSA cSx esa ls ,d xsna ;kn`PN;k fudkyh tkrh gSA bldh izkf;drk D;k gS fd xsna (i) yky gks (ii) yky ugha gks\ ¼2 ½ A bag contains 3 red balls and 5 black balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is (i) red (ii) not red? vFkok@ vFkok@OR 20 cYckas ds ,d lewg esa 4 cYc [kjkc gaSA bl lewg esa ls ,d cYc ;kn`PN;k fudkyk tkrk gSA bldh D;k izkf;drk gS fd cYc [kjkc gksxk\ A lot of 20 bulbs contain 4 defective ones. One bulb is drawn at random from the lot. What is the probability that this bulb is defective? iz-16 xq.ku[kaM fof/k ls f}?kkr lehdj.k x2 – 3x – 10 = 0 ds ewy Kkr djksA ¼3 ½ Find the roots of the quadratic equation x2 – 3x – 10 = 0 by factorisation. vFkok@ vFkok@OR f}?kkr lehdj.k 2x2 – 4x + 3 = 0 dk fofoDrdj Kkr dhft, vkSj ewyksa dh izÑfr Hkh Kkr djksA Find the discriminant of the quadratic equation 2x2 – 4x + 3 = 0 and hence also find the nature of its roots. iz-17 ml A.P. dk 31oka in Kkr djks ftldk 11oka in 38 gS vkSj 16oka in 73 gSA ¼3 ½ Find the 31st term of an A.P. whose 11th term is 38 and the 16th term is 73. vFkok@ vFkok@OR nks vadksa okyh fdruh la[;k,a 3 ls foHkkT; gSa\ How many two digit numbers are divisible by 3? R-10 Page 11 of 15  iz-18 fuEu vkÑfr es]a DE | | BC gSA (i) EC dk eku Kkr djksA (ii) AD dk eku Kkr djks & ¼3 ½ A A 1.5 lseh. 1 lseh. 1.8 lseh. D E D E 3 lseh. 7.2 lseh. 5.4 lseh. B C B C (i) (ii) In figure (i) and (ii), DE | | BC. Find EC in (i) and AD in (ii) - A A 1.5 cm 1 cm 1.8 cm D E D E 3 cm 7.2 cm 5.4 cm B C B C (i) (ii) vFkok@ vFkok@OR +, +- vkÑfr es]a DE | | AC vkSj DF | | AE gS] fl) dhft, fd ,- = -. gS & A D B F E C +, +- In figure, DE | | AC and DF | | AE, prove that = - ,- -. A D B F E C R-10 Page 12 of 15  iz-19 fl) djks fd cká fcanq ls o`Ùk ij [khaph xbZ nks Li’kZ js[kkvksa dh yackbZ cjkcj gksrh gSA ¼3 ½ Prove that the length of two tangents drawn from an external point to a circle are equal. vFkok@ vFkok@OR ,d fcanq Q ls] tks o`Ùk ds dsUnz ls 25 lseh- dh nwjh ij gS o`Ùk ij fcanq Q ls [khaph Li’kZ js[kk dh yackbZ 24 lseh- gS] rks o`Ùk dh f=T;k Kkr dhft,A From a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the centre is 25 cm, then find the radius of the circle. iz-20 izfrLFkkiu fof/k dk iz;ksx dj fuEu jSf[kd lehdj.k ;qXe dks gy dhft, & ¼4 ½ x + y = 14 x–y=4 Solve the following pair of linear equations by substitution method – x + y = 14 x–y=4 vFkok@ vFkok@OR foyksiu fof/k ds iz;ksx ls fuEu jSf[kd lehdj.k ;qXe dks gy dhft, & 3x + 4y = 10 2x – 2y = 2 Solve the following pair of linear equations by elimination method – 3x + 4y = 10 2x – 2y = 2 R-10 Page 13 of 15  iz-21 7.6 lseh- yack ,d js[kk[k.M [khafp, vkSj bls 5:8 vuqikr esa foHkkftr dhft,A jpuk ds pj.k Hkh fy[ksAa ¼4 ½ Draw a line segment of length 7.6 cm and divide in the ratio 5:8. Write the step of construction. vFkok@ vFkok@OR 5 lseh-] 6 lseh- vkSj 7 lseh- Hkqtkvksa okys ,d f=Hkqt dh jpuk dhft, vkSj fQj ,d vU; / f=Hkqt dh jpuk dhft,] ftldh Hkqtk,a fn, x, f=Hkqt dh laxr Hkqtkvksa dh xquh gSA 0 jpuk ds pj.k Hkh fy[ksaA Construct a triangle with sides 5 cm, 6 cm and 7 cm and then other triangle, / whose sides are of the corresponding sides of the first triangle. Write the 0 step of construction. iz-22 Øe’k% 6 lseh-] 8 lseh- vkSj 10 lseh- f=T;kvksa okys /kkrq ds rhu Bksl xksyksa dks fi?kykdj ,d cM+k Bksl xksyk cuk;k tkrk gS] bl xksys dh f=T;k Kkr djksA ¼4 ½ Metallic spheres of radii 6 cm, 8 cm and 10 cm respectively are melted to form a single solid sphere, find the radius of resulting sphere. vFkok@ vFkok@OR f=T;k 4.2 lseh- okys /kkrq ds ,d xksys dks fi?kykdj] f=T;k 6 lseh- okys ,d csyu ds :Ik esa