Board Paper 10th Delhi (2014)

Q1The first three terms of an AP respectively are 3y 1, 3y+ 5 and 5y+ 1. Thenyequals:(A) 3(B) 4(C) 5(D) 2Solution:The first three terms of an AP are3y1,3y+5and5y+1, respectively.

We need to find the value of y.We know that if a, b and c are in AP, then:b a = c b 2b = a + c

2(3y+5)=3y1+5y+16y+10=8y10=8y6y2y=10y=5

Hence, the correct option is C. Q2In Fig. 1, QR is a common tangent to the given circles, touching externally at the point T. The tangent at T meets QR at P. If PT = 3.8 cm, then the length of QR (in cm) is :

(A) 3.8(B) 7.6(C) 5.7(D) 1.9Solution:

It is known that the length of the tangents drawn from an external point to a circle are equal.

QP = PT = 3.8 cm ...(1)

PR = PT = 3.8 cm ...(2)

From equations (1) and (2), we get:

QP = PR = 3.8 cm

Now, QR = QP + PR = 3.8 cm + 3.8 cm = 7.6 cm

Hence, the correct option is B. Q3In Fig. 2, PQ and PR are two tangents to a circle with centre O. If QPR = 46, then QOR equals:

(A) 67(B) 134(C) 44(D) 46Solution:Given: QPR = 46PQ and PR are tangents.Therefore, the radius drawn to these tangents will be perpendicular to the tangents.So, we have OQ PQ and OR RP. OQP = ORP = 90So, in quadrilateral PQOR, we haveOQP +QPR + PRO + ROQ = 360 90+ 46 + 90+ ROQ = 360 ROQ = 360 226= 134

Hence, the correct option is B. Q4A ladder makes an angle of 60 with the ground when placed against a wall. If the foot of the ladder is 2 m away from the wall, then the length of the ladder (in metres) is:(A)43

(B)43

(C)22

(D) 4Solution:

In the figure, MN is the length of the ladder, which is placed against the wall AB and makes an angle of60with the ground.

The foot of the ladder is at N, which is 2 m away from the wall.

BN = 2 m

In right-angled triangle MNB:

cos60=BNMN=2mMN12=2mMNMN=4m

Therefore, the length of the ladder is 4 m.

Hence, the correct option is D.

(Very similar question :http://www.meritnation.com/ask-answer/question/a-ladder-makes-an-angle-of-60degree-with-the-floor-is-lower/triangles/3012427) Q5If two different dice are rolled together, the probability of getting an even number on both dice, is:(A)136(B)12(C)16(D)14Solution:Possible outcomes on rolling the two dice are given below:{(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6),(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6),(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6),(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6),(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6),(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}

Total number of outcomes = 36

Favourable outcomes are given below:{(2, 2), (2, 4), (2, 6),(4, 2), (4, 4), (4, 6), (6, 2), (6, 4), (6, 6)}

Total number of favourable outcomes = 9

Probability of getting an even number =TotalnumberoffavourableoutcomesTotalnumberofoutcomes=936=14

Hence, the correct option is D. Q6A number is selected at random from the numbers 1 to 30. The probability that it is a prime number is:(A)23

(B)16

(C)13

(D)1130Solution:Total number of possible outcomes = 30

Prime numbers between 1 to 30 are 2, 3, 5, 7, 11, 13, 17, 19, 23 and 29.


Probability of selecting a prime number =TotalnumberoffavourableoutcomesTotalnumberofoutcomes=1030=13

Hence, the correct option is C.

(Very similar question :http://www.meritnation.com/ask-answer/question/find-the-probability-that-a-number-selected-at-random-from/probability/3285990) Q7If the points A(x, 2), B(3, 4) and C(7, 5) are collinear, then the value ofxis:(A) 63(B) 63(C) 60(D) 60Solution:It is given that the three points A(x, 2), B(3, 4) and C(7, 5) are collinear.

Area of ABC = 0

12[x1(y2y3)+x2(y3y1)+x3(y1y2)]=0Here,x1=x,y1=2,x2=3,y2=4andx3=7,y3=5

x[4(5)]3(52)+7[2(4)]=0x(4+5)3(52)+7(2+4)=0x3(7)+76=0x+21+42=0

x+63=0x=63

Thus, the value ofxis 63.

Hence, the correct option is A.

(Very similar question :http://www.meritnation.com/cbse/class10/board-papers/math/math/math-2008-set-1/starttest/qkpbOVC$jkYJ2mhMSLxvjw!!) Q8The number of solid spheres, each of diameter 6 cm that can be made by melting a solid metal cylinder of height 45 cm and diameter 4 cm, is:(A) 3(B) 5(C) 4(D) 6Solution:Letrandhbe the radius and the height of the cylinder, respectively.Given:Diameter of the cylinder = 4 cm Radius of the cylinder,r= 2 cmHeight of the cylinder,h= 45 cmVolume of the solid cylinder =r2h=(2)245cm3=180cm3

Suppose the radius of each sphere beRcm.Diameter of the sphere = 6 cm Radius of the sphere,R= 3 cm

Letnbe the number of solids formed by melting the solid metallic cylinder.

n Volume of the solid spheres = Volume of the solid cylindern43R3=180

n43(3)3=180

n=1803427=5

Thus, the number of solid spheres that can be formed is 5.

Hence, the correct option is B.

(Exactly same question :http://www.meritnation.com/ask-answer/question/how-many-solid-spheres-of-diameter-6cm-are-required-to-b/surface-areas-and-volumes/769880) Q9Solve the quadratic equation 2x2+axa2= 0 forx.Solution:We have:2x2+axa2= 0Comparing the given equation with the standard quadratic equation (ax2+bx+c= 0), we get:a=2,b=aandc=a2Usingthequadraticformula,x=bb24ac2a,weget:x=aa242(a2)22=a9a24=a3a4x=a+3a4=a2orx=a3a4=aSo, the solutions of the given quadratic equation arex=a2orx=a. Q10The first and the last terms of an AP are 5 and 45 respectively. If the sum of all its terms is 400, find its common difference.Solution:Letabe the first term anddbe the common difference.Given:a= 5Tn= 45Sn= 400

We know:Tn=a+ (n 1)d 45 = 5 + (n 1)d 40 = (n 1)d ... (i)

and Sn=n2(a+Tn)400=n2(5+45)n2=40050n=28=16

On substitutingn =16 in (i), we get:40 = (16 1)d 40 = (15)dd=4015=83Thus, the common difference is83.

(Exactly same question :http://www.meritnation.com/ask-answer/question/the-first-term-of-an-ap-is-5-the-last-term-is-45-and-the-s/arithmetic-progressions/6294895) Q11Prove that the line segment joining the points of contact of two parallel tangents of a circle, passes through its centre.Solution:Let XBY and PCQ be two parallel tangents to a circle with centre O.Construction:Join OB and OC.Draw OAXY

Now, XBAOXBO +AOB = 180 (sum of adjacent interior angles is 180)

Now,XBO = 90 (A tangent to a circle is perpendicular to the radius through the point of contact) 90+AOB = 180AOB = 180 90= 90

Similarly,AOC = 90AOB +AOC = 90+ 90= 180Hence, BOC is a straight line passing through O.Thus, the line segment joining the points of contact of two parallel tangents of a circle, passes through its centre.

(Exactly same question :http://www.meritnation.com/ask-answer/question/prove-that-the-line-segment-joining-the-points-of-contact/circles/3416886) Q12If from an external point P of a circle with centre O, two tangents PQ and PR are drawn such that QPR = 120, prove that 2PQ = PO.Solution:Let us draw the circle with external point P and two tangents PQ and PR.

We know that the radius is perpendicular to the tangent at the point of contact.

OQP=90

We also know that the tangents drawn to a circle from an external point are equally inclined to the segment, joining the centre to that point.

QPO=60

Now, inQPO:

cos60=PQPO12=PQPO2PQ=PO Q13Rahim tosses two different coins simultaneously. Find the probability of getting at least one tail.Solution:Rahim tosses two coins simultaneously. The sample space of the experiment is {HH, HT, TH, TT}.


Outcomes in favour of getting at least one tail on tossing the two coins ={HT, TH, TT}

Number ofoutcomes in favour of getting at least one tail = 3

Probability of getting at least one tail on tossing the two coins

=NumberoffavourableoutcomesTotalnumberofoutcomes=34114=34 Q14In fig. 3, a square OABC is inscribed in a quadrant OPBQ of a circle. If OA = 20 cm, find the area of the shaded region. (Use = 3.14)Solution:Let us join OB.

In OAB:OB2= OA2+ AB2= (20)2+ (20)2= 2 (20)2

OB = 20 2Radius of the circle,r= 20 2 cmArea of quadrant OPBQ =360r2=903603.14(202)2cm2=143.14800cm2= 628 cm2Area of squareOABC = (Side)2= (20)2cm2= 400 cm2 Area of the shaded region = Area of quadrant OPBQ Area of square OABC = (628 400) cm2 = 228 cm2(Exactly same question :http://www.meritnation.com/ask-answer/question/in-the-given-figure-a-square-oabc-is-inscribed-in-a-quadra/areas-related-to-circles/6294553) Q15Solve the equation4x3=52x+3;x0,32, forx.Solution:Given equation:4x3=52x+3;x0,32

4x3=52x+343xx=52x+3(43x)(2x+3)=5x6x2+8x9x+12=5x6x2+6x12=0x2+x2=0x2+2xx2=0(x+2)(x1)=0x+2=0orx1=0x=2orx=1

Thus, the solutions of the given equation is 2 or 1. Q16If the seventh term of an AP is19and its ninth term is17, find its 63rdterm.Solution:Letabe the first term anddbe the common difference of the given A.P.

Given:a7=19a9=17

a7=a+(71)d=19

a+6d=19...(1)

a9=a+(91)d=17

a+8d=17...(2)

Subtracting equation (1) from (2), we get:

2d=263d=163

Puttingd=163in equation (1), we get:

a+(6163)=19a=163

a63=a+(631)d=163+62(163)=6363=1

Thus, the63rdterm of the given A.P. is 1. Q17Draw a right triangle ABC in which AB = 6 cm, BC = 8 cm and B = 90. Draw BD perpendicular from B on AC and draw a circle passing through the points B, C and D. Construct tangents from A to this circle.Solution:Follow the given steps to construct the figure.Step 1Draw a line BC of 8 cm length.Step 2Draw BX perpendicular to BC.Step 3Mark an arc at the distance of 6 cm on BX. Mark it as A.Step 4Join A and C. Thus, ABC is the required triangle.Step 5With B as the centre, draw an arc on AC.Step 6Draw the bisector of this arc and join it with B. Thus, BD is perpendicular to AC.Step 7Now, draw the perpendicular bisector of BD and CD. Take the point of intersection as O.Step 8With O as the centre and OB as the radius, draw a circle passing through points B, C and D.Step 9Join A and O and bisect it. Let P be the midpoint of AO.Step 10Taking P as the centre and PO as its radius, draw a circle which will intersect the circle at point B and G. Join A and G.Here, AB and AG are the required tangents to the circle from A.

Q18If the point A(0, 2) is equidistant from the points B(3,p) and C(p, 5), findp. Also find the length of AB.Solution:The given points are A(0, 2), B(3,p) and C(p, 5).

It is given that A is equidistant from B and C.

AB = AC

AB2= AC2

(3 0)2+ (p 2)2= (p 0)2+ (5 2)2

9 +p2+ 4 4p=p2+ 9

4 4p= 0

4p= 4

p= 1

Thus, the value ofpis 1.

Length of AB =(30)2+(12)2=32+(1)2=9+1=10units Q19Two ships are there in the sea on either side of a light house in such a way that the ships and the light house are in the same straight line. The angles of depression of two ships as observed from the top of the light house are 60 and 45. If the height of the light house is 200 m, find the distance between the two ships. [Use3=1.73]Solution:Letdbe the distance between the two ships. Suppose the distance of one of the ships from the tower isxmetres, then the distance of the other ship from the tower isdxmetres.

In right-angled ADO, we have:

tan45=ODAD=200x1=200xx=200...(1)

In right-angled BDO, we have:

tan60=ODBD=200dx3=200dxdx=2003Puttingx=200,wehave:d200=2003d=2003+200d=200(3+13)d=2001.58d=316

Thus, the distance between two ships is 316 m. Q20If the points A(2, 1), B(a,b) and C(4, 1) are collinear andab= 1, find the values ofaandb.Solution:The given points are A (2, 1), B (a,b) and C (4, 1).

Since the given points are collinear, the area of the triangle ABC is 0.12[x1(y2y3)+x2(y3y1)+x3(y1y2)]=0Here,x1=2,y1=1,x2=a,y2=bandx3=4,y3=1

12[2(b+1)+a(11)+4(1b)]=0

2b22a+44b=0

2a+6b=2

a+3b=1...(1)

Given:ab=1...(2)

Subtracting equation (1) from (2), we get:

4b=0b=0

Substitutingb= 0 in (2), we get:

a= 1

Thus, the values ofaandbare 1 and 0, respectively. Q21In Fig 4, a circle is inscribed in an equilateral triangle ABC of side 12 cm. Find the radius of inscribed circle and the area of the shaded region. [Use = 3.14 and3=1.73]

Solution:It is given that ABC is an equilateral triangle of side 12 cm.Construction:Join OA, OB and OC.

Draw:OPBCOQACORAB

Let the radius of the circle bercm.Area ofAOB + Area ofBOC + Area ofAOC = Area ofABC12 AB OR +12 BC OP +12 AC OQ =34 (Side)212 12 r+12 12 r+12 12 r=34 (12)2 3 12 12 r=34 12 12r= 23= 2 1.73 = 3.46Therefore, the radius of the inscribed circle is 3.46 cm.

Now, area of the shaded region = Area ofABC Area of the inscribed circle = [34 (12)2(23)2] cm2 = [363 12] cm2 = [36 1.73 12 3.14] cm2 = [62.28 37.68] cm2 = 24.6cm2Therefore, the area of the shaded region is24.6cm2. Q22In Fig.5, PSR, RTQ and PAQ are three semicircles of diameters 10 cm, 3 cm and 7 cm respectively. Find the perimeter of the shaded region. [Use = 3.14]Solution:Radius of semicircle PSR =1210cm=5cmRadius of semicircle RTQ =123cm=1.5cmRadius of semicircle PAQ =127cm=3.5cm

Perimeter of the shaded region = Circumference of semicircle PSR + Circumference of semicircle RTQ + Circumference of semicircle PAQ = [12 2(5) +12 2(1.5) +12 2(3.5)] cm =(5 + 1.5 + 3.5) cm= 3.14 10 cm = 31.4 cm Q23A farmer connects a pipe of internal diameter 20 cm from a canal into a cylindrical tank which is 10 m in diameter and 2 m deep. If the water flows through the pipe at the rate of 4 km per hour, in how much time will the tank be filled completely?Solution:For the given tank:Diameter = 10 mRadius,R = 5 mDepth, H = 2 m

Internal radius of the pipe =r=202cm=10cm=110mRate of flow of water =v=4km/h=4000m/hLettbe the time taken to fill the tank.So, the water flown through the pipe inthours will be equal to the volume of the tank.

r2vt=R2H(110)24000t=(5)22

t=2521004000=114

Hence, the time taken is114hours. Q24A solid metallic right circular cone 20 cm high and whose vertical angle is 60, is cut into two parts at the middle of its height by a plane parallel to its base. If the frustum so obtained be drawn into a wire of diameter112cm, find the length of the wire.Solution:Let ACB be the cone whose vertical angle ACB = 60. LetRandxbe the radii of the lower and upper end of the frustum.Here, height of the cone, OC = 20 cmLet us consider P as the mid-point of OC.After cutting the cone into two parts through P,OP =202=10cmAlso, ACO and OCB =1260o=30oAfter cutting cone CQS from cone CBA, the remaining solid obtained is a frustum.

Now, in triangle CPQ:tan30=x1013=x10x=103cm

In triangle COB:

tan30=RCO13=R20R=203cm

Volumeofthefrustum,V=13R213x2V=13(R2x2)

V=13((203)2(103)2)=13(40031003)=13(3003)=13100=1003

The volumes of the frustum and the wire formed are equal.

(124)2l=1003l=10032424l=19200cm=192m

Hence, the length of the wire is 192 m. Q25The difference of two natural numbers is 5 and the difference of their reciprocals is110. Find the numbers.Solution:Let one natural number bex.

Given:Difference between the natural numbers = 5

Other natural number =x+ 5

Difference of their reciprocals =110(given)

1x1x+5=110x+5xx(x+5)=1105x2+5x=110x2+5x=50x2+5x50=0x2+10x5x50=0x(x+10)5(x+10)=0(x+10)(x5)=0x+10=0orx5=0x=10orx=5x=5(xisanaturalnumber)

One natural number = 5

Other natural number =x+ 5 = 5 + 5 = 10

Thus, the two natural numbers are 5 and 10. Q26Prove that the length of the tangents drawn from an external point to a circle are equal.Solution:Let AP and BP be the two tangents to the circle with centre O.

To prove:AP = BPProof:InAOP andBOP:OA = OB (radii of the same circle)OAP =OBP =90 (since tangent at any point of a circle is perpendicular to the radius through the point of contact)OP = OP (common)AOPBOP (by R.H.S. congruence criterion)

AP = BP (corresponding parts of congruent triangles)

Hence, the length of the tangents drawn from an external point to a circle are equal.

(Exactly same question :http://www.meritnation.com/ask-answer/question/plzzz-i-want-an-proof-abtthe-length-of-tangents-drawn-from/circles/1590650) Q27The angles of elevation and depression of the top and the bottom of a tower from the top of a building, 60 m high, are 30 and 60 respectively. Find the difference between the heights of the building and the tower and the distance between them.Solution:Let AB be the building and CD be the tower.

In right ABD:

ABBD=tan60o60BD=3BD=603BD=203

In right ACE:

CEAE=tan30oCEBD=13(AE=BD)CE=2033=20

Height of the tower = CE + ED = CE + AB = 20 m + 60 m = 80 m

Difference between the heights of the tower and the building = 80 m 60 m = 20 m

Distance between the tower and the building = BD =203m Q28A bag contains cards numbered from 1 to 49. A card is drawn from the bag at random, after mixing the cards thoroughly. Find the probability that the number on the drawn card is:(i) an odd number(ii) a multiple of 5(iii) a perfect square(iv) an even prime numberSolution:Total number of cards = 49

(i)Total number of outcomes = 49

The odd numbers form 1 to 49 are 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47 and 49.


Required probability =TotalnumberoffavourableoutcomesTotalnumberofoutcomes=2549

(ii)


The numbers 5, 10, 15, 20, 25, 30, 35, 40 and 45 are multiples of 5.


Required probability =TotalnumberoffavourableoutcomesTotalnumberofoutcomes=949

(iii)


The numbers 1, 4, 9, 16, 25, 36 and 49 are perfect squares.


Required probability =TotalnumberoffavourableoutcomesTotalnumberofoutcomes=749=17

(iv)


We know that there is only one even prime number, which is 2.


Required probability =TotalnumberoffavourableoutcomesTotalnumberofoutcomes=149 Q29Find the ratio in which the point P(x, 2) divides the line segment joining the points A(12, 5) and B(4, 3). Also find the value ofx.Solution:Let the point P (x, 2) divide the line segment joining the points A (12, 5) and B (4, 3) in the ratiok:1.

Then, the coordinates of P are(4k+12k+1,3k+5k+1).

Now, the coordinates of P are (x, 2).

4k+12k+1=xand3k+5k+1=2

3k+5k+1=23k+5=2k+25k=3k=35Substitutingk=35in4k+12k+1=x, we get:x=435+1235+1

x=12+603+5

x=728

x= 9

Thus, the value ofxis 9.

Also, the point P divides the line segment joining the points A(12, 5) and (4, 3) in the ratio35:1, i.e. 3:5. Q30Find the values ofkfor which the quadratic equation (k+ 4)x2+ (k+ 1)x+ 1 = 0 has equal roots. Also find these roots.Solution:Given quadratic equation:(k+4)x2+(k+1)x+1=0.

Since the given quadratic equation has equal roots, its discriminant should be zero.

D = 0(k+1)24(k+4)1=0k2+2k+14k16=0k22k15=0k25k+3k15=0(k5)(k+3)=0k5=0ork+3=0k=5or3Thus, the values ofkare 5 and 3.

Fork= 5:(k+4)x2+(k+1)x+1=09x2+6x+1=0(3x)2+2(3x)+1=0(3x+1)2=0x=13,13Fork= 3:(k+4)x2+(k+1)x+1=0x22x+1=0(x1)2=0x=1,1

Thus, the equal root of the given quadratic equation is either 1 or13. Q31In an AP of 50 terms, the sum of first 10 terms is 210 and the sum of its last 15 terms is 2565. Find the A.P.Solution:Letaanddbe the first term and the common difference of an A.P., respectively.

nthterm of an A.P.,an=a+(n1)d

Sum ofnterms of an A.P.,Sn=n2[2a+(n1)d]

We have:Sum of the first 10 terms =102[2a+9d]210=5[2a+9d]42=2a+9d...(1)

15thterm from the last = (5015 + 1)th= 36thterm from the beginning

Now,a36=a+35d

Sumofthelast15terms=152(2a36+(151)d)=152[2(a+35d)+14d]=15[a+35d+7d]

2565=15[a+42d]171=a+42d...(2)

From (1) and (2), we get:

d= 4a= 3

So, the A.P. formed is 3, 7, 11, 15 . . . and 199. Q32Prove that a parallelogram circumscribing a circle is a rhombus.Solution:Given:ABCD be a parallelogram circumscribing a circle with centre O.To prove:ABCD is a rhombus.

We know that the tangents drawn to a circle from an exterior point are equal in length. AP = AS, BP = BQ, CR = CQ and DR = DS.AP + BP + CR + DR = AS + BQ + CQ + DS(AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ) AB + CD = AD + BC or 2AB = 2BC (since AB = DC and AD = BC) AB = BC = DC = AD.Therefore, ABCD is a rhombus.Hence, proved.

http://www.meritnation.com/cbse/class10/board-papers/math/math/math-2013-set-1:-solutions/starttest/UzAIT1cah1fy2h7a1@HzSA!! Q33Sushant has a vessel, of the form of an inverted cone, open at the top, of height 11 cm and radius of top as 2.5 cm and is full of water. Metallic spherical balls each of diameter 0.5 cm are put in the vessel due to which25th of the water in the vessel flows out. Find how many balls were put in the vessel. Sushant made the arrangement so that the water that flows out irrigates the flower beds. What value has been shown by Sushant?Solution:

Height (h) of the conical vessel = 11 cmRadius (r1) of the conical vessel = 2.5 cmRadius (r2) of the metallic spherical balls =0.52=0.25cmLetnbe the number of spherical balls that were dropped in the vessel.Volume of the water spilled = Volume of the spherical balls dropped25Volumeofcone=nVolumeofonesphericalball2513r21h=n43r32r21h=n10r32(2.5)211=n10(0.25)368.75=0.15625nn=440

Hence, the number of spherical balls that were dropped in the vessel is 440.Sushant made the arrangement so that the water that flows out, irrigates the flower beds. This shows the judicious usage of water. Q34From a solid cylinder of height 2.8 cm and diameter 4.2 cm, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid.[Take=227]Solution:The following figure shows the required cylinder and the conical cavity.

Given:Height (h) of the conical part = Height (h) of the cylindrical part = 2.8 cmDiameter of the cylindrical part = Diameter of the conical part = 4.2 cm Radius (r) of the cylindrical part = Radius (r) of the conical part = 2.1 cmSlant height (l) of the conical part =r2+h2=(2.1)2+(2.8)2cm=4.41+7.84cm=12.25cm=3.5cm

Total surface area of the remaining solid = Curved surface area of the cylindrical part + Curved surface area of the conical part + Area of the cylindrical base=2rh+rl+r2=(22272.12.8+2272.13.5+2272.12.1)cm2=(36.96+23.1+13.86)cm2=73.92cm2

Thus, the total surface area of the remaining solid is 73.92 cm2.meritnation.comUpgradeCall Us at011-40705050orClick to CallDownload our Mobile App

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Board Paper 10th Delhi (2014)