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This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 26001. |
How many cubes of 10 cm edge can be put in a cubical box of 1m edge |
| Answer» Given Edge = 1 m = 100 cmVolume = side3= (100)3= 1000000 cm3Volumes of cubes of 10 cm edge = 103 = 1000 cm3Number of cubes\xa0{tex}= \\frac { 1000000 } { 1000 }{/tex}= 1000 cubes | |
| 26002. |
How to find underoot |
| Answer» | |
| 26003. |
If cosecA-cotA =1/4Then cosecA+contact =? |
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Answer» cosecA-cotA = 1/4 {given}Now we multiply and divide cosecA + cotA on L.H.SSo, we get{cosecA-cotA × cosecA+cotA} / cosecA + cotA = 1/4cosec^2A - cot^2A / cosecA+cotA =1/4We know that cosec^2A-cot^2A=1So, we get1/cosecA +cotA =1/4Therefore,cosecA+cotA =4 cosecA- cotA= 1/4(cosecA-cotA)^2=(1/4)^2(cosec^2)A+(cot^2)A-2cosecAcotA=1/161+(cot^2)A+(cot^2)A-2cosecAcotA=1/162(cot^2)A-2cosecAcotA=1-1/162cotA(cotA-cosecA)=-15/16. -1)Now,cosecA-cotA=1/4cotA-cosecA=-1/4Now putting this in -1)2cotA(-1/4)=-15/16cotA=15/8now cosecA=(1/4)+15/8 cosecA=17/8To find out: cosecA+cotA =(17/8)+(15/8) =32/8 =4 Ans Sorry there is cotA not contact |
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| 26004. |
8+9 |
| Answer» 17 | |
| 26005. |
the 4 th term of AP is 0 . prove that the 25th term of the A.P is three times its 11th term |
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Answer» Did uh get dat? Given, a+3d=025th term,a+24d11th term,a+10d24d can b written as 3d+21dsame as(24d)10d can b written as 3d+7d (10d)So,A+24dA+3d+21d,A+10dA+3d+7d,Atq..A+3d=0Soo A+3d+21d= 0+21=21A+3d+7d=0+7=721= 3(7)Proved .. I love you |
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| 26006. |
details of cards |
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Answer» Page no. 301 total crds is 52 It is in Ncert. |
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| 26007. |
Whose board preparation has been completed? |
| Answer» Aap Apni bato humne to abhi start bhi nhi Ki | |
| 26008. |
If sinα = 1/2, prove that 3cosα - 4cos3α = 0 |
| Answer» Sinα=p/h so p=1and h=2 by applying Pythagoras theoram we get b here so you have to put the value of cosα that is b/h. | |
| 26009. |
which sample paper is best for preparing board exam |
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Answer» Oswaal... Oswal U like |
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| 26010. |
Very hard chapter in mathematics anybody know..? |
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Answer» No any Surface area and volume ?? Sahi bole Triangles |
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| 26011. |
General term of ap |
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Answer» One More is sum A+D A+(N-1)D It is also known as nth term of an AP is written as an |
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| 26012. |
Solve for x : 1/x+1 + 2/x+2 = 4/x+4 , x not equal 1,-2-4. |
| Answer» {tex}\\frac { 1 } { x + 1 } + \\frac { 2 } { x + 2 } = \\frac { 4 } { x + 4 }{/tex}or,\xa0{tex}\\frac { x + 2 + 2 ( x + 1 ) } { ( x + 1 ) ( x + 2 ) } = \\frac { 4 } { x + 4 }{/tex}or,{tex}\\frac { 3 x + 4 } { x ^ { 2 } + 3 x + 2 } = \\frac { 4 } { x + 4 }{/tex}or,{tex}( 3 x + 4 ) ( x + 4 ) = 4 \\left( x ^ { 2 } + 3 x + 2 \\right){/tex}or,{tex}3 x ^ { 2 } + 16 x + 16 = 4 x ^ { 2 } + 12 x + 8{/tex}or,{tex}x ^ { 2 } - 4 x - 8 = 0{/tex}or,\xa0{tex}x = \\frac { - b \\pm \\sqrt { b ^ { 2 } - 4 a c } } { 2 a }{/tex}or,\xa0{tex}x = \\frac { - ( - 4 ) \\pm \\sqrt { ( - 4 ) ^ { 2 } - 4 ( 1 ) ( - 8 ) } } { 2 \\times 1 }{/tex}or,\xa0{tex}x = \\frac { 4 \\pm \\sqrt { 16 + 32 } } { 2 }{/tex}or,{tex}x = \\frac { 4 \\pm \\sqrt { 48 } } { 2 } = \\frac { 4 \\pm 4 \\sqrt { 3 } } { 2 }{/tex}or,\xa0{tex}x = 2 \\pm 2 \\sqrt { 3 }{/tex}Hence,\xa0{tex}x = 2 + 2 \\sqrt { 3 } \\text { or } 2 - 2 \\sqrt { 3 }{/tex} | |
| 26013. |
How we make 10 root 3 by 30/ root 3 |
| Answer» | |
| 26014. |
Find the value of k for which x =1 is a root of x^2 +kx+3=0 |
| Answer» (1)^2+k1+3=01+k+3=04+k=0K= -4 | |
| 26015. |
Prove that underoot P +underoot Q is irrational number |
| Answer» Suppose that {tex} \\sqrt { p } + \\sqrt { q }{/tex} is a rational number equal to {tex} \\frac { a } { b }{/tex}, where a and b are integers having no common factor.Now,\xa0{tex} \\sqrt { p } + \\sqrt { q } = \\frac { a } { b }{/tex}{tex} \\Rightarrow \\sqrt { p } = \\frac { a } { b } - \\sqrt { q }{/tex} (squaring both side){tex} \\Rightarrow \\quad ( \\sqrt { p } ) ^ { 2 } = \\left( \\frac { a } { b } - \\sqrt { q } \\right) ^ { 2 }{/tex}{tex} \\Rightarrow \\quad p = \\frac { a ^ { 2 } } { b ^ { 2 } } - 2 \\left( \\frac { a } { b } \\right) \\sqrt { q } + q{/tex}{tex} \\Rightarrow \\quad 2 \\left( \\frac { a } { b } \\right) \\sqrt { q } = \\frac { a ^ { 2 } } { b ^ { 2 } } + q - p{/tex}{tex} \\Rightarrow \\quad 2 \\frac { a } { b } \\sqrt { q } = \\frac { a ^ { 2 } + b ^ { 2 } ( q - p ) } { b ^ { 2 } }{/tex}{tex} \\Rightarrow \\quad \\sqrt { q } = \\frac { a ^ { 2 } + b ^ { 2 } ( q - p ) } { 2 a b }{/tex}{tex} \\Rightarrow \\sqrt { q }{/tex}\xa0is a rational number. (because sum of two rational numbers is always rational)This is a contradiction as\xa0{tex} \\sqrt { q }{/tex}\xa0is an irrational number.Hence,\xa0{tex} \\sqrt { p } + \\sqrt { q }{/tex}\xa0is an irrational number. | |
| 26016. |
The value of tan A is always less than 1. True or false |
| Answer» Hiii!!!! | |
| 26017. |
The lengths of the diagonals of a rhombus are 30cm and 40cm.Find the side of rhombus |
| Answer» 25cm | |
| 26018. |
Alpha |
| Answer» Particle | |
| 26019. |
X×3/y-3 = 18/11...........(1). X+8/y×2=2/5...............(2)Find x/y |
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Answer» X=12and y=25 Sidhe sidhe bol na nai ban raha hai Abe unpadh hai kya wrong question bol raha hai Wrong question... stupid What is this????? **** |
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| 26020. |
What is Orrd number |
| Answer» Heyyy ,,,atul how r u.?? | |
| 26021. |
How to solve quadratic equations involving varisble a b and x. |
| Answer» | |
| 26022. |
in the figure given alongside, I find the measure of angle ACD. |
| Answer» | |
| 26023. |
Find the positive root of the equations √3x^2 +6=9 |
| Answer» Root under 3 | |
| 26024. |
If points (p,q),(m,n) & (p-m,q-n) are collinear than proof that , pn = qm. |
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Answer» Okay!!!!!!!!! So easy apply the area formula and bcoz points are collinear....area = 0 |
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| 26025. |
If sec(they)= x+1/4x prove sec(the)+tan(there)= 2x or 1/2x |
| Answer» By the given condition of question\xa0{tex}\\sec \\theta = x + \\frac { 1 } { 4 x }{/tex}{tex}\\therefore \\quad \\tan ^ { 2 } \\theta = \\sec ^ { 2 } \\theta - 1{/tex}{tex}\\Rightarrow \\quad \\tan ^ { 2 } \\theta = \\left( x + \\frac { 1 } { 4 x } \\right) ^ { 2 } - 1 = x ^ { 2 } + \\frac { 1 } { 16 x ^ { 2 } } + \\frac { 1 } { 2 } - 1 = x ^ { 2 } + \\frac { 1 } { 16 x ^ { 2 } } - \\frac { 1 } { 2 } = \\left( x - \\frac { 1 } { 4 x } \\right) ^ { 2 }{/tex}{tex}\\Rightarrow \\quad \\tan \\theta = \\pm \\left( x - \\frac { 1 } { 4 x } \\right){/tex}{tex}\\Rightarrow \\quad \\tan \\theta = \\left( x - \\frac { 1 } { 4 x } \\right) \\text { or, } \\tan \\theta = - \\left( x - \\frac { 1 } { 4 x } \\right){/tex}CASE 1: When\xa0{tex}\\tan \\theta = - \\left( x - \\frac { 1 } { 4 x } \\right) :{/tex}\xa0In this case,{tex}\\sec \\theta + \\tan \\theta = x + \\frac { 1 } { 4 x } + x - \\frac { 1 } { 4 x } = 2 x{/tex}CASE 2: When\xa0{tex}\\theta = - \\left( x - \\frac { 1 } { 4 x } \\right) :{/tex}\xa0In this case,{tex}\\sec \\theta + \\tan \\theta = \\left( x + \\frac { 1 } { 4 x } \\right) - \\left( x - \\frac { 1 } { 4 x } \\right) = \\frac { 2 } { 4 x } = \\frac { 1 } { 2 x }{/tex}Hence,\xa0{tex}\\sec \\theta + \\tan \\theta = 2 x \\text { or } , \\frac { 1 } { 2 x }{/tex} | |
| 26026. |
If d is the HCF of 468 and 222, find the value of integers x and y which satisfy d=468x+222y. |
| Answer» Given integers are 468 and 222, where 468 > 222By applying Euclid’s division lemma, we get468 = 222 {tex}\\times{/tex}\xa02 + 24.222 = 24 {tex}\\times{/tex}\xa09 + 6.24 = 6 {tex}\\times{/tex}\xa04 + 0.We observe that remainder is 0. So the last divisor 6 is the H.C.F. of 468 and 222 .6 = 222 - 24 {tex}\\times{/tex}\xa096 = 222 - (468 - 222 {tex}\\times{/tex}\xa02) {tex}\\times{/tex}\xa09 [Substituting 24 = 468 - 222 {tex}\\times{/tex}\xa02]6 = 222 - 468 × 9 + 222 {tex}\\times{/tex}\xa0186 = 222(1 + 18) - 468 × 96 = 222 {tex}\\times{/tex}\xa019 - 468 {tex}\\times{/tex}\xa096 = 468 ×\xa0(-9) + 222 ×\xa0196 = 468x +\xa0222y where x = - 9 and y = 19.Hence, obtained. | |
| 26027. |
Two roots of polynomial is -2,-3 |
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Answer» Pls.Complete the que Yes. What?? So what?? |
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| 26028. |
I less than 1000 is the |
| Answer» greatest 3- digit number | |
| 26029. |
What are your hours |
| Answer» Kaha ke ho tum | |
| 26030. |
1+tan2a |
| Answer» is equal to sec2a | |
| 26031. |
Prove that a quadrilateral drawn outside a circle is square |
| Answer» See in RD Sharma | |
| 26032. |
2+3+36+365554896+9*72687528842 |
| Answer» 654553314515 | |
| 26033. |
Ac |
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Answer» Alternating current What Alternating current |
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| 26034. |
if Sn denotes sum of n terms then prove that S12=3(S8 - S4) |
| Answer» Let a be the first term and the common difference be d.Sn\xa0=\xa0{tex}\\frac n2{/tex}[2a + (n - 1)d]S12 = {tex}\\frac {12}2{/tex}[2a + (12 - 1)d]= 6[2a + 11d]= 12a + 66dS8 =\xa0{tex}\\frac 82{/tex}[2a + (8 - 1)d]= 4[2a + 7d]= 8a + 28dS4 =\xa0{tex}\\frac 42{/tex}[2a + (4- 1)d]= 2[2a + 3d]= 4a + 6d3(S8 - S4) = 3[(8a + 28d) - (4a + 6d)]= 3[8a + 28d - 4a - 6d]= 3[4a + 22d]= 12a + 66d= S12 | |
| 26035. |
SinA-CosA=0 ,find Sin^4A+ Cos^4A |
| Answer» | |
| 26036. |
1-3x-1÷56×87=? |
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Answer» Ku Shame on you |
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| 26037. |
4x²+4bx-(a²-b²)=0 |
| Answer» We have the following equation,4x2 + 4bx - (a2 - b2) = 0Now,4x2 + 4bx - (a2 - b2) = 0{tex}\\Rightarrow{/tex}\xa04x2 - 2(a - b)x + 2(a + b)x - (a2 - b2) = 0{tex}\\Rightarrow{/tex} 2x[2x - (a - b)] + (a + b)[2x - (a - b)] = 0{tex}\\Rightarrow{/tex} [2x - (a - b)] [2x + (a + b)] = 0{tex}\\Rightarrow{/tex} 2x - (a - b) = 0 or 2x + (a + b) = 0{tex}\\Rightarrow{/tex}\xa02x = a - b or 2x = -a - b{tex} \\Rightarrow x = \\frac{{a - b}}{2}{/tex} or {tex}x = \\frac{{ - a - b}}{2}{/tex} | |
| 26038. |
Prove that root3×root5 is irrational |
| Answer» If possible, let\xa0{tex}\\sqrt{3}{/tex}\xa0+\xa0{tex}\\sqrt{5}{/tex}\xa0be a rational number equal to x. Then,x =\xa0{tex}\\sqrt{3}{/tex}\xa0+\xa0{tex}\\sqrt{5}{/tex}{tex}\\Rightarrow{/tex}\xa0x2 = ({tex}\\sqrt{3}{/tex}\xa0+\xa0{tex}\\sqrt{5}{/tex})2\xa0{tex}\\Rightarrow{/tex}\xa0x2 = ({tex}\\sqrt{3}{/tex})2 + ({tex}\\sqrt{5}{/tex})2 + 2\xa0{tex}\\times{/tex}\xa0{tex}\\sqrt{3}{/tex}\xa0{tex}\\times{/tex}\xa0{tex}\\sqrt{5}{/tex}= 3 + 5 + 2{tex}\\sqrt{15}{/tex}= 8 + 2{tex}\\sqrt{15}{/tex}{tex}\\Rightarrow{/tex}\xa0x2 - 8 = 2{tex}\\sqrt{15}{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}\\frac{x^2 -8}{2}{/tex}\xa0=\xa0{tex}\\sqrt{15}{/tex}Now, x is rational{tex}\\Rightarrow{/tex}\xa0x2 is rational{tex}\\Rightarrow{/tex}\xa0{tex}\\frac{x^2 - 8}{2}{/tex}\xa0is rational{tex}\\Rightarrow{/tex}\xa0{tex}\\sqrt{15}{/tex}\xa0is rationalBut,\xa0{tex}\\sqrt{15}{/tex}\xa0is irrational.Thus, we arrive at a contradiction. So, our supposition that\xa0{tex}\\sqrt{3}{/tex}\xa0+\xa0{tex}\\sqrt{5}{/tex}\xa0is rational is wrong.Hence,\xa0{tex}\\sqrt{3}{/tex}\xa0+\xa0{tex}\\sqrt{5}{/tex}\xa0is an irrational number. | |
| 26039. |
1- cos square |
| Answer» Sin square | |
| 26040. |
What is the square root of 3 |
| Answer» 1.732 | |
| 26041. |
Prove that √2 + √3 are irrational number |
| Answer» Let us assume that\xa0{tex}\\sqrt 2 + \\sqrt 3{/tex}\xa0is a rational numberLet {tex}\\sqrt2+\\sqrt3=\\frac{\\mathrm a}{\\mathrm b}{/tex} Where a and b are co-prime positive integersOn squaring both sides, we get{tex}(\\sqrt2+\\sqrt3)^2=\\frac{\\mathrm a^2}{\\mathrm b^2}{/tex}{tex}2+3+2\\sqrt6=\\frac{\\mathrm a^2}{\\mathrm b^2}{/tex}{tex}5 + 2\\sqrt 6 =\\frac{a^2}{b^2}{/tex}\xa0{tex}2\\sqrt6=\\frac{\\mathrm a^2}{\\mathrm b^2}-5{/tex}{tex}2\\sqrt6=\\frac{\\mathrm a^2-5\\mathrm b^2}{\\mathrm b^2}{/tex}{tex}\\sqrt6=\\frac{\\mathrm a^2-5\\mathrm b^2}{2\\mathrm b^2}{/tex}Now\xa0{tex}\\frac{\\mathrm a^2-5\\mathrm b^2}{2\\mathrm b^2}{/tex}\xa0is a rational number.This shows that\xa0{tex}\\sqrt 6{/tex}\xa0is a rational number.But this contradicts the fact that\xa0{tex}\\sqrt 6{/tex}\xa0is an irrational number.This contradiction has raised because we assume that\xa0{tex}\\left( {\\sqrt 2 + \\sqrt 3 } \\right){/tex}\xa0is a rational number.Hence, our assumption is wrong and\xa0{tex}\\left( {\\sqrt 2 + \\sqrt 3 } \\right){/tex}\xa0is an irrational number. | |
| 26042. |
Find the co ordinate of point which is nearest to the point (-2,5) |
| Answer» The point on y-axis that is nearest to the point(-2,5) is (0,5). | |
| 26043. |
Prove that √2 and √3 are irrational number |
| Answer» \tSuppose\xa0{tex}\\sqrt2{/tex}\xa0is a rational number. That is ,\xa0{tex}\\sqrt2{/tex}\xa0=\xa0{tex}\\frac{p}{q}{/tex}\xa0for some p{tex}\\in{/tex}Z and q {tex}\\in{/tex}Z.\xa0We can assume the fraction is in lowest fraction, That is p and q shares no common factors.\tThen {tex}\\sqrt2q=p{/tex}\xa0\tSquaring both side we get,\xa0\t{tex}2q^2=p^2{/tex}\tSo\xa0{tex}p^2{/tex}\xa0is a multiple of 2,\tlet\'s assume\xa0{tex}p=2m{/tex}\xa0\tThen,\xa0{tex}2q^2=\\left(2m\\right)^2{/tex}\xa0\t{tex}2q^2=4m^2{/tex}\tOr {tex}q^2=2m^2{/tex}\tSo {tex}q^2{/tex}\xa0is a multiple of 2,\t{tex}\\therefore{/tex} q is multiple of 2\tThus p and q shares a common factor.This is contradiction.\t{tex}\\Rightarrow {/tex}{tex}\\sqrt { 2 }{/tex}\xa0is an irrational number.\t\t\tLet us assume that\xa0{tex}\\sqrt 3{/tex}\xa0be a rational number.\t{tex}\\sqrt { 3 } = \\frac { a } { b }{/tex}, where a and b are integers and co-primes and b{tex} \\neq{/tex}0\tSquaring both sides, we have\t{tex}\\frac { a ^ { 2 } } { b ^ { 2 } } = 3{/tex}\tor,\xa0{tex}a ^ { 2 } = 3 b ^ { 2 }{/tex}--------(i)\ta2\xa0is divisible by 3.\tHence a is divisible by 3..........(ii)\tLet a = 3c ( where c is any integer)\tsquaring on both sides we get\t(3c)2\xa0= 3b2\t9c2\xa0= 3b2\tb2\xa0= 3c2\tso b2\xa0is divisible by 3\thence, b is divisible by 3..........(iii)\tFrom equation(ii) and (iii), we have\t3 is a factor of a and b which is contradicting the fact that a and b are co-primes.\tThus, our assumption that\xa0{tex}\\sqrt 3{/tex} is rational number is wrong.\tHence,\xa0{tex}\\sqrt 3{/tex}\xa0is an irrational number.\t | |
| 26044. |
What is the probability that a leap year selected at random will contain 53 Thursday |
| Answer» 2\\7 | |
| 26045. |
Prove that √3 is irrational |
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Answer» See the answer in rs aggrawal class 10 pg no. 28 yha prove kse kre?? |
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| 26046. |
Prove that root3 is irrational |
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Answer» Let us assume that |
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| 26047. |
If x=3+root 7 then find the value of x+1/× |
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Answer» 9+root 7÷2 9+root7/2 39-13root3/6 |
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| 26048. |
The sum of two numbers is 17 and the sum of their squars is 157. Find the numbes |
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Answer» May I know the process 11 and 6 |
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| 26049. |
How to find value of k in given quadratic equation |
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Answer» Where is equation According to given question |
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| 26050. |
Show that the product of three consecutive natural numbers is divisible by 6 |
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Answer» Skb* Yes kab is Sahil.. 3,5,7 Ok Bataoge tum sahil ho na Pujara? I m Cheteshwar ? nahi mazaa lene Ya DB Skb are you sahil SKB TIME PASS KERNE AYE HO KYA? Very hard but chota sa h You can take help from rd sharma Very easy but quite long to |
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