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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.
| 23201. |
Someone know where is Sm ??? |
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| 23202. |
Oye tough question bahubali 3 kab aa rhi h yaaro? |
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| 23203. |
55+53 |
| Answer» 108 | |
| 23204. |
Hello friends??? |
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| 23205. |
Hey guys...!! |
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| 23206. |
Divide 39 into two pats such that their product is 324 |
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Answer» 27×12=324 and 27+12=39 ..SO THE TWO NUMBERS IS 27 and 12 ..??? 12 & 27 |
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| 23207. |
Divide 39 into parts such that. Their product is 324 |
| Answer» 12 & 27 | |
| 23208. |
Rd sharma solution |
| Answer» Refer google or download app from play store. | |
| 23209. |
If cosα/cosβ=m and cosα/sinβ=n, then show that (m²+n²)cos²β=n². |
| Answer» Please let alpha=a bit=b(Cos^2a/cos*2b+cos*2a/sin*b)cos*2b=cos*2a/sin*b(Cos*2a sin*2b+cos*a cos*b)cos*2b÷÷÷÷÷÷÷÷÷÷÷÷÷÷÷÷÷÷÷÷÷÷÷Cos*2bsin*2bCos*2a(sin*2b+cos*2b)÷÷÷÷÷Sin*2bCos*2a. Sin*2b+cos*2b=1÷÷÷÷÷÷Sin*2b | |
| 23210. |
Prove that 3 root3 divieded by 5 is irrational |
| Answer» Suppose\xa0{tex}\\frac { 3 \\sqrt { 3 } } { 5 }{/tex}\xa0be a\xa0rational number{tex}\\therefore \\frac { 3 \\sqrt { 3 } } { 5 } = \\frac { a } { b }{/tex}, a and b are co-prime, b{tex}\\ne{/tex}\xa00{tex}\\Rightarrow \\sqrt { 3 } = \\frac { 5 a } { 3 b }{/tex}{tex}5a\\ and\\ 3b{/tex} are integers and\xa0{tex}\\sqrt3{/tex}\xa0is irrational.{tex}\\frac { 5 a } { 3 b }{/tex}\xa0is rational.{tex}\\therefore \\sqrt { 3 } \\neq \\frac { 5 a } { 3 b }{/tex}{tex}\\therefore{/tex}\xa0Our supposition is wrong{tex}\\Rightarrow \\frac { 3 \\sqrt { 3 } } { 5 }{/tex}\xa0is an\xa0irrational number. | |
| 23211. |
Excercise 8.1 10 |
| Answer» Refer this app. | |
| 23212. |
Prove that a parallogram circumscribing a circle is a rhombus |
| Answer» Given ABCD is a parallelogram in which all the sides touch a given circleTo prove:- ABCD is a rhombusProof:-{tex}\\because{/tex}\xa0ABCD is a parallelogram{tex}\\therefore{/tex}\xa0AB = DC and AD = BCAgain AP, AQ are tangents to the circle from the point A{tex}\\therefore{/tex}\xa0AP = AQSimilarly, BR = BQCR = CSDP = DS{tex}\\therefore{/tex}(AP + DP) + (BR + CR) = AQ + DS + BQ + CS = (AQ + BQ) + (CS + DS){tex}\\Rightarrow{/tex}\xa0AD + BC = AB + DC{tex}\\Rightarrow{/tex}\xa0BC + BC = AB + AB [{tex}\\because{/tex} AB = DC, AD = BC]{tex}\\Rightarrow{/tex}\xa02BC = 2AB{tex}\\Rightarrow{/tex}\xa0BC = ABHence, parallelogram ABCD is a rhombus | |
| 23213. |
Find coordinates of point which |
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| 23214. |
Show that the product of three consecutive natural numbers is divisible by 6. |
| Answer» Let three consecutive positive integers be, n, n + 1 and n + 2. When a number is divided by 3, the remainder obtained is either 0 or 1 or 2. ∴ n = 3p or 3p + 1 or 3p + 2, where p is some integer.If n = 3p, then n is divisible by 3. If n = 3p + 1, ⇒ n + 2 = 3p + 1 + 2 = 3p + 3 = 3(p + 1) is divisible by 3. If n = 3p + 2, ⇒ n + 1 = 3p + 2 + 1 = 3p + 3 = 3(p + 1) is divisible by 3. So, we can say that one of the numbers among n, n + 1 and n + 2 is always divisible by 3. ⇒ n (n + 1) (n + 2) is divisible by 3. Similarly, when a number is divided 2, the remainder obtained is 0 or 1. ∴ n = 2q or 2q + 1, where q is some integer. If n = 2q ⇒ n and n + 2 = 2q + 2 = 2(q + 1) are divisible by 2. If n = 2q + 1 ⇒ n + 1 = 2q + 1 + 1 = 2q + 2 = 2 (q + 1) is divisible by 2. So, we can say that one of the numbers among n, n + 1 and n + 2 is always divisible by 2. ⇒ n (n + 1) (n + 2) is divisible by 2. Hence n (n + 1) (n + 2) is divisible by 2 and 3.∴ n (n + 1) (n + 2) is divisible by 6. | |
| 23215. |
Prove that "root 2 + root 3" is an irrational number |
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Answer» solve for both differentlylet root 2 in the form (a/b) where a and b are coprimeroot2= a/bsquaring both sides2=a square/b squarenow cross multiplywe get 2*bsquare=a [email\xa0protected]=>2 devides a square=>2 devides atherefore [email\xa0protected]solving @1 and @2 we get2*b square=(2p)square2*b square=4*( p square)this impliseb square =2* p squaretherfore a and b have a common factor 2therefore it contradictshence root3 is an irrational no.solve similarly for root2 or simply writeany no. added to irrational no. is irrational It is in R.D.Sharma I have searched..its not there in RS AGGARWAL Is it there in RS..?? See in RS agarwal |
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| 23216. |
Find coordinates of point on y-axis which is nearest to point (-2,5) |
| Answer» 2 point (-2,5) and (0,y) solve | |
| 23217. |
How to take out zeroes |
| Answer» By splitting d middle term | |
| 23218. |
Find the roots of 6x^2 - root2x - root2=0 by factorization method. |
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| 23219. |
If the sum of the roots of the equation x2 -x= alpha (2x-1) is zero , find alpha |
| Answer» {tex}\\frac{{ - 1}}{2}{/tex} | |
| 23220. |
sahil khi mp ka to nhi kyu ki mera sahil boiddha name phle classmate tha |
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| 23221. |
TAN 30 |
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| 23222. |
sahil apni fb id ka neme ya number btao aur vaise kha se ho |
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| 23223. |
Sahil please na think se bolo n |
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| 23224. |
Hlo any one online |
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| 23225. |
If angle B and angle Q are acute angles such that sinB= SinQ them prove that Angle B = Angle Q |
| Answer» Side lekar Pythagoras lgade answer as Jaye ga ncert example | |
| 23226. |
How can we learn prove type question |
| Answer» Figures of such questions help a lot | |
| 23227. |
Find the value of m if HCF of 65 and 117is expressible in the form 65m - 117 . |
| Answer» 117 = 13 {tex}\\times{/tex}\xa03 {tex}\\times{/tex}\xa0365 = 13 {tex}\\times{/tex}\xa05HCF (117, 65) = 13LCM(117,65) = 13 {tex}\\times{/tex}\xa05 {tex}\\times{/tex}\xa03 {tex}\\times{/tex}\xa03 = 585Here is given that:{tex}HCF =65m-117{/tex}{tex}13=65m-117{/tex}{tex}65m=130{/tex}m =\xa0{tex}\\frac { 130 } { 6 5 } ={/tex}2 | |
| 23228. |
Can anybody give answes of the 2 sets of ftre... |
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| 23229. |
Find x =?x*x+(a+1/a)x+1=0 |
| Answer» X=-a,X=-1/a | |
| 23230. |
Hey RIMSHA kesi ho yrr |
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| 23231. |
Gygftg |
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| 23232. |
Rani L your commet is wrong |
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| 23233. |
Which term of AP 92, 88, 84, 80........ 0 |
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Answer» Yes right n=22 22th term |
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| 23234. |
triangle ABC DE||BC, if AD:DB=3:5, AC=4.8cm, find AE. |
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Answer» AE = 1.8 1.8 |
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| 23235. |
Message I did not know Hindi please talk with me in English |
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| 23236. |
Eat all this thing specially for u ?????????????????????????????????? |
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| 23237. |
Join this dj anyone |
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| 23238. |
x square - 4 x + 1 polynomial |
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| 23239. |
how many terms of the ap : 9,17,25,... must be taken to give a sum of 636 |
| Answer» a=9a2=17 difference =a2-a1 =17-9=8sum=636 put the formula n get ans | |
| 23240. |
2cot=tan5 find the value of 4 tan + tan |
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| 23241. |
6/x+y =7/x-y + 3 and 1/2(x+y)=1/3(x-y) |
| Answer» According to question given system of equations are{tex}\\frac { 6 } { x + y } = \\frac { 7 } { x - y } + 3{/tex}Let {tex}\\frac{1}{{x + y}} = u{/tex}\xa0and {tex}\\frac{1}{{x - y}} = v{/tex}. Then, the given system of equations becomes6u = 7v + 3{tex}\\Rightarrow{/tex}\xa06u - 7v = 3 ...........(i)Given equation,{tex}\\frac{u}{2} = \\frac{v}{3}{/tex}{tex}\\Rightarrow{/tex}\xa03u = 2v{tex}\\Rightarrow{/tex}\xa03u - 2v = 0 .......(ii)Multiplying equation (ii) by 2, we get6u - 4v = 0 ....(iii)Subtracting equation (i) from equation (iii), we get-7v + 4v = 3{tex}\\Rightarrow{/tex}\xa0-3v = 3{tex}\\Rightarrow{/tex}\xa0v = -1Put the value of v = -1 in equation (ii), we get3u - 2 {tex}{/tex}× (-1) = 0{tex}\\Rightarrow{/tex}\xa03u + 2 = 0{tex}\\Rightarrow{/tex}\xa03u = -2{tex}\\Rightarrow u = \\frac{{ - 2}}{3}{/tex}Now,{tex}u = \\frac{{ - 2}}{3}{/tex}{tex}\\Rightarrow \\frac{1}{{x + y}} = \\frac{{ - 2}}{3}{/tex}{tex}\\Rightarrow x + y = \\frac{{ - 3}}{2}{/tex}\xa0.......(iv)and, v = -1{tex}\\Rightarrow \\frac{1}{{x - y}} = -1{/tex}{tex}\\Rightarrow{/tex}\xa0x - y = -1 ........(v)Adding equation (iv) and equation (v), we get{tex}2x = \\frac{{ - 3}}{2} - 1{/tex}{tex}\\Rightarrow 2x = \\frac{{ - 3 - 2}}{2}{/tex}{tex}\\Rightarrow 2x = \\frac{{ - 5}}{2}{/tex}{tex}\\Rightarrow x = \\frac{{ - 5}}{4}{/tex}Put the value of\xa0{tex}x = \\frac{{ - 5}}{4}{/tex}\xa0in equation (v), we get{tex}\\frac{{ - 5}}{4} - y = - 1{/tex}{tex}\\Rightarrow \\frac{{ - 5}}{4} + 1 = y{/tex}{tex}\\Rightarrow \\frac{{ - 5 + 4}}{4} = y{/tex}{tex}\\Rightarrow \\frac{{ - 1}}{4} = y{/tex}{tex}\\Rightarrow y = \\frac{{ - 1}}{4}{/tex}Hence, the solution of the system of equation is {tex}x = \\frac{{ - 5}}{4},y = \\frac{{ - 1}}{4}{/tex}. | |
| 23242. |
Koi h..???? |
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| 23243. |
Number of honourable cards or cards of honour |
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| 23244. |
What is hypothenious |
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| 23245. |
If x=-4 is a root of equation xsquare +2x+4p=0 find the value of p |
| Answer» P=-2 | |
| 23246. |
search of a princess anyone want to chat |
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| 23247. |
Proof root 3 is irrational |
| Answer» If possible let root 3 be rational.Then we can write root 3 = p/ q where p and q are co primes and q is not = 0.Now root 3 = p/ q Psquare = 3 qsquare P sq is a multiple of 3P is a multiple of 3 Let p = 3mP sq = 9m sq Q sq = 3m sq Q is a multiple of 3 Thus 3 is a common multiple of p and q This is a contradiction, since p and q are Co primes Therefore root 3 is a irrational | |
| 23248. |
1040root long division method process |
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| 23249. |
find the discriminant of the quadratic equation 6x^2-7x+2=0 |
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| 23250. |
By geometrical construction ,it is possible to divide a line segment in ratio √3:1/√3 |
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