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InterviewSolution
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.
| 39201. |
Prove that √2 is irrational? |
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Answer» Ho jaega par bra ans hai Sorry in complete Let.assume that root 2 is rationala and b be the positive integers where they have factors other |
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| 39202. |
Additional sub ke marks math me fail me replace honge ya paas hone ke baad |
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| 39203. |
Write the quadratic polynomial having zeroes 1 and -2 |
| Answer» x²-x-2 | |
| 39204. |
Ex 1.4 problem |
| Answer» | |
| 39205. |
If the product of the two number is 18 one integer is -72 find the oher |
| Answer» - 0.25May be..... | |
| 39206. |
Sin + cos thita=1 |
| Answer» Sin square thita +cos square thitha =1 | |
| 39207. |
Prove that 2*2=(2)2 by using identiy |
| Answer» | |
| 39208. |
Cos theta - sin theta \\cos theta +sin theta=1- under root3/1+under root 3 |
| Answer» 60o | |
| 39209. |
Sallabus of class 10 |
| Answer» Check syllabus here :\xa0https://mycbseguide.com/cbse-syllabus.html | |
| 39210. |
Two candles of equal height but different thickness are lighted.the first burns of in 6 hrs |
| Answer» Let height of each candle = {tex}x\\ unit.{/tex}First candle burns off in 6 hours.Second candle burns off in\xa08 hours.Height of 1st candle after burning for\xa01 hr =\xa0{tex}\\frac{x}{6}{/tex}{tex}unit{/tex}and height of 2nd candle after burning for\xa01 hr\xa0=\xa0{tex}\\frac{x}{8}{/tex}{tex}unit{/tex}Let the required time {tex}= y\\ hrs{/tex}.Length of 1st candle burnt after y hrs =\xa0{tex}\\frac{y \\times x}{6}{/tex}{tex}unit{/tex}Height of 1st candle left =\xa0{tex}\\left(x-\\frac{x y}{6}\\right){/tex}Length of 2nd candle burnt after y hrs =\xa0{tex}\\left(\\frac{y \\times x}{8}\\right){/tex}{tex}unit{/tex}Height of 2nd candle left =\xa0{tex}\\left(x-\\frac{x y}{8}\\right){/tex}According to the question,Height of 1st candle =\xa0{tex}\\frac{1}{2} \\times{/tex}Height of 2nd candle{tex}\\Rightarrow \\quad x-\\frac{x y}{6}=\\frac{1}{2}\\left(x-\\frac{x y}{8}\\right){/tex}{tex}\\Rightarrow \\quad x\\left(1-\\frac{y}{6}\\right)=\\frac{1}{2} x\\left(1-\\frac{y}{8}\\right){/tex}{tex}1-\\frac{y}{6}=\\frac{1}{2}\\left(1-\\frac{y}{8}\\right){/tex}{tex}\\Rightarrow \\quad 2-\\frac{y}{3}=1-\\frac{y}{8}{/tex}{tex}2 -1 ={/tex}\xa0{tex}\\frac{y}{3}-\\frac{y}{8}{/tex}1 =\xa0{tex}\\frac{8 y-3 y}{24}{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}24 = 5y{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}y ={/tex}\xa0{tex}\\frac{24}{5}{/tex}{tex}y = 4.8 hours = 4\\ hours\\ 48\\ minutes.{/tex} | |
| 39211. |
XzgejurZj?? |
| Answer» ?????? | |
| 39212. |
Sin 60° cos 30°+ sin 30° cos 60° |
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Answer» Put the values and do the needful 1 |
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| 39213. |
Ramkali would need 1800 for admission fee and books etc, for her daughter to start |
| Answer» Since, the difference between the savings of two consecutive months is Rs.\xa020,\xa0therefore the series is an A.P.Here, the savings of the first month is Rs. 50First term, a = 50, Common difference, d = 20No. of terms = no. of monthsNo. of terms, n = 12{tex}S _ { n } = \\frac { n } { 2 } [ 2 a + ( n - 1 ) d ]{/tex}{tex}= \\frac { 12 } { 2 } [ 2 \\times 50 + ( 12 - 1 ) 20 ]{/tex}= 6[100 + 220]= 6(320)= 1920After a year,\xa0Ramakali\xa0will save Rs. 1920.Yes,\xa0Ramakali\xa0will be able to fulfill her dream of sending her daughter to school. | |
| 39214. |
8-2 ( 10-2 under root 5) whole under root bracket . Solve this to a square +b square |
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| 39215. |
Hcf of 196 and 38220 |
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Answer» 195 Bro LCM is 196. .38220=196×195+0 195 is correct answer 195 It is 196 only |
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| 39216. |
Express 140 as its product of prime factore |
| Answer» 2* 2* 5* 7 = 140 | |
| 39217. |
Prove:CotA - TanA +1÷CotA +TanA - 1=1+CosA÷sinA |
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| 39218. |
the numrator of a fraction |
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| 39219. |
tan a -cot a÷sin a× cos a=tan^2 a- cot^2a |
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| 39220. |
1+sin= |
| Answer» Cosec+1upon cosec | |
| 39221. |
Prove the area of somilar triangle is proportional to the area of square present at its sides |
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Answer» Similar*May be it\'s typing mistake. What is somilar |
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| 39222. |
Wh what is the solution of the question is if d = HCF of 48 and 72 then find the value of D |
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Answer» 24 is the hcf 24 24 |
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| 39223. |
Evaluate Sin 60 cos 30 + sin30 cos 60 |
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Answer» 2 Ans = 1 Sin90=Sin60Cos30+Sin30Cos60Sin90=1 sin(90_60) sin30+sin30sin(90_60)sin30sin30+sin30sin30=0 |
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| 39224. |
What is condition that the pair of linear equations kx+2y=5and 3x+y=1have a unique solution |
| Answer» According to question the given system of equations are3x + y = 1.......(1)and kx + 2y = 5..........(2)Since we know that,The given equations are of the forma1x + b1y + c1= 0 anda2x + b2y + c2 = 0has a unique solution if\xa0{tex}\\frac { a _ { 1 } } { a _ { 2 } } \\neq \\frac { b _ { 1 } } { b _ { 2 } }{/tex}Thus, {tex}\\frac { 3 } { k } \\neq \\frac { 1 } { 2 } \\Rightarrow k \\neq 6{/tex}Thus, k can take any real values except 6 | |
| 39225. |
23564+252627-822772/2626727827177171717 |
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| 39226. |
Angle |
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| 39227. |
What is completing the square |
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| 39228. |
What is nature of root |
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| 39229. |
Prove Sin90=1 |
| Answer» 0° 30 45 60 90sin root0 root1 root2 root3 root4 Dividing all numbers by 2=sin 0÷2 1÷2 root2÷2 root3÷2 2÷2 0 1/2 1/root2 root3/2 1Understood you fool!! | |
| 39230. |
determine the 2nd term and rth term of AP.whow 6th term 12 and 8th term is 22. |
| Answer» Given a6\xa0= 12{tex}\\Rightarrow{/tex}\xa0a + (6 - 1)d = 12{tex}\\Rightarrow{/tex}\xa0a + 5d = 12 ............(i)and, a8\xa0= 22{tex}\\Rightarrow{/tex}\xa0a + (8\xa0- 1)d = 22{tex}\\Rightarrow{/tex}\xa0a + 7d = 22 ............(ii)Subtracting equation (i) from (ii), we get(a + 7d) - (a + 5d ) = 22 - 12{tex}\\Rightarrow{/tex}\xa0a + 7d - a - 5d = 10{tex}\\Rightarrow{/tex}2d - 10{tex}\\Rightarrow \\quad d = \\frac { 10 } { 2 } = 5{/tex}Using value of d in equation (i), we geta + 5\xa0{tex}\\times{/tex}\xa05 = 12{tex}\\Rightarrow{/tex}\xa0a = 12 - 25 = -13nth term(an) = a + (n - 1)a= -13 + (n - 1)(5)= 5n - 18 | |
| 39231. |
if nth term of ap is 2x-1 .find sum of first n term. |
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Answer» Or n=?? means 1 N=1 |
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| 39232. |
Find largest number which divides 224 and 250 and 302 and leaves remainder 3 in each case |
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| 39233. |
If a and b are the zeros of ax2+bx+c then find a3+b3 |
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| 39234. |
Check whether 15^n can end with 0 for any natural number |
| Answer» N | |
| 39235. |
find roots of 45x+ 50 |
| Answer» There will be 1root. X=_10/9 | |
| 39236. |
Find the greatest number that will divide 385, 509 and 636 leaving remainder 4, 5 and 6 respectively |
| Answer» Pleass frnd answer me fast please ☺? | |
| 39237. |
7th class math ex=5.1 complete |
| Answer» this is for tenth class.... | |
| 39238. |
2×3+4÷5/6% |
| Answer» | |
| 39239. |
×=2,×=6 |
| Answer» Which ques is this?? | |
| 39240. |
Express the trigonometric ratios sinA secA and tanA in terms of cotA |
| Answer» For sin A,By using identity {tex}cosec ^ { 2 } A - \\cot ^ { 2 } A = 1 \\Rightarrow \\cos e c ^ { 2 } A = 1 + \\cot ^ { 2 } A{/tex}{tex}\\Rightarrow \\frac { 1 } { \\sin ^ { 2 } A } = 1 + \\cot ^ { 2 } A{/tex}{tex}\\Rightarrow \\sin A = \\frac { 1 } { \\sqrt { 1 + \\cot ^ { 2 } A } }{/tex}For secA,\xa0By using identity {tex}\\sec ^ { 2 } A - \\tan ^ { 2 } A = 1 \\Rightarrow \\sec ^ { 2 } A = 1 + \\tan ^ { 2 } A{/tex}{tex}\\Rightarrow \\sec ^ { 2 } A = 1 + \\frac { 1 } { \\cot ^ { 2 } A } = \\frac { \\cot ^ { 2 } A + 1 } { \\cot ^ { 2 } A } \\Rightarrow \\sec ^ { 2 } A = \\frac { 1 + \\cot ^ { 2 } A } { \\cot ^ { 2 } A }{/tex}{tex}\\Rightarrow \\sec A = \\frac { \\sqrt { 1 + \\cot ^ { 2 } A } } { \\cot A }{/tex}For tanA,{tex}\\tan A = \\frac { 1 } { \\cot A }{/tex} | |
| 39241. |
Irrational number under root 5 + under root 3 prove |
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| 39242. |
Can i have maths of 10th class in hindi |
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| 39243. |
Fx3 |
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| 39244. |
formula of all chapters |
| Answer» See in your book?????? | |
| 39245. |
Please give ans urgent |
| Answer» | |
| 39246. |
Write first form of AP of a=-1,d=1/2 |
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Answer» -1,-1/२,0,1/2..... -1/2 and 0 and 1/2 and 1 |
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| 39247. |
All formula in the book |
| Answer» Which chapter | |
| 39248. |
Which is greater 4/8 and 2/4 |
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Answer» Both are same Equal Same |
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| 39249. |
If i buy any sample paper from you so how i will pay money |
| Answer» Paytm | |
| 39250. |
Pytagoras theorem |
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Answer» A square + b square is equal to C square in which A and B are the perpendicular sides and C is the hypotenuse H2=B\u200b\u200b\u200b\u200b\u200b\u200b2+P2H stands for hypotenuse, B stands for base and P stands for perpendicular of the right angled triangle. H^2 = b^2 + p^2Here,H=hypoteusb=baseAnd..p=perpendicular.. A^2=B^2+C^2Here A is hypotaneous B and C are other 2 sides of right angle triange. |
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