Saved Bookmarks
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.
| 5251. |
Cf value |
|
Answer» I think It means calorific value |
|
| 5252. |
1+sin ×n |
| Answer» | |
| 5253. |
A number x is chosen from these number -5,-4,-3,-2,-1,0,1,2,3,4,5 what is the probability of [x] |
|
Answer» 9/11 ?? 9/11 |
|
| 5254. |
S.I 500 time 2 year rate 5% p=5000 |
| Answer» | |
| 5255. |
If cosA=√3/2 and tanB=1/√3.find the value of sin(A+B) |
|
Answer» ?? Root 3/2 √3/2 |
|
| 5256. |
How to attempt the question paper so that all questions would be answered |
|
Answer» Don\'t read read paper because it may disturb your mind of don\'t know it so attempt first if u don\'t know about it move to next.. this way can get best of your preparation Don \'t be so tensed Always do paper with cool mind but don\'t take it lightly..and please don\'t leave any question attempt all... try to watch clock at every interval of time and set in your mind that i can do this ques. in 5min.or even more ..step marking will be there therefore attempt all and if u will not able to solve any quest. Or u think that it is wrong then please do not cut it becaude there is step marking for every step ....fast your speed of writing and try to think fast but currectly and then write as u think in this way u can complete whole paper ..or u canDivide the time by section of paper.Hope it will help u.((## I do same)))##All the best. First you have to solve those questions which you are confident that you are able to do. After that you have to divide your time according to the number of questions and how much marks they carry. If your mind is concentrate then you is able to attempt all questions |
|
| 5257. |
I want the prove of trigonometric ratios |
| Answer» | |
| 5258. |
(x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)=0 |
| Answer» Given,(x - a)(x - b) + (x - b)(x - c) + (x - c)(x - a) = 0{tex}\\Rightarrow x ^ { 2 } - a x - b x + a b + x ^ { 2 } - b x - c x + b c + x ^ { 2 } - c x - a x+ a c = 0{/tex}{tex}\\Rightarrow 3 x ^ { 2 } - 2 a x - 2 b x - 2 c x + a b + b c + c a = 0{/tex}For equal roots\xa0{tex}B ^ { 2 } - 4 A C = 0{/tex}or,\xa0{tex}\\{ - 2 ( a + b + c ) \\} ^ { 2 } = 4 \\times 3 ( a b + b c + c a ){/tex}or,\xa0{tex}4 ( a + b + c ) ^ { 2 } - 12 ( a b + b c + c a ) = 0{/tex}or,\xa0{tex}a ^ { 2 } + b ^ { 2 } + c ^ { 2 } + 2 a b + 2 b c + 2 a c - 3 a b - 3 b c - 3 a c = 0{/tex}or,\xa0{tex}\\frac { 1 } { 2 } \\left[ 2 a ^ { 2 } + 2 b ^ { 2 } + 2 c ^ { 2 } - 2 a b - 2 a c - 2 b c \\right] = 0{/tex}or,\xa0{tex}\\frac { 1 } { 2 } \\left[ \\left( a ^ { 2 } + b ^ { 2 } - 2 a b \\right) + \\left( b ^ { 2 } + c ^ { 2 } - 2 b c \\right) + \\left( c ^ { 2 } + a ^ { 2 } - 2 a c \\right) \\right] = 0{/tex}or,\xa0{tex}\\frac { 1 } { 2 } \\left[ \\left( a ^ { 2 } + b ^ { 2 } - 2 a b \\right) + \\left( b ^ { 2 } + c ^ { 2 } - 2 b c \\right) + \\left( c ^ { 2 } + a ^ { 2 } - 2 a c \\right) \\right] = 0{/tex}or,\xa0{tex}( a - b ) ^ { 2 } + ( b - c ) ^ { 2 } + ( c - a ) ^ { 2 } = 0 \\text { if } a \\neq b \\neq c{/tex}Since\xa0{tex}( a - b ) ^ { 2 } > 0 , ( b - c ) ^ { 2 } > 0 ( c - a ) ^ { 2 } > 0{/tex}\xa0Hence,\xa0{tex}( a - b ) ^ { 2 } = 0 \\Rightarrow a = b{/tex}{tex}( a - c ) ^ { 2 } = 0 \\Rightarrow b = c{/tex}{tex}( c - a ) ^ { 2 } = 0 \\Rightarrow c = a{/tex}{tex}\\therefore a = b = c{/tex}\xa0Hence Proved. | |
| 5259. |
045 code qestion paper |
| Answer» | |
| 5260. |
If the ratio of the first n terms of two AP is (7n+1):(4n+27),find the ratio of their mth term |
| Answer» Let a, and A be the first terms and d and D be the common difference of two A.PsThen, according to the question,{tex}\\frac { S _ { n } } { S _ { n } ^ { \\prime } } = \\frac { \\frac { n } { 2 } [ 2 a + ( n - 1 ) d ] } { \\frac { n } { 2 } [ 2 A + ( n - 1 ) D ] } = \\frac { 7 n + 1 } { 4 n + 27 }{/tex}or,\xa0{tex}\\frac { 2 a + ( n - 1 ) d } { 2 A + ( n - 1 ) D } = \\frac { 7 n + 1 } { 4 n + 27 }{/tex}or,{tex}\\frac { a + \\left( \\frac { n - 1 } { 2 } \\right) d } { A + \\left( \\frac { n - 1 } { 2 } \\right) D } = \\frac { 7 n + 1 } { 4 n + 27 }{/tex}Putting,\xa0{tex}\\frac { n - 1 } { 2 } = m - 1{/tex}{tex}n-1 = 2m - 2{/tex}{tex}n= 2m - 2 + 1{/tex}or, {tex}n = 2m - 1{/tex}{tex}\\frac { a + ( m - 1 ) d } { A + ( m - 1 ) D } = \\frac { 7 ( 2 m - 1 ) + 1 } { 4 ( 2 m - 1 ) + 27 }{/tex}{tex}\\frac { a + ( m - 1 ) d } { A + ( m - 1 ) D } = \\frac { 14 m - 7 + 1 } { 8 m - 4 + 27 }{/tex}{tex}\\frac { a + ( m - 1 ) d } { A + ( m - 1 ) D } = \\frac { 14 m - 6 } { 8 m + 23 }{/tex}Hence,\xa0{tex}\\frac { a _ { m } } { A _ { m } } = \\frac { 14 m - 6 } { 8 m + 23 }{/tex} | |
| 5261. |
Please do study friends our board exams coming very soon you also study in mobile but few hour |
|
Answer» ? Ok Yeah Right |
|
| 5262. |
In ∆ABC, DE||BC,find the value of ×.If, AD=× DB=×+1 EC=×+3 AE=×+5 |
|
Answer» X=3 ×=3 |
|
| 5263. |
Tell quadratic formula |
|
Answer» x=(-b±√b²-4ac)/2a -b+roots/2a - b+-root d / 2 a D = bsquare - 4ac |
|
| 5264. |
5×10+20÷20+10 |
|
Answer» (50*10)+(20/20)+(10)50+1+10=51+10=61So,Ans=61 61 61 |
|
| 5265. |
(a+b)cube= |
|
Answer» a^3+b^3+3abb+3aabAs it is easy to learn acube+ bcube+3asquareb+3absquare a³+b³+3ab(a+b) |
|
| 5266. |
Find the values of y for which the distance between the points p(2,-3) and q(10,y) is 10 units |
| Answer» Is y is (-1) | |
| 5267. |
The perimeter of rectangle is 60cm and its hypotenuse is 25 cm .find area of triangle. |
| Answer» Given: a + b + 25 = 60\xa0{tex}\\Rightarrow{/tex}\xa0a + b = 35a2 + b2 = 252ab\xa0{tex}=\\frac{1}{2}\\left[(a+b)^{2}-\\left(a^{2}+b^{2}\\right)\\right]{/tex}{tex}=\\frac{1}{2}[1225-625]{/tex}{tex}=\\frac{1}{2}[600]{/tex}= 300{tex}\\therefore{/tex}\xa0area of\xa0{tex}\\triangle{/tex}\xa0= 150cm2 | |
| 5268. |
How to find middle term of an ap |
| Answer» N+1/2 | |
| 5269. |
CosecA+cotA=11/5.then find the cosa |
| Answer» | |
| 5270. |
Prove tana/1-cota+cota/1-tana=1+sec a×coseca |
| Answer» We have,{tex}\\mathrm { LHS } = \\frac { \\tan A } { 1 - \\cot A } + \\frac { \\cot A } { 1 - \\tan A }{/tex}{tex}\\Rightarrow \\quad \\mathrm { LHS } = \\frac { \\tan A } { 1 - \\frac { 1 } { \\tan A } } + \\frac { \\frac { 1 } { \\tan A } } { 1 - \\tan A }{/tex}{tex}\\Rightarrow \\quad \\text { LHS } = \\frac { \\tan A } { \\frac { \\tan A -1 } { \\tan A } } + \\frac { 1 } { \\tan A ( 1 - \\tan A ) }{/tex}{tex}\\Rightarrow \\quad \\mathrm { LHS } = \\frac { \\tan ^ { 2 } A } { \\tan A - 1 } + \\frac { 1 } { \\tan A ( 1 - \\tan A ) }{/tex}{tex}\\Rightarrow \\quad \\text { LHS } = \\frac { \\tan ^ { 2 } A } { \\tan A - 1 } - \\frac { 1 } { \\tan A ( \\tan A - 1 ) }{/tex}{tex}\\Rightarrow \\quad \\mathrm { LHS } = \\frac { \\tan ^ { 3 } A - 1 } { \\tan A ( \\tan A - 1 ) }{/tex}\xa0[Taking LCM]{tex}\\Rightarrow \\quad \\mathrm { LHS } = \\frac { ( \\tan A - 1 ) \\left( \\tan ^ { 2 } A + \\tan A + 1 \\right) } { \\tan A ( \\tan A - 1 ) }{/tex}\xa0[{tex}\\because{/tex}\xa0a3\xa0- b3\xa0= ( a - b )(a2\xa0+ ab + b2)]{tex}\\Rightarrow \\quad \\mathrm { LHS } = \\frac { \\tan ^ { 2 } A + \\tan A + 1 } { \\tan A }{/tex}{tex}\\Rightarrow \\quad \\mathrm { LHS } = \\frac { \\tan ^ { 2 } A } { \\tan A } + \\frac { \\tan A } { \\tan A } + \\frac { 1 } { \\tan A }{/tex}{tex}\\Rightarrow{/tex}\xa0LHS = tanA + 1 + cotA [ since (1/tanA) =cotA ].= (1 + tanA + cotA){tex}\\therefore \\quad \\frac { \\tan A } { 1 - \\cot A } + \\frac { \\cot A } { 1 - \\tan A }{/tex}\xa0= 1 + tanA + cotA ...........(1)Now, 1 + tanA + cotA = 1 +\xa0{tex}\\frac { \\sin A } { \\cos A } + \\frac { \\cos A } { \\sin A }{/tex}\xa0= 1 +\xa0{tex}\\frac { \\sin ^ { 2 } A + \\cos ^ { 2 } A } { \\sin A \\cos A }{/tex} = 1 +\xa0{tex}\\frac { 1 } { \\sin A \\cos A }{/tex}\xa0[{tex}\\because{/tex}Sin2A + Cos2 A = 1 ]\xa0= 1 + cosecAsecASo, 1 + tanA + cotA = 1+ cosecAsecA.......(2)From (1) and (2), we obtain{tex}\\frac { \\tan A } { 1 - \\cot A } + \\frac { \\cot A } { 1 - \\tan A }{/tex}\xa0= 1 + tanA + cotA = 1 + cosecAsecA | |
| 5271. |
2x²+4x+6 |
| Answer» | |
| 5272. |
How we get oswal solutions of 6-10 sample paper |
|
Answer» Yaha kaise du pdf me dwnld kar loo. Oswallbook.com website se loo link http://www.oswaalbooks.com/download-sample-question-papers.php?id=10. Www.oswallbookssample.in From oswaal website it is given |
|
| 5273. |
X cube minus 6 x square plus 11 x minus 6 |
| Answer» | |
| 5274. |
What is the distance between the points (a cos 45 ,0 )and ( 0,a cos 45) |
| Answer» | |
| 5275. |
Prove that 6 is irrational. |
|
Answer» 1 min.... Yha √6 hoga naa... Same ques in ncert.... Yha prove kre??? |
|
| 5276. |
If the circumference of a circle is pie units more than the diameter of circle find the diameter |
| Answer» | |
| 5277. |
Find the area of ∆ABCwith A (1,-4) and midpoint of sides through A being (2,-1)and (0,-1). |
| Answer» Let E be the midpoint of AB.{tex}\\therefore \\quad \\frac { x + 1 } { 2 } = 2{/tex}\xa0or x = 3and\xa0{tex}\\frac { y + ( - 4 ) } { 2 } = - 1{/tex} or, y = 2or, B(3, 2)Let F be the mid-point of AC.Then,{tex}0=\\frac{x_1+1}{2}{/tex}\xa0or\xa0{tex}x_1=-1{/tex}and {tex}\\frac { y _ { 1 } + ( - 4 ) } { 2 }{/tex}\xa0= -1 or, y1\xa0= 2or, C= (-1, 2)Now the co-ordinates are A(1, - 4), B(3,2), C (-1,2)Area of triangle{tex}= \\frac { 1 } { 2 } \\left[ x _ { 1 } \\left( y _ { 2 } - y _ { 3 } \\right) + x _ { 2 } \\left( y _ { 3 } - y _ { 1 } \\right) + x _ { 3 } \\left( y _ { 1 } - y _ { 2 } \\right) \\right]{/tex}{tex}= \\frac { 1 } { 2 } [ 1 ( 2 - 2 ) + 3 ( 2 + 4 ) - 1 ( - 4 - 2 ) ]{/tex}{tex}= \\frac { 1 } { 2 } [ 0 + 18 + 6 ]{/tex}= 12 sq units. | |
| 5278. |
If sintheta + 2costheta = 1 the prove that 2sintheta - costheta = 2 |
| Answer» | |
| 5279. |
What value for cos180°? |
| Answer» -1 | |
| 5280. |
Who is going to score full marks in maths |
|
Answer» How??? i thin k we all |
|
| 5281. |
Latest cbse news. |
| Answer» What?? | |
| 5282. |
What is homologous argans |
|
Answer» same structure different function Having same basic structure performed different function ex.....wings of bat,paw of a cat ?? The organ which r same in appearance and different in function By mistake clicked maths |
|
| 5283. |
Show that exactly one of the no n,n+2,n+4 is divisible by 3 |
|
Answer» Sol :We applied Euclid Division algorithm on n and 3.a = bq +r on putting a = n and b = 3n = 3q +r , 0 |
|
| 5284. |
Prove that the sum of the squares of the sides of a rhombus is equal to the squares of its diagnol |
| Answer» Given: Let, ABCD is a rhombus and since diagonals of a rhombus bisect each other at\xa0{tex} 90 ^ { \\circ }{/tex}To Prove: {tex} \\therefore A B ^ { 2 } + B C ^ { 2 } + C D ^ { 2 } + A D ^ { 2 } = A C ^ { 2 } + B D ^ { 2 }{/tex}Proof :\xa0{tex}\\therefore {/tex}{tex}A O = O C {/tex}\xa0{tex}\\Rightarrow A O ^ { 2 } = O C ^ { 2 }{/tex}{tex}B O = O D {/tex}\xa0{tex}\\Rightarrow B O ^ { 2 } = O D ^ { 2 }{/tex}and\xa0{tex}\\angle A O B = 90 ^ { \\circ }{/tex}\xa0{tex}\\therefore {/tex}\xa0{tex} A B ^ { 2 } = O A ^ { 2 } + B O ^ { 2 }{/tex}Similarly,\xa0{tex} A D ^ { 2 } = x ^ { 2 } + y ^ { 2 } {/tex}{tex}BC ^ { 2 } = x ^ { 2 } + y ^ { 2 } {/tex}{tex}C D ^ { 2 } = x ^ { 2 } + y ^ { 2 } {/tex}{tex} \\therefore A B ^ { 2 } + B C ^ { 2 } + C D ^ { 2 } + D A ^ { 2 } = 4 A O ^ { 2 } + 4 D O ^ { 2 }{/tex}\xa0{tex} = ( 2 A O ) ^ { 2 } + ( 2 D O ) ^ { 2 }{/tex}\xa0{tex} = ( 2 x ) ^ { 2 } + ( 2 y ) ^ { 2 }{/tex}\xa0{tex} \\therefore A B ^ { 2 } + B C ^ { 2 } + C D ^ { 2 } + A D ^ { 2 } = A C ^ { 2 } + B D ^ { 2 }{/tex}\xa0Hence proved. | |
| 5285. |
Preparation of all the subject within 1 month. How? |
|
Answer» 1 month v nahi reh gaya half to pre board mein nikal jayega OK thanks Aur kuch Read all subject per day |
|
| 5286. |
Which is best amongst :Arihant,super20,exam guru and Oswaal. Dont answer"NCERT" |
|
Answer» oswal Oswaal All in one |
|
| 5287. |
Which formula we can use to find area of a given segment?? |
|
Answer» Only mind Ha h n iska formula |
|
| 5288. |
Which formula we can use to find area of a give sector? |
| Answer» (Thita /360)πr² | |
| 5289. |
Long type |
| Answer» Long type ageeee? | |
| 5290. |
Find the coordinates of the centroid of a triangle whose vertices are (0, 6) (8 ,12) and (8, 0) |
|
Answer» ???☠????? No this is wrong answer 8,9 |
|
| 5291. |
Basic proportional theorem |
|
Answer» Example It there in text book See in NCERT |
|
| 5292. |
I m confused about centroid of a triangle in coordinate geometry.plzz explain |
| Answer» The mid point of triangle , at this point all the median\'s of triangle intersect each other | |
| 5293. |
What is the centroid of triangle |
|
Answer» Median of a triangle Centroid of a triangle is a point where all medians of triangle intersect Medium Point where all 3 media\'s intersects Oh |
|
| 5294. |
What is the centroid of a triangle called;plzzz explain me. |
| Answer» X1+x2+x3/2 | |
| 5295. |
If SecA +TanA =p So prove that SinA =p2-1/p2+1 |
| Answer» | |
| 5296. |
What is the distance between two parallel tangents of a circle radius 14cm |
|
Answer» 28 cm 28 cm 28cm 28cn |
|
| 5297. |
Any type of questions |
| Answer» | |
| 5298. |
Find the value of x+y; if 3x-2y=5 and 3y-2x=3 |
|
Answer» 3x-2y=5 _13y-2x=3. =-2x+3y=3 _2Use _1 3x-2y=53x=5+2yX=5+2y/3Put the value of x in_2-2x+3y=3-2(5+2y/3)+3y=3-10-4y/3+3y=3 -10-4y+9y/3=3-10+5y=95y=9+10Y=19/5Put the vlue of y in_13x-2(19/5)=5 3x=5+38/53x=25+38/53x=63/5x= 63/5/3x=63/5×1/3x= 21/5 Which is correct answer guys 13/2 Might be x+y=(-4) =8 |
|
| 5299. |
-0.12 divide the following number |
| Answer» | |
| 5300. |
Prove that 3+5 |
|
Answer» ❤❤fattu What we have to prove rational aur irrational |
|