InterviewSolution
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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.
| 40101. |
Given an=4 , d=2 , Sn= -14 , find n and a |
| Answer» n=7, a=-8 | |
| 40102. |
Given l=28, S=144,and there are total 9 terms. Find a |
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Answer» Sn=n/2(a+l)144=9/2(a+28)288=9(a+28)32=a+28a=32-28a=4 a=4 |
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| 40103. |
6(cos^10theta+sin^10theta)-15(cos^8theta+sin^8theta)+10(cos^6theta+sin^6theta)=1 ,prove that |
| Answer» | |
| 40104. |
If ΔABC~ΔPQR, |
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Answer» arre congruent prove kr do na aisa ek example bhi toh h book me Then what ? |
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| 40105. |
Find out the value of k equation x minus. 2y=3 and 3x+ky=1 |
| Answer» 5 | |
| 40106. |
F(x)=2ycube - 3kxsquare + 4x -5 is 6 |
| Answer» | |
| 40107. |
If tan +sec=l then prove that sec square =L square +1divide by |
| Answer» Given, tan θ + sec θ =\xa0{tex}l{/tex}........(1)We know that, sec2\xa0θ – tan2\xa0θ = 1.......(2)Now, sec θ + tan θ = {tex}l{/tex}\xa0[ from (1) ]⇒ (sec θ + tan θ)\xa0{tex}\\frac { ( \\sec \\theta - \\tan \\theta ) } { \\sec \\theta - \\tan \\theta } = 1{/tex}⇒\xa0{tex}\\frac { \\sec ^ { 2 } \\theta - \\tan ^ { 2 } \\theta } { \\sec \\theta - \\tan \\theta } = l{/tex}⇒\xa0{tex}\\frac { 1 } { \\sec \\theta - \\tan \\theta } = l{/tex}\xa0[ from equation (2) ]or, sec θ – tan θ =\xa0{tex}\\frac { 1 } { l }{/tex}\xa0........(3)Now, to get sec θ , eliminating tan θ from (1) and (3)adding (1) and (3) we get :-⇒ 2 sec θ =\xa0{tex}l + \\frac { 1 } { l }{/tex}⇒ 2 sec θ =\xa0{tex}\\frac { l ^ { 2 } + 1 } { l }{/tex}⇒ sec θ =\xa0{tex}\\frac { l ^ { 2 } + 1 } { 2 l }{/tex}Hence, proved. | |
| 40108. |
X-1=xspuare +x+1 |
| Answer» | |
| 40109. |
X (3+4) = 2 (4_5) |
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| 40110. |
What is sin25.cos65 and how? |
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Answer» 1 sin90-65.cos65 Cos65.cos65 =1 1 |
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| 40111. |
In the figure, angle BED=angle BDE and E divides BC in the ratio 2:1. Prove that AF×BE = 2AD×CF. |
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| 40112. |
If sin A -- cos A then tanA--? |
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Answer» TanA----- Cot A CotA |
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| 40113. |
If bac is 90degree AD is it\'s bisector. If DE is perpendicular to AC. Prove that DE*(AB+AC)=AB*AC |
| Answer» To prove the given result,we will use the following theorm.The internal bisector of an angle of a triangle divides the opposite side internally in the ratio of the sides containing the angleSince AD is the bisector of {tex}\\angle{/tex}A of {tex}\\Delta{/tex}ABC.{tex}\\therefore \\quad \\frac { A B } { A C } = \\frac { B D } { D C }{/tex}\xa0[by above theorm]{tex}\\Rightarrow \\quad \\frac { A B } { A C } + 1 = \\frac { B D } { D C } + 1{/tex}[Adding 1 on both sides]{tex}\\Rightarrow \\quad \\frac { A B + A C } { A C } = \\frac { B D + D C } { D C }{/tex}{tex}\\Rightarrow \\quad \\frac { A B + A C } { A C } = \\frac { B C } { D C }{/tex}\xa0... (i)In {tex}\\Delta{/tex}\'s CDE and CBA, we have{tex}\\angle{/tex}DCE = {tex}\\angle{/tex}BCA = {tex}\\angle{/tex}C [Common]{tex}\\angle{/tex}BAC = {tex}\\angle{/tex}DEC [Each equal to 90°]So, by AA-criterion of similarity, we have{tex}\\Delta{/tex}CDE ~ {tex}\\Delta{/tex}CBA{tex}\\Rightarrow \\quad \\frac { C D } { C B } = \\frac { D E } { B A }{/tex}{tex}\\Rightarrow \\quad \\frac { A B } { D E } = \\frac { B C } { D C }{/tex}\xa0...(ii)From (i) and (ii), we obtain{tex}\\frac { A B + A C } { A C } = \\frac { A B } { D E } \\Rightarrow D E \\times ( A B + A C ) = A B \\times A C{/tex} | |
| 40114. |
Tan theta upon 1 - cos theta + cot theta upon 1 - tan theta is equals to 1 + sec theta cosec theta |
| Answer» Please check the question it should be tan theta upon 1 -cot thetha .... then put the identity of tan theta =sin thetha upon cos thetha after that take their LCM\'S your answer will come . | |
| 40115. |
Find the area of the quadrilateral, the coordinates of whose vertices are (-3,2),(5,4),(7,6),(-5-4) |
| Answer» draw any of its digonals to form 2 diffrent triangles and find their area then add both area so you can find area of quadrilateral | |
| 40116. |
If xa+y6and z5 find xyz |
| Answer» Not understandable | |
| 40117. |
Which refresher is best for maths? |
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Answer» R.D. SHARMA All are good but ncert is the best....? |
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| 40118. |
Solve:- x^2+5x-900 by trinomial method |
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| 40119. |
Reappear compartment Ka paper 10 students ke saath hoga Ya private hoga 2019 march mai |
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Answer» Reapper cmprtmnt wlo se alg hge.... March mw hgw core subjects ke nd vocational ke febuary me hge |
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| 40120. |
cbse blue print of maths ,science and sst chapterwise not unitwise |
| Answer» Check the pattern here :\xa0https://mycbseguide.com/cbse-syllabus.html | |
| 40121. |
when is cbse calss10 2019 board exam |
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Answer» 3rd week of the february 6 march to 22 march but yet discssion not a confirm date. B/w feb. And march. |
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| 40122. |
What is rayional no. |
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Answer» By the way rational no.is which is in the form of p/q. What do you mean by rayional? Rational numbers are the numbers that can be written in the form of p/q A Rational Number of the form p/q or a number which can be expressed in the form of p/q , where p and q are integers and q ≠ zero, is called a Rational - Number.Example : 2 / 3 , -5 / 7, -10 / -3 are Rational Number. Rational ya rayional no. I dont know |
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| 40123. |
Find distance pair of points (-5,7),(-1,3) |
| Answer» (-5, 7), (-1, 3)Required distance{tex}= \\sqrt {([- 1 - {{(-5)}^2]} + {{(3- 7)}^2}}{/tex}{tex}= \\sqrt {16 + 16} = \\sqrt {32}{/tex}{tex}= 4\\sqrt 2{/tex} | |
| 40124. |
Hcf of 196 and398220 |
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Answer» The factors of 196 are: 1, 2, 4, 7, 14, 28, 49, 98, 196The factors of 398220 are: 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60, 6637, 13274, 19911, 26548, 33185, 39822, 66370, 79644, 99555, 132740, 199110, 398220Then the highest common factor is 4. Answer |
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| 40125. |
Ncert ch 7 solutions |
| Answer» Check NCERT solutions here :\xa0https://mycbseguide.com/ncert-solutions.html | |
| 40126. |
For what value of K, pair of linear equations Kx+3y+2=0,2x+y+3=0 has no solution |
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Answer» K=6 The value of K =6 |
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| 40127. |
If two zeroes are given so how can find other zeroes |
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Answer» If two zeroes are given than we can find the sum of zeroes and the product of zeroes after that we can put the values of sum and product of zeroes in Quadratic formula. By help of -b/a and c/a formula |
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| 40128. |
Solve x2+6x+13>0 |
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Answer» It is wrong question It\'s wrong question |
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| 40129. |
Solve for positive numbers x,y. if 2x3+3y3=246 and 6x3-7y3=-286 |
| Answer» | |
| 40130. |
Ax+by=0 |
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| 40131. |
How many 3 digit number is divisible by 7 |
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Answer» 127 105,112,119,....... 996a=105d=7an=196996=105(n_1)×7N=128 |
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| 40132. |
In given figure ,st||rq ,ps=3&Sr= 4 . find the ratio of the area of ∆ pst to the area ∆ Prq. |
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Answer» Kaise pura nikle gaa answer suru se likh kar batao Ar(PST) : Ar(PRQ) = 9 : 49. Koi answer batao Ek triangle bana hai isme pqr aur usi triangle st|| qr |
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| 40133. |
1is the prime number or not |
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Answer» No No, 1 is not a prime no. because if we divide 1 in any number it will give same answer. So,1 is not a prime number. Yes 1is a prime no. No 1 is not a prime number Yes |
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| 40134. |
Lcm of 3825 |
| Answer» 1955 | |
| 40135. |
Find the locus of the centres of circles which touch a given line at a given point. |
| Answer» Let APB be the given line, and let a circle with centre O touch APB at P.Then, {tex}\\angle O P B{/tex}= 90°. Let there be another circle with centre O\' which touches the line APB at P.Then, {tex}\\angle O ^ { \\prime } P B{/tex}=90°.This is possible only when O and O\' lie on the same line O\'OP. Hence,the required locus is a line perpendicular to the given line at the point of contact. | |
| 40136. |
Find the distance of a points A(-3/2.,-1/3) from the origin |
| Answer» √85/6 | |
| 40137. |
a=2d=5Find the 7th term of an AP. |
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Answer» a7 =a+6d =2+6×5 =2+30 =32 32 a+6d=2+6*5=32 32 32 32 32 |
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| 40138. |
3+7 |
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Answer» 10 Sorry i questions my mistak -1-(-1) 10 |
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| 40139. |
If the sum of first n term of an A.P. is given by Sn= 3n²+ 5n , find the n th term of the A.P.? |
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Answer» 6(n-1) n th term = 14n - 6 6n+2 is correct answer |
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| 40140. |
When exam of board 2019 |
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Answer» Starting Feb 2019, cause election will be conducted in march Starting in March. But vocational subject\'s exam is starting in Feb.? Starting Feb.....There is an update by CBSE for class 10 boards 2019 Starting march |
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| 40141. |
Board me trigonomenty kitne marks ki aayegi |
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Answer» I conform 12 numbers Approx 5 marks ki from ch 8 and 3 - 4 marks ki from ch 9. |
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| 40142. |
Final exams mai Trigonometry kitne marks ki aaigi |
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Answer» 12 18 12 marks |
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| 40143. |
If the roots of the equation (a-b)x²+(b-c)x+(c-a) are equal ,prove that 2a=b+c. |
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Answer» Without copy i cannot solve this question Tere hath kis din kam aaenge Bhai kaise solve karu without copy |
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| 40144. |
What is discriminent ? |
| Answer» D=b2-4ac where D=O (means roots are real and same) D>0(means roots are real but distinct) D<0(means no real roots or imaginary roots) | |
| 40145. |
If underoot3 tan=3sin,find the value of sin^2-cos^2 |
| Answer» | |
| 40146. |
Sin0 ki value |
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Answer» 1 0 0 1 0 1 |
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| 40147. |
If ax^2+bx+c=0 and c=b^2/4a. Find nature of roots |
| Answer» | |
| 40148. |
a1 =2an =an-1n is greater than =2 |
| Answer» | |
| 40149. |
an =3^n+1 |
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| 40150. |
Solve this by completing squar method 2x+7y-9=0 |
| Answer» | |