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.
| 35551. |
Traingle a=12\'b=34;c=45,m/ |
| Answer» Maths ispe nhi ho skti | |
| 35552. |
I have problem to ch - Triangle plzz help me |
| Answer» Bhai aise kaise help karo | |
| 35553. |
2 7 12.......... 10an=? |
| Answer» a=2d=5n=10an=?an=a+(n-1)dan=2+(10-1)5an=2+45an=47 | |
| 35554. |
6th class math book |
| Answer» To problem kya hai | |
| 35555. |
The area of a square field is 6050m2 .The length of its diagonal is |
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Answer» February...2nd weak... Febuary 20 approx Febuary ke ...1 ya 2 nd weak mw start hjayege.... Kisi ko pta hai kya board exam kab hoga February ya march se? |
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| 35556. |
1/2*22/7*7*7=187 how???? |
| Answer» πr²/2 (area of semi circle) | |
| 35557. |
5x-6=0+2+1=1x |
| Answer» | |
| 35558. |
1 and 1 |
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Answer» Gud sgun...?? 11?? |
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| 35559. |
Solve x and y 217x+113y=973113x+217y=887 |
| Answer» Fhkvzuvxg | |
| 35560. |
Showthat one and only one out of n,n+2 ,n+4 is divisible by 3 where n is any positive integer |
| Answer» Bhak | |
| 35561. |
5x(3_2) |
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Answer» 5x 5x |
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| 35562. |
If two zeroes of polynomials are 2,3 and the polynomials are 3x3-4x+3 find the one zero |
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Answer» _1/6 Hzbuss |
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| 35563. |
For some integers m, every even integer is of the form:-(i) m (ii) m+1 (iii) 2m (iv) 2m+1 |
| Answer» 1 | |
| 35564. |
Easiest way to find trigonometric identities |
| Answer» | |
| 35565. |
Can it can be prove that 5=0 |
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Answer» Yes, by multiplying to both side with zero a 5×0 =0Simple No |
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| 35566. |
2+75312 |
| Answer» 75314 | |
| 35567. |
x¹+x²+x³+x⁴+.........x¹°° is a polynomial or notGive reason |
| Answer» | |
| 35568. |
Prive that: tan /(1+ tan^2)^2 +cot/(1+cot^2)^2 =sin×cos |
| Answer» LHS\xa0{tex}=\\frac{\\tan A}{\\left(1+\\tan ^{2} A\\right)^{2}}+\\frac{\\cot A}{\\left(1+\\cot ^{2} A\\right)^{2}}{/tex}{tex}=\\frac{\\tan A}{\\left(\\sec ^{2} A\\right)^{2}}+\\frac{\\cot A}{\\left(\\ cosec ^{2} A\\right)^{2}}{/tex}{tex}=\\frac{\\sin A}{\\cos A}{/tex}\xa0. cos4\xa0A\xa0{tex}+\\frac{\\cos A}{\\sin A}{/tex}\xa0.\xa0sin4\xa0A= sin A cos3\xa0A + cos A sin3\xa0A= cos A sin A (cos2\xa0A + sin2\xa0A)= sin A cos A = RHS\xa0 | |
| 35569. |
Sin 90= |
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Answer» 1 obviously 1 LOL 1 Ncert mai clearly diya hai dude! 1 |
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| 35570. |
sin-cos+1÷sin+cos-1=1÷sec-tan |
| Answer» We have to prove that,{tex}\\frac{{\\sin \\theta - \\cos \\theta + 1}}{{\\sin \\theta + \\cos \\theta - 1}} = \\frac{1}{{\\sec \\theta - \\tan \\theta }}{/tex} using identity\xa0{tex}sec^2\\theta=1+tan^2\\theta{/tex}LHS = {tex}\\frac{{\\sin \\theta - \\cos \\theta + 1}}{{\\sin \\theta + \\cos \\theta - 1}} {/tex}{tex} = \\frac{{\\tan \\theta - 1 + \\sec \\theta }}{{\\tan \\theta + 1 - \\sec \\theta }}{/tex} [ dividing the numerator and denominator by {tex}\\cos{\\theta}{/tex}.]{tex} = \\frac{{(\\tan \\theta + \\sec \\theta)-1 }}{{(\\tan \\theta - \\sec \\theta )+1}}{/tex}{tex}=\\frac{\\{{(\\tan\\theta+\\sec\\theta)-1\\}}(tan\\theta-\\sec\\theta)}{\\{{(\\tan\\theta-\\sec\\theta)+1\\}}(\\tan\\theta-\\sec\\theta)}{/tex} [ Multiplying and dividing by {tex}(\\tan{\\theta}-\\sec{\\theta}){/tex}]{tex}=\\frac{{(\\tan^2\\theta-\\sec^2\\theta)-}(tan\\theta-\\sec\\theta)}{\\{{(\\tan\\theta-\\sec\\theta)+1\\}}(\\tan\\theta-\\sec\\theta)}{/tex} [{tex}\\because (a-b)(a+b)=a^2-b^2{/tex}]{tex} = \\frac{{-1-\\tan \\theta + \\sec \\theta }}{{(\\tan \\theta - \\sec \\theta+1)(\\tan{\\theta}-\\sec{\\theta}) }}{/tex}[{tex}\\because \\tan^2\\theta-\\sec^2\\theta=-1{/tex}]{tex}=\\frac{-(\\tan\\theta-\\sec\\theta+1)}{(\\tan\\theta-\\sec\\theta+1)(\\tan\\theta-\\sec\\theta)}{/tex}{tex}=\\frac{-1}{\\tan{\\theta}-\\sec{\\theta}}{/tex}{tex} = \\frac{1}{{\\sec \\theta - \\tan \\theta }}{/tex}=RHSHence Proved. | |
| 35571. |
What is the value of alpha minus beta |
| Answer» √b²-4ac/a | |
| 35572. |
Write the quadratic polynomial whose roots are 2 and -3 |
| Answer» Roots of polynomial are 2 and -3 . So we get the sum of roots = -1 and product of roots =-6. So polynomial is x²+1x-6. | |
| 35573. |
Sides of the similar triangle 4cm and 9cm are in the ratio of? |
| Answer» 2/3 | |
| 35574. |
Find the value of -8-(10) using no. Line |
| Answer» - 18 | |
| 35575. |
Can anybody tell me that what is da difference between a tangent to a circle and a secant |
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Answer» Tanget touch at a single point whereas secant touch at two points.... Tangent touches only one point of the perimeter of cirle (eg football on the ground ) and secant touches two points of the perimeter of circle.... Tangent touch at only one point but secant at two point |
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| 35576. |
2:3 |
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Answer» 2/3 It means 2/3 What is the question ? Clarify your question . |
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| 35577. |
Prove that 6-root3 is irrational. |
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Answer» Yes Let 6√3 be a rational number , so it can be written in the form of a/b where b≠0 and a and b are co prime numbers ......... .......... 6√3=a/b .. now √3 = a/b-6 , as we know that √3 is an irrational number , therefore our assumption is wrong. Hence due to contradiction 6√3 is an irrational number .hence proved . |
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| 35578. |
For what value of k the quadratic equation x^2 -2(k+1)x+k^2=0 has real and equal roots |
| Answer» | |
| 35579. |
What is the total surface area of the frustum? |
| Answer» πl(R ➕ r) ➕πR^2 ➕πr^2 | |
| 35580. |
X =3 is one root of quadratic x² -2kn-6=0 find the value of k |
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Answer» x2-2kn-6=0(3)2-2k(3)-6=09-6k-6=03-6k=0-6k=3K=-3/6=-1/2 x2 - 2kx - 6 = 0(3)2 - 2k(3) - 6 = 09 - 6k - 6 = 0-6k - 3 = 0- 6k = 3k = 3/-6k = -1/2 Put x=3 and solve |
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| 35581. |
Cosec = 5/3 evaluate 4sec - 2tan + 5 sin .÷ 20 cosec - 3 cosec + 9 cot |
| Answer» Ans of the question is 13/49 | |
| 35582. |
What is meant by concentric circles? |
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Answer» Many circles having same radii A circle inscribed in an same circle is known as concentric circle |
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| 35583. |
Find the HCFof of 65 & 117 and express it in the form 65m+117n |
| Answer» First find the HCF of 65 and 117 by Using Euclid\'s division algorithm,117 = 65{tex}\\times{/tex} 1 + 5265 = 52{tex}\\times{/tex} 1 + 1352 = 13{tex}\\times{/tex} 4 + 0So, HCF of 117 and 65 = 13HCF = {tex}65m + 117n{/tex}For, {tex}m= 2{/tex} and {tex}n = -1{/tex},HCF = 65{tex}\\times{/tex} 2 + 117{tex}\\times{/tex} (-1)= 130 - 117= 13Hence, the integral values of m and\xa0n are 2 and -1 respectively and the HCF of 117 and 65 is 13. | |
| 35584. |
Ncert book need for cbse board exam ? |
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Answer» Yes,offcourse Yes why not Obviously yes! Yes or no |
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| 35585. |
What is a composit number ? |
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Answer» Wo no.jinke one hi factor ho . This is called composite number. Composite numbera have mor than 2 factors. For example:- 4 ,6....... Which has more than two factor |
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| 35586. |
√3tantheta=1 then find the value of sin²theta-cos²theta |
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Answer» -1/2 1 |
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| 35587. |
5x+2y=2k, 2(k+1)x+ky=(3k+4) |
| Answer» {tex}5x + 2y = 2k{/tex}{tex}2(k + 1)x + ky = (3k + 4){/tex}These are of the form{tex}a_1x + b_1y + c_1\xa0= 0 , a_2x + b_2y + c_2\xa0= 0{/tex}where,{tex}a_1= 5\xa0,b_1= 2, c_1\xa0= -2k,{/tex}{tex}a_2= 2(k +1)\xa0,b_2= k\xa0,c_2\xa0= -(3k + 4){/tex}For infinitely many solutions, we must\xa0have{tex}\\frac { a _ { 1 } } { a _ { 2 } } = \\frac { b _ { 1 } } { b _ { 2 } } = \\frac { c _ { 1 } } { c _ { 2 } }{/tex}This holds only when{tex}\\frac { 5 } { 2 ( k + 1 ) } = \\frac { 2 } { k } = \\frac { - 2 k } { - ( 3 k + 4 ) }{/tex}{tex}\\Rightarrow \\frac { 5 } { 2 ( k + 1 ) } = \\frac { 2 } { k } = \\frac { 2 k } { ( 3 k + 4 ) }{/tex}Now, the following cases arises:Case 1:{tex}\\frac { 5 } { 2 ( k + 1 ) } = \\frac { 2 } { k }{/tex}[Taking I and II]{tex}\\Rightarrow 5 k = 4 ( k + 1 ) \\Rightarrow 5 k = 4 k + 4{/tex}k = 4Case 2:{tex}\\frac { 2 } { k } = \\frac { 2 k } { ( 3 k + 4 ) }{/tex}[Taking II and III]{tex}\\Rightarrow{/tex}2(3k +\xa04) = 2k2\xa0{tex}\\Rightarrow{/tex}6k + 8 = 2k2{tex}\\Rightarrow{/tex}{tex}2k^2\xa0-\xa06k + 8 =\xa00{/tex}{tex}\\Rightarrow{/tex}{tex}2(k^2\xa0- 3k + 4)=\xa00{/tex}{tex}\\Rightarrow{/tex}{tex}k^2\xa0- 3k - 4 = 0{/tex}{tex}\\Rightarrow{/tex}{tex}k^2\xa0- 4k + k - 4 = 0{/tex}{tex}\\Rightarrow{/tex}k(k - 4) + 1(k - 4) = 0{tex}\\Rightarrow{/tex}(k - 4)(k +\xa01) = 0(k - 4) = 0 or k + 1 = 0{tex}\\Rightarrow{/tex}\xa0k = 4 or k = -1Case 3:{tex}\\frac { 5 } { 2 ( k + 1 ) } = \\frac { 2 k } { ( 3 k + 4 ) }{/tex}[Taking I and II]{tex}\\Rightarrow 15 k + 20 = 4 k ^ { 2 } = 4 k{/tex}{tex}\\Rightarrow{/tex}4k2\xa0+ 4k - 15k - 20 = 0{tex}\\Rightarrow{/tex}\xa04k2\xa0- 11k - 20 = 0{tex}\\Rightarrow{/tex}4k2\xa0- 16k + 5k - 20 =0{tex}\\Rightarrow{/tex}4k(k - 4) + 5(k - 4) = 0{tex}\\Rightarrow{/tex}(k - 4)(4k + 5) = 0{tex}\\Rightarrow k = 4 \\text { or } k = \\frac { - 5 } { 4 }{/tex}Thus, k = 4, is the common value of which there are infinitely many solutions. | |
| 35588. |
If the points (-1.5,3),(k,2),(-3,4)are collinear.find k? |
| Answer» 0 | |
| 35589. |
HCf of 625 |
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Answer» 5×5×5×5= 625 Its the number itself i.e. 625 5 power 4 5 is the hcf |
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| 35590. |
Formula of area of square |
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Answer» (Side)^2 OR side × side Side × side Side *side |
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| 35591. |
AP is 2,6,4,8,5 .find common difference. |
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Answer» This is not AP It is not in the form of an A.P No common diference6-2=4 4-6=-2then no c..d This is not in the form of AP 4 |
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| 35592. |
How to learn trigonometry formulas and identities ? How it apply in quesrions ? |
| Answer» Sin cosec cos=sec tan =cot | |
| 35593. |
How many digits will be there in decimal part of 73÷2power4 × 5 power 3 |
| Answer» | |
| 35594. |
Find root of the equation:-* 4x2+4bx-(a2-b2)=0* ax2+bx+c=0 |
| Answer» 16a2 | |
| 35595. |
TanA +TanB/ CotA + CotB =tanA.tanB |
| Answer» | |
| 35596. |
150+(-40)×20 |
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Answer» =150 - 800= - 650 -650 |
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| 35597. |
[email\xa0protected]+(+88+=( |
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Answer» What type of question us this .And from which chapter does it belong What |
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| 35598. |
If the median of the series exceeds the mean by 3, find by what number the mode exceeds its mean? |
| Answer» 3(median) = 2(mean) + modeLet the mean be \'x\'So, median will be x + 3Then, according to the question3(x+3) = 2(x) + mode3x + 9 = 2x + mode3x - 2x + 9 - modex+9 = modeNow, mode - mean⇒ x+9 - x= 9 So, mode exceeds the mean by 9. \xa0 | |
| 35599. |
What happens to a value of tanA when A increases from 0°to90° |
| Answer» Value of tan A increases when A increases from {tex}0^o{/tex} to {tex}90^o{/tex}{tex}(0,\\frac{1}{\\sqrt{3}},1,\\sqrt{3},{/tex}\xa0not defined at\xa0{tex}90^{\\circ}{/tex}) | |
| 35600. |
Sin theta value |
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Answer» Perpendicular /hypotenuse Sin theta=p/h Length/ hypotenuse 0 |
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