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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.
| 40001. |
State and prove phythagoras theorem |
| Answer» It is given in book | |
| 40002. |
Sin(A-B) =1/2,Cos(A+B)=1/2. Find the value of A and B. |
| Answer» A=45B=15 | |
| 40003. |
If the sum of first 7 terms of an ap is 49 and that of 17 terms is 289 find the sum of n terms |
| Answer» 49=7/2[2a+6d]14=2a+6d7=a+3d_________134=2a+16d17=a+8d__________2a+3d=7a+8d=17______________on sub-5d=-10d=2a=1Sn=n/2{2+(n-1)2}Sn=n×n | |
| 40004. |
How to do cross multiplication method |
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Answer» First shift the terms of RHS to LHS then sign will change , then use the formula Its very easy see pg 60 of your ncert |
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| 40005. |
Teri ma\'am kids chuf |
| Answer» What the hell??? | |
| 40006. |
Find the centre of a circle passing through(6,-6),(3,-7),(3,3) |
| Answer» Let\xa0A → (6, –6), B\xa0→ (3, –7) and C\xa0→ (3, 3).Let the centre of the circle be I(x, y)Then, IA = IB = IC [By definition of a circle]{tex}\\Rightarrow{/tex} IA2 = IB2 = IC2{tex}\\Rightarrow{/tex} (x - 6)2 + (y + 6)2 = (x - 3)2 + (y + 7)2 = (x - 3)2 + (y - 3)2Taking first two, we get(x - 6)2 + (y + 6)2 = (x - 3)2 + (y + 7)2{tex}\\Rightarrow{/tex} x2 - 12x + 36 + y2 + 12y + 36 = x2 - 6x + 9 + y2 + 14y + 49{tex}\\Rightarrow{/tex} 6x + 2y = 14{tex}\\Rightarrow{/tex} 3x + y = 7 ......(1) ....[Dividing throughout by 2]Taking last two, we get(x - 3)2 + (y + 7)2 = (x - 3)2 + (y - 3)2{tex}\\Rightarrow{/tex} (y + 7)2 = (y - 3)2{tex}\\Rightarrow{/tex} (y + 7) = {tex}\\pm{/tex}(y-3)taking +e sign, we gety + 7 = y - 3{tex}\\Rightarrow{/tex} 7 = -3which is impossibleTaking -ve sign, we gety + 7 = -(y - 3){tex}\\Rightarrow{/tex} y + 7 = -y + 3{tex}\\Rightarrow{/tex} 2y = -4{tex}\\Rightarrow y = \\frac{{ - 4}}{2} = - 2{/tex}Putting y = -2 in equation (1), we get{tex}\\Rightarrow{/tex} 3x - 2 = 7{tex}\\Rightarrow{/tex} 3x = 9{tex}\\Rightarrow{/tex} x = 3Thus, I {tex}\\rightarrow{/tex} (3, -2)Hence, the centre of the circle is (3, -2). | |
| 40007. |
X square + bx + c equal to zero |
| Answer» What we have to find | |
| 40008. |
What is the externalSection formula |
| Answer» X=m1x2-m2x1÷m2-m1 similar for y put x to y | |
| 40009. |
How much exam paper taken math |
| Answer» | |
| 40010. |
Prove that cotA-cosA/cotA+cosA = cosecA-1/cosecA+1 |
| Answer» cotA -cosA)/(cotA + cosA) = (cosecA-1)/(cosecA+1) (cotA -cosA)/(cotA + cosA)((cosA/sinA)-cosA)/((cosA/sinA)-cosA) cotA=cosA/sinAtaking cosA common from numretor nd denominatr nd cancelling it we get ((1/sinA)-1)/((1/sinA)+1) since 1/sinA=cosecA therfore (cosecA-1)/(cosecA+1) | |
| 40011. |
@ is called.... |
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Answer» @ is called at or at the rate. At the rate At the rate At the rate |
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| 40012. |
how to make prsentation on fraction |
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Answer» It is not possible How is it possible no not class 10 i am from class 6 |
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| 40013. |
If (1.7)^x=(0.17)^y=10^4 then find the value of:Xy/(y-x) |
| Answer» | |
| 40014. |
2ab=cd |
| Answer» | |
| 40015. |
How many number can be divisible by 7 |
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Answer» Infinite but if you are asking the ques of ncert then ans will be a= 105. an=994 d=7 105 to 904 Infinite Infinite |
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| 40016. |
In an ap if sn=n(4n+1),find first term,common difference and a10 |
| Answer» | |
| 40017. |
if x plus 1by x is 7then xcube plus 1by x cube is equal to |
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Answer» x+1/x=7 ( taking cube on both sides) then (x+1/x )cube=7cube ,xcube+1/xcube+3 (x+1/x)=343 then we get xcube +1/x cube= 343-21so we get xcube +1/xcube=322 217/36 |
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| 40018. |
Q.E how to solve x2-3x+2=0 |
| Answer» x2 - 3x + 2 = 0x2 - 2x - 1x + 2= 0x(x - 2) - 1( x - 2) = 0(x -1) ( x -2) = 0x - 1 = 0 or x - 2 = 0x = 1 or x = 2 | |
| 40019. |
Sin-cos+1/sin+cos-1 =1/sec-tan prove |
| 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. | |
| 40020. |
If alpha and beta are the zeroes of the polynomial x^2-4x+1Find the value of α+1/α |
| Answer» | |
| 40021. |
Sample papers of 2019 latest |
| Answer» Buy apc sample paper | |
| 40022. |
188 |
| Answer» | |
| 40023. |
x°2-x-p(p+1) |
| Answer» x2 - x - p (p + 1)= x2 - x - p2 - p | |
| 40024. |
How to learn statistic in few minutes? |
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Answer» Gather all the formulas of statistics and their meaning. Also do the examples of each formula Get those formulas memorised. |
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| 40025. |
If Sn denotes the sum of the first n terms of an AP, prove that S30=3(S20-S10). |
| Answer» Sn=n/2(2a+(n-1)d)S30=30/2(2a+29d)S30=30a+435d... (1)S20=20/2(2a+19d)S20=20a+190d....(2)S10=10/2(2a+9d)S10=10a+45d....(3)3(S20-S10)=3(20a+190d-10a-45d) =3(10a+145d) =30a+435d=S30 (from 1) So, S30=3(S20-S10) Hence proved. | |
| 40026. |
What is the formula of sin |
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Answer» Perpendicular /hypotenuse SinA=tanA.cosA P/H Perpendicular/hypotenuse |
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| 40027. |
Math ka sllyebus |
| Answer» Chapter 1st to chapter 9th | |
| 40028. |
Find three consecutive terms which are in AP whose sum is 24 and product is 440 |
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Answer» (a+d) +a+(a+d) =24 and(a-d)(a) (a+d) =440 (a-d)+a+(a+d) =24 and (a-d)(a)(a+d)=440 |
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| 40029. |
2(2) |
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Answer» 4 4 |
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| 40030. |
Find the value of tan A, where A is an acute angle and cosec A= 2/√3 |
| Answer» Under root 3 | |
| 40031. |
Prove that ac +Ab + bc greater 2ap + pq |
| Answer» | |
| 40032. |
Sin (A+B)=? |
| Answer» Let A =B=30 Sin(A+B) =sin(30 +30)=sin60=√3/2 | |
| 40033. |
AP 3,8,13-----253 |
| Answer» What we have to find here | |
| 40034. |
How to find the zeroes of cubic polynomialActually I want exact procedure ...pls help |
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Answer» Acccha ....zyada gyan mat baato jitna poocha h utna batao agar pta h to Itna bhi nahi aaata hamare esh ki kab pragati hogi gawaaar |
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| 40035. |
What is tangent and circumferance in circle? |
| Answer» Tangent is line draw outside of circle and get in contact with circle on a certain one point...Circumferance of circle is 2πr | |
| 40036. |
1÷(x-1)(x-2)+1÷(x-2)(x-3)+1÷(x-3)(x-4)=1/6 solve the value of x |
| Answer» We have,{tex}\\frac{1}{{(x - 1)(x - 2)}} + \\frac{1}{{(x - 2)(x - 3)}}{/tex}{tex}+ \\frac{1}{{(x - 3)(x - 4)}} = \\frac{1}{6}{/tex}{tex}\\Rightarrow{/tex}{tex} (x - 3)(x - 4) + (x - 1)(x - 4) + (x - 1)(x - 2) = {/tex}{tex}\\frac{1}{6}{/tex}{tex}(x - 1)(x - 2)(x - 3)(x - 4){/tex}[{tex}\\because{/tex} Multiplying both sides by (x -1)(x - 2)(x - 3)(x - 4)]{tex}\\Rightarrow{/tex}\xa0{tex}x^2 - 4x - 3x + 12 + x^2 - 4x - x + 4 + x^2 - 2x - x + 2 ={/tex}{tex}\\frac{1}{6}{/tex}{tex}[(x - 1)(x - 2)(x - 3)(x - 4)]{/tex}{tex}\\Rightarrow{/tex}{tex}3x^2 - 15x + 18 ={/tex}{tex}\\frac{1}{6}{/tex}{tex}(x - 1)(x - 2)(x - 3)(x - 4){/tex}{tex}\\Rightarrow{/tex}{tex}3(x^2 - 5x + 6) =\u200b\u200b\u200b\u200b\u200b\u200b\u200b{/tex}{tex}\\frac{1}{6}{/tex}{tex}(x - 1)(x - 2)(x - 3)(x - 4){/tex}{tex}\\Rightarrow{/tex}{tex}18[x^2 - 3x - 2x + 6] = (x - 1)(x - 2)(x - 3)(x - 4){/tex}{tex}\\Rightarrow{/tex}{tex}18[x(x - 3) - 2(x - 3)] = (x - 1)(x - 2)(x - 3)(x - 4){/tex}{tex}\\Rightarrow{/tex}{tex}18(x - 3)(x - 2) = (x - 1)(x - 2)(x - 3)(x - 4){/tex}{tex}\\Rightarrow{/tex} 18 = (x - 1)(x - 4){tex}\\Rightarrow{/tex} 18 = x2 - 4x - 1x + 4{tex}\\Rightarrow{/tex} x2 - 5x + 4 - 18 = 0{tex}\\Rightarrow{/tex} x2 - 5x - 14 = 0In order to factorize x2 - 5x - 14, we have to find two numbers \'a\' and \'b\' such that.a + b = - 5 and ab = -14Clearly, -7 + 2 = -5 and (-7)(2) = -14{tex}\\therefore{/tex}\xa0a = -7 and b = 2Now,x2 - 5x - 14 = 0{tex}\\Rightarrow{/tex} x2 - 7x + 2x - 14 = 0{tex}\\Rightarrow{/tex} x(x - 7) + 2(x - 7) = 0{tex}\\Rightarrow{/tex} (x - 7)(x + 2) = 0{tex}\\Rightarrow{/tex} x - 7 = 0 or x + 2 = 0{tex}\\Rightarrow{/tex} x = 7 or x = -2 | |
| 40037. |
Sin - 2sin power 3 / 2cos power 3 - cos |
| Answer» Sin/cos | |
| 40038. |
If A,BandC are interior angles of triangle ABC,then show thatSin(B+C by 2)=cosAby2 |
| Answer» \xa0A, B, C, are interior angles of a\xa0{tex}\\Delta {/tex}{tex}\\because A + B + C = 180 ^ { 0 }{/tex}{tex}\\Rightarrow B + C = 180 ^ { 0 } - A \\Rightarrow \\frac { B + C } { 2 } = 90 ^ { 0 } - \\frac { A } { 2 }{/tex}{tex}\\Rightarrow \\sin \\frac { \\mathrm { B } + \\mathrm { C } } { 2 } = \\sin \\left( 90 ^ { \\circ } - \\frac { \\mathrm { A } } { 2 } \\right) \\left[ \\because \\sin \\left( 90 ^ { 0 } - \\theta \\right) = \\cos \\theta \\right]{/tex}{tex}\\Rightarrow \\sin \\frac { \\mathrm { B } + \\mathrm { C } } { 2 } = \\cos \\frac { \\mathrm { A } } { 2 } \\text { proved }{/tex}LHS = RHS | |
| 40039. |
If Sin(A-B) =0, Cos(A+B) =0, 0 degree |
| Answer» Sin(A-B) =0 Sin(A-B) =Sin0A-B=0A=BCos(A+B)=0Cos(A+B)=Cos90A+B=902A=90A=B=45Sin(A+B)=Sin(45+45)Sin90=1Cos(A-B)=Cos(45-45)Cos0=1 | |
| 40040. |
Tan(90-a)cot |
| Answer» | |
| 40041. |
The first term of ap is 5 the last term is 45 and the sum is 400 find number of terms |
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Answer» S = N / 2 ( A + L ) .....400= N / 2 ( 5 + 45 ) = 400 = N/ 2 ( 50 ) 400 = 25n .....N = 16 16 |
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| 40042. |
1_2 |
| Answer» I have taken arihant math sample paper | |
| 40043. |
What will the value for 7/cot square theta - 7/cos squate theta |
| Answer» 7 | |
| 40044. |
Math important chapter for class10 board |
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Answer» All are imp. But most imp is the frog and nightangle ,patol babu and both drama all are important |
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| 40045. |
How to prove root5 irrational |
| Answer» Let √5 be a rational number then it can be written in the form of a/b where b≠0 , a and b are co prime numbers ... So √5= a/b , now squaring on both sides gives 5=a²/b². 5b² = a² . Now we can say that a² is divisible by 5 then a is also divisible by 5. Let a =5c , squaring on both sides gives a²=25 c². Putting value of a² ... 5b² =25 c²... b²=5c².. so we can say that b² is divisible by 5 then b is also divisible by 5. So our assumption is wrong , so due to contradiction √5 is an irrational number as it a common factor other than a and b . Hence proved. | |
| 40046. |
Cross multipleplication |
| Answer» | |
| 40047. |
Find the value of k sothat the lines 2x-3y=9 and kx- 9 y=18 will be parallel |
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Answer» K=6 Condition for parallel lines or no solution ---> a1/a2 = b1/b2 ≠ c1/c2..... So 2/k = 1/3 ≠1/2. So from here by solving them we get , k=6 and k≠4. |
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| 40048. |
If the perimeter of a sector of a circle of radius 6.5cm is 29cm what is its area?? |
| Answer» Maths ke ques. Ispe nhi hote | |
| 40049. |
6+5 |
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Answer» 11 11 |
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| 40050. |
(sin A+cosecA)square +(cos A+secA)square =7+tan square A +cot square A. |
| Answer» this can be donevby applying a²+b²then then sin²a+cos²a become 1 and other can be done by its recuprocal | |