InterviewSolution
Saved Bookmarks
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.
| 2001. |
Find the area of a quadrant of a circle whose circumference is 44cm |
| Answer» Circumferences of a circle\xa0{tex}= 2 \\pi r{/tex}{tex}\\Rightarrow 2 \\pi r = 44{/tex}{tex}\\Rightarrow 2 \\times \\frac { 22 } { 7 } \\times r = 44{/tex}{tex}\\Rightarrow{/tex}\xa0r = 7 cm{tex}\\therefore{/tex}\xa0Area of the quadrant{tex}= \\left( \\frac { 1 } { 4 } \\times \\frac { 22 } { 7 } \\times 7 \\times 7 \\right) \\mathrm { cm } ^ { 2 }{/tex}{tex}= \\frac { 77 } { 2 } \\mathrm { cm } ^ { 2 }{/tex}{tex}= 38.5 cm^2{/tex} | |
| 2002. |
Kx?+2x+1 Find value of x |
| Answer» | |
| 2003. |
Express each number as a product of its prime factor 140 |
| Answer» So, the prime factors of 140 = 2 {tex}\\times{/tex}\xa02 {tex}\\times{/tex}\xa05 {tex}\\times{/tex}\xa07 = 22\xa0{tex}\\times{/tex}\xa05 {tex}\\times{/tex}\xa07. | |
| 2004. |
In A.P the sum of first "n" terms is 3n^2/2+13n/2. Find the 25 term of A.P |
|
Answer» Sum of n terms = 3n2/2 +13n/2Put n = 1, we get sum of 1 term = first term = 3/2+13/2= 16/2= 8So first term =8\xa0put n = 2, we get S2 =\xa03(2)2/2 +13*2/2 = 3*4/2 +13*2/2 = 6+13= 19So second term = 19-8 = 11So A.P is 8 ,11,14,17.........\xa0a= 8 and d=3a25 =\xa0a+(25-1)d = 8+24*3= 8+72 = 80\xa0 Put the value of n Ans will be 80 |
|
| 2005. |
Related number system |
| Answer» | |
| 2006. |
Cosec / cosec - 1 + cosec / cosec + 1=2sec |
| Answer» Cross multiply first then you will get,Cosec2+cosec+cosec2-cosec/cosec2-1= 2cosec.squard theta/cot squard thetaNext expand it,2× 1/sin× 1/sin ÷ cos/ sinSin theta will be cancelled.You will get, 2×1/cos= 2Sec theta | |
| 2007. |
Horrible |
| Answer» | |
| 2008. |
x+h/x-h =tanA/tanB;=2x/2h=tanA+tanB/tanA-tanB how |
| Answer» tan A +tan B/ tan A-tan BDivide numerator and denominator by tan B, we gettan A/tanB +tanB/tanB/tanA/tanB - tan B/tanBx+h/x-h+1/x+h/x-h-1x+h+x-h/x+h-x+h2x/2hhence proved\xa0 | |
| 2009. |
Completing the square |
| Answer» | |
| 2010. |
Kya Hindi mein Kabhi Parichay ayege |
|
Answer» Nhi kabhi nhi anwser me |
|
| 2011. |
Sin square theta + cos square theta equal to 1 |
| Answer» | |
| 2012. |
Board exam date |
| Answer» | |
| 2013. |
Which books I have follow for getting 95% in 10the board exam results |
| Answer» NCERT because it\'s our Bible and to score more consult S.chan for science and R.S. Aggarwal or R.D.Sharma for maths and solve as many sample paper as possible to score more. | |
| 2014. |
If in a ΔABC,AD is median and AM is perpendicular to BC ,then prove that AB^2+AC^2=2AD^2+1/2BC^2 |
| Answer» | |
| 2015. |
Exam me kis strategy se likhe jisse 30question ka paper 3hours me complete ho jaye |
| Answer» Always start from back so that u won\'t miss a good wheightage of marks !! | |
| 2016. |
The mth term of an Ap is n and the nth term is n . Find the rth term |
| Answer» Let a be the first term and d be the common difference of the given AP. Then, in general the mth and nth terms can be written as\xa0Tm = a + (m - 1) d and Tn = a + (n - 1)d respectively.According to the question,we are given that,(m.Tm) = (n.Tn){tex}\\Rightarrow{/tex}\xa0m.{a + (m - 1)d} = n.{a + (n - 1)d}{tex}\\Rightarrow{/tex}\xa0a.(m - n) + {(m2 - n2) - (m - n)} . d = 0{tex}\\Rightarrow{/tex}\xa0(m - n).{a + (m + n - 1)}d.{tex}\\Rightarrow{/tex}\xa0(m - n).Tm+n = 0{tex}\\Rightarrow{/tex}Tm+n = 0 [{tex}\\because{/tex}\xa0(m-n){tex}\\neq{/tex}0].Hence, the (m + n)th term is zero. | |
| 2017. |
If a, b, c, d, e, f are in A.P., then find the value of e - c |
| Answer» According to the question, we have\xa01st term = a and common difference = 3{tex}\\therefore{/tex}\xa0b = a + 3, c = a + 2(3) = a + 6d = a + 3(3) = a + 9,e = a + 4(3) = a + 12Now, e - c = (a + 12) - (a + 6) = 6 | |
| 2018. |
Find the sum of 3 digit number which are not Divisiable by 7 |
| Answer» | |
| 2019. |
The product of two successive integral multiples by5is300 determine the multiple |
| Answer» Let the successive multiples of 5 be 5x and 5x + 5. Then according to question we have,\xa05x {tex}\\times{/tex} (5x + 5) = 300{tex}\\Rightarrow{/tex}\xa025x2 + 25x = 300{tex}\\Rightarrow{/tex}25x2 + 25x - 300 = 0{tex}\\Rightarrow{/tex}25(x2 + x - 12) = 0{tex}\\Rightarrow{/tex}x2 + x - 12 = 0Solve by factorization method we have,x2 + 4x - 3x - 12 = 0{tex}\\Rightarrow{/tex}x(x + 4) - 3(x + 4) = 0{tex}\\Rightarrow{/tex}(x + 4)(x - 3) = 0Therefore, either x = -4 or x = 3 If x = -4, we have 5x = 5 {tex}\\times{/tex} (-4) = -20 and 5x + 5 = 5 {tex}\\times{/tex} (-4) + 5 = -15 Aslo If x = 3, we have 5x = 5 {tex}\\times{/tex} 3 = 15 and 5x + 5 = 5 {tex}\\times{/tex} 3 + 5 = 20 Hence, the required multiples of 5 are 15, 20. or\xa0-20, -15 | |
| 2020. |
How to prove root 2 irrational |
| Answer» Suppose\xa0{tex}\\sqrt2{/tex}\xa0is a rational number. That is ,\xa0{tex}\\sqrt2{/tex}\xa0=\xa0{tex}\\frac{p}{q}{/tex}\xa0for some p{tex}\\in{/tex}Z and q {tex}\\in{/tex}Z.\xa0We can assume the fraction is in lowest fraction, That is p and q shares no common factors.Then {tex}\\sqrt2q=p{/tex}\xa0Squaring both side we get,\xa0{tex}2q^2=p^2{/tex}So\xa0{tex}p^2{/tex}\xa0is a multiple of 2,let\'s assume\xa0{tex}p=2m{/tex}\xa0Then,\xa0{tex}2q^2=\\left(2m\\right)^2{/tex}\xa0{tex}2q^2=4m^2{/tex}Or {tex}q^2=2m^2{/tex}So {tex}q^2{/tex}\xa0is a multiple of 2,{tex}\\therefore{/tex} q is multiple of 2Thus p and q shares a common factor.This is contradiction.{tex}\\Rightarrow {/tex}{tex}\\sqrt { 2 }{/tex}\xa0is an irrational number. | |
| 2021. |
How sin is equal tp cos |
| Answer» | |
| 2022. |
4x-5 |
| Answer» | |
| 2023. |
Prove that sin theata /1-1/sintheata |
| Answer» LHS = {tex}\\sqrt { \\frac { 1 + \\sin \\theta } { 1 - \\sin \\theta } } + \\sqrt { \\frac { 1 - \\sin \\theta } { 1 + \\sin \\theta } }{/tex}{tex}= \\sqrt { \\frac { ( 1 + \\sin \\theta ) } { ( 1 - \\sin \\theta ) } \\times \\frac { ( 1 + \\sin \\theta ) } { ( 1 + \\sin \\theta ) } }{/tex}+\xa0{tex}\\sqrt { \\frac { ( 1 - \\sin \\theta ) } { ( 1 + \\sin \\theta ) } \\times \\frac { ( 1 - \\sin \\theta ) } { ( 1 - \\sin \\theta ) } }{/tex}{tex}= \\sqrt { \\frac { ( 1 + \\sin \\theta ) ^ { 2 } } { 1 - \\sin ^ { 2 } \\theta } } + \\sqrt { \\frac { ( 1 - \\sin \\theta ) ^ { 2 } } { 1 - \\sin ^ { 2 } \\theta } }{/tex}{tex}= \\sqrt { \\frac { ( 1 + \\sin \\theta ) ^ { 2 } } { \\cos ^ { 2 } \\theta } } + \\sqrt { \\frac { ( 1 - \\sin \\theta ) ^ { 2 } } { \\cos ^ { 2 } \\theta } }{/tex}\xa0{tex}[\\because sin^2\\theta+cos^2\\theta=1]{/tex}{tex}= \\frac { 1 + \\sin \\theta } { \\cos \\theta } + \\frac { 1 - \\sin \\theta } { \\cos \\theta }{/tex}\xa0{tex}= \\frac { 1 + \\sin \\theta + 1 - \\sin \\theta } { \\cos \\theta }{/tex}{tex}= \\frac { 2 } { \\cos \\theta }{/tex}=\xa0{tex}2sec\\theta{/tex}= RHS | |
| 2024. |
Their are any important question which will come in class 10 board exam in2018 |
|
Answer» solve cbse sample paper Check it on official website of CBSE |
|
| 2025. |
2+3-536+ |
| Answer» | |
| 2026. |
If the point A (K+1,2k), B (3K,2k+3) and C (5K-1,5K) are collinear, then find the value of K |
| Answer» Given points will be collinear, if area of the triangle formed by them is zero.Area =\xa0{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}\\Rightarrow 0 = \\frac { 1 } { 2 } {/tex}[(k + 1)(2k + 3 - 5k) + 3k(5k - 2k) + (5k - 1) (2k - 2k - 3)]{tex}\\Rightarrow {/tex}\xa00 = (k + 1)(3 - 3k) + 3k(3k) +(5k - 1)(-3){tex}\\Rightarrow {/tex}\xa00 =\xa03k - 3k2\xa0+ 3 - 3k + 9k2\xa0- 15k + 3{tex}\\Rightarrow {/tex}0 = 6k2\xa0- 15k + 6{tex}\\Rightarrow {/tex}0 = 2k2\xa0- 5k + 2{tex}\\Rightarrow {/tex}0 = 2k2\xa0- 4k - k + 2\xa0{tex}\\Rightarrow {/tex}0 = 2k(k - 2) - 1(k - 2){tex}\\Rightarrow {/tex}0 = (2k - 1)(k - 2){tex}\\Rightarrow {/tex}2k - 1 = 0 or k - 2 = 0{tex}\\Rightarrow k = \\frac { 1 } { 2 }{/tex} Or k = 2 | |
| 2027. |
Under root 660 whole square + 660 whole square |
| Answer» | |
| 2028. |
Under root 660 is whole square + 660 whole square |
| Answer» | |
| 2029. |
Which value of n in 2^n× 5^n make the unit place as 5 |
| Answer» {tex}\\begin{array}{l}(2\\times5)^{\\mathrm n}=2^{\\mathrm n}\\times5^{\\mathrm n}\\end{array}{/tex}{tex}\\text{=10}^n{/tex}{tex}\\text{If n=0 then 10}^0\\text{=1}{/tex}{tex}\\text{If n>0 then 10}^n\\text{ will end with 0 }{/tex}{tex}\\mathrm{If}\\;\\mathrm n<0\\;\\mathrm{then}\\;10^{\\mathrm n}\\;\\mathrm{ends}\\;\\mathrm{with}1\\;(\\mathrm e.\\mathrm g.\\;0.1,0.01,0.001){/tex}Hence for all values of n, {tex}2^n\\times 5^n{/tex} can never end with 5. | |
| 2030. |
Px^2-2√5px+15=0 |
| Answer» We have, px2\xa0-\xa0{tex}2\\sqrt{5}{/tex}px + 15 = 0\xa0Since the roots of equation are equal. Then,Discriminant, D = 0{tex}\\Rightarrow{/tex}\xa0({tex}2\\sqrt{5}{/tex}p)2\xa0- 4(p)(15) = 0{tex}\\Rightarrow{/tex}\xa020p2\xa0- 60p = 0{tex}\\Rightarrow{/tex}\xa020p(p -3) = 0{tex}\\Rightarrow{/tex}\xa020p = 0 or p - 3 = 0{tex}\\Rightarrow{/tex}\xa0p = 3 [since p\xa0{tex}\\neq{/tex}\xa00] | |
| 2031. |
What is the volome of wire? |
| Answer» V = πr2L, where "r" is the wire radius and "L" is its length.Remembering that radius is half of diameter, the expression becomes\xa0V = (πd2L)/4. | |
| 2032. |
Please tell some ncert important question of ch trigonometry |
| Answer» Solve the last exercise | |
| 2033. |
Statics |
| Answer» | |
| 2034. |
Mid point theorem of triangle |
| Answer» | |
| 2035. |
Prove that S12 =3 (S8-S4) |
|
Answer» So simple On Lhs. 3(s8-s4)=3(s4)=s12So, s12=s12OkOk So simple On Lhs. 3(s8-s4)=3(s4)=s12So, s12=s12Ok |
|
| 2036. |
On +de + fear +kf+6545 = |
| Answer» | |
| 2037. |
Areas related to circle |
| Answer» | |
| 2038. |
If a+c+e =0 and b+d=0 then find zeroes of ax4+bx3+cx2+dx+e |
| Answer» | |
| 2039. |
Chapter. Circle |
| Answer» Related with tangent | |
| 2040. |
If sin theta= 2/5 , then find the value of 4+4cot² theta. |
| Answer» | |
| 2041. |
If sin theta=2/5, then find the value of 4+4tan² theta. |
| Answer» Sin A= 2/5 it means P = 2 and Hypotenuse = 5 . So base = √52\xa0-22\xa0= √25-4 = √21. So tan A= P/B = 2/√21So 4+4tan2A = 4+4(2/√21)2 = 4+4*4/21 = 4+16/21 = 84+16/21 = 100/21 | |
| 2042. |
If tan30\' -- |
| Answer» | |
| 2043. |
√3 is rational number?explain |
| Answer» | |
| 2044. |
2x2+2x+5=0 |
| Answer» | |
| 2045. |
Similar triangle had same ratio of similar sides |
| Answer» Yes | |
| 2046. |
How to find area of segment when angle of theta is 110? |
| Answer» | |
| 2047. |
When we used the formula b2-4ac and when we used b2+-(b2-4ac)/2a |
|
Answer» Hmko ye kuch v samaj ni aa rha kya h yeee It is =0.5748 |
|
| 2048. |
Give me All the formulas in chapter surface area and Volume accept the volume |
| Answer» (a+b)sq==????(a-b)sq==??.?? | |
| 2049. |
The sum of 4 th and 8th term is 24 and the sum of 6th and 10 th term is 44 .Find AP |
| Answer» Let the first term and the common difference of the AP be a and d respectively.Then,4th term = a + (4 - 1)d = a + 3d\xa0{tex}\\because {a_n} = a + (n - 1)d{/tex}8th term = a + (8 - 1)d = a + 7d\xa0{tex}\\because {a_n} = a + (n - 1)d{/tex}6th term = a + (6 - 1)d = a + 5d\xa0{tex}\\because {a_n} = a + (n - 1)d{/tex}and 10th term = a + (10 - 1)d = a + 9d\xa0{tex}\\because {a_n} = a + (n - 1)d{/tex}According to the question,4th term + 8th term = 24{tex} \\Rightarrow {/tex}\xa0(a + 3d) + (a + 7d) = 24{tex} \\Rightarrow {/tex}\xa02a + 10d = 24{tex} \\Rightarrow {/tex}\xa0a + 5d = 12 .......... (1) (Dividing throughout by 2)6th term + 10th term = 24{tex} \\Rightarrow {/tex}\xa0(a + 5d) + (a + 9d) = 44{tex} \\Rightarrow {/tex}\xa02a + 14d = 44{tex} \\Rightarrow {/tex}\xa0a + 7d = 22 .......... (2) (Dividing throughout by 2)Solving (1) and (2), we geta = -13d = 15So, First term = -13Second term = -13 + 5 = -8Third term = -8 + 5 = -3Hence, the first three terms of the given AP are -13, -8 and -3. | |
| 2050. |
X+1/2 +y-1/2=4 find x and y |
| Answer» | |