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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.
| 34851. |
Is class 10 maths board exam standered level paper will be from ncert book |
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Answer» 100% 75 percent |
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| 34852. |
Is sal ncert me se kitna ane wala ha |
| Answer» Tuzhe jo ata hai vo | |
| 34853. |
2*6 |
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Answer» So tough question ??? 12 12 |
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| 34854. |
How solve ap sums easily |
| Answer» By applying the formulas seeing the question wisely | |
| 34855. |
boat and stream problem |
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Answer» This one is important question and you get ans in textbook in example boat and stream problem |
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| 34856. |
What is formula of major sector in terms of π,r, theta,sin,cos, and theta. |
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Answer» Also area of major segment= area of circle - area of minor segment For area of segment you need to find the area of sector and area of triangle formed in circle with chord. And subtract them. Sorry not area of sector i mean to say segment on Area of sector = (theta/360° )×(πr^2) |
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| 34857. |
I have lots of problems while solving AP. I get confused ? that which is\' n\' and which is an |
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Answer» Thanks you ☺ N=no.of termsan=last term Sn= sum of termsD=common difference ? \'n \' means number of terms in a series and \'an\' is the specification to a particular nth term. Also an can be considered as last term having n terms.Example : please come ahead the fourth student from of the line having 12 students. In this question n=12 ( total number of students)an =a4 that mean 4th student. |
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| 34858. |
Fhdghdh |
| Answer» | |
| 34859. |
More than ogive and less than ogive for given data with mean = 23 meet at (20,22) . Find mode |
| Answer» | |
| 34860. |
The ratio of the sum of n terms of two A.P\'s is (7n+1) : (4n+27) find the ratio of their mth terms |
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Answer» Sn=n/2(2A+(n-1)D)/n/2(2a+(n-1)d=7n+1/4n+272A+(n-1)D/2a+(n-1)d=7n+1/4n+27N=22A+(17-1)D/2a+(17-1)d=120/952A+16D/2a+16d=120/95A+8D/a+8d=120/95Ratio 24:19 Given ratio of sum of n terms of two AP’s = (7n+1):(4n+27)Let’s consider the ratio these two AP’s mth terms as am : a’m\xa0Recall the nth term of AP formula, an = a + (n – 1)dHenceam : a’m = a + (m – 1)d : a’ + (m – 1)d’On multiplying by 2, we getam : a’m = [2a + 2(m – 1)d] : [2a’ + 2(m – 1)d’]= [2a + {(2m – 1) – 1}d] : [2a’ + {(2m – 1) – 1}d’]= S2m – 1 : S’2m – 1= [7(2m – 1) + 1] : [4(2m – 1) +27] [from (1)]= [14m – 7 +1] : [8m – 4 + 27]= [14m – 6] : [8m + 23]Thus the ratio of mth terms of two AP’s is [14m – 6] : [8m + 23] |
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| 34861. |
What os the formula for major segment |
| Answer» Area of circle -Area of minor segment | |
| 34862. |
Sin30-2tan45+cos60/sin45cos45+2sin30cos60 |
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Answer» Sin60 3tan30+cos90/sin60tan45+2cos45tan60 Nfldngakm |
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| 34863. |
The probability of the getting 53 Sunday in non leap year |
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Answer» The answer is 1/52 bcoz there is only one odd day in 52weeks and accordig to that equation favourable outcomes /total no of outcomes=1/52 or 1/365 A non leap year has 365 days or 52 weeks and1 odd day , the odd day can be sunday, monday, tuesday, wednesday, thursday, friday, saturday ,So there are 7 possiblities out of which 1 is favourable,So the probablity of 53 sunday in non leap year is 0 because in a non leap year there will only 52 sunday and probability of impossible outcome is 0 In a non-leap year there will be 52 Sundays and 1day will be left.\xa0This 1 day can be Sunday, Monday, Tuesday, Wednesday, Thursday,friday,Saturday, Sunday.Of these total 7 outcomes, the favourable outcomes are 1.\xa0Hence the probability of getting 53 sundays =\xa01 / 7. |
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| 34864. |
use Euclid division of algorithm to find the HCF of 867 &255 |
| Answer» H.C.F=51 | |
| 34865. |
(1+sinA/1-sinA) |
| Answer» | |
| 34866. |
Ap main sirf nth term,sum of first n term hee pucha jatavh kya |
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| 34867. |
AP kitne no.ka atta hai |
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Answer» 9 or 10 marks 10 no ka |
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| 34868. |
Which book has the most no of chance that its question can come in exam |
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Answer» Ncert Algebra and maths For maths element Always ncert NECERT |
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| 34869. |
Write the value of cot square theta minus 1 by sin squared theta |
| Answer» 1 | |
| 34870. |
If sec¢ + tan¢ = p, show that p²-1 ÷ p²+1 = sin¢ |
| Answer» We have,{tex}\\mathrm { LHS } = \\frac { p ^ { 2 } - 1 } { p ^ { 2 } + 1 } = \\frac { ( \\sec \\theta + \\tan \\theta ) ^ { 2 } - 1 } { ( \\sec \\theta + \\tan \\theta ) ^ { 2 } + 1 }{/tex}{tex}\\Rightarrow \\quad \\mathrm { LHS } = \\frac { \\sec ^ { 2 } \\theta + \\tan ^ { 2 } \\theta + 2 \\sec \\theta \\tan \\theta - 1 } { \\sec ^ { 2 } \\theta + \\tan ^ { 2 } \\theta + 2 \\sec \\theta \\tan \\theta + 1 }{/tex}{tex}\\Rightarrow \\quad \\mathrm { LHS } = \\frac { \\left( \\sec ^ { 2 } \\theta - 1 \\right) + \\tan ^ { 2 } \\theta + 2 \\sec \\theta \\tan \\theta } { \\sec ^ { 2 } \\theta + 2 \\sec \\theta \\tan \\theta + \\left( 1 + \\tan ^ { 2 } \\theta \\right) }{/tex}{tex}\\Rightarrow \\quad \\mathrm { LHS } = \\frac { \\tan ^ { 2 } \\theta + \\tan ^ { 2 } \\theta + 2 \\sec \\theta \\tan \\theta } { \\sec ^ { 2 } \\theta + 2 \\sec \\theta \\tan \\theta + \\sec ^ { 2 } \\theta }{/tex}{tex}\\Rightarrow \\quad \\mathrm { LHS } = \\frac { 2 \\tan ^ { 2 } \\theta + 2 \\tan \\theta \\sec \\theta } { 2 \\sec ^ { 2 } \\theta + 2 \\sec \\theta \\tan \\theta } = \\frac { 2 \\tan \\theta ( \\tan \\theta + \\sec \\theta ) } { 2 \\sec \\theta ( \\sec \\theta + \\tan \\theta ) } = \\frac { \\tan \\theta } { \\sec \\theta } = \\frac { \\sin \\theta } { \\cos \\theta \\cdot \\sec \\theta } = \\sin \\theta{/tex}=RHS | |
| 34871. |
2sin A =cos A |
| Answer» Karna kya hi isme | |
| 34872. |
What is the hcf of 3sq3 ×5 and 3sq3×5sq2 |
| Answer» 15 sq3 | |
| 34873. |
(2x-3) (3x+1) |
| Answer» (2x -3) (3x + 1)= 2x (3x + 1) - 3(3x + 1)= 6x2 + 2x - 9x - 3= 6x2 - 7x - 3 | |
| 34874. |
Prove that sin square tita plus cos square tita is equals to one |
| Answer» as per puthgoras\xa0theoren\xa0in right angle triangle:base2\xa0+ height\xa02\xa0= hypo 2devide\xa0both sides by hypo2(base / hypo)2 +( height/ hypo)2 =1sin2\xa0Θ + cos\u200b\u200b\u200b\u200b\u200b\u200b\u200b2\xa0Θ\u200b\u200b\u200b\u200b\u200b\u200b\u200b = 1\xa0\xa0 | |
| 34875. |
Formula of hemisphere |
| Answer» A hemisphere is the half part of a sphere. We can find many of the real-life examples of the hemispheres such as our planet Earth can be divided into two segments the southern & northern hemispheres and much more.Curved surface area of hemisphere = 2πr2total surface area= curved surface area + area of the base circle = 3πr2Volume-Hemisphere = 2/3 π r3 | |
| 34876. |
how many questions have in board exams |
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| 34877. |
Which book is best for maths class 10 |
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Answer» RD sharma NCERT and R.S AGRAWAL Xam idea R.s aggrawal and Ncert book. Ncert and Evergreen self study R.s agarwal |
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| 34878. |
Tan A+ Sin A=m and tan A- SinA = n Prove that (m2-n2)2=16 mn |
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| 34879. |
Theorem 6.4 |
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Answer» Ch 6 Dghgfg |
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| 34880. |
0 divide 0 =2 |
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Answer» Are you kidding ?? 0/0100-100/100-100(10) square-(10) square/10(10-10)(10-10)(10+10)/10(10-10)10+10/1020/102 answer It is infinity Wrong question How . |
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| 34881. |
How to solve qestions simply |
| Answer» Eaaaaaaasy | |
| 34882. |
If the mean of 1,2,3......,x is 6x/11,find the value of x. |
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Answer» How Answer is 11 |
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| 34883. |
Cosec theta + cot theta equal to one by cosec theta minus cot theta prove it |
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| 34884. |
What angle subtended at the centre of a circle of radius 6cm by an arc of length 3π cm? |
| Answer» | |
| 34885. |
Tanget at any point of a circle is perpendicular to the radius through the point of contact |
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Answer» Mr.Vansh that is theorem 10.1 This is theorm 10.2 of math |
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| 34886. |
Distance between the point p(a+b, a-b) q(a-b, a+b) |
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Answer» Let distance between joining of two points (x1,y1) and (x2, y2)= √(x2-x1)²+(y2-y1)²(x1,y1) = P(a+b ,a-b)(x2, y2) = Q(a-b, a+b)PQ =√[a-b -(a+b)]²+ [a+b-(a-b)]²= √(a-b-a-b)²+ (a+b-a+b)²=√(-2b)²+ (2b)²=√4b²+4b²=√8b²=√(2*2)*2*b*b=2b√2 Ab |
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| 34887. |
Rhombus shape |
| Answer» A Rhombus is a flat shape with 4\xa0equal\xa0straight sides.A rhombus looks like a diamond | |
| 34888. |
Why maths is so difficult |
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Answer» It is not so difficult but less practice make it It is npt difficult, people make it difficult Make it more interesting |
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| 34889. |
In a triangle ABC, write cos (b+c/2) in terms of angle A |
| Answer» Sum of three angles of triangle is\xa0{tex}180^{\\circ}{/tex}A + B + C = 180°or, B + C = 180° - ADivide both sides by 2{tex}\\frac{B+C}{2}=\\frac{180^{\\circ}}{2}-\\frac{A}{2}{/tex}{tex}\\frac{B+C}{2}=90-\\frac{A}{2}{/tex}{tex}\\therefore \\quad \\cos \\left( \\frac { B + C } { 2 } \\right) {/tex}{tex}= \\cos \\left( 90 - \\frac { A } { 2 } \\right){/tex}{tex}= \\sin \\frac { A } { 2 }{/tex} | |
| 34890. |
Why the solution of trignomentry are solve by trignometric formula |
| Answer» Because trigonometry can be solved by trigonometry and not by the theorems of triangles | |
| 34891. |
Difference between nth term and last term |
| Answer» nth term is the number of terms in AP series and the last term is the last term of AP series | |
| 34892. |
Proof that circumference/ diameter=π |
| Answer» 2πr/d=2πr/2r=π | |
| 34893. |
If P(E) =0.05,what is the probability of \'not E\' |
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Answer» 1-0.05=0.95. P(not e)=1-P(E) P(E)+P(not E)=1 Therefore , 0.05+ P(not E)=1P(not E ) =1-0.05= 0.95 |
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| 34894. |
Find the largest number which divides 2053 and 967 and leaves a remainder of 5 and 7 respectively |
| Answer» 2053 leaves remainder 5So,2053-5=2048Now, 967 leaves remainder 7So,967-7=960The no. Are 2048 and 960And HCF of 2048 and 960 is 64Thereforev the largest no. Which is divisible by 2053 and 967 and leaves remainder 5 and 7 is 64 | |
| 34895. |
Prove that sinΦ-cosΦ+1/sinΦ+cosΦ-1=1/secΦ-tanΦ, |
| 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. | |
| 34896. |
What is the shape of point |
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Answer» I think it has no shape because it has no diametre and no length and height I guess it\'s circular |
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| 34897. |
Solve by factorization method:-9x^2-9(a+b)x+(2a^2+5ab+2b^2)=0 |
| Answer» 9x^2-9 (a+b)x+(2a^2+5ab+2b^2)=0= 9x^2-3(a+2b)x-3(2a+b)x+(a+2b)(2a+b)=0= 3x[3x-(a+2b)]-(2a+b)[3x-(a+2b)]=0= x= a+2b/3 and x= 2a+b/3 | |
| 34898. |
What is the formula of object distance |
| Answer» f = focal lengthu = object distancev = image distance1/f = 1/u + 1/v | |
| 34899. |
When two lines are parllel then what thing happen |
| Answer» There corresponding angles are equal , there alternare interior angle are equal and alternate exterior angle are equal | |
| 34900. |
If I\'m the term of an ap is 2n + 1 what is the sum of its first 3 terms |
| Answer» n = 1 2(1)+1=3 n=2 2(2)+1=5 a=3 d=2n=3Ab sum vala formula laga | |