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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.
| 36701. |
What is the formulae of circumference of a circle |
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Answer» 2πR 2π r 2 pi r only correct pi r square is the area of circle 2 pie r I don\'t know how to type pie symbol Pi r square 2 pie r |
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| 36702. |
1/(cosecA--cotA) --1/sinA =1/sinA -- 1/(cosecA+cotA) |
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Answer» For this question you have to change the location of LHS and RHS..... LHS:= 1/(CosecA-CotA) - 1/(cosecA+CotA) =CosecA+CotA+CosecA-CotA/Cosec^2A - Cot^2A=2cosecARHS:=1/sinA + 1/sinA=2/sinA=2CosecAProved!!!! 1 |
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| 36703. |
What do you mean by incenter |
| Answer» Point where all angle bisectors meet within a triangle | |
| 36704. |
The ratio od sum of n terms of two A.P.s is (7n +1):(4n+27).Find the ratio of \'m\'th terms. |
| Answer» Given ratio of sum of n terms of two AP’s = (7n+1):(4n+27)....(1)Let’s consider the ratio these two AP’s mth terms as am : a’m →(2)The nth term of AP formula, an = a + (n – 1)dHence equation (2) becomes,am : 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 {tex}\\frac{14m-6}{8m+23}{/tex} | |
| 36705. |
Wetage of all chapter |
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| 36706. |
After how many decimal places will the decimal expansion of 147 upon 120 terminate |
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Answer» 3 places of decimal120=2*2*2*3*5147*3/2*2*2*3*5Now, 147/2*2*2*5147*5*5/2*2*2*5*5*5147*25/2³*5³3675/(2*5)³3675/10003.675 Convert it into the form of 5^m ×2^n..You will get the answer |
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| 36707. |
Cbse board the maths exam will very difficult or not |
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Answer» Prepare well, do well and other leave on teachers on teachers, it will be really easy. Don\'t know... No... Yes it will very difficult |
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| 36708. |
Can we use black pen in board exam please tell me anybody?????? |
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Answer» by cbse rule you should carry only Blue/Royal blue ball point/Gel and fountain pens only you can use No harsh ....only blue pen use in exam ,this is written in admit card Please |
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| 36709. |
What does D means in trignometry |
| Answer» | |
| 36710. |
Find hcf of 225 and 925 |
| Answer» 25 | |
| 36711. |
chapter-13 all formulae |
| Answer» Volume of cylinde | |
| 36712. |
Plzz tell me any important sums in linear equations in two variables.?? |
| Answer» | |
| 36713. |
6x+4=10 6x+3y=2 Find elimination method |
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Answer» If it is 4y thenSubtract 6x+3y=2 from 6x +4y =10 In this y will be 8 then put y value in any equation. U will get x is it 6x + 4 = 10 or 6x + 4y = 10 |
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| 36714. |
2(sin⅔ + cos⅔) - 3(sin⁴ + cos⁴) + 1 |
| Answer» Btao | |
| 36715. |
Is math paper consite only ncert or others for exam target for getting 80 above |
| Answer» | |
| 36716. |
1+cot square alpha/1+cosec alpha= cosec alpha |
| Answer» LHS\xa0{tex} = 1 + \\frac{{{{\\cot }^2}\\theta }}{{1 + \\cos ec\\theta }}{/tex}{tex} = 1 + \\frac{{\\cos e{c^2}\\theta - 1}}{{1 + \\cos ec\\;\\theta }}{/tex}\xa0{tex}\\left[ {\\because {{\\cot }^2}\\theta = \\cos e{c^2}\\theta - 1} \\right]{/tex}{tex} = 1 + \\frac{{(\\cos ec\\theta + 1)(\\cos ec\\theta - 1)}}{{(1 + \\cos ec\\theta )}}{/tex}\xa0{tex}\\left[ {\\because {a^2} - {b^2} = (a + b)(a - b)} \\right]{/tex}{tex} = 1 + \\cos ec\\;\\theta - 1{/tex}{tex} = \\cos ec\\theta {/tex}= RHSHence proved. | |
| 36717. |
Plz give me notes of construction |
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| 36718. |
(2,3) (4,1) |
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| 36719. |
find the centre of circle passing through the points (6 -6), (3 -7) and (3 3). |
| Answer» U can find by mid point formula | |
| 36720. |
secA+tanA=p find the value of cosecA |
| Answer» Psq. + 1 upon psq._ 1 | |
| 36721. |
If sec theta + tan theta = P find value of cosec theta |
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Answer» LHS1/cosA + sinA/cosA =p1+sinA/cosA =p1+sinA=pcosA1+sinA=p√1-sin sqA Squaring both side(1+sinA )sq = p sq. (1 -sin sq.A)(1+sinA) (1+sinA) =p sq. (1-sin sq.A)(1+sinA) (1+sinA)/(1-sinA)(1+sinA) =p sq. {:.a sq.+b sq =(a+b)( a-b)}(1+sinA)/(1-sinA) = p sq.(1+sinA) =p sq.( 1 - sin A) 1 + sinA = p sq. -p sq sinASinA + p sq.sinA = p sq-1 sin A (1 + p sq.) = p sq-1 Sin A = (p sq. -1)/(p sq+1) CosecA = (p sq.+1) / (p sq.-1) {sinA = 1/ cosecA} H.P SecA+tanA =p |
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| 36722. |
If A,B and C are interior angles of triangle ABC,then show that cos(B+C/2)=sinA/2 |
| Answer» A+B+C = 180°B+C=180°-AB+C/2=90°-A/2Cos(B+C/2)=cos(90°-A/2)Cos(B+C/2)=sin(A/2)Hence proved | |
| 36723. |
A boy observes that the angle of elevation |
| Answer» Let O be the position of the bird and B be the position of the boy. Let FG be the building and G be the position of the girl.In {tex}\\triangle{/tex}OLB,{tex}\\frac { OL } { B O }= \\sin 30 ^ { \\circ }{/tex}{tex}\\Rightarrow \\quad \\frac { O L } { 100 } = \\frac { 1 } { 2 }{/tex}{tex}\\Rightarrow{/tex}\xa0OL = 50 mOM = OL - ML= OL - FG= 50 - 20 = 30 mIn {tex}\\triangle{/tex}OMG{tex}\\frac {OM}{OG}=sin45°=\\frac{1}{√2}{/tex}OG = OM√2 = 30 √2 = 42.3 meter | |
| 36724. |
Construct a trinagle with sides 5 cm 6cm and 7 cm then another triangle whos sides 7/5 of thia |
| Answer» To construct: To construct a triangle of sides 5 cm, 6 cm and 7 cm and then a triangle similar to it whose sides are of\xa0{tex}\\frac{7}{5}{/tex} the corresponding sides of the first triangle.Steps of construction:\tDraw a triangle ABC of sides 5 cm, 6 cm and 7 cm.\tFrom any ray BX, making an acute angle with BC on the side opposite to the vertex A.\tLocate 7 points B1, B2, B3, B4, B5, B6 and B7 on BX such that BB1 = B1 B2 = B2 B3 = B3 B4 = B4 B5 = B5 B6 = B6 B7.\tJoin B5 C and draw a line through the point B7, draw a line parallel to B5 C intersecting BC at the point C\'.\tDraw a line through C\' parallel to the line CA to intersect BA at A\'.\tThen, A\'BC\' is the required triangle.\tJustification :\t{tex}\\because CA||C\'A\'{/tex} [By construction]\t{tex}\\therefore {/tex}\xa0{tex}\\triangle ABC \\sim \\triangle A\'BC\'{/tex}\xa0[AA similarity]\t{tex}\\therefore \\frac{{A\'B}}{{AB}} = \\frac{{A\'C\'}}{{AC}} = \\frac{{BC\'}}{{BC}}{/tex}\xa0[By Basic Proportionality Theorem]\t{tex}\\because {B_7}C\'||{B_5}C{/tex}\xa0[By construction]\t{tex}\\therefore \\triangle B{B_7}C\' \\sim \\triangle B{B_5}C{/tex}\xa0[AA similarity]\tBut {tex}\\frac{{B{B_5}}}{{B{B_7}}} = \\frac{5}{7}{/tex}\xa0[By construction]\tTherefore, {tex}\\frac{{BC}}{{BC\'}} = \\frac{5}{7} \\Rightarrow \\frac{{BC\'}}{{BC}} = \\frac{7}{5}{/tex}\t{tex}\\therefore \\frac{{AB}}{{AB}} = \\frac{{A\'C\'}}{{AC}} = \\frac{{BC\'}}{{BC}} = \\frac{7}{5}{/tex} | |
| 36725. |
If m term is 1/n and n term is 1/m ,find mn term |
| Answer» Answer is 1 | |
| 36726. |
What are basics and important concepts in class 10 1st chapter Real Numbers |
| Answer» Proof of irrational, and show that any positive integer can be written in 3m, 3m + 1, these type of question are important | |
| 36727. |
I m the last year CBSE topper so guys u ask any question ❓❓❓ |
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| 36728. |
O is any point inside a rectangle ABCD .prove that OB2+OA2+OC |
| Answer» I don\'t know | |
| 36729. |
How can we rectify that wether we have to find the csa or tsa of a solid |
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| 36730. |
In elimination method when we to add or subtract |
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| 36731. |
If X=2sin , Y=2cos +1then find the value of x+y |
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| 36732. |
Mathematica |
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| 36733. |
Prove that √7 be irrational |
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Answer» Sry yrr glti se 7 ki jgh 3 ho gya aap 7 kr lena 3 ki jgh baki same ese hi h Let us assume to the contrary ,that √3 is rational.That is,we can find integers a and b(#0 )such that√3=a/b.Suppose a and b have a common factor other tha 1,then we can divide by the common factor , and assume that a and b are coprime.So,b√3=aSquaring both sides,a2 is divisible by 3 that a is also divisible by 3So,we can write a=3c for some integer c.Substituting for a,we get 3b2= 9c2.This means that b2 is divisible by 3,and so b is also divisible by 3 ..a and b have at least 3 as a common factor.But this contradicts the fact that a and b are coprime ..This contradiction has arisen because of our incorrect assumption that √3 is rational..So,we conclude that √3 is irrational..?? |
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| 36734. |
Is verification in construction necessary |
| Answer» No if asked then write the way of construction | |
| 36735. |
How to prove pythagoras therom ? |
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Answer» one side of triangle = sum of other two sides Use ncert bro |
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| 36736. |
Step of construction is needed or not? |
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Answer» Very imp needed Yes in 4 and 3 marks question Yes ; its provide u full marks |
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| 36737. |
find the peobability that a leap year has 53 sumdays |
| Answer» 2\\7 | |
| 36738. |
Pgt therom |
| Answer» Use rs aggarwal bro | |
| 36739. |
SARCASM AHEAD!The most important question that will come in exam...prove that 1 2 ka 4 and 4 2 ka 1 |
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| 36740. |
Yeh linear combination kya hota hai....??Anybody... |
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| 36741. |
Tangent drawn to a circle is 90 degree angle proof?? |
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Answer» Let there be any other point on tangent that would be perpendicular...but its distance from center is greater than that of point of contact. This will be the case with any other point. Thus distance between center and point of contact of tangent is shortest. Hence it is perpendicular. Theorem 10.1 pg.208 |
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| 36742. |
200logs are thousand in following number 20logs in the box |
| Answer» Answer is 16 | |
| 36743. |
Which of the followingHas the sum of its roots is 3? |
| Answer» ?...incomplete question! | |
| 36744. |
135 ka angle kisko bisect krke banta hai?? |
| Answer» Pehle 150° ka angle banae aata h Aata h toh uske baad 120° aur 150° ke beech ek arc cut karo phir tumhara 135° ka angle ban jayega I hope this will help u | |
| 36745. |
What is unit digit of 9 raise to power n |
| Answer» 9? | |
| 36746. |
Let the three no are x-1,x-7 ,x-7 are in ap find the value of x |
| Answer» Correct ur ques plz | |
| 36747. |
For what value of k so that kx+3y=k-3 & 12k+ky=k infinite solution |
| Answer» 0 | |
| 36748. |
Two coins are tossed simultaneously. Find the probability of exactly one head |
| Answer» When two coins are tossed simultaneouslyTotal number of outcomes = {HH, HT, TH, TT}Total number of outcomes = 4Favourable outcomes = {HT, TH} = 2Probability of getting exactly one head ={tex}\\frac { 2 } { 4 } = \\frac { 1 } { 2 }{/tex} | |
| 36749. |
Bpt theorem? |
| Answer» | |
| 36750. |
Chapter 10 theorem 10.2 |
| Answer» | |