Explore topic-wise InterviewSolutions in .

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.

10851.

Prove that sum of angle of triangle is180

Answer»
10852.

What is discriminate and their formula

Answer» B2-4ac
D is used to denote discriminate and I think its formula is d=x-A
10853.

49log10

Answer» 49log10 = 49*1= 49
Your school is MPboard or not
10854.

Find the sum of all 11 terms of an A. P whose middle most term is 30?

Answer» Let a be the first term and d be the common difference of the given A.P.Clearly, in\xa0an A.P. consisting of 11 terms,\xa0{tex} \\left( \\frac { 11 + 1 } { 2 } \\right) ^ { t h }{/tex}\xa0i.e. 6th term is the middle term.{tex}\\text{ it is given that the middle term =30}{/tex} So a+5d=30 .....(1).{tex}S_{11}=\\frac{11}{2}(2a+10d){/tex}= 11(a+5d) But a+5d=30 from (1)Hence S\u200b\u200b\u200b\u200b\u200b\u200b11\xa0= 11 × 30= 330
10855.

Find missing number in series 4,13,31,58--------

Answer» the answer is 94
the miss number for fraction equivalent 4÷13=
it is 944+9=1313+18=3131+27=5858+36=94
10856.

Sn=n(n+1).....find A20

Answer»
10857.

If x =secA + sinA and y= secA - sinA then prove (2 /x+y)+(x-y/2)=1

Answer»
10858.

If secA– tanA = a &. Cosec A+ cotA = b , then show that ab+a–b+1=0.

Answer»
10859.

Tignametry

Answer»
10860.

(m)square +(m –)s

Answer»
10861.

If sin (A-B)=1/2 COS (A+B)=1/2 FIND A AND B

Answer» Thanks this is very much helpful for me
As we know , Value of sin 30°= 1/2 Value of cos 60°= 1/2 Sin(A—B)= sin 30° A—B=30°---------------------------©Cos(A+B)=cos 60° A+B=60°---------------------------®From equation © and ® A+B=60° A–B=30° ______________ 2A= 90° A=45° &. B=45°
10862.

If thita,bita are zeroes of the polynomial axsquare bx+c,than evaluate thita - bita

Answer»
10863.

Area ratio them

Answer»
10864.

What is prime numbers

Answer» The number which are not divisible by any number except only one and own
Which not divisible by 2
Dnt know
10865.

HCF of 140

Answer» 140
Sorry 140
149
10866.

Tell us about all trigonometric ratios

Answer» Trigonometry is very important chapter for competitive exams Mainly it is used in the chapter calculus of class 11and12 Trigonometry mean Tri = threeGono = side Metry = measurement
10867.

Say about old sin theta

Answer»
10868.

Introduction to financial market

Answer»
10869.

2+4

Answer» Very tough question
yeh to aryabhatta ko hi Mallom hoga!!!
the to aryabhatta no hi Mallom hoga!!!!!
10870.

I want chapter wise question paper of all chapter

Answer»
10871.

How to find co-ordinate of co-ordinate geometry

Answer»
10872.

Which term of an AP 121,117,113,....... is its first negative term?

Answer» n<0 0=121+(n-1)(-4)-121=-4n+4-125=-4nn=125/4 =31.25So 32 will be the first negative term
10873.

Which term of AP 14,11,8.... is -1

Answer» 6
10874.

Graph of x^2-6x+9

Answer»
10875.

Complete the statement:Probability of event E + probability of event "not E"= _______

Answer» 1 or 1÷1
10876.

What is compliment event

Answer»
10877.

2%1.865490

Answer»
10878.

1/3x+y +1/3x-y=3/41/2 (3x+y)-1/2 (3x-y)=1/8

Answer»
10879.

Find the least positive integer which is divisible by first five natural number

Answer» 60
10880.

Sqare of any number

Answer» Multiply the number two timesEx 2×2=4
10881.

Ratio for unique solution ,infinitely many solution,no solution

Answer»
10882.

If sec theta×sin theta=0 ,find the value of theta

Answer» Angle 0
10883.

Sin65cos25cos65sin25=?

Answer»
10884.

8+3=

Answer» 8+3=11
11
11
10885.

1234567890×1234567890

Answer» 1494157345019043010
1.52415788e18
0
10886.

What is the new pattern of class 10th maths question paper

Answer»
10887.

In adjoining figure, find tan P-cot R

Answer»
10888.

What is the sum of nth term of an AP

Answer» It means last term.As for example- 5,8,9,7In this nth term is 7.
10889.

TP and TQ are tangents. TP=TQ. Angle PTQ=2angle OPQ.

Answer» Given A circle with centre O and an external point T and two tangents TP and TQ to the circle, where P, Q are the points of contact.To Prove: {tex}\\angle{/tex}PTQ = 2{tex}\\angle{/tex}OPQProof: Let {tex}\\angle{/tex}PTQ = {tex}\\theta{/tex}Since TP, TQ are tangents drawn from point T to the circle.TP = TQ{tex}\\therefore{/tex}\xa0TPQ is an isoscles triangle{tex}\\therefore{/tex}\xa0{tex}\\angle{/tex}TPQ = {tex}\\angle{/tex}TQP = {tex}\\frac12{/tex} (180o - {tex}\\theta{/tex}) = 90o - {tex}\\fracθ2{/tex}Since, TP is a tangent to the circle at point of contact P{tex}\\therefore{/tex}\xa0{tex}\\angle{/tex}OPT = 90o{tex}\\therefore{/tex}\xa0{tex}\\angle{/tex}OPQ = {tex}\\angle{/tex}OPT - {tex}\\angle{/tex}TPQ = 90o - (90o -\xa0{tex}\\frac12{/tex}{tex}\\theta{/tex}) = {tex}\\fracθ2{/tex}= {tex}\\frac12{/tex}{tex}\\angle{/tex}PTQThus, {tex}\\angle{/tex}PTQ = 2{tex}\\angle{/tex}OPQ
10890.

. In right angle triangle one acute is double The Other find the acute angles

Answer» Acute angles are 30 degree and 60 degree [Let one acute angle be x then other angle =2x By angle sum property x + 2x +90=1803x = 90x=302x = 60]
10891.

What is the construction and working of thermal power plant?

Answer»
10892.

If sec =17/8 ,verify that 3-4sin^2/4cos^2-3= 3-tan^2/1-3tan^2

Answer»
10893.

Chapter 6 thearom 6.5and 6.6

Answer»
10894.

Cos (cosx)+sin (cosx)what is least and greatest value?

Answer»
10895.

The decimal represention of 6/1250 will terminate after how many places of decimal

Answer» First we get prime factorization of 1250 we got 6/2×1×5×5×5×5 so it is terminating decimal no.
10896.

If cosec(3x -15°)=2 find x

Answer» Cosec(3x-15) =cosec30(. :cosec30=2) 3x=45. : x=15. answer
Cosec (3x-15)=2Cosec (3x-15)=cosec303x-15=303x=45Therefore x=45/3=15
X=15
10897.

Show that positive integer to be a perfect square. It should be of the form8k+1

Answer» Since, any odd positive integer n is of the form 4m\xa0+ 1 or 4m + 3.if n = 4m + 1n2 = (4m + 1)2= 16m2 + 8m + 1= 8(2m2 + m) + 1So n2 = 8q + 1 ........... (i)) (where q = 2m2 + m is a positive integer)If n = (4m + 3)n2 = (4m + 3)2= 16m2 + 24m + 9= 8(2m2 + 3m + 1) + 1So n2 = 8q + 1 ...... (ii) (where q = 2m2 + 3m + 1 is a positive integer)From (i) and (ii) we conclude that the square of an odd positive integer is of the form 8q + 1, for some integer q.
10898.

Show that positive integer to be a perfect square. It should be of the form 8k+1

Answer» Since, any odd positive integer n is of the form 4m\xa0+ 1 or 4m + 3.if n = 4m + 1n2 = (4m + 1)2= 16m2 + 8m + 1= 8(2m2 + m) + 1So n2 = 8q + 1 ........... (i)) (where q = 2m2 + m is a positive integer)If n = (4m + 3)n2 = (4m + 3)2= 16m2 + 24m + 9= 8(2m2 + 3m + 1) + 1So n2 = 8q + 1 ...... (ii) (where q = 2m2 + 3m + 1 is a positive integer)From (i) and (ii) we conclude that the square of an odd positive integer is of the form 8q + 1, for some integer q.
10899.

What type of decimal expantion does 29by 2square into 5 into 7have?

Answer»
10900.

How to express linear combination

Answer»