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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 |
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| 10852. |
What is discriminate and their formula |
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Answer» B2-4ac D is used to denote discriminate and I think its formula is d=x-A |
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| 10853. |
49log10 |
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Answer» 49log10 = 49*1= 49 Your school is MPboard or not |
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| 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-------- |
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Answer» the answer is 94 the miss number for fraction equivalent 4÷13= it is 944+9=1313+18=3131+27=5858+36=94 |
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| 10856. |
Sn=n(n+1).....find A20 |
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| 10857. |
If x =secA + sinA and y= secA - sinA then prove (2 /x+y)+(x-y/2)=1 |
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| 10858. |
If secA– tanA = a &. Cosec A+ cotA = b , then show that ab+a–b+1=0. |
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| 10859. |
Tignametry |
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| 10860. |
(m)square +(m –)s |
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| 10861. |
If sin (A-B)=1/2 COS (A+B)=1/2 FIND A AND B |
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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° |
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| 10862. |
If thita,bita are zeroes of the polynomial axsquare bx+c,than evaluate thita - bita |
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| 10863. |
Area ratio them |
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| 10864. |
What is prime numbers |
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Answer» The number which are not divisible by any number except only one and own Which not divisible by 2 Dnt know |
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| 10865. |
HCF of 140 |
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Answer» 140 Sorry 140 149 |
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| 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 |
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| 10868. |
Introduction to financial market |
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| 10869. |
2+4 |
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Answer» Very tough question yeh to aryabhatta ko hi Mallom hoga!!! the to aryabhatta no hi Mallom hoga!!!!! |
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| 10870. |
I want chapter wise question paper of all chapter |
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| 10871. |
How to find co-ordinate of co-ordinate geometry |
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| 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 |
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| 10875. |
Complete the statement:Probability of event E + probability of event "not E"= _______ |
| Answer» 1 or 1÷1 | |
| 10876. |
What is compliment event |
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| 10877. |
2%1.865490 |
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| 10878. |
1/3x+y +1/3x-y=3/41/2 (3x+y)-1/2 (3x-y)=1/8 |
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| 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 |
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| 10882. |
If sec theta×sin theta=0 ,find the value of theta |
| Answer» Angle 0 | |
| 10883. |
Sin65cos25cos65sin25=? |
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| 10884. |
8+3= |
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Answer» 8+3=11 11 11 |
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| 10885. |
1234567890×1234567890 |
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Answer» 1494157345019043010 1.52415788e18 0 |
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| 10886. |
What is the new pattern of class 10th maths question paper |
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| 10887. |
In adjoining figure, find tan P-cot R |
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| 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? |
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| 10892. |
If sec =17/8 ,verify that 3-4sin^2/4cos^2-3= 3-tan^2/1-3tan^2 |
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| 10893. |
Chapter 6 thearom 6.5and 6.6 |
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| 10894. |
Cos (cosx)+sin (cosx)what is least and greatest value? |
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| 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 |
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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 |
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| 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? |
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| 10900. |
How to express linear combination |
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