InterviewSolution
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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.
| 1801. |
Pyhtagoraes theoram |
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Answer» Not pythogoras theorem but it\'s proof is important for examination point of view Hypoteneous ^2=perpendicular ^2+base^2 |
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| 1802. |
What is difference between angle of elevation and angle of depression???? |
| Answer» The angle at which we see something from bottom to top is called angle of elevationEg- when we see the building from the ground. The angle at which we see something from top to bottom is called angle of depressionEg- when we see the ground from the building. | |
| 1803. |
If SecA + TanA =P,then SecA in terms of P is |
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Answer» (psquare+1)/2p P-TanA |
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| 1804. |
Can u plz rell me marking scheme of mathematics |
| Answer» | |
| 1805. |
prove that. tanΦ+sinΦ \\ tanΦ-sinΦ = secΦ+1 \\ secΦ-1 |
| Answer» after simplifying the lhs we get sin+sincos /sin-sincosAfter simplifying rhs we get 1+cos/1-cosNow take sin common from lhs and cancel it then it will be proved lhs=1+cos/1-cos=rhs | |
| 1806. |
Is sum of irrational number always irrational number |
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Answer» No Yss...... |
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| 1807. |
Area related to circle |
| Answer» | |
| 1808. |
What is the formula of cos |
| Answer» B/h | |
| 1809. |
How to find the sqare root of12.25 |
| Answer» | |
| 1810. |
If sinA+cosA=√2,then evaluate tanA+cotA |
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Answer» let A=45,sin45+cos45=1/root2+1/root2,root2 common,so1+1/root2=》2/root2 BT RATIONALISATION,2/root2 ×root2/root2 =2*root2/2=root2these tells A=45tanA+cotA=tan45+cot45 =》1+1=2hence tanA+cotA=2 SinA + cosB =\xa0{tex}\\sqrt2{/tex}squaring both sides,1+2sinA*sinB = 2sin2A = 1{tex}\\Rightarrow{/tex} 2A = 90°A=45°tanA + cotA = tan45° + cot45°= 1 + 1 = 2 (ans) |
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| 1811. |
Explain any three methods for conservation of water resources |
| Answer» Do not waste water. Repair leaky tapsHarvest rain water. | |
| 1812. |
How can I find vertices of triangle if coordinates of mid points are given |
| Answer» | |
| 1813. |
What is theeta |
| Answer» | |
| 1814. |
Name the line drawn from the eye of an observer to the point in the object viewed by the observer |
| Answer» line of sight | |
| 1815. |
Find the median of the following wages |
| Answer» | |
| 1816. |
If mean of 1,2,3, _ _ _ _ x is 6 |
| Answer» | |
| 1817. |
Can you please tell me the syllabus of maths |
| Answer» | |
| 1818. |
CosA÷1- tanA + sinA÷ 1- cotA |
| Answer» The answr of this question | |
| 1819. |
Why centroid is denoted by g |
| Answer» | |
| 1820. |
What is "Pi" |
| Answer» | |
| 1821. |
Cos2 A+sin2A |
| Answer» =1 | |
| 1822. |
1386- 346.5 |
| Answer» 1039.5 | |
| 1823. |
Equation of origin in co ordinate geometry |
| Answer» | |
| 1824. |
Area of rhombus |
| Answer» 1/2*product of its diagonals | |
| 1825. |
What is the marking scheme of board paper |
| Answer» You can check marking scheme here in the syllabus :\xa0https://mycbseguide.com/cbse-syllabus.html | |
| 1826. |
The volume of hemisphere is 4851/2 . Find its curved surface area. |
| Answer» Volume of the sphere = 4851{tex}\\Rightarrow \\frac { 4 } { 3 } \\pi r ^ { 3 } = 4851{/tex}{tex}\\Rightarrow 4 \\times \\frac { 22 } { 7 } \\times r ^ { 3 } = 4851{/tex}{tex}\\Rightarrow r ^ { 3 } = \\frac { 4851 \\times 7 } { 88 }{/tex}{tex}\\Rightarrow r ^ { 3 } = \\sqrt [ 3 ] { \\frac { 4851 \\times 7 } { 88 } }{/tex}{tex}\\Rightarrow r ^ { 3 } = \\sqrt [ 3 ] { 385.875 }{/tex}{tex}\\Rightarrow{/tex}\xa0r = 7.28 cm3Curved surface area of a sphere =\xa0{tex}4 \\pi r ^ { 2 }{/tex}{tex}= 4 \\times \\frac { 22 } { 7 } \\times ( 7.280 ) ^ { 2 }{/tex}{tex}= \\frac { 4663.8592 } { 7 }{/tex}= 666.2656 cm2 | |
| 1827. |
(2x+1/x+1)^4-10(2x+1/x+1)^2+9=0 |
| Answer» | |
| 1828. |
Two dice are rolled what Is the probability that the product is less than 9? |
| Answer» 4/9 is the right or wrong | |
| 1829. |
Please show me the figure of question 9 of ch 13.3 |
| Answer» | |
| 1830. |
2+2=5 prove That |
| Answer» 2+2=4+9/2-9/2 (4-9/2)sq.ka sq. Root+9/2(16+81/4-36)sq.root+9/2(81/4-20)sq . root +9/2(81/4+25-45)sq.root +9/2(5-9/2) square kasq. Root +9/25-9/2+9/25proved | |
| 1831. |
what happens to the value of tan theta when theta increases from 0` to 90 |
| Answer» Value of tan theta increases | |
| 1832. |
(a-b)x^2 +(b-c)x+(c-a)=0 has equal roots. prove that 2a=b+c. |
| Answer» Use the relation, Bsq. - 4ac=0 | |
| 1833. |
Plz solve question number 31 of RS Agarwal of exersice 19 B |
| Answer» | |
| 1834. |
Sum of 3rd and 7th term of A.P. is 6 and product is 8. Find the sum of First 16 terms. |
| Answer» Here according to question the sum of the 3rd and 7th terms of an A.P. is 6 and their product is 8. We have to\xa0find the sum of the first 20 terms of the A.P.According to question,\xa0a3 + a7= 6 (where a3,a7 represents 3rd\xa0and 7th\xa0terms)And\xa0,a3{tex}\\times{/tex}\xa0a7\xa0=8{tex}{/tex}.Now ,a3= a+2d\xa0and a7= a+6d, are 3rd\xa0\xa0and 7th terms respectively.So,(a+2d)+(a+6d)=6{tex}\\implies{/tex} 2a + 8d = 6{tex}\\implies{/tex}2 (a + 4d)= 6.{tex}\\implies a+4d=\\frac {6 }{2}{/tex}{tex}\\implies a=3-4d{/tex}. And ( a3×a7) = (a + 2d)(a + 6d) = 8.......(i).Substituting the value of a = (3 - 4d) in (i) we get( 3 - 4d + 2d)(3- 4d + 6d) =8{tex}\\Rightarrow{/tex}(3 + 2d)(3 - 2d) = 8{tex}\\Rightarrow{/tex} 9 - 4d2 = 8{tex}\\therefore{/tex}4d2\xa0= 1,d2\xa0=\xa0{tex}\\frac 14{/tex}d =\xa0{tex}\\pm\\frac 12{/tex}Case (i):When\xa0d =\xa0{tex}\\frac 12{/tex}S20\xa0=\xa0{tex}\\frac n2{/tex}[2a+ (n - 1)d]{tex}\\implies{/tex}S20{tex}= \\frac { 20 } { 2 } \\left[ 2 + \\frac { 19 } { 2 } \\right]{/tex}S20\xa0= 115Now, Case(ii):When\xa0d = -{tex}\\frac 12{/tex}S20\xa0=\xa0{tex}\\frac { 20 } { 2 } \\left[ 2 \\times 5 + 19 \\times \\left( - \\frac { 1 } { 2 } \\right) \\right]{/tex}{tex}= 10 \\left[ 10 - \\frac { 19 } { 2 } \\right] {/tex}{tex}=5{/tex}\xa0 | |
| 1835. |
Real numbers Prime factorisation 7529 |
| Answer» | |
| 1836. |
How i can get print out of sylabus and other material |
| Answer» Yes, you can print syllabus from here :\xa0https://mycbseguide.com/cbse-syllabus.html | |
| 1837. |
Has anyone downloaded all sample papers of cbse guide of anyone subject |
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Answer» yes me Yes |
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| 1838. |
Explain checker board |
| Answer» | |
| 1839. |
Date of Annual examination 2017-18 |
| Answer» Sterts from 9 march 2018 and finished on 10 april 2018 | |
| 1840. |
11×11=4 12×12=913×13=?14×14=? |
| Answer» 16 & 25. | |
| 1841. |
ex 1.1 que 5 |
| Answer» | |
| 1842. |
My math is weak , how i strong? Please tell. |
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Answer» Do the homework given to you and practise previous exercises daily Daily practice with full attention and always solving previous year question paper |
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| 1843. |
Prove that 1 +1 =2 |
| Answer» 1 + 1 = 2 2 = 2Therefore , 1 + 1 = 2 | |
| 1844. |
Sin 3€+cos2€ |
| Answer» | |
| 1845. |
Sin^2A +cos^2A=1 |
| Answer» | |
| 1846. |
Find the area of an angle a° of a circle with radius R |
| Answer» Area of sector =\xa0{tex}\\frac { \\theta } { 360 } \\times \\pi r ^ { 2 }{/tex}where\xa0{tex}\\theta{/tex}= angle, r = radius of circleHere, we have{tex}\\theta {/tex}= p and radius = RPutting these in formula,Area of sector =\xa0{tex}= \\frac { p } { 360 } \\times \\pi R ^ { 2 }{/tex} | |
| 1847. |
Kiya class 10 ki board datesheet net pr aa gaye h |
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Answer» No No |
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| 1848. |
To verify that sum of first n natural number is n(n+1)By 2 by graphical method |
| Answer» sum of n natural nos are 1+2+3+4+.......+.na = 1d=1we know that Sn=n/2(2a+(n-1)d)Sn=n/2(2+n-1) =\xa0{tex}n(n+1) \\over 2{/tex} | |
| 1849. |
Find the smallest number that if divided by 7,9,11 leaves a remainder of 1,2, and 3 respectively |
| Answer» | |
| 1850. |
√2 is irrationalb |
| Answer» | |