InterviewSolution
Saved Bookmarks
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.
| 34701. |
Congurent symbol |
| Answer» ≅ | |
| 34702. |
Area of squar |
|
Answer» Thanks Side into side |
|
| 34703. |
Find the zeroes of the following polynomial-5√5x^+30x+8√5 |
| Answer» The given quadratic polynomial is\xa0{tex}5 \\sqrt{5} x^{2}+30 x+8 \\sqrt{5}.{/tex}In order to factorize it, we have to find two numbers / and m such thatl + m = 30 and lm\xa0{tex}=5 \\sqrt{5} \\times 8 \\sqrt{5}=200{/tex}Clearly, 10 + 20 = 30 and 10\xa0{tex}\\times{/tex}20=200. Therefore, l = 10 and m = 20Now,\xa0{tex}5 \\sqrt{5} x^{2}+30 x+8 \\sqrt{5}{/tex}{tex}=5 \\sqrt{5} x^{2}+10 x+20 x+8 \\sqrt{5}{/tex}{tex}=\\left(5 \\sqrt{5} x^{2}+10 x\\right)+(20 x+8 \\sqrt{5}){/tex}{tex}=\\left(5 \\sqrt{5} x^{2}+10 x\\right)+(4 \\sqrt{5} \\times \\sqrt{5} x+8 \\sqrt{5}){/tex}{tex}=5 x(\\sqrt{5} x+2)+4 \\sqrt{5}(\\sqrt{5} x+2){/tex}{tex}=(5 x+4 \\sqrt{5})(\\sqrt{5} x+2){/tex}{tex}=\\sqrt{5}(\\sqrt{5} x+4)(\\sqrt{5} x+2){/tex} | |
| 34704. |
Find distance between pair of points: (2,3),(4,1) |
|
Answer» 2√2 2^2 2 roort 2 4√2 2√2 |
|
| 34705. |
Irrational prove karna |
| Answer» Kisko | |
| 34706. |
Maths circles ruls |
| Answer» | |
| 34707. |
Root 5 rational no. Hai ya nhi |
|
Answer» Nahi Nahi |
|
| 34708. |
75+89 |
|
Answer» Ye addition apse nahi huee Board exam main kya karoge???? 164 164 164 164 164 |
|
| 34709. |
Prove that root 5 is a rational number |
|
Answer» Let √5 is rational no.Let two integers[ p ] [ q ]√5=p/qWe consume wrong that √5is rational but,it is irrarional Root5 is an irrational number. |
|
| 34710. |
76+65.79 |
|
Answer» 141.79 141.79 141.799 141.79????? 141.79 |
|
| 34711. |
60/360 × 3.14 × 15 × 15 |
|
Answer» 117.75 117.75???? 117.75 |
|
| 34712. |
Construct an obtuse angle and bisect it |
| Answer» By using compass or ......??? | |
| 34713. |
What is snow flakes of Mathematics ??? ??? |
| Answer» | |
| 34714. |
Write an AP whose first term is 10 and common difference is 3 |
|
Answer» 10, 13, 16, 19, 22,............ The first term,a = 10common difference,d = 3So, the next two terms would be a +d, a +2d.That is, the next two terms are 13, 16, 19, 21,........ |
|
| 34715. |
For what the value of k ,10 , k-2 are in AP |
|
Answer» How Yes its 11 11 |
|
| 34716. |
Test of lesson 1 and 2 |
|
Answer» Matlab kya ha Ok |
|
| 34717. |
Is hindi a vocational subject? |
|
Answer» Hindi is not a vocational subject. Hindi is a main subject. No it\'s not a vocational subject. |
|
| 34718. |
What is the value of sin |
|
Answer» sin 30=1/2, sin 45 =1 / √2sin 60=\xa0√3/2 and sin 90 =1 Also sin=1/cosec Perpendicular /hypotenuse |
|
| 34719. |
If angle b and angle q are acute such that sinb =sin q then prove that angle b = angle q |
| Answer» When sinb=sinq So sin angle b=sin angle qThen angle b = angle q | |
| 34720. |
any girl talk to me please |
|
Answer» Kyu nhi Why?? |
|
| 34721. |
Solve for x√2^(2)+7x+5√2=0 |
|
Answer» -√2 answer -5√2-2/7 18root 2 answer |
|
| 34722. |
Sin60-cot45 |
|
Answer» √3-2/2 Sin60=√3/2Cot45 =1Sin60-cot45=√3/2-1After Taking LCM =(√3-2)/2 root 3/2 -1 ? |
|
| 34723. |
100-50 |
|
Answer» This app is for helping in a logic questions not that what is 100-50...faltu mat bhejo 50 |
|
| 34724. |
the circumference of a circle exceeds the daimeter by 16.8cm find the circumference of the circle |
|
Answer» 2 x 3.14 x r = 2r + 16.8r(6.28-2)=16.8r= 16.8/4.28So circumference = 2 x 3.14 x 16.8 / 4.28 = 24.65 cm\xa0 but how to solve this question 24.64 Edjfjxesidic |
|
| 34725. |
What is the HCF of smallest prime number and the smallest composite number? Help me friends....... |
|
Answer» Smallest prime number is 2Smallest composite number is 4H.C.F(2,4) is 2 1 |
|
| 34726. |
question paper design 2018-19 mathematics |
| Answer» Check Question Papers here :\xa0https://mycbseguide.com/cbse-question-papers.html | |
| 34727. |
Convert each of the following per cent into a fraction in the lowest term means?? |
|
Answer» Where is the percentage?? Plz answer quickly |
|
| 34728. |
Solve for x and y37x+43y=123 43x+ 37y=117 |
|
Answer» 37x + 43y = 123 ……..(1)43x + 37y = 117 ……..(2)Adding (1) and (2)80x + 80y = 24080(x + y) = 240x + y = Try with elimination method Ans 1 and 2 This will be done by addition and subtraction 1&2 |
|
| 34729. |
We can say that 1,1,1... is an AP |
|
Answer» yes Yes because the common diffrence is same t2-t1=1-1=0 Can say No |
|
| 34730. |
a cot theta + b cosec theta =p, b cot theta +a cosec theta =q, then find value of p^2 –q^2 |
|
Answer» a cotA + b cosecA =pb cotA + a cosecA=qso p2 -q2\xa0= (\xa0a cotA + b cosecA)2 - (b cotA + a cosecA)2=cot2A( a2-b2) +cosec2A (b2-a2) + 2ab cotA cosecA- 2ab cotA cosecA=( a2-b2) (cot2A-cosec2A)=\xa0( a2-b2)(-1)=b2-a2 a2 + b2 |
|
| 34731. |
Sec A + tanA =P, find the value of cosec A? |
| Answer» {tex}sec\\ \\theta+ tan\\ \\theta = p{/tex} ...(i)Also {tex}sec^2\xa0\\theta - tan^2 \\theta = 1{/tex}{tex}\\Rightarrow{/tex}\xa0(sec\xa0{tex}\\theta{/tex}\xa0- tan\xa0{tex}\\theta{/tex}) (sec\xa0{tex}\\theta{/tex}\xa0+ tan\xa0{tex}\\theta{/tex}) = 1{tex}\\Rightarrow{/tex}\xa0p(sec\xa0{tex}\\theta{/tex}\xa0-\xa0tan\xa0{tex}\\theta{/tex}) = 1[using equation (i)]{tex}\\Rightarrow{/tex}\xa0sec\xa0{tex}\\theta{/tex}\xa0-\xa0tan\xa0{tex}\\theta{/tex}\xa0{tex}=\\frac{1}{p}{/tex}\xa0...(ii)(ii) - (i) we get{tex}-2 tan{/tex}\xa0{tex}\\theta{/tex}\xa0{tex}=\\frac{1-p^{2}}{p}{/tex}{tex}\\Rightarrow{/tex}- tan\xa0{tex}\\theta{/tex}\xa0{tex}=\\frac{1-p^{2}}{2 p}{/tex}{tex}\\Rightarrow{/tex}- cot\xa0{tex}\\theta{/tex}\xa0{tex}=\\frac{2 p}{1-p^{2}}{/tex}cot\xa0{tex}\\theta{/tex}\xa0{tex}=\\left(\\frac{2 p}{1-p^{2}}\\right)^{2}{/tex}{tex}=\\frac{-4 p^{2}}{\\left(1-p^{2}\\right)^{2}}{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}cosec^2{/tex}\xa0{tex}\\theta{/tex}\xa0- 1\xa0{tex}=\\frac{-4 p^{2}}{\\left(1-p^{2}\\right)^{2}}{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}cosec^2{/tex}\xa0{tex}\\theta{/tex}\xa0{tex}=\\frac{-4 p^{2}}{\\left(1-p^{2}\\right)^{2}}+1=\\frac{-4 p^{2}+\\left(1-p^{2}\\right)^{2}}{\\left(1-p^{2}\\right)^{2}}{/tex}{tex}cosec^2{/tex}\xa0{tex}\\theta{/tex}\xa0{tex}=\\frac{-4 p^{2}+1+p^{4}-2 p^{2}}{\\left(1-p^{2}\\right)^{2}}{/tex}{tex}=\\frac{\\left(1+p^{2}\\right)^{2}}{\\left(1-p^{2}\\right)^{2}}{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}cosec\\\xa0\\theta{/tex}\xa0{tex}=\\frac{1+p^{2}}{1-p^{2}}{/tex} | |
| 34732. |
Cos theta + sin theta =root 2 cos theta prove that cos theta - sin theta =root 2 sin theta |
| Answer» | |
| 34733. |
1/a+1/b+1/y=1/a+b+ya≠0 b≠0 & y=-a+b |
| Answer» What find | |
| 34734. |
How to draw perpendicular in right triangle ABC on BC? |
| Answer» Draw altitude AD on BC. | |
| 34735. |
find the length of median AD if vertices are a(1,-1) B (-4,6) C (-3,-5 ) |
|
Answer» Firstly find mid point of BC by mid point formula, then find length of AD by distance formula. (1;-1) |
|
| 34736. |
Hi friends.... I am new here .,can u all help me in my doubts |
|
Answer» Yes Ofcourse ? Yes ?? Ok, Osama Bin Laden Zinda Hein?? Ofcourse Yes☺️ |
|
| 34737. |
In an AP is the sum of its first n terms is 3n^2+5n and its kth term is 164 find the value of k. |
|
Answer» S = 3n2\xa0+ 5n\xa0S1\xa0= a1\xa0= 3 + 5 = 8\xa0S2\xa0= a1\xa0+ a2= 12 + 10 = 22 ⇒ a2\xa0=\xa0S2\xa0-\xa0S1\xa0= 22 - 8 = 14\xa0S3\xa0= a1\xa0+ a2\xa0+ a3 = 27 + 15 = 42⇒a3\xa0=\xa0S3\xa0-\xa0S2\xa0= 42 - 22 = 20\xa0∴ Given AP is 8,14,20,.....\xa0Thus a = 8 , d = 6 Given tk= 164.\xa0164 = [a + (k -1)d]\xa0164 = [(8) + (k-1)6]\xa0164 = [8 + 6k - 6]\xa0164 = [2 + 6k]162= 6k, k= 162 / 6.\xa0∴ k = 27 |
|
| 34738. |
7, 10, 8, 11, 9, 12, ... What number should come next? |
|
Answer» 10 will be the next no. Here uh can use the formula an=a+(n-1)d 10 15 10 10 |
|
| 34739. |
Q.2. 3,5, 9,17,33,.....? |
|
Answer» 65 65 can be found by same formula 65 67 65 32 |
|
| 34740. |
How I find last 5 years question papers which are came in bord exams |
| Answer» Buy oswal sample papers book | |
| 34741. |
How to solve 576=60n-2nsquare-576 |
| Answer» | |
| 34742. |
Why the area of cylinder is added in examplé 14 of chapter surface area and volume |
|
Answer» Because the metal sheet is also used in making the cylinder. ncert ,rs aagrawal ya rd sharma book ya koi others books |
|
| 34743. |
Writa 10 th term of ap 2,3333 |
|
Answer» a=2d=3333-2=333110th term=?10th term=a+9d " " =2+9×3331 " " =2+29979 " " =29981So, the 10th term of the APs=29981 29981 |
|
| 34744. |
If diameter of a circle is 7cm. Find perimeter of semi circle |
|
Answer» 11 D=7,r=7/2 Perimeter =2πr/2 =πr=22/7*7/2=22/2=11 11 Circumference of semicircle ___ 1/2*2TT r |
|
| 34745. |
Which term of the AP :121,117,113,......................., is its first negative term |
|
Answer» Please solve this step by step 32nd term |
|
| 34746. |
what we will use in application of trignometrey what we take tan,cot,sin,cos,sec,cosec |
| Answer» It depends on angle given in figure If you have to find hypotenuse then you can use both sin and cos For base,use cos tan For perpendicular, use sin, tan | |
| 34747. |
Explain how to solve with completing the square method |
|
Answer» Steps\tStep\xa01 Divide all terms by a (the coefficient of x2).\tStep\xa02 Move the number term (c/a) to the right side of the equation.\tStep\xa03\xa0Complete the square\xa0on the left side of the equation and balance this by adding the same value to the right side of the equation (1/2.x)2 |
|
| 34748. |
ABC is an equilateral triangle D is a point on BC such that BD=1/3BC then prove that 9AB^2 = 7AD^2 |
|
Answer» Sorry i can not draw figure here\xa0Just drop perpendicular AE from A to BC\xa0suppose side of triangle is 6x thenAD=2x and DE=x and AB = 3x and BE=3xso AE2= AB2 -BE2= (6x)2 -(3x)2 =36x2-9x2 =27x2Now in triangle ADEAD2=AE2 + DE2 =27 x2\xa0+ x2\xa0 = 28 x2AD2\xa0= 28/36 *36x2 = 28/36 *AB2 = 7/9*AB27 AD2\xa0= 9 AB2\xa0\xa0\xa0\xa0 Sorry , show that 9AD^2 = 7AB^² |
|
| 34749. |
X/a +y/b=a+b |
| Answer» | |
| 34750. |
Find the quadratic equations the sum and product of whose zeroes are -3and 2,respectively |
| Answer» x^2+3x+2 | |