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.
| 39751. |
Plz divide 3 by 7 |
|
Answer» Its not completely divisible...Answer is :-0.4285 approx 0.428571428571428 |
|
| 39752. |
Prove that Thales theorem |
| Answer» Fast | |
| 39753. |
What is the compound interest of p=18000 for 2.5years at 10% per annum |
| Answer» By using year by year calculationS.I. on Rs. 18000 at\xa010%\xa0per annum for 1 year{tex} = \\frac{{18000 \\times 10 \\times 1}}{{100}} = Rs.1800{/tex}∴ Amount at the end of 1st year= Rs. 18000 + Rs. 1800= Rs. 19800= Principle for 2nd year.S.I. on Rs. 19800 at\xa010% per annum for 1 year{tex} = \\frac{{19800 \\times 10 \\times 1}}{{100}}{/tex}= Rs. 1980∴ Amount at the end of 2nd year= Rs. 19800 + Rs. 1980= Rs. 21780= Principle for 3rd yearS.I. on Rs. 21780 at\xa010%\xa0per annum for {tex}\\frac{1}{2}{/tex}\xa0year{tex} = \\frac{{21780 \\times 10 \\times 1}}{{2 \\times 100}}{/tex}= Rs. 1089∴ Amount at the end of\xa0{tex}2\\frac{1}{2}{/tex}\xa0years= Rs. 21780 + Rs. 1089\xa0= Rs. 22869this is the required amount.Now,C.I. = Rs. 22869 – Rs. 18000= Rs. 4869. | |
| 39754. |
How to represent three upon four on the number line. |
|
Answer» Divide 0-1into four equal parts and third part will denote3/4 Question Thats not the answer for his queston Hello ayush |
|
| 39755. |
Give some tips about 10th CBSE board papers |
|
Answer» Nice thought Radhika Write clearly and properly , if you don\'t know the correct answer than write anything related to that chapter , if u writing any definition than u have to write it\'s formula also, lastly if you don\'t know the correct answer so you can get marks if your formula is correct. That\'s all |
|
| 39756. |
The angle of cyclic quadrilateral ABCD are: |
|
Answer» sum of opposite angles = 180sum of all the interior angles= 360 360 |
|
| 39757. |
To prove sin^4 + cos ^4 = 1 |
| Answer» LhsSin4 +cos4(Sin2)2+(cos2)2(sin2 +cos2) 2(1)21Rhs1 Verified | |
| 39758. |
Raja dashrath ram ko uvraj kyu banana chahte the |
| Answer» kyuki ram dashrath ke bade putra the | |
| 39759. |
In an AP the sum of 1st n term Sn=3n^2/2+5n/2 find 25th term |
| Answer» According to the question,we are given that,\xa0{tex}S _ { n } = \\frac { 3 n ^ { 2 } } { 2 } + \\frac { 5 n } { 2 } = \\frac { 3 n ^ { 2 } + 5 n } { 2 }{/tex}{tex}\\Rightarrow S _ { n - 1 } = \\frac { 3 ( n - 1 ) ^ { 2 } + 5 ( n - 1 ) } { 2 }{/tex}{tex}= \\frac { 3 \\left( n ^ { 2 } - 2 n + 1 \\right) + 5 n - 5 } { 2 }{/tex}{tex}= \\frac { 3 n ^ { 2 } - 6 n + 3 + 5 n - 5 } { 2 }{/tex}{tex}= \\frac { 3 n ^ { 2 } - n - 2 } { 2 }{/tex}Now,nth term = Tn=Sn-Sn-1={tex}= \\frac { 3 n ^ { 2 } + 5 n } { 2 } - \\frac { 3 n ^ { 2 } - n - 2 } { 2 } = \\frac { 3 n ^ { 2 } + 5 n - 3 n ^ { 2 } + n + 2 } { 2 }{/tex}={tex}\\frac{{6n + 2}}{2}{/tex}=3n+125th term=T25=3(25)+1=75+1=76. | |
| 39760. |
Percatage of 7/8 |
| Answer» 87.5% | |
| 39761. |
Ch8trignomatry |
|
Answer» Spelling : Trigonometry Chapter -8 |
|
| 39762. |
RD sharma solutions of ch 2 |
| Answer» All but how | |
| 39763. |
Wah wah wah |
| Answer» | |
| 39764. |
1+2-3×4÷5 |
|
Answer» 0.6 0.6 2.4 0 3/5 7/12 712 |
|
| 39765. |
What is a unique\u200b solution , coincident solution and parellel??? |
|
Answer» Unique solution is a single solution is there for the quadratic equation and it intersects in graph . Coincident is that the graph is a line which cannot intersects and always coincides . Parallel is that the graph has two lines which is parallel to eachother when we get a single solution of different variable than thisbis called as unique solution. |
|
| 39766. |
Solution of 5th question of maths exercise 5.2 |
|
Answer» Now we find that the common difference d=13-7=6 , d=6 And first term A = 7 And , last term An = 205 By using formula we have An = a + (n -1) d = = 205 = 7 + (n -1) 6 = ( n -1 ) 6 = 205 - 7 = n - 1= 198/6 = n = 33 + 1 = 34 Ans Hence , 34 th term is 205 I hope this is the helpful for u .......... a=7,d=6,an=205an=a+(n-1)×d205=7+(n-1)×6205=7+6n-6205=1+6n6n=205-16n=204n=204÷6n=34My name is sarthak Bhardwaj |
|
| 39767. |
Class 7 chapter -6 6.3 que no 2 |
| Answer» | |
| 39768. |
Write all the formulas of chapter 3 linear equations in two variables. Anyone please?? |
| Answer» Graphical method substitution elimination method cross multiply method | |
| 39769. |
The first term of AP is -13, the last term is 13 find the common difference |
| Answer» It is wrong question sum bhi de rakha hai | |
| 39770. |
Which term of the AP 48,43,38,33,....,is it\'s first negative term |
| Answer» 11th term will be negative (-2) | |
| 39771. |
Who is Pythagoras |
|
Answer» Pythagoras was a great mathematician who modified the theirem which is given by baudhayan fro india . This theorem is also known as baudhayan Pythagoras theorem Pythagoras is mathamathesion .He give theorem base square + perpendicular square= hypotenuse square |
|
| 39772. |
5+-4y+8=0 |
| Answer» | |
| 39773. |
0=2f-40÷44+f |
| Answer» F=20 | |
| 39774. |
9573÷474 |
| Answer» 20.196 | |
| 39775. |
CosA-sinA+1/cosA+sinA-1=1/cosecA-cotA |
| Answer» | |
| 39776. |
Prove that the sum of infinity is -1/12 |
| Answer» | |
| 39777. |
How to prove AAA similarity theorem? |
|
Answer» What if it comes in exam? Jai bhole ki?? Kkh Dont do that It is not imp. |
|
| 39778. |
A~b mean |
|
Answer» A is similar to B It means triangle A is similar to triangle B |
|
| 39779. |
2,34,84,68,______. |
|
Answer» Now You tell me how 130 come 2=1ki power 6 +134=2ki power 5 +284=3ki power 4 +368=4ki power 3 +4Similarly5ki power 2 +5=30 How come 30 is he right ans . Can u tell me Right answer is 30 130 |
|
| 39780. |
8.3 |
| Answer» | |
| 39781. |
Find the distance bw (h,k) where for x axis |
| Answer» The distance of the point (h, k) from x-axis i.e. distance between (h,k) and (x,0) is{tex}Distance=\\sqrt{(x-h)^2 +(0-k)^2}{/tex}\xa0[By using distance formula]{tex}\\Rightarrow Distance=\\sqrt{x^2+h^2-2xh+k^2}{/tex}{tex}\\Rightarrow Distance=\\sqrt{x^2+h^2+k^2-2xh}{/tex} | |
| 39782. |
If sec 5A = cosec (A +30) what is the value of A |
|
Answer» Value of A=10 10 A= 10 |
|
| 39783. |
2 + 5 |
|
Answer» 7 7 7 |
|
| 39784. |
32x4_500 |
| Answer» -372 | |
| 39785. |
Two consucative no. 265 |
| Answer» | |
| 39786. |
A + b ka hole square |
|
Answer» Bhai a+b me hole kaise kiya (a+b)2\xa0= a2\xa0+ b2 + 2ab a2+ b2+2 ab |
|
| 39787. |
If tan theta +1/tantheta =5,find sin theta and cos theta |
| Answer» tanQ+1/tanQ=5tan^2Q+1/tanQ=5Sec^2Q/tanQ=5CosQ/cos^2Q×sinQ=5CotQ=5 | |
| 39788. |
Parallegoram prove square |
|
Answer» And have to prove all sides and angles is 90 as parallelogram s opposite sides are equal it is a hint for u You have to do construction in it Copy le aa |
|
| 39789. |
If (x+a) is a factor of 2x^2+2ax+5x+10 then find a |
| Answer» p(x) = 2x²+2ax+5x+10 ............(i),....................take, x+a = 0 =› x = -asince x+a is a factor of (i) .............................therefore, .........................................................p(-a) = 0 ............................................................=› 2(-a)² + 2a(-a)+5(-a)+10 = 0 ......................=› 2a² - 2a² - 5a + 10 = 0 .................................=› -5a = -10 =› a = 2 | |
| 39790. |
If sin Tita-cos tita=√2 prove that cos tita+sin tita =√2 |
| Answer» | |
| 39791. |
Express cot 85°+cos45°in term of trigonometric ratio |
| Answer» | |
| 39792. |
solve for x =1/2a+b+2x=1/2a+1/b+1/2x,x. 0 |
| Answer» {tex}\\frac{1}{2a + b + 2x}{/tex}\xa0=\xa0{tex}\\frac{1}{2a}{/tex}\xa0+\xa0{tex}\\frac{1}{b}{/tex}\xa0+\xa0{tex}\\frac{1}{2x}{/tex}\xa0{tex}\\Rightarrow{/tex}\xa0{tex}\\frac{1}{2a + b + 2x}{/tex}\xa0-\xa0{tex}\\frac{1}{2x}{/tex}\xa0=\xa0{tex}\\frac{1}{2a}{/tex}\xa0+\xa0{tex}\\frac{1}{b}{/tex}\xa0{tex}\\Rightarrow{/tex}{tex}\\frac { 2 x - 2 a - b - 2 x } { ( 2 a + b + 2 x ) ( 2 x ) }{/tex}\xa0=\xa0{tex}\\frac{b + 2a}{2a \\times b}{/tex}\xa0{tex}\\Rightarrow{/tex}\xa0{tex}\\frac { - ( 2 a + b ) } { ( 2 a + b + 2 x ) 2 x }{/tex}\xa0=\xa0{tex}\\frac{b + 2a}{2ab}{/tex}{tex}\\Rightarrow{/tex}{tex}\\frac { - 1 } { 4 a x + 2 b x + 4 x ^ { 2 } }{/tex}\xa0=\xa0{tex}\\frac{1}{2ab}{/tex}\xa0{tex}\\Rightarrow{/tex}\xa0{tex}4x^2 + 2bx + 4ax = -2ab{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}4x^2 + 2bx + 4ax + 2ab = 0{/tex}\xa0{tex}\\Rightarrow{/tex}\xa0{tex}2x(2x + b) + 2a(2x + b) = 0{/tex}{tex}\\Rightarrow{/tex}\xa0(2x + b)(2x + 2a) = 0{tex}\\Rightarrow{/tex}\xa0x = -{tex}\\frac{b}{2}{/tex} or x = -a | |
| 39793. |
Find the area of triangle with A(1,-4) and mid point of sides through being (2,-1)and (0,-1). |
| Answer» Let E be the midpoint of AB.{tex}\\therefore \\quad \\frac { x + 1 } { 2 } = 2{/tex}\xa0or x = 3and\xa0{tex}\\frac { y + ( - 4 ) } { 2 } = - 1{/tex} or, y = 2or, B(3, 2)Let F be the mid-point of AC.Then,{tex}0=\\frac{x_1+1}{2}{/tex}\xa0or\xa0{tex}x_1=-1{/tex}and {tex}\\frac { y _ { 1 } + ( - 4 ) } { 2 }{/tex}\xa0= -1 or, y1\xa0= 2or, C= (-1, 2)Now the co-ordinates are A(1, - 4), B(3,2), C (-1,2)Area of triangle{tex}= \\frac { 1 } { 2 } \\left[ x _ { 1 } \\left( y _ { 2 } - y _ { 3 } \\right) + x _ { 2 } \\left( y _ { 3 } - y _ { 1 } \\right) + x _ { 3 } \\left( y _ { 1 } - y _ { 2 } \\right) \\right]{/tex}{tex}= \\frac { 1 } { 2 } [ 1 ( 2 - 2 ) + 3 ( 2 + 4 ) - 1 ( - 4 - 2 ) ]{/tex}{tex}= \\frac { 1 } { 2 } [ 0 + 18 + 6 ]{/tex}= 12 sq units. | |
| 39794. |
One line prove it 180 |
| Answer» | |
| 39795. |
Show 4cos geometrically |
| Answer» | |
| 39796. |
How I will get good marks in exams |
|
Answer» Do study before 2month of exarm and you are topper of 2019 arun kumar national topper of CBSE Board 2019. Understand the concept and try to recollect the points.Create ur strategy and follow it.Have patience. Do hard work Study well...... study well |
|
| 39797. |
Underroot 3sin x=cos x ;find x |
|
Answer» Then answer is 30 ° Do you mean √3 sin x = cos x |
|
| 39798. |
Prove that :CosA-sinA+1/cosa+sinA-1=cosecA+cotA |
| Answer» | |
| 39799. |
Proof that the parallelogram cricumscribing a circle is a rhombus |
| Answer» Let ABCD be the llgm circumscribing a circle wid centre O. As ABCD is a llgm.AB= CD AND BC =AD -----> Eq.1Also, the tangents to a circle frm an external point are equal in length.Therefore, AM= AP, BM= BN, CO= CN, DO= DP.=> (AM+ BM) +(CO+ DO)= AP+BN+CN+DP=> AB+CD=(AP+BC) +( BN+ NC) =AD+BC ---->Eq.2From Eq 1 and 2, we get,AB=BC=CD=ADHence, ABCD is a rhombus. | |
| 39800. |
If Power of 9 is x+2 and power of 6.3 is x+1 then find the value of x |
| Answer» | |