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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.
| 38851. |
If the product of two numbers 396×576 and their LCM is 6336.Find their HCF |
| Answer» 36 | |
| 38852. |
Explain some one question without using pythagoras theroem |
| Answer» Let A (4, 4), B (3, 5) and C(-1, -1) be three vertices of a {tex}\\Delta {\\rm A}{\\rm B}C{/tex}.{tex}\\therefore{/tex}\xa0Slope of AB {tex} = \\frac{{5 - 4}}{{3 - 4}} = \\frac{1}{{ - 1}} = - 1{/tex}{tex}\\therefore{/tex}\xa0Slope of BC {tex}= \\frac{{ - 1 - 5}}{{ - 1 - 3}} = \\frac{{ - 6}}{{ - 4}} = \\frac{3}{2}{/tex}{tex}\\therefore{/tex}\xa0Slope of AC {tex} = \\frac{{ - 1 - 4}}{{ - 1 - 4}} = \\frac{{ - 5}}{{ - 5}} = 1{/tex}Now slope of AB {tex}\\times{/tex} slope of AC = -1{tex}\\times{/tex}1 = -1This shows thatAB{tex}\\bot{/tex}AC. Thus {tex}\\Delta {\\rm A}{\\rm B}C{/tex} is right angled at point A. | |
| 38853. |
Obtain all other zeroes of 2x(cube)+5x(square)-28x_15,if one of its zeroes is 3 |
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Answer» Here polynomial is 2x^3+5x^2-28x-15One zero=3Therefore it\'s first factor is (x-3)Now, divide the polynomial from it\'s first factor.We obtain the quotient (2x^2+11x+5) and remainder=0Now, polynomial = 1st factor + QuotientNow factorise it\'s quotient(2x^2+11x+5)=2x^2+10x+1x+5 = 2x(x+5)+1(x+5) = (x+5)(2x+1)Now it\'s factors are minus 5 and minus 1÷2 . -4.5 and -0.5 |
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| 38854. |
Show that the quadratic polynomial 4 x square + 12 x + 15 has no real zeros |
| Answer» Use the formula of discriminant | |
| 38855. |
If 3sin +5cos =5 prove that 5sin - 3cos = +- 3 |
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| 38856. |
Sin theta=12/13then the value of the 2cos theta+3tan theta/sin theta+ tan theta.sin theta |
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| 38857. |
how to find the area of a segment wrire the formula also |
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| 38858. |
Syllabus of periodic test 1 |
| Answer» Ch-1,2,3,4,5,6,7,14,15 | |
| 38859. |
If x+2 is a factor of polynomial 5x cube +(k+2)x square - 3kx + 2 then find the value of k. |
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Answer» K=3 X=-2,(5×-2)cube +(k+2)-2-3k(-2)-1000-2k-2+6k4k=1000+24k=1002K= 1002/4K=250.5 |
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| 38860. |
Solve x and y . Ax upon b minus byupon a is equall to a plus b . Ax minus by is equal to 2ab |
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| 38861. |
Solve the foll system of equation 3x-4y-12x-8÷3y+5 |
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| 38862. |
What is the value of cos square 67 - sin square 23 |
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Answer» Thnks 0 |
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| 38863. |
Show that any number of the form (6)x, x€N can never end with the digit 0 |
| Answer» {tex}6 ^ { n } = ( 2 \\times 3 ) ^ { n } = 2 ^ { n } \\times 3 ^ { n }{/tex}{tex}\\therefore \\quad 5 \\text { is not a factor of } 6 ^ { n }{/tex}{tex}\\therefore {/tex}\xa0It never ends with 0. | |
| 38864. |
Show that the square of any positive integer cannot be of form 5q+2 or 5q +3 for any integer q |
| Answer» Let n be any positive integer. Applying Euclids division lemma with divisor = 5, we get{tex}\\style{font-family:Arial}{\\begin{array}{l}n=5q+1,5q+2,5q+3\\;and\\;5q+4\\;\\\\\\end{array}}{/tex}Now (5q)2 = 25q2 = 5m, where m = 5q2, which is an integer;{tex}\\style{font-family:Arial}{\\begin{array}{l}(5q\\;+\\;1)^{\\;2}\\;=\\;25q^2\\;+\\;10q\\;+\\;1\\;=\\;5(5q^2\\;+\\;2q)\\;+\\;1\\;=\\;5m\\;+\\;1\\\\where\\;m\\;=\\;5q^2\\;+\\;2q,\\;which\\;is\\;an\\;integer;\\\\\\;(5q\\;+\\;2)^2\\;=\\;25q^2\\;+\\;20q\\;+\\;4\\;=\\;5(5q^2\\;+\\;4q)\\;+\\;4\\;=\\;5m\\;+\\;4,\\\\\\;where\\;m\\;=\\;5q^2\\;+\\;4q,\\;which\\;is\\;an\\;integer;\\\\\\;(5q\\;+\\;3)^{\\;2}\\;=\\;25q^2\\;+\\;30q\\;+\\;9\\;=\\;5(5q^2\\;+\\;6q+\\;1)\\;+\\;4\\;=\\;5m\\;+\\;4,\\\\\\;where\\;m\\;=\\;5q^2\\;+\\;6q\\;+\\;1,\\;which\\;is\\;an\\;integer;\\\\\\;(5q\\;+\\;4)^2\\;=\\;25q^2\\;+\\;40q\\;+\\;16\\;=\\;5(5q^2\\;+\\;8q\\;+\\;3)\\;+\\;1\\;=\\;5m\\;+\\;1,\\;\\\\where\\;m\\;=\\;5q^2\\;+\\;8q\\;+\\;3,\\;which\\;is\\;an\\;integer\\\\\\end{array}}{/tex}Thus, the square of any positive integer is of the form 5m, 5m + 1 or 5m + 4 for some integer m.It follows that the square of any positive integer cannot be of the form 5m + 2 or 5m + 3 for some integer m. | |
| 38865. |
If tan = root 1-a^2 then prove that sec+tan^3cosec =( 2-a^2)^3/2 |
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| 38866. |
If a polynomial x•4-3x•3-6x•2+kx-16 is exactly divisible by x•3-3x+2, find the value of K. |
| Answer» Given\xa0p(x) = x4\xa0- 3x3\xa0- 6x2 + kx - 16and g(x) =\xa0x2\xa0- 3x\xa0+ 2On long division of f(x) by g(x) we getNow, remainder = (k - 24)xWhen p(x) is exactly divisible by g(x), then remainder = 0{tex}\\Rightarrow{/tex}\xa0(k - 24)x = 0k - 24 = 0k = 24 | |
| 38867. |
Round off 593702164 to the nearest ten |
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Answer» 593702160 593702165 |
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| 38868. |
X2 -20 =0 |
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| 38869. |
For what value of x are the points A (_3,12),B (7,6),C (x,9) collinear ? |
| Answer» According to the question, A(-3, 12), B(7, 6) and C(x, 9).(x1 = -3, y1 = 12), (x2 = 7, y2 = 6) and\xa0(x3 = x, y3 = 9)The given points A,B,C are collinear. So, area of triangle is zero.{tex}\\Rightarrow{/tex}\xa0x1(y2 - y3) + x2(y3 - y1)+ x3(y1 - y2)=0{tex}\\Rightarrow{/tex}\xa0{tex}(-3)(6 - 9) + 7 (9 - 12) +x(12 - 6) =0{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}9 - 21 + 6x = 0{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}6x = 12\xa0{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}x = 2{/tex} | |
| 38870. |
Kya hai yeh trigonometry jara vistar me batai |
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| 38871. |
Find the distance between the following pairs of points |
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| 38872. |
Can every positive campasite numbers be represent as product at prime number if yes explain how? |
| Answer» Yes every composite number can be represent as product of prime number . As 16 can be expressed in the form of 2 to the power 4 which is a prime number product and other example 42 which can be expressed in the form form of 2 x 3 x 7 thus this shows that every composite number can be expressed in the form of product of prime numbers | |
| 38873. |
What is the geographyclly representation of arithmetic progression |
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| 38874. |
If the zeros of the polynial f(x);x3 -3x2+x+1are a-b;a;a+bfind aand b |
| Answer» Let alpha beta and gamma are the three zeros of the given polynomial FX and Alpha is equals to a minus b beta is equals to a and gamma is equals to A + B by sum of zeros we have alpha + beta + Gamma is equals to minus minus 3 by 1 so we get 3 a is equals to 3 then a is equals to 1 by this we get alpha × beta × Gama is equals to -1 and putting the value of a in this equation we get -b^2 =-1 thus a is equals to 1 and b is equals to 1 | |
| 38875. |
How to define chemical equation. |
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Answer» The method of representing a chemical reaction with the help of symbols and formulas is called chemical eqations Change in states , product , colour , etc are defiend the reaction chemical The way to represent chemical reaction is called chemical equation. |
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| 38876. |
If a+b equals to c then what is c |
| Answer» Real number if a,b are integers | |
| 38877. |
3 cos 55°/7 sin35°=4cos70° cosec20/7(tan5° tan25° tan45° tan65° tan85°) |
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| 38878. |
2√x+3√y=2, 4√x+9√y=-1 then find x,y |
| Answer» . | |
| 38879. |
Write quadritic formula |
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Answer» Minus B plus minus under root b squared minus 4 AC / 2a ax2 + bx + c |
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| 38880. |
5+5+5+5=555 |
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Answer» 545+5+5 =555, just strike a line in the first \'+\' to make it 4. 555 prove it |
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| 38881. |
State euclid\'s divison lemma |
| Answer» Let a,b be any two positive integers (a>b) . When there exists two unique whole number q and R such that a =bq ±r , o≤ r <6 | |
| 38882. |
Find the medium value |
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| 38883. |
Prove that 3+√2is an irrational number. |
| Answer» Here p/q q equal to not zeroHere p and q are some integer3+√2=p/qHere √2=p/q=3P-3q/2Here p is also divisible by 29u square =3qU=3Here 3+√2 is a irrational | |
| 38884. |
CIRCIE 1.4 ANSWERS |
| Answer» DRAW A LABELLED DIAGRAM TO SHOW THE PROCESS OFREPARATION OF SODIUM HYDROXIDE FROM COMMON SALT | |
| 38885. |
6d=54+d |
| Answer» 6d=54+d=6d-d =54=5d =54 = d = 1.8 | |
| 38886. |
What are The formula for all algebric identities |
| Answer» Solve the questions eeleated to identuties from different books | |
| 38887. |
What is 10² |
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Answer» 100 100 is 10 square 100 How silly u are...?? 100 100 |
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| 38888. |
How to make trigonometry easy and interesting? |
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Answer» Try to use the concepts in Real - Life.......... Practice it regularly at home Due to practice any things can make easy Please help |
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| 38889. |
How many sides does a regular polygon have if each of its interior angles is 165 |
| Answer» Its each interior angle is 165 degree then to obtain exterior angle we have to subtract 165 degree to 180 degrees then we have 15 degree as an exterior angle and so to obtain its side divide 360 by 15 which is equal to 24 | |
| 38890. |
Draw a line segment AB= 7cm.Take a point P on AB such that AP:PB= 3:4 |
| Answer» Steps of construction:\tDraw a line segment AB = 7 cm}\tDraw a ray AX making an acute angle BAX with AB\tAlong AX mark (3 + 5) = 8 points.\tA1, A2, A3, A4, A5, A6, A7, A8\tsuch that AA1\xa0=\xa0A1A2\xa0= A2A3\xa0= A3A4\xa0= A4A5\xa0= A5A6\xa0= A6A7\xa0= A7A8\tJoin A8B.\tThrough the point A3, draw a line parallel to A8B by making an angle equal to\xa0{tex}\\angle{/tex}AA8B at A3.Suppose this line meets AB at a point P.Thus, P is the required point such that\xa0{tex}\\frac { A P } { P B } = \\frac { 3 } { 5 }{/tex}. | |
| 38891. |
Is the syllabus of maths half course or full course? |
| Answer» Obviously, full course in finals | |
| 38892. |
Draw a graph of polynomial f(x)=-3x2+2x-1 |
| Answer» | |
| 38893. |
What is LCM of P and Q where P = a3b2 and Q = b3a2?\xa0 |
| Answer» LCM(P,Q) = a2b2 | |
| 38894. |
Hcf of 75 and 160 |
| Answer» 5 | |
| 38895. |
Solve 99x + 101y = 499 |
| Answer» Given, 99x + 101y = 499 ....(i)101x + 99y = 501... (ii)Adding eqn. (i) and (ii),( 99x + 101y ) + (101x + 99y ) = 499 + 50199x + 101y + 101x + 99y = 1000200x + 200y = 1000x + y = 5 ...(iii)Subtracting eqn. (ii) from eqn. (i), we get( 99x + 101y ) - (101x + 99y ) = 499 - 50199x + 101y - 101x - 99y = -2-2x + 2y = - 2or, x - y= 1 ........ (iv)Adding equations (iii) and (iv)x + y + x - y = 5 + 12x = 6{tex}\\therefore {/tex}\xa0x = 3Substituting the value of x in eqn. (iii), we get3 + y = 5y = 2Hence the value of x and y of given equation are 3 and 4 respectively. | |
| 38896. |
prove that underoot 3is rational |
| Answer» Let us assume on the contrary that root 3 is a Rational Number. Then , there exists co-prime positive integers a and b such thatUnder root 3 =a/bOn Squaring both sides, We get a^2= 3b^2---------------(¡)=> 3|a^2=>3|a-------------------(ii)Let a= 3c , for some integer ca^2=9c^2 [ On Sq. Both sides ]=>3b^2 =9c^2 [From (¡) ]=>b^2= 3c^2=> 3|b^2=>3|b-------------------(¡¡¡)From (ii) and (¡¡¡ ), We get , that 3 is a common factor of both a and b.This contradicts the fact that a and b are co-primes [i.e. their H.C.F. is 1].So , Our Assumption is incorrect. Hence,Underroot 3 is an irrational number. [PROVED] | |
| 38897. |
Math lab manual experimant 7 |
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| 38898. |
In mathematics I have my doubt in optional exercise 5th question |
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| 38899. |
Prove that one of every three consecutive positive integers is divisble by 3 |
| Answer» Let the three consecutive positive integers be n, n + 1 and n + 2, where n is any integer.By Euclid’s division lemma, we havea = bq + r; 0 ≤ r < bFor a = n and b = 3, we haven = 3q + r ...(i)Where q is an integer and 0 ≤ r < 3, i.e. r = 0, 1, 2.Putting r = 0 in (i), we get{tex}n = 3q{/tex}∴ n is divisible by 3.{tex}n + 1 = 3q + 1{/tex}∴ n + 1 is not divisible by 3.{tex}n + 2 = 3q + 2{/tex}∴ n + 2 is not divisible by 3.Putting r = 1 in (i), we get{tex}n = 3q + 1{/tex}∴ n is not divisible by 3.{tex}n + 1 = 3q + 2{/tex}∴ n + 1 is not divisible by 3.{tex}n + 2 = 3q + 3 = 3(q + 1){/tex}∴ n + 2 is divisible by 3.Putting r = 2 in (i), we get{tex}n = 3q + 2{/tex}∴ n is not divisible by 3.{tex}n + 1 = 3q + 3 = 3(q + 1){/tex}∴ n + 1 is divisible by 3.{tex}n + 2 = 3q + 4{/tex}∴ n + 2 is not divisible by 3.Thus for each value of r such that 0 ≤ r < 3 only one out of n, n + 1 and n + 2 is divisible by 3. | |
| 38900. |
Prove that n2-n is divisble by 2 for every positive integers n |
| Answer» Any positive integer is of the form 2q or 2q + 1, for some integer q.{tex}\\therefore{/tex} When n = 2q{tex}\\style{font-family:Arial}{n^2\\;-\\;n\\;=\\;n(n\\;-\\;1)\\;=\\;2q(2q\\;-\\;1)=\\;2m,}{/tex}where m = q(2q - 1) ( m is any integer)This is divisible by 2When n = 2q + 1{tex}\\style{font-family:Arial}{\\begin{array}{l}n^2\\;-\\;n\\;=\\;n(n\\;-\\;1)\\;=\\;(2q\\;+\\;1)(2q+1-1)\\\\=2q(2q+1)\\end{array}}{/tex}= 2m, when m = q(2q + 1) ( m is any integer)which is divisible by 2.Hence, n2 - n is divisible by 2 for every positive integer n. | |