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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.
| 34651. |
Define H.O.E in Euclid\'s lemma |
| Answer» | |
| 34652. |
Chapter wise identities |
| Answer» Get identities in the notes :\xa0https://mycbseguide.com/cbse-revision-notes.html | |
| 34653. |
49-113-64-113 |
| Answer» -241 | |
| 34654. |
What u mean by euclid\'s division algorithm |
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Answer» According to Euclid’s Division Lemma if we have two positive integers a and b, then there exist unique integers q and r which satisfies the condition a = bq + r where 0 ≤ r ≤ b. Given positive integers a and b,their exist unique integers q and r satisfying a=bq+r, where 0 is equl to or smaller than r is smaller than b |
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| 34655. |
X + ✅y =7 and ✔x + y = 11 |
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Answer» Given equations are{tex}\\frac{x}{4}{/tex}\xa0+\xa0{tex}\\frac{2y}{3} = 7{/tex}...............(i)and\xa0{tex}\\frac{x}{6}{/tex}\xa0+\xa0{tex}\\frac{3y}{5} = 11{/tex} .................(ii)From equation (i), we get{tex}\\frac{x}{4}{/tex}\xa0+\xa0{tex}\\frac{2y}{3} = 7{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}\\frac{x}{4}{/tex}\xa0= 7 -\xa0{tex}\\frac{2y}{3}{/tex}{tex}\\Rightarrow{/tex}\xa0{tex} x = 4(7 -\\frac{2y}{3}){/tex}{tex}\\Rightarrow{/tex}\xa0x = 28 -\xa0{tex}\\frac{8y}{3}{/tex}.................(iii)substituting x = 28 -\xa0{tex}\\frac{8y}{3}{/tex}\xa0in equation (ii), we get{tex}\\frac { 28 - \\frac { 8 y } { 3 } } { 6 } + \\frac { 3 y } { 5 } = 11{/tex}\xa0{tex}\\Rightarrow{/tex}\xa0{tex}\\frac { 84 - 8 y } { 18 } + \\frac { 3 y } { 5 } = 11{/tex}{tex}\\Rightarrow{/tex}{tex}\\frac{420 - 40y + 54y}{18 \\times 5}{/tex}\xa0= 11{tex}\\Rightarrow{/tex}\xa0420 + 14y = 990{tex}\\Rightarrow{/tex} 14y = 570{tex}\\Rightarrow{/tex} y =\xa0{tex}\\frac{570}{14}{/tex}\xa0=\xa0{tex}\\frac{285}{7}{/tex}When y =\xa0{tex}\\frac{285}{7}{/tex}, equation (iii) becomes{tex}x = 28 - \\frac { 8 \\times \\frac { 285 } { 7 } } { 3 }{/tex}{tex}\\Rightarrow{/tex}\xa0{tex} x = 28 -\\frac{8 \\times 285}{21}{/tex}{tex}\\Rightarrow{/tex}{tex}x = \\frac{588 - 2280}{21}{/tex}\xa0=\xa0{tex}\\frac{-1692}{21}{/tex}{tex}\\Rightarrow{/tex} -{tex}x =\\frac{564}{7}{/tex}{tex}\\therefore{/tex}\xa0x = -{tex}\\frac{564}{7}{/tex}, y =\xa0{tex}\\frac{285}{7}{/tex} is the solution of given system of equations. Sorry nahi samajh me aya . Pic |
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| 34656. |
Explain terminating and non terminating in long |
| Answer» The rational number for which the long division terminates after a finite number of steps is known as terminating decimal. Examples: The rational number for which the long division does not terminate after any number of steps is known as non-terminating decimal. | |
| 34657. |
Prove that (2 + ROOT under 3 )is irrational |
| Answer» We will prove this result by contradiction method.Let assume that\xa0{tex}( 2 + \\sqrt { 3 } ){/tex} be rational.Then 2 and {tex}\\sqrt { 3 }{/tex} are rational.{tex}\\Rightarrow 2 + \\sqrt { 3 } - 2{/tex} is rational ..... ({tex}\\because{/tex} difference of two rational numbers\xa0is rational){tex}\\Rightarrow \\sqrt { 3 }{/tex} is rationalThis contradicts the fact that {tex}\\sqrt { 3 }{/tex} is irrational.The Contradiction arises because\xa0we assume that {tex}( 2 + \\sqrt { 3 } ){/tex} is rational.Hence, {tex}2 + \\sqrt { 3 }{/tex} is not a rational but an irrational number. | |
| 34658. |
Class 10 ch 4 ncert book example with solution |
| Answer» What | |
| 34659. |
Define terms |
| Answer» | |
| 34660. |
Prime factorisation of 7429 |
| Answer» 7429 = 17 × 19 × 23 | |
| 34661. |
What do you mean by Euclid\'s division lema |
| Answer» According to Euclid’s Division Lemma if we have two positive integers a and b, then there exists unique integers\xa0q\xa0and\xa0r\xa0which satisfies the condition\xa0a = bq + r\xa0where 0\xa0≤ r ≤ b.The basis of Euclidean division algorithm is Euclid’s division lemma. To calculate the Highest Common Factor (HCF) of two positive integers\xa0a\xa0and\xa0b\xa0we use Euclid’s division algorithm. HCF is the largest number which exactly divides two or more positive integers. By exactly we mean that on dividing both the integers\xa0a\xa0and\xa0b\xa0the remainder is zero. | |
| 34662. |
Prove that √3 is irrational??? |
| Answer» Let us assume that √3 is a rational numberThat is, we can find integers a and b (≠ 0) such that √3 = (a/b)Suppose a and b have a common factor other than 1, then we can divide by the common factor, and assume that a and b are coprime.√3b = a⇒\xa03b2=a2 (Squaring on both sides) → (1)Therefore, a2 is divisible by 3Hence \x91a\x92 is also divisible by 3.So, we can write a = 3c for some integer c.Equation (1) becomes,3b2 =(3c)2⇒\xa03b2 = 9c2∴ b2 = 3c2This means that b2 is divisible by 3, and so b is also divisible by 3.Therefore, a and b have at least 3 as a common factor.But this contradicts the fact that a and b are coprime.This contradiction has arisen because of our incorrect assumption that √3 is rational.So, we conclude that √3 is irrational. | |
| 34663. |
Peove that root 5 is irrational no |
| Answer» let √5 be rationalthen it must in the form of p/q [q is not equal to 0][p and q are co-prime]√5=p/q=> √5 × q = psquaring on both sides=> 5×q×q = p×p ----- 1p×p is divisible by 5p is divisible by 5p = 5c [c is a positive integer] [squaring on both sides ]p×p = 25c×c --------- 2sub p×p in 15×q×q = 25×c×cq×q = 5×c×c=> q is divisible by 5thus q and p have a common factor 5there is a contradictionas our assumtion p &q are co prime but it has a common factorso\xa0√5 is an irrational | |
| 34664. |
A2 \\a2 |
| Answer» | |
| 34665. |
Are there matks for....steps writing...in construction? ?? |
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Answer» They are important only if asked in the question...if not then writing them is a waste of time...but if let\'s say it\'s asked to write the steps of construction and you don\'t do so...then you will definitely loose your marks... Yes it is necessary. But don\'t waste time for learning the words for the steps just write in our own words because there is rarely any teacher who will read the steps they will directly see the construction but if you don\'t write it then you will surely lose 1 or 1.5 marks. Yes ...sometimes. But justification is must |
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| 34666. |
3 x - 2y + 3 = 04 x + 3 y - 47 equals to zeroSolve these equations by cross multiplication method |
| Answer» {tex}3x - 2y + 3 = 0{/tex}........(i){tex}4x + 3y - 47 = 0{/tex}......(ii)By cross multiplication, we have{tex}\\therefore \\frac { x } { [ ( - 2 ) \\times ( - 47 ) - ( 3 \\times 3 ) ] }{/tex}{tex}= \\frac { y } { [ ( 3 \\times 4 ) - ( - 47 ) \\times 3 ] }{/tex}{tex}= \\frac { 1 } { [ 3 \\times 3 - ( - 2 ) \\times 4 ] }{/tex}{tex}\\Rightarrow \\quad \\frac { x } { ( 94 - 9 ) } = \\frac { y } { ( 12 + 141 ) } = \\frac { 1 } { ( 9 + 8 ) }{/tex}{tex}\\Rightarrow \\quad \\frac { x } { 85 } = \\frac { 1 } { 17 } , \\frac { y } { 153 } = \\frac { 1 } { 17 }{/tex}{tex}17x = 85, \\ 17y = 153{/tex}{tex}\\Rightarrow \\quad x = \\frac { 85 } { 17 } , y = \\frac { 153 } { 17 }{/tex}Therefore, the solution is {tex}x = 5,\\ y = 9{/tex} | |
| 34667. |
Can you solve all questions of last year question paper |
| Answer» Done | |
| 34668. |
Find all the zeroes of x cube +11x square +23x-35, if two of its zeroes are 1 and -5 |
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Answer» Hence sum of zeroes = -b/aLet the another zero is x. Hence x + 1 +(-5) = -(11)/1 =x + (-4) =. (-11) x =-11+4 x=-7 Vjgig |
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| 34669. |
Show that 7-sq. root 5 is irrational |
| Answer» Let us assume that {tex}7 - \\sqrt 5{/tex}\xa0is a rational number.Then, there must exist positive co primes a and b such that{tex}\\begin{array}{l}7 - \\sqrt 5=\\frac{\\mathrm a}{\\mathrm b}\\\\\\end{array}{/tex}{tex}-\\sqrt5=\\frac{\\mathrm a}{\\mathrm b}-7{/tex}{tex}\\begin{array}{l}\\sqrt5=7-\\frac{\\mathrm a}{\\mathrm b}\\\\\\end{array}{/tex}{tex}\\begin{array}{l}\\sqrt 5=\\frac{7\\mathrm b-\\mathrm a}{\\mathrm a}\\\\\\end{array}{/tex}The right side\xa0{tex}\\begin{array}{l}\\frac{7\\mathrm b-\\mathrm a}{2\\mathrm a}\\\\\\end{array}{/tex}\xa0is a rational numbers so\xa0{tex}\\sqrt5{/tex} is a rational numberThis contradicts the fact that {tex}\\sqrt 5{/tex}\xa0is an irrational numberHence our assumption is incorrect and {tex}7 - \\sqrt 5{/tex}\xa0is an irrational number. | |
| 34670. |
Find the value of k so that the pair of linear equations x+2y =5 and 3x+ky+5=0 have unique solution |
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Answer» The given pair of linear equations isx + 2y - 5= 0 and\xa03x + ky + 5 = 0Here, a1\xa0= 1, b1\xa0= 2,a2\xa0= 3, b2\xa0= kFor having a unique solution, we must have{tex}\\frac { a _ { 1 } } { a _ { 2 } } \\neq \\frac { b _ { 1 } } { b _ { 2 } }{/tex}{tex}\\Rightarrow \\quad \\frac { 1 } { 3 } \\neq \\frac { 2 } { k } \\Rightarrow k \\neq 6{/tex} Unique solution=a1/a2 not equal to b1/b2here, a1=1; a2=3; b1=2; b2=kthen, 1/3=1/kcross multiplythn k =3 |
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| 34671. |
X plus root under y =7, root under x plus y =11 . Find the value of x and y |
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| 34672. |
What is pressure group |
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Answer» Another defination A\xa0group that tries to influence public policy in the interest of a particular cause |
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| 34673. |
Check whether the equation is quadratic or not? x*2+1/x*2 =2 |
| Answer» We have,x2 +\xa0{tex}\\frac{1}{x^2}{/tex}\xa0= 0{tex} \\Rightarrow{/tex}\xa0x4\xa0- 1 = 0{tex} \\Rightarrow{/tex}\xa0x4\xa0- 1 = 0and (x4\xa0- 1) is a polynomial of degree 4. It is also not in the form of\xa0{tex}ax^2+bx+c=0{/tex}{tex}\\therefore{/tex}\xa0Given equation\xa0is not a quadratic equation. | |
| 34674. |
10y+15x =0 |
| Answer» 15x = - 10yX = 10y/15Ok but what you have to find | |
| 34675. |
Find the value of c in the eq.systemcx+3y+(3-c)=0 eq.112x+cy-c=0 eq.2 |
| Answer» cx + 3y + ( 3 - c ) = 0 and 12x +\xa0cy - c = 0Condition for infintely many solutions,{tex}\\frac { a _ { 1 } } { a _ { 2 } } = \\frac { b _ { 1 } } { b _ { 2 } } = \\frac { c _ { 1 } } { c _ { 2 } }{/tex}From given system of equation,a1\xa0= c, b1\xa0= 3, c1\xa0= 3 - cand a2\xa0= 12, b2\xa0= c, c2\xa0= -cPutting these values in condition,we get{tex}\\frac { c } { 12 } = \\frac { 3 } { c } = \\frac { 3 - c } { - c }{/tex}Considering first equality, i.e{tex}\\frac { c } { 12 } = \\frac { 3 } { c }{/tex}{tex}( c ) \\times ( c ) = 3 \\times 12{/tex}c2\xa0= 36{tex}c = \\pm \\sqrt { 36 }{/tex}c = ± 6here c = -6 is rejected as it does not satisfy the 2nd equality.Therefore, c = 6 | |
| 34676. |
X=1+√2+√3Find 2x^4-8x^3-5x^2+26x-29 |
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Answer» Give me answer only not derivation It is very long question send your number I will send you on WhatsApp |
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| 34677. |
Which math best |
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Answer» Advance if you want to go in the field of math or basic is you are not interested in maths Me? |
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| 34678. |
What is math best for 10 |
| Answer» Write proper question | |
| 34679. |
How many questions in 3.1 10 class |
| Answer» 3 | |
| 34680. |
√10 |
| Answer» Value 3.162 | |
| 34681. |
If cosec theta=x+1/4x prove that cosec theta+cot theta=2x or 1/2x |
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Answer» Question:\xa0cosec theta=x +1/4x prove that cosec theta+cot theta=2x or 1/2xSolution: Its 2ooo long |
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| 34682. |
If cosec theta + cot theta = X, find the value of cosec theta - cot theta\xa0 |
| Answer» X^2-1/ 2X | |
| 34683. |
after how many places with the decimal expansion of 147/120 terminate ? |
| Answer» {tex} \\frac{{147}}{{120}}{/tex}={tex} \\frac{{49}}{{40}}{/tex}= {tex} \\frac{{122.5}}{{100}}{/tex}= 1.225 thus the decimal expansion has three digits after the decimal point. | |
| 34684. |
Prove that the following numbers are irrational |
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Answer» Root 2 Prove that the following numbers are irrational |
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| 34685. |
22/7×22/7×3/11 |
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Answer» २.६९३८७७५५१० 2.69387755 Oo sorry ?? 2.69387755102041 Are iss question ka answer .....nikala hai maine Ya ky ha 2.69...... 2.69.......... ?? Naam kya hai bhai tera ??? |
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| 34686. |
1*1+1-1 |
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Answer» 1?? 1 1 Kidding aren\'t you 1 1 1 |
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| 34687. |
Find a rational number between /2 and /5 |
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Answer» May be Value of √2and√5in between any no.write |
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| 34688. |
The ratio of the volumes of two spheres is 8 is to 27 what is the ratio of their surface areas |
| Answer» Let the radius of 1st sphere be \'r1\' and the radius of 2nd sphere be \'r2\'According to question,Ratio\xa0of the volume of the given spheres is,{tex}\\frac { \\text { Volume of } 1 ^ { \\text { st } } \\text { sphere } } { \\text { Volume of } \\Pi ^ { \\text { nd } } \\text { sphere } } = \\frac { \\frac { 4 } { 3 } \\pi r _ { 1 } ^ { 3 } } { \\frac { 4 } { 3 } \\pi r _ { 2 } ^ { 3 } } = \\frac { 8 } { 27 }{/tex}{tex}\\therefore \\quad \\quad \\frac { r _ { 1 } ^ { 3 } } { r _ { 2 } ^ { 3 } } = \\frac { 8 } { 27 }{/tex}{tex} \\frac { r _ { 1 } } { r _ { 2 } } = \\frac { 2\\sqrt2} { 3 }{/tex}The ratio\xa0of the radius of the given spheres, r1\xa0: r2\xa0= 2{tex}\\sqrt 2{/tex}:3Now,Ratio of the\xa0surface areas of the spheres\xa0{tex}= \\frac { \\text { Surface area of } 1 ^ { \\text { st } } \\text { sphere } } { \\text { Surface area of } \\Pi ^ { \\text { nd } } \\text { sphere } }{/tex}{tex}\\frac { 4 \\pi r _ { 1 } ^ { 2 } } { 4 \\pi r _ { 2 } ^ { 2 } } = \\left( \\frac { r _ { 1 } } { r _ { 2 } } \\right) ^ { 2 }{/tex}{tex}=\\left( \\frac { 2\\sqrt2 } { 3 } \\right) ^ { 2 } = \\frac { 8 } { 9 }{/tex}= 16 : 9 | |
| 34689. |
I want to get the topper answer sheet of 2019 |
| Answer» You can check the topper\'s answer sheet in each chapter\'s category. | |
| 34690. |
2/✓x+3/✓y=2 4/✓x-9/✓y=-1 |
| Answer» | |
| 34691. |
A two digit number 4 time the some of its digit if 18 is added to the number the digit are reversed |
| Answer» | |
| 34692. |
Polynomial. Chapter. Of10class |
| Answer» So what can we do then | |
| 34693. |
Show that one and only one out of n, (n+1), (n+2) is divisible by 3, where n is any positive integer |
| Answer» If n=2 then (n+1) is divisibleIf n=3 then (n) is divisibleIf n=4 then (n+2) is divisible | |
| 34694. |
What is the difference between algebraic and geometrical geographically in linear equation |
| Answer» In algebra the x term can continue | |
| 34695. |
How to convert 4hour 10 min in hour |
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Answer» 1 hr= 60 min Therefore 1 min = 1/60 hrs.... 10 min = 10/60 hrs = 1/6 hrs..... It implies 4 hr 10 min = 4 hr + 1/6 hr = 4+1/6 = (24+1)/6= 25/6= 4.1 hrs.. 4.1 hours |
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| 34696. |
4x-5x-20=03x+5y-15=0 |
| Answer» On a graph paper, draw a horizontal line X\'OX and a vertical line YOY\' as the x-axis and the y-axis respectively.Graph of {tex}4x - 5y - 20 = 0{/tex}{tex}4x - 5y - 20 = 0 {/tex}{tex}\\Rightarrow{/tex}\xa0{tex}5y = (4x - 20){/tex}{tex}\\Rightarrow \\quad y = \\frac { ( 4 x - 20 ) } { 5 }{/tex}.....(i)Table for {tex}4x - 5y - 20 = 0.{/tex}\tx025y-4-2.40\tNow, plot the points {tex}A(0, -4),\\ B(2, -2.4)\\ and\\ C(5, 0){/tex} on the graph paper.Join AB and BC to get the graph line ABC. Extend it on both ways.Thus, the line ABC is the graph of {tex}4x - 5y - 20 = 0.{/tex}{tex}3x\xa0+ 5y -15 = 0{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}5y = (15 -3x){/tex}{tex}\\Rightarrow \\quad y = \\frac { ( 15 - 3 x ) } { 5 }{/tex}.........(ii)Table for {tex}3x\xa0+ 5y -15 = 0.{/tex}\tx-505y630\tOn the same graph paper as above, plot the points P (-5, 6) and Q(0, 3).The third point C(5, 0) has already been plotted.Join PQ and QC to get the graph line PQC. Extend it on both ways.Thus, the line PQC is the graph of 3x + 5y -15 = 0.The two graph lines intersect at the point C(5,0).{tex}\\therefore{/tex}\xa0x = 5, y = 0 is the solution of the given system of equations.Clearly, the given equations are represented by the graph lines ABC and PQC respectively.The vertices of {tex}\\triangle{/tex}AQC formed by these lines and the y-axis are A(0, -4), Q(0,3) and C(5,0). | |
| 34697. |
15×28 |
| Answer» 420 | |
| 34698. |
Find the roots by the method of completing the square 2x/2+x+4=0 |
| Answer» X=-8 | |
| 34699. |
Find HCF. Of 867&255 |
| Answer» On applying division lemma to 867 and 255We get867=255×3+102255=102×2+51102=51×2+0Hence HCF(867,255)=51 | |
| 34700. |
Formula for l in frustum |
| Answer» [H^2+{R-r}^2]^(1/2) | |