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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.
| 23701. |
2(2x-1/x+3)-3(x+3/2x-1)=5 |
| Answer» x= -10 & - 1/5 | |
| 23702. |
When was exam.start class 10 cbse |
| Answer» 2 march 2018 | |
| 23703. |
class 10 latest mock paper solution |
| Answer» | |
| 23704. |
Volume of cone |
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Answer» 1/3 py r square h 1/3πr2h |
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| 23705. |
Find the middle term of the AP 6,13,20......... 216 |
| Answer» The given AP is 6,13,20,...,216a = 6 and l = 216So the middle term will be average of a and lHence middle term = {tex}=\\frac{\\mathrm a+\\mathrm l}2{/tex}{tex}=\\frac{6+216}2=\\frac{222}2{/tex}\xa0=111So the middle term of this AP is 111 | |
| 23706. |
What is tangant ? |
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Answer» A line touches the circle externally and make 90 degree with the center A line touch the circle externally at one point A line touching circle at only one point externaly. A line meeting a circle only in one point is called tangent. Tangent is a line which touch a circle only at one point |
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| 23707. |
How to find frequency for finding mode |
| Answer» The largest frequency help to find the mode | |
| 23708. |
Verify that 3,-1,-1/3,are the zeroes of a cubic polynomial p(x) =3x²- 5x²-11x²-3 |
| Answer» Given polynomial is p(x) = 3x3 -5x2 - 11x - 33Put x = 3 in p(x), we get{tex} \\therefore{/tex}p(3) = 3 {tex} \\times{/tex}\xa033 - 5 {tex} \\times{/tex}\xa032 - 11 {tex} \\times{/tex}\xa03 - 3 = 81 - 45 - 33 - 3 = 0Put x = -1 in p(x), we getp(-1) = 3 {tex} \\times{/tex}\xa0(-1)3 - 5\xa0{tex} \\times{/tex} (-1)2 -11 {tex} \\times{/tex}\xa0(-1) - 3 = - 3 - 5 + 11 - 3 = 0Put x = {tex}- \\frac { 1 } { 3 }{/tex} in p(x), we getand,\xa0{tex} p \\left( - \\frac { 1 } { 3 } \\right){/tex}={tex} 3 \\times \\left( - \\frac { 1 } { 3 } \\right) ^ { 3 } - 5 \\times \\left( - \\frac { 1 } { 3 } \\right) ^ { 2 } - 11 \\times \\left( - \\frac { 1 } { 3 } \\right) - 3{/tex}={tex} - \\frac { 1 } { 9 } - \\frac { 5 } { 9 } + \\frac { 11 } { 3 } - 3{/tex}= 0\xa0So, 3,-1 and\xa0{tex} -\\frac 13{/tex}\xa0are the zeros of polynomial p(x).Let,\xa0{tex} \\alpha = 3 , \\beta = - 1 \\text { and } \\gamma = - \\frac { 1 } { 3 }{/tex} .Then,{tex} \\alpha + \\beta + \\gamma{/tex} ={tex} 3 - 1 - \\frac { 1 } { 3 } = \\frac { 5 } { 3 }{/tex}, and\xa0{tex} - \\frac { \\text { Coefficient of } x ^ { 2 } } { \\text { Coefficient of } x ^ { 3 } } = - \\left( \\frac { - 5 } { 3 } \\right) = \\frac { 5 } { 3 }{/tex}{tex} \\therefore \\quad \\alpha + \\beta + \\gamma = - \\frac { \\text { Coefficient of } x ^ { 2 } } { \\text { Coefficient of } x ^ { 3 } }{/tex}{tex} \\alpha \\beta + \\beta \\gamma + \\gamma \\alpha{/tex} = {tex} 3 \\times ( - 1 ) + ( - 1 ) \\times \\left( - \\frac { 1 } { 3 } \\right) + \\left( - \\frac { 1 } { 3 } \\right) \\times 3{/tex}=\xa0{tex} - 3 + \\frac { 1 } { 3 } - 1 = - \\frac { 11 } { 3 }{/tex}and,\xa0{tex} \\frac { \\text { Coefficient of } x } { \\text { Coefficient of } x ^ { 3 } } = - \\frac { 11 } { 3 }{/tex}{tex} \\therefore \\quad \\alpha \\beta + \\beta \\gamma + \\gamma \\alpha = \\frac { \\text { Coefficient of } x } { \\text { Coefficient of } x ^ { 3 } }{/tex}{tex} \\alpha \\beta \\gamma = 3 \\times ( - 1 ) \\times \\left( - \\frac { 1 } { 3 } \\right) = 1{/tex}......... (i)And,\xa0{tex} - \\frac { \\text { Constant term } } { \\text { Coefficient of } x ^ { 3 } } = - \\left( \\frac { - 3 } { 3 } \\right) = 1{/tex} ........ (ii)From (i) and (ii){tex} \\therefore \\quad \\alpha \\beta \\gamma = - \\frac { \\text { Constant term } } { \\text { Coefficient of } x ^ { 3 } }{/tex} | |
| 23709. |
Find the circumference if the area of the circle is 38.5cm. |
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Answer» 22 22 22 Circumference=77 |
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| 23710. |
64 +10 |
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Answer» 74 74 |
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| 23711. |
what is SN formula |
| Answer» Sn=n/2(2a+(n-1)d) | |
| 23712. |
In the given figure AC=24cm, AB=7cm and angle BOD =90°. Find the area of the shaded region. |
| Answer» Where is the figure | |
| 23713. |
Will the questions in optional exercise will come in exams? |
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Answer» 50 -50 chances. No Yes,it\'s very important exercise Less chances. Only two questions can come like that |
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| 23714. |
Cosec =3/5 then find the value of (5 cosec theeta -4 tan theeta /sec theeta +cot theeta ) |
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Answer» Correct answer is 11/25 May be 40/31.. |
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| 23715. |
2 cos2 theeta .1 2/1+cot2 theta =2 |
| Answer» | |
| 23716. |
Paper comes from ncert or another material |
| Answer» | |
| 23717. |
Ncert question and answers of circle |
| Answer» Check NCERT Questions and Solutions here :\xa0https://mycbseguide.com/ncert-solutions.html | |
| 23718. |
4x/2 |
| Answer» x=0.5 | |
| 23719. |
Ncert question of circle |
| Answer» Check NCERT Questions and Solutions here :\xa0https://mycbseguide.com/ncert-solutions.html | |
| 23720. |
Can two no. have 18 and 380 their hcf and lcm respectively. Give reason |
| Answer» No, because HCF is always the Factor Of LCM | |
| 23721. |
Circle invert solution |
| Answer» | |
| 23722. |
What is eculid lemma |
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Answer» For every two positive integers a,b there exists another two integers b and r such that a=bq+r Euclid\'s lemma is a lemma that captures a fundamental property of prime numbers, namely: Euclid\'s lemma - If a prime p divides the product ab of two integers a and b, p must divide at least one of those integers a and b. \u200eThe basis of Euclidean division algorithm is Euclid\'s division lemma. To calculate the Highest Common Factor (HCF) of two positive integers a and b ,we use Euclid\'s division algorithm. HCF is the largest number which exactly divides two or more positive integers. a=bq+r |
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| 23723. |
Find 32 percent of425 |
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Answer» 32 % of 42532 x425/10032 x 85/2032 x 17/48 x17= 136\xa0 136 |
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| 23724. |
Divide a line segment of length 8cm internally in the ratio 3:4 |
| Answer» Steps of construction STEP I Draw the line segment AB of length 8 cm.STEP II Draw any ray AX making an acute angle {tex}\\angle{/tex}BAX with AB.STEP III Draw a ray BY parallel to AX by making\xa0{tex}\\angle A B Y{/tex}\xa0equal to {tex}\\angle B A X.{/tex}STEP IV Mark the three point A1, A2, A3 on AX and 4 points B1, B2, B3, B4 on BY such thatAA1 = A1A2 = A2A3 = BB1 = B1B2 = B2B3 = B3B4.STEP V Join A3 B4. Suppose it intersects AB at a point P.Then, P is the point dividing AB internally in the ratio 3:4. | |
| 23725. |
If the point (p, q), (m, n) and(p-n, q-m) are Colliner show that pn=qm |
| Answer» Given points are collinear. Therefore[p {tex}\\times{/tex}\xa0n + m(q - n) + (p - m) q] - [m {tex}\\times{/tex}\xa0q + (p - m) n + p (q - n)] = 0{tex}\\Rightarrow{/tex}\xa0(pn + qm - mn + pq - mq) - (mq + pn - mn + pq - pn) = 0{tex}\\Rightarrow{/tex}\xa0(pn + p q - mn) - (mq - mn + pq) = 0{tex}\\Rightarrow{/tex}\xa0pn - mq = 0{tex}\\Rightarrow{/tex}\xa0pn = qm | |
| 23726. |
Value of Sin90° |
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Answer» Yrr ye San to book m hi mil jayega Sin90=1 1 1 The value of Sin 90° =1 according to trigonometry table |
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| 23727. |
If A,B and C are the interior angles of triangle ABC, find that tan(B+C)/2 |
| Answer» In\xa0{tex} \\Delta A B C{/tex}, we have,A + B + C = 180°{tex} \\Rightarrow{/tex}\xa0B + C =180° - ADividing 2 on both sides, we get\xa0{tex} \\Rightarrow \\quad \\frac { B + C } { 2 } = 90 ^ { \\circ } - \\frac { A } { 2 }{/tex}Applying Tan on both sides\xa0{tex} \\Rightarrow \\tan \\left( \\frac { B + C } { 2 } \\right){/tex}{tex} = \\tan \\left( 90 ^ { \\circ } - \\frac { A } { 2 } \\right){/tex}{tex} \\Rightarrow \\tan \\left( \\frac { B + C } { 2 } \\right) = \\cot \\frac { A } { 2 }{/tex} | |
| 23728. |
2 sin 68/cos 22 -2 cot 15/5tan 75 =5tan45 tan 20 tan 40 tan 5 tan 7/5 |
| Answer» {tex}\\frac{{2\\sin 68^\\circ }}{{\\cos 22^\\circ }} - \\frac{{2\\cot 15^\\circ }}{{5\\tan 75^\\circ }} - \\frac{{3\\tan 45^\\circ \\tan 20^\\circ \\tan 40^\\circ \\tan 50^\\circ \\tan 70^\\circ }}{5}{/tex}{tex} = \\frac{{2\\sin \\left( {90^\\circ - 22^\\circ } \\right)}}{{\\cos 22^\\circ }} - \\frac{{2\\cot \\left( {90^\\circ - 75^\\circ } \\right)}}{{5\\tan 75^\\circ }}{/tex}\xa0{tex} - \\frac{{3\\tan 45^\\circ \\tan \\left( {90^\\circ - 70^\\circ } \\right)\\tan \\left( {90^\\circ - 50^\\circ } \\right)\\tan 50^\\circ \\tan 70^\\circ }}{5}{/tex}{tex} = \\frac{{2\\cos 22^\\circ }}{{\\cos 22^\\circ }} - \\frac{{2\\tan 75^\\circ }}{{5\\tan 75^\\circ }} - \\frac{{3 \\times 1 \\times \\cot 70^\\circ \\times \\tan 50^\\circ \\times \\frac{1}{{\\cot 50^\\circ }} \\times \\frac{1}{{\\cot 70^\\circ }}}}{5}{/tex}{tex} = 2 - \\frac{2}{5} - \\frac{3}{5}{/tex}{tex} = \\frac{{10 - 2 - 3}}{5} = \\frac{5}{5} = 1{/tex} | |
| 23729. |
I want to know the question were arranged form ncert and examoler or from every book |
| Answer» | |
| 23730. |
Find the sum of the first 40 positive integers divisible by 6 |
| Answer» The first 40 positive integers divisible by 6 are 6, 12, 18, 24, .....Here, a2 - a1 = 12 - 6 = 6a3 - a2 = 18 - 12 = 6a4 - a3 = 24 - 18 = 6i.e. ak+1 - ak is the same everytime.So, the above list of numbers from an AP.Here, a = 6d = 6n = 40{tex}\\therefore {/tex}\xa0Sum of the first 40 positive integers = S40{tex} = \\frac{{40}}{2}\\left[ {2a + (40 - 1)d} \\right]{/tex}\xa0.........{tex}\\because {S_n} = \\frac{n}{2}\\left[ {2a + (n - 1)d} \\right]{/tex}= 20[2a + 39d]{tex} = (20)[2 \\times 6 + 39 \\times 6]{/tex}= (20) (246)= 4920 | |
| 23731. |
What is tha passing marks in board exam |
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Answer» 27 you have to gain in all subjects if you gain it then out of 20 marks your teacher will giving you 33% 27/80 |
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| 23732. |
2017 paper board exam |
| Answer» Check last year papers here :\xa0https://mycbseguide.com/cbse-question-papers.html | |
| 23733. |
sinA _ 2cosA = 1Prove that2cosA _ sinA = 2 |
| Answer» | |
| 23734. |
Ch 6 triangle exs 6.5 question 11 |
| Answer» | |
| 23735. |
What is the maximum number of parallel tangent a circle can have on a diameter? |
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Answer» 2 PAIR of tangents from both vertically and horizontally Only 1 pair of // tangents can be drawn on a diameter of a circle |
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| 23736. |
What number should be added to polynomial x² + 7 x - 35 show that 3 is the Zero of the polynomial |
| Answer» - 5 | |
| 23737. |
What is cube root of 74 |
| Answer» 405224 | |
| 23738. |
If the perimeter of a semi circular protractor is 36 cm,find the diameter of the protractor |
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Answer» πr=3622/7 ×r= 36r= 36×7/22r=18×7/11d=2×18×7/11d=22.9 Perimeter of semi circular is given by = π r+2r36 = πr+2r36 = r( π+2)36 = r(22/7 +2)36 = r(22+14/7)36= r(36/7)1 = r/7r= 7 cm\xa0d = 2r = 2*7 = 14 cm SorRy but Wright answer is 11.45 is it 22.90 |
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| 23739. |
SOLVE BY SUBSTITUTION METHOD 3x+y=3 And 9x_3y =9SOLVE THIS PROBLEM |
| Answer» 3x+y = 3y= 3-3xSubstituting the value of y in second equation , we get\xa09x - 3(3-3x) = 99x -9 +9x = 918x-9= 918x = 9+9 = 18x= 18/18 = 1y= 3-3*1= 3-3= 0x= 1 and y= 0 | |
| 23740. |
For what value of n, are the nth terms of two A. Ps 63, 65, 67,.... and 3, 10, 17,.... equal? |
| Answer» First APs63, 65, 67, ......Here, a = 63d = 65 - 63 = 2{tex}\\therefore {/tex}\xa0nth term = 63 + (n - 1)2\xa0{tex}\\because {/tex}\xa0an = a + (n - 1)dSecond APs3, 10, 17, .....Here, a = 3d = 10 - 3 = 7{tex}\\therefore {/tex}\xa0nth term = 3 + (n - 1)7\xa0{tex}\\because {/tex}\xa0an = a+(n - 1)dIf the n th terms of two APs are equal then63 + (n - 1)2 = 3 + (n - 1)7{tex} \\Rightarrow {/tex}\xa0(n - 1)2 - (n - 1)7 = 3 - 63{tex} \\Rightarrow {/tex}\xa0(n - 1) (2 - 7) = -60{tex} \\Rightarrow {/tex}\xa0(n - 1) (-5) = -60{tex} \\Rightarrow n - 1 = \\frac{{ - 60}}{{ - 5}}{/tex}{tex} \\Rightarrow {/tex}\xa0n - 1 = 12{tex} \\Rightarrow {/tex}\xa0n = 12 + 1{tex} \\Rightarrow {/tex}\xa0n = 13Hence, for n = 13th terms of the two APs are equal | |
| 23741. |
The surface area of a sphere is 616 cm ki square find its radius |
| Answer» 4π r2\xa0=6164*22/7*r2\xa0= 616r2\xa0= 616*7/4*22r2\xa0= 154*7/22r2\xa0= 7*7r= 7 cm | |
| 23742. |
Hello come here...ishi |
| Answer» Ishi sorry I don\'t hate u but hate another ishi | |
| 23743. |
Which teem of the AP 3,15,27,39will be 132 more than its 54th term |
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Answer» First term of A.P. a = 3 and common difference d = 12a54 =\xa0a+(54-1)d = 3+53*12 = 3+636 = 639New term is 132 more than 639. So nth term = 132+639= 771771 = 3+(n-1)*12771= 3+12 n-12771 = 12n-912n= 771+9=780n= 780/12=65So 65 th term will be 132 more than its 54 th term. 65th |
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| 23744. |
Area of square is equal to the area of triangle why? |
| Answer» | |
| 23745. |
How to remember trigonometric ratios? |
| Answer» | |
| 23746. |
Sorry guys |
| Answer» | |
| 23747. |
Find HCF of (x^3-3x+2) and (x^2-4x+3) |
| Answer» | |
| 23748. |
What is nuclear atomic size of Radium and benzene ecine |
| Answer» Atomic Radius of Radium is\xa0200\xa0pmThe six-membered ring in\xa0benzene\xa0is a perfect hexagon (all carbon-carbon bonds have an identical length of 1.40 Å).\xa0 | |
| 23749. |
The 4th vertix D of a llgm ABCD whose 3vertices are A(-2,3)B(6,7)and C(8,3) is_________. |
| Answer» | |
| 23750. |
If the nth term of an AP is 7+4n then find the common difference |
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Answer» D = 4 D=7 |
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