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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.
| 35051. |
Sin²x+1-cos²x/cos²x+1-sin²x=1+tanx-cotx |
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| 35052. |
Which is the most important chapter for cbse exams |
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Answer» 2,3,5,6,8,10,13,12 All chaprers are important Trignometery Hii 6 and 13 Chapter 2,5,8,9,12,13 |
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| 35053. |
How much marks do i get if i do ncert properly 3 times? |
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Answer» Full But 80 marks ka to paper hi aayega 80 marks Because 20 marks come from out of book 80 marks |
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| 35054. |
प्रोवे पाइथागोरस theorem |
| Answer» Hypotenuse square =square of one side+the other side | |
| 35055. |
Ncert book exercise 13.3 qno. 3 and 9 |
| Answer» Ans3. Height of well ( cylinder) = 20 mDiameter 7 m So radius 7÷ 2mLength of platform 22 mBreadth 14 mVol. Of platform = vol. of cylinder L× b× h= π r2 h22×14× h = 22÷7×7÷2×7÷2×20Solve it | |
| 35056. |
What is thales theorem? |
| Answer» Basic Proportionality Theorem or Thales Theorem.If a line is drawn parallel to one side of a triangle, to interest the other two sides indistinct points, the other two sides are divided in the same ratio. | |
| 35057. |
Sin²35+Sin²55 ÷Sec²15-Cot²75+(tan5. tan15. tan60. tan75. tan85) |
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Answer» But how 1 1 |
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| 35058. |
8 cgepter |
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| 35059. |
एक वृत के चतुथॉश का area मान निकालीए जिसकी परिधि २२सेमी. है? |
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Answer» 9.625 Question in English |
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| 35060. |
The wheels of a car are of diameter 80 cm each..........66km per hour? |
| Answer» Diameter of the wheel = 80 cm{tex}\\therefore {/tex} Radius of the wheel =\xa0{tex}\\frac { 80 } { 2 } = 40 \\mathrm { cm }{/tex}{tex}\\therefore {/tex}Circum ference of the wheel =\xa0{tex}2 \\pi ( 40 ) \\mathrm { cm } = 80 \\pi \\mathrm { cm }{/tex}{tex}\\therefore {/tex}\xa0Distance covered by the wheel in one complete revolution{tex}= 80 \\pi \\mathrm { cm } = 80 \\times \\frac { 22 } { 7 } \\mathrm { cm } = \\frac { 1760 } { 7 } \\mathrm { cm }{/tex}\xa0Distance covered by the wheel in 1 hour\xa0( = 60 minutes) = 66 km = 6600000 cm{tex}\\therefore {/tex}\xa0Distance covered by the wheel in 10 minutes{tex}= \\frac { 6600000 } { 60 } \\times 10 = 1100000 \\mathrm { cm }{/tex}\xa0{tex}\\therefore {/tex} Required number of complete revolution{tex}= \\frac { 1100000 } { \\frac { 1760 } { 7 } } = \\frac { 1100000 \\times 7 } { 1760 } = 4375{/tex} | |
| 35061. |
Ax+by=a-b;bx-ay=a+b by elimination |
| Answer» Given\xa0pair of linear equation is\xa0{tex}ax+by-a+b=0{/tex} .....(i)and {tex}bx-ay-a-b=0{/tex} ........... (ii)Multiplying\xa0{tex}ax+by-a+b=0{/tex}\xa0by a and\xa0{tex}bx-ay-a-b=0{/tex}\xa0by b, and adding them, we geta2x + aby - a2 + ab = 0 and b2x - aby - ab - b2 = 0(a2x + aby - a2 + ab ) + (b2x - aby - ab - b2 ) = 0a2x + aby - a2 + ab + b2x - aby - ab - b2 = 0a2x + b2x - a2- b2 = 0{tex} \\Rightarrow ({a^2} + {b^2})x = ({a^2} + {b^2}){/tex}{tex} \\Rightarrow x = \\frac{{({a^2} + {b^2})}}{{({a^2} + {b^2})}} = 1{/tex}On putting x =1 in first equation, we get{tex}ax+by-a+b=0{/tex}{tex}a + by = a - b {/tex}{tex}\\Rightarrow y = - \\frac{b}{b} = - 1{/tex}Hence, x=1 and y=-1, which is the required unique solution. | |
| 35062. |
1000*4878978*6 |
| Answer» 29274768000 | |
| 35063. |
Is question came from NCERT optional exercises |
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Answer» It does but what matters if we do extra we could do that also? Questions will definitely come from optional exercises No becoj that not from examination point of view? No as they are not from examination point of view Sometimes it does so be prepared |
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| 35064. |
SinA(1+tanA)+cosA(1+cotA)= secA + cosecA . Prove it |
| Answer» =SinA (1 + sinA/cosA) + cosA (1 + cosA/sinA)=sinA + sinA×sinA/cosA + cosA + cosA×cosA/sinA=sinA + sin2a/cosA + cosA + cos2A/sinA=sinA + cos2A/sinA + sin2A/cosA + cosA =Sin2A+cos2A/sinA + sin2A+cos2A/cosA=1/sinA + 1/cosA=cosecA + secAHENCE PROVED... | |
| 35065. |
WhY is the real numbers |
| Answer» Write full questions | |
| 35066. |
State the arithmetic thorem of ch 1 |
| Answer» a=bq+r. | |
| 35067. |
If seven times the 7th terms of ap in equal to elven times the 11th term then what will be the 18 th |
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Answer» 7(a7) = 11(a11) Find a187(a+6d) = 11(a+10d)7a + 42d = 11a + 110d 7a - 11a = 110d - 42d-4a = 68d a= -68/4a = -17da+17d = 0a18 = 0 0 Zero |
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| 35068. |
If (n-2),(4n-1),(5n+1) are consecutive terms of an AP. Then find n. |
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Answer» Formula=2b=a+c,so 2(4n-1)=n-2+5n+1,so, 8n-2=6n-1,8n-6n=-1+2,so,2n=1,so,n=1/2 ans .... d=a2-a1d=a3-a2(4n-1)-(n-2)=(5n+1)-(4n-1)3n+1=n+2n=1/2 Batado koi |
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| 35069. |
State the fundamental theorem of arthmeticsof chapter 1 |
| Answer» The fundamental theorem of arithmetic states that every number\xa0can be represented as the\xa0product of the primes numbers.{tex}Eg: 404=2\\times2\\times101{/tex} | |
| 35070. |
R we have to get (99)$ |
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| 35071. |
Is ko kaise improve aur fast kre |
| Answer» Keise ko | |
| 35072. |
Is all papers of board exams comes from ncert books only |
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Answer» Hi Not all but some questions 100% all questions come from ncert book. Only 33%...!!:-) Mostly Yes. |
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| 35073. |
Find the value of k ,for which one root of the quadratic equation kx^2-14x-8 =0 is 2 |
| Answer» Kx2-14x-8= 0 where x=2 as given so, k(2)2 -14×2-8=0,4K-28-8=0,4k=36,where k=9 . Ok | |
| 35074. |
Find the value of a for which point |
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| 35075. |
How to score good in SSC & Mathematics? |
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Answer» Do again again again Daliy study of mathematics and ssc is the best way. Practice practice and practice |
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| 35076. |
If HCF of 408 & 1032 is expresible in 1032p - 408 * 5 find p |
| Answer» If the HCF of 408 and 1032 can be written as\xa0By Euclid \'s division algorithm,1032 = 408×2 + 216408 = 216×1 + 192216 = 192×1 + 24192 = 24×8 + 0Since the remainder becomes 0 here, so HCF of 408 and 1032 is 24Now,\xa01032x - 408*5 = HCF of these numbers⇒ 1032x - 2040 = 24⇒ 1032x = 24+2040⇒ x = 2064/1032⇒ x = 2So value of x is 2. | |
| 35077. |
How derive the formula for frustrum of cone |
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| 35078. |
The circumference of a circle exceeds its diameter by 120cm, then its radius is? |
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Answer» 28 cm 60 28 cm. |
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| 35079. |
There are 3 circles all are touching each other r1=1.5cm,r2 = 2cm and r3= 3 cm. Find xy |
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| 35080. |
Konse sample paper best h |
| Answer» Bhai koi bhi ek badiya publication ki badiya question bank buy kar lo or fir uska revision kar lo 5-6 baar best h | |
| 35081. |
SecA(1-sinA) (secA+tanA=1 |
| Answer» 1/cosA(1-sinA)(1/cosA+sinA/cosA)=((1-sinA)/cosA)((1+sin)/cosA)=((1-sinA)(1+sinA))/cos2A=(1-sin2A)/cos2A=Cos2a/cos2A=1 | |
| 35082. |
Value of √i^3 |
| Answer» Value of √3 is 1.73 approx | |
| 35083. |
15/5 |
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Answer» 3 3 |
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| 35084. |
πβα |
| Answer» | |
| 35085. |
On which side new sample papers are in maths |
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Answer» Vedantu also Latest books of sample paper are provided to students by teachers |
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| 35086. |
(X+1) =2(x_3) |
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Answer» x-7 X-4 |
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| 35087. |
Prove,Angles in alternate segments of a circle are equal . |
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| 35088. |
Area of the segment formed by the corresponding chord |
| Answer» l/2*r-r^2sin thieta/2*cos thieta/2or area of sector- area of triangle | |
| 35089. |
Rs Aggrawal probability solutions |
| Answer» | |
| 35090. |
find the zero\'s of the polynomials are root 2 & -root 3(x- root2) (x + root2)=0 |
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| 35091. |
All the geometrical figure along with the Relation |
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| 35092. |
Determinants questions |
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| 35093. |
In equation x-2y=0 y= |
| Answer» X/2 | |
| 35094. |
Determined +3 2nd year 4th semester |
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Answer» (a-b)(b-c)(c-a)(ab+bc+ca) Miscellaneous illustration questions |
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| 35095. |
Prive that the area ofcircular path of uniform width h sourrounded a region of radius r πh(2r+h) |
| Answer» Given,radius of inner circle = rwidth of path = hThen radius of outer circle = R= r + h{tex}\\therefore {/tex}\xa0Area of path = Area of outer circle - Area of inner circle{tex} = \\pi {(R)^2} - \\pi {(r)^2}{/tex}{tex} = \\pi {(r + h)^2} - \\pi {r^2}{/tex}{tex} = \\pi ({r^2} + {h^2} + 2rh) - \\pi {r^2}{/tex}{tex} = \\pi {r^2} + \\pi {h^2} + 2\\pi rh - \\pi {r^2}{/tex}{tex} = \\pi {h^2} + 2\\pi rh{/tex}{tex}= \\pi h(h + 2r){/tex} | |
| 35096. |
Prove that the area ofcircular last la all larkspur |
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| 35097. |
Divide 12 into two number such that sum of their square is 74 |
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Answer» Let the two number be a and ba + b = 12a2 + b2 = 74If a = 7 and b = 572 + 52 = 49+ 25 =74 5 and 7 |
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| 35098. |
Exercise 11c in question no_15 |
| Answer» | |
| 35099. |
Math exercise 10.2 question no 10 |
| Answer» Draw a figure as tangent is perpendicular to the radius therefore the sum of those two angles will be 180. Subtract it from 360(ASP Of Quadrilateral) . Your answer is 180. What is required | |
| 35100. |
If a cos-b sin=c, prove that a sin+b cos=+-√a×a+b×b-c×c . |
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