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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.
| 36151. |
Prove that:(1/4-3)³= 4 raised to the power 9 |
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| 36152. |
Cos2A/1-tanA + Sin3A/sinA - cosA = 1 + sinAcosA |
| Answer» Bhai question wrong h . | |
| 36153. |
Alpha + beta equal to dash |
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Answer» -Cofficient of x / cofficient of xsquare -b/a |
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| 36154. |
Formula of sum of AP |
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Answer» Sn=n/2[2a+(n-1)d] Sn= n/2(2a+(n -1)d) Sn = n/2(a+l) If the last term is given we should use sn=n/2(2a+(n-1)d) Sn =n/2(2a+(n-1))d |
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| 36155. |
Find the greatest number of six digits exactly divisible by 18,24 and 36 |
| Answer» LCM of 18,24 and 3618 = 2 {tex}\\times{/tex}\xa03224 = 23{tex}\\times{/tex}\xa0336 = 22{tex}\\times{/tex}\xa032LCM (18,24,36) = 23\xa0{tex}\\times{/tex}\xa032 = 72Hence if any number is divisible by 72 that will be divisible by 18,24 and 36 also.Now the largest 6 digit number is 999999On dividing 999999 by 72 we get{tex}\\;999999=13888\\times72+\\;63=999936+63{/tex}Hence 999936 is the greatest 6 digit number divisible by 18,24 and 36. | |
| 36156. |
How to draw 6tracks of 200mtrs and place the staggers |
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| 36157. |
In coordinate geometry ex 7.2 2ndOne how the answer will come 39 ans |
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| 36158. |
if x = asec + btan and y = atono + bsec Prove that x2 -y2 = a2-b2 |
| Answer» We\xa0have,x = asec{tex}\\theta{/tex}\xa0+ btan{tex}\\theta{/tex}\xa0and y = atan{tex}\\theta{/tex}\xa0+ bsec{tex}\\theta{/tex}LHS = (x2\xa0- y2) = (asec{tex}\\theta{/tex}\xa0+ btan{tex}\\theta{/tex})2\xa0- (atan{tex}\\theta{/tex}\xa0+ bsec{tex}\\theta{/tex})2= (a2sec2{tex}\\theta{/tex}\xa0+ b2tan2{tex}\\theta{/tex}\xa0+ 2absec{tex}\\theta{/tex}tan{tex}\\theta{/tex}) -\xa0(a2tan2{tex}\\theta{/tex}\xa0+ b2sec2{tex}\\theta{/tex}\xa0+ 2absec{tex}\\theta{/tex}tan{tex}\\theta{/tex})= a2(sec2{tex}\\theta{/tex}\xa0- tan2{tex}\\theta{/tex}) - b2(sec2{tex}\\theta{/tex}\xa0- tan2{tex}\\theta{/tex})= a2\xa0- b2\xa0= RHSTherefore, LHS = RHS | |
| 36159. |
3+11---------803 |
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Answer» ?? AP find karna hai easy hai ye Dekho (a) diya hua hai 3 aur (d) ho jaiye ga 11-3 = 8 Aur Tn = 803 hai bas formula pe rakhoo Tn = a + (n-1)d803 = 3 + (n-1)8 |
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| 36160. |
What is the formula of a square + b square |
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Answer» (a+b) ka hole square-2ab (a+b)ka hole square (a+b)(a+b) |
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| 36161. |
D is a point on the side BC of a triangle ABC suct that angle ADC=angle BAC. Show that CA2=CB.CD |
| Answer» ∆ADC and ∆BACAngle ADC=angle BACAngle C =angle CHence ∆ADC~∆BACCA/CD=CB/CAHence CA2=CB.CDProved | |
| 36162. |
What is the value of sin45 |
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Answer» 1/√2 1/√2 1/√2 1/√2 |
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| 36163. |
Prove that √3 is an irrational number |
| Answer» Find it by assuming it to be rational in the form p/q. Prove that the assumption is wrong by substituting and equating wherever necessary! Simple!!!☺️ | |
| 36164. |
What is the value of infinity |
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Answer» Or bhi aage khud dhund le 1000009o00098990000000000000000000000000000000000000000 Not defined Infinite |
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| 36165. |
Can two numbers have 15 as their HCF and175 as their LCM? Give reason |
| Answer» {tex}\\frac{175}{15}=11.667{/tex}Hence 175 is not divisible by 15But LCM of two numbers should be divisible by their HCF.{tex}\\therefore{/tex}\xa0Two numbers cannot have their HCF as 15 and LCM as 175. | |
| 36166. |
A number when divided by 53 gives 33 as quotient and 19 as ramainder. find the number |
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Answer» 1768 First of all we must know thatDivisior×Quotent+Remainder=A53×33+19=1768 1768 |
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| 36167. |
2x+9x=244x_6x=25 |
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| 36168. |
Prove that *√(sec theta -1/sec theta +1) +√(sec theta -1/sec theta +1) = 2cosec theta * |
| Answer» LHS ={tex}\\sqrt { \\frac { \\sec \\theta - 1 } { \\sec \\theta + 1 } } + \\sqrt { \\frac { \\sec \\theta + 1 } { \\sec \\theta - 1 } }{/tex}\xa0Rationalise the denominator and we get,\xa0{tex}= \\sqrt{\\frac{(sec\\theta-1)^2}{sec^2\\theta-1}}+ \\sqrt{\\frac{(sec\\theta+1)^2}{sec^2\\theta-1}}{/tex}=\xa0{tex}\\frac { ( \\sec \\theta - 1 ) + ( \\sec \\theta + 1 ) } { \\sqrt { ( \\sec \\theta + 1 ) ( \\sec \\theta - 1 ) } }{/tex}=\xa0{tex}\\frac { 2 \\sec \\theta } { \\sqrt { \\sec ^ { 2 } \\theta -1 } } = \\frac { 2 \\sec \\theta } { \\sqrt { \\tan ^ { 2 } \\theta } } = \\frac { 2 \\sec \\theta } { \\tan \\theta }{/tex}=\xa0{tex}2 \\times \\frac { 1 } { \\cos \\theta } \\times \\frac { \\cos \\theta } { \\sin \\theta }{/tex}=\xa0{tex}2 \\times \\frac { 1 } { \\sin \\theta }{/tex}= 2 cosec{tex}\\theta{/tex}= RHSHence Proved | |
| 36169. |
cos0-sin0 |
| Answer» -1 | |
| 36170. |
If CosecA+cotA=mShow that - M square - 1 / 1 + m square =cosA |
| Answer» Given: cosec A + cot A = m{tex}\\Rightarrow{/tex}\xa0(cosec A + cot A)2 = (m)2 \xa0[squaring both sides ]{tex}\\Rightarrow{/tex}\xa0cosec2A + cot2A + 2 cosec A cot A = m2\xa0\xa0.......(1)Now, LHS\xa0{tex}=\\frac{m^{2}-1}{m^{2}+1}{/tex}{tex}=\\frac{\\ cosec ^{2} A+\\cot ^{2} A+2 \\ cosec A \\cot A-1}{\\ cosec ^{2} A+\\cot ^{2} A+2 \\ cosce A \\cdot \\cot A+1}{/tex}. [ From (1) ]{tex}=\\frac{\\cot ^{2} A+\\cot ^{2} A+2 \\ cosec A \\cdot \\cot A}{\\ cosec ^{2} A+\\ cosec ^{2} A+2 \\ cosec A \\cdot \\cot A}{/tex} [Since, Cosec2A - Cot2A = 1]{tex}=\\frac{2 \\cot ^{2} A+2 \\ cosec A \\cot A}{2 \\ cosec ^{2} A+2 \\ cosec A \\cot A}{/tex}{tex}=\\frac{2 \\cot A(\\cot A+\\ cosec A)}{2 \\ cosec A(\\ cosec A+\\cot A)}{/tex}{tex}=\\frac{\\cot A}{cosec A}{/tex}{tex}=\\frac{\\frac{\\cos A}{\\sin A}}{\\frac{1}{\\sin A}}{/tex}{tex}=\\frac{\\cos A}{\\sin A} \\times \\frac{\\sin A}{1}{/tex}= cos A = RHSHence, Proved. | |
| 36171. |
What is tha percentag of 250 out of 180 |
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Answer» ye kaisa question h 60% 72% There can be no prcentage.....if u want,then, the percentage would be 138.8 |
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| 36172. |
What is the percentag 100 out of 38 |
| Answer» 38% | |
| 36173. |
If (tan theta+sin theta)=m and (tan theta -sin theta )=n.prove that (m^2-n^2)^2=16mn |
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Answer» Theta is angle Tan +theta Plzzz solve this ques |
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| 36174. |
What is the value of x for X + 4 is equals to 10 and minus 4 is equal to 5 |
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| 36175. |
1root of the equstion 2x square - 8x - m is5/2 find the other root s the value of m |
| Answer» {tex}2x^2 - 8x - m = 0{/tex}{tex}\\because{/tex}{tex}\\frac{5}{2}{/tex}\xa0is one root2({tex}\\frac{5}{2}{/tex})2 - 8{tex}\\times{/tex}\xa0{tex}\\frac{5}{2}{/tex}\xa0{tex}- m = 0{/tex}\xa0{tex}\\Rightarrow{/tex}\xa0{tex}\\frac{2 \\times 25}{4}{/tex}\xa0{tex}- 20 - m = 0{/tex}{tex}\\Rightarrow{/tex}{tex}\\frac{25 - 40}{2}{/tex}\xa0{tex}- m = 0{/tex}\xa0{tex}\\Rightarrow{/tex}\xa0{tex}\\frac{-15}{2}{/tex}\xa0= m\xa0{tex}\\Rightarrow{/tex}m =\xa0{tex}\\frac{-15}{2}{/tex}.Also, sum of roots =\xa0-{tex}\\frac{b}{a}{/tex}\xa0=\xa0{tex}\\frac{-(-8)}{2}{/tex}\xa0= 4One root =\xa0{tex}\\frac{5}{2}{/tex}\xa0[given]{tex}\\therefore{/tex}\xa0Other root = 4 -\xa0{tex}\\frac{5}{2}{/tex}\xa0=\xa0{tex}\\frac{3}{2}{/tex} | |
| 36176. |
Example 7 of chapter 4 page no 80 |
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| 36177. |
tan2A + cot2A = sec2A cosec2A - 2 prove it. |
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| 36178. |
2.2 quation no.2 |
| Answer» 2.(i) 1/4,1According to master(suhail) FormulaX2-(Some Of zeroes)X + (Product of Zeroes)Note=X2 where 2 is power of X X2-(1/4)x + (-1)Now LCM 4X2-X-4ansNote X is a Veriableand 2 is his Power | |
| 36179. |
Please solve for x where x≠-1,-2,-4[1/x+1]+[2/x+2]=4/x+4 |
| Answer» After The lots of TrialYour Answer Is2x2+4x+10-3xyNot->x is a veriable And 2x2 means 2x(2 is power of x) | |
| 36180. |
A(5,1);B(1,5)andC(-3,-1)are the vetices of ∆ABC.find the length of median AD. |
| Answer» √1 | |
| 36181. |
What is an |
| Answer» Last term of ap | |
| 36182. |
Find a cubic polynomial whose zeroes are 3,1/2and-1 |
| Answer» Let\xa0{tex}\\alpha{/tex} = 3, {tex}\\beta{/tex}\xa0=\xa0{tex}\\frac{1}{2}{/tex} and {tex}\\gamma{/tex}\xa0= -1. Then,{tex}( \\alpha + \\beta + \\gamma ) = \\left( 3 + \\frac { 1 } { 2 } - 1 \\right) = \\frac { 5 } { 2 }{/tex},{tex}( \\alpha \\beta + \\beta \\gamma + \\gamma \\alpha ) = \\left( \\frac { 3 } { 2 } - \\frac { 1 } { 2 } - 3 \\right) = \\frac { - 4 } { 2 }{/tex}\xa0= -2and\xa0{tex}\\alpha \\beta y = \\left\\{ 3 \\times \\frac { 1 } { 2 } \\times ( - 1 ) \\right\\} = \\frac { - 3 } { 2 }{/tex}The polynomial with zeros α,β and\xa0{tex}\\gamma{/tex} is:{tex}\\mathrm x^3-(\\mathrm\\alpha+\\mathrm\\beta+\\mathrm\\gamma)\\mathrm x^2+(\\mathrm{αβ}+\\mathrm{βγ}+\\mathrm{γα})\\mathrm x-\\mathrm{αβγ}{/tex}{tex}=\\mathrm x^3-\\frac52\\mathrm x^2-2\\mathrm x+\\frac32{/tex}Thus,\xa02x3- 5x2- 4x + 3 is the desired polynomial. | |
| 36183. |
solve :- sin 63° + cos 27°cos17°+ sin 73° |
| Answer» 2 | |
| 36184. |
Hi Everyone,Koi aisa hai Jo mujhey Math Padha sakey ? |
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Answer» Yes I am Ye app |
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| 36185. |
20+20 |
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Answer» 20+20. 40 20+20=40 |
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| 36186. |
2-3=? |
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Answer» —1 -1 -1 |
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| 36187. |
2 x = -5y |
| Answer» X-0 ;y-0 | |
| 36188. |
2x/x-3 + 1/2x+3 + 3x+9/(x-3)(2x+3) |
| Answer» Given, {tex}\\frac{2x}{x-3}+\\frac{1}{2x+3}+\\frac{3x+9}{(x-3)(2x+3)}=0{/tex}{tex}\\Rightarrow\\frac{{2x(2x + 3) + x - 3 + 3x + 9}}{{(x - 3)(2x + 3)}} = 0{/tex}{tex} \\Rightarrow \\frac{{4{x^2} + 6x + x - 3 + 3x + 9}}{{2{x^2} + 3x - 6x - 9}} = 0{/tex}{tex} \\Rightarrow \\frac{4{x^2} + 10x +6 }{{2{x^2} - 3x - 9}} = 0{/tex}Cross multiplying equation , we get ,{tex} \\Rightarrow 4{x^2} + 10x +6 = 0{/tex}{tex} \\Rightarrow 4{x^2} + 6x+4x +6 = 0{/tex}{tex} \\Rightarrow x(4{x} + 6)+1(4x +6) = 0{/tex}{tex}\\Rightarrow(x+1)=0{/tex} or {tex} (4x+6)=0{/tex}{tex}\\Rightarrow x=-1{/tex}{tex}\\Rightarrow x=-\\frac{3}{2}{/tex}Hence , the roots of the given quadratic equation are -1 and {tex}-\\frac{3}{2}{/tex} | |
| 36189. |
If x=rsinAcosB,y=rsinAsinB andz=rcosA, then prove that r sq=x sq+y sq+z sq |
| Answer» Since, x2 = r2sin2Acos2By 2 = r2sin2A sin2Band z2 = r2cos2AL.H.S. = x2 + y2 + z2{tex}=r^2sin^2Acos^2B+r^2sin^2Asin^2B+r^2cos^2A{/tex}=\xa0{tex}r^2sin^2A(cos^2B+sin^2B)+r^2cos^2A{/tex}= r2sin2 A + r2cos2A= r2(sin2 A + cos2A)= r2.= R.H.SHence Proved. | |
| 36190. |
Solve x+1/x=3 by completing square method |
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| 36191. |
(2x+3y)-(2x-3y)=4 |
| Answer» 2x+3y-2x+3y=4 3y+3y=46y=4Y=2/3Now, put value of y in this eq 2x+3y=02x+3x2/3=02x=-2X=-1 | |
| 36192. |
If m = cosecA-sinA and n= secA -tanA prove that (m^2 n)^2/3+(mn^3)^2/3=1 |
| Answer» We have,cosec{tex}\\theta{/tex}\xa0- sin{tex}\\theta{/tex}\xa0= m and sec{tex}\\theta{/tex}\xa0- cos{tex}\\theta{/tex}\xa0= n{tex}\\Rightarrow \\quad \\frac { 1 } { \\sin \\theta } - \\sin \\theta = m \\text { and } \\frac { 1 } { \\cos \\theta } - \\cos \\theta{/tex}\xa0= n{tex}\\Rightarrow \\quad \\frac { 1 - \\sin ^ { 2 } \\theta } { \\sin \\theta } = m \\text { and } \\frac { 1 - \\cos ^ { 2 } \\theta } { \\cos \\theta }{/tex}\xa0= n{tex}\\Rightarrow \\quad \\frac { \\cos ^ { 2 } \\theta } { \\sin \\theta } = m \\text { and } \\frac { \\sin ^ { 2 } \\theta } { \\cos \\theta }{/tex}\xa0= n{tex}\\therefore \\quad \\left( m ^ { 2 } n \\right) ^ { 2 / 3 } + \\left( m n ^ { 2 } \\right) ^ { 2 / 3 } = \\left( \\frac { \\cos ^ { 4 } \\theta } { \\sin ^ { 2 } \\theta } \\times \\frac { \\sin ^ { 2 } \\theta } { \\cos \\theta } \\right) ^ { 2 / 3 } + \\left( \\frac { \\cos ^ { 2 } \\theta } { \\sin \\theta } \\times \\frac { \\sin ^ { 4 } \\theta } { \\cos ^ { 2 } \\theta } \\right) ^ { 2 / 3 }{/tex}= (cos3{tex}\\theta{/tex})2/3\xa0+ (sin3{tex}\\theta{/tex})2/3\xa0= cos2{tex}\\theta{/tex}\xa0+ sin2{tex}\\theta{/tex}\xa0= 1Hence, (m2n)2/3 + (mn2)2/3 = 1 | |
| 36193. |
X/a+y/b=a+bX/a²+y/b²=ab |
| Answer» | |
| 36194. |
Nnxnnd |
| Answer» What do you want? | |
| 36195. |
Prove that,(secA_cosecA)(1+tanA+cotA)=tanA×secA_cotA×cosecA |
| Answer» | |
| 36196. |
What is curved surface of cone |
| Answer» C.S.A of cone is πrl, here π is 22/7 or 3.142 and r is radius of cone and l is slant height of cone. | |
| 36197. |
In triangle ABC DE parallel to BC, BE and CD intersect at F, AD:BD = 5:4, find Ar(def)/ar(cfb) |
| Answer» Let AD = 5x cm and DB = 4x cm. Then AB = AD + DB = (5x + 4x) cm = 9x cm.In\xa0{tex}\\triangle ADE{/tex}\xa0and\xa0{tex}\\triangle ABC{/tex}, we have{tex}\\angle A D E = \\angle A B C{/tex}\xa0[corresponding angles]{tex}\\angle A E D = \\angle A C B{/tex}\xa0[corresponding angles]{tex}\\angle A = \\angle A {/tex}\xa0[common in both triangles]\xa0{tex}\\therefore \\quad \\triangle A D E \\sim \\triangle A B C{/tex}\xa0[By AAA - Similarity Criteria ]{tex}\\Rightarrow \\frac { D E } { B C } = \\frac { A D } { A B }{/tex}[ As corresponding sides of similar triangles are proportionate to each other]\xa0{tex}= \\frac { 5 x } { 9 x } = \\frac { 5 } { 9 }{/tex}In\xa0{tex}\\triangle DFE{/tex}\xa0and\xa0{tex}\\triangle CFB{/tex}\xa0, we have{tex}\\angle E D F = \\angle B C F{/tex}\xa0[alternate interior angles as DE|| BC ]and\xa0{tex}\\angle D E F = \\angle C B F{/tex}\xa0[alternate interior angles as DE || BC].\xa0{tex}\\therefore \\quad \\triangle D F E \\sim \\triangle C F B{/tex}\xa0( By AA similarity Criteria)\xa0{tex}\\Rightarrow \\quad \\frac { \\operatorname { ar } ( \\Delta D F E ) } { \\operatorname { ar } ( \\Delta C F B ) } = \\frac { D E ^ { 2 } } { C B ^ { 2 } } = \\frac { D E ^ { 2 } } { B C ^ { 2 } }{/tex}{tex}= \\left( \\frac { D E } { B C } \\right) ^ { 2 }{/tex}{tex}= \\left( \\frac { 5 } { 9 } \\right) ^ { 2 } = \\frac { 25 } { 81 }{/tex}{tex}\\Rightarrow \\quad \\operatorname { ar } ( \\triangle D F E ) : \\operatorname { ar } ( \\triangle C F B ){/tex}= 25 : 81. | |
| 36198. |
If sin(2x+3y)=1and cos(2x-3y)=root 3÷2..find the value of x and y |
| Answer» 2x+3y=90. (Sin 90°=1). (1)2x-3y=√3/2 (cos 30°=√3/2). (2) Solving both the equation, we get 2x+3y=902x-3y=30= 4x=120= x=120/4= x= 30Now, Putting the value of x=30 in (1)2×30+3y=9060+3y=903y=30Y=10Therefore, the value of x=30 and y=10. | |
| 36199. |
X+y=5and 2x-3y=4 |
| Answer» X+y=5. (1)2x-3y=4. (2) Multiplying (1) by (2) 2(x+y)=2×5 2x+2y=10. (3) Subtracting (3) from (2) 2x+2y=10 2x-3y=4 = -5y=-6 = y=6/5Putting y=6/5 in (1) X+6/5=5 5x+6=25 5x=25-6 X= 19/5Therefore, the value of x is 19/5 and y is 6/5. | |
| 36200. |
In figure l || m ,angle OAC=80° ,angle ODB =70° . Is ∆ OCA SIMILAR ∆ODB |
| Answer» | |