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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.
| 39151. |
Prove that (3+2√5)^2 is irrational . |
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| 39152. |
If a triangle ABC is an equilateral triangle such that AD is perpendicular to BC, then find AD^2 |
| Answer» According to the question,\xa0{tex}\\triangle ABC {/tex}\xa0is an equilateral triangle.In {tex}\\triangle{/tex}ABD, using Pythagoras theorem,{tex}\\Rightarrow{/tex} AB2 = AD2 + BD2{tex}\\Rightarrow{/tex} BC2 = AD2 + BD2, (as AB = BC = CA){tex}\\Rightarrow{/tex} (2 BD)2 = AD2 + BD2, (perpendicular is the median in an equilateral triangle){tex}\\Rightarrow{/tex} 4BD2 - BD2 = AD2{tex}\\therefore{/tex}\xa0AD2\xa0=3BD2 | |
| 39153. |
Show that x=bc/ad is a solution of the quadratic equation ad2(ax/b+2c/d)x+bc2=0. |
| Answer» We have, {tex}ad^2x ({a \\over b}x + {2c \\over d}) + c^2b =0{/tex}{tex}\\implies {a^2d^2 \\over b}x^2 + 2acdx + c^2b = 0{/tex}{tex}\\implies {a^2d^2 \\over b}x^2 + acdx +acdx + c^2b = 0{/tex}{tex}\\implies adx({ad \\over b}x+c) + bc({ad \\over b}x + c) = 0{/tex}{tex}\\implies (adx+ bc) ({ad \\over b}x + c) = 0{/tex}Either adx + bc =0 or {tex}({ad \\over b}x + c) = 0{/tex}{tex}\\implies x = -{bc \\over ad}{/tex}Hence, {tex}x = -{bc \\over ad}{/tex} is the requirreed solution. | |
| 39154. |
5x-7x |
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Answer» _2x -2x -2x 5x-7x = 0 -2x = 0 x = 1/2 |
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| 39155. |
How we know that in 3 case aur tw0 case r in the positive integer |
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| 39156. |
1/4,4 zeros find |
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| 39157. |
What is middle term spilting |
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Answer» When we multiple term (ac) we got number after this we find two number whose sum or substraction is the middle term whose multiplication is multiply is ac Middle term splitting means that where we first multiply a and c then compare with b. Then we form the factors of ac like ex 30=15*2 where b=17 and equation is x2+17x+30 and the we form the zeros |
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| 39158. |
(5+7) |
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Answer» 12.... Its not a class 10 question anyway 12 12 |
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| 39159. |
L.C.M of 32 and 24 |
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Answer» 32/1,24/1 L C M is 1 3 |
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| 39160. |
If the zeros of the polynomial X³-3x²+x+1are a-b, a+B, find a and b |
| Answer» Given polynomial is f(x) = x3\xa0- 3x2\xa0+ x + 1Let\xa0{tex} \\alpha{/tex}\xa0= (a - b),\xa0{tex} \\beta{/tex}\xa0= a and\xa0{tex} \\gamma{/tex}\xa0= (a + b)Now,\xa0{tex} \\alpha + \\beta + \\gamma{/tex}\xa0=\xa0{tex} - \\frac { ( - 3 ) } { 1 }{/tex}⇒\xa0(a - b) + a + ( a + b ) = 3⇒ a - b + a + a+ b = 3⇒ a + a + a = 3⇒\xa03a = 3⇒ a = 3/3⇒\xa0a = 1Also,\xa0{tex} \\alpha \\beta + \\beta y + \\gamma \\alpha = \\frac { 1 } { 1 }{/tex}⇒\xa0(a - b)a + a (a + b) + (a + b)(a - b) = 1\xa0⇒\xa0a2\xa0- ab + a2\xa0+ab + a2\xa0- b2\xa0= 1⇒\xa03a2\xa0- b2\xa0= 1 ( ∵ a = 1)⇒\xa03(1)2\xa0- b2\xa0= 1( ∵ a = 1)⇒ 3 - b2 = 1⇒\xa0b2\xa0= 2⇒\xa0b =\xa0{tex} \\pm \\sqrt{2}{/tex}Hence, a = 1 and b =\xa0{tex} \\pm \\sqrt{2}{/tex} | |
| 39161. |
2+5-6 |
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Answer» 1 is answer 3 1 1 1 is answerAre you really a student of class 10 |
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| 39162. |
5 +09 |
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Answer» +14 14 14 14 |
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| 39163. |
Use Eucli\'s algorithm to find the H.C.F. of 420 and 130 |
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Answer» Therefore HCF=10 420=130×3+30130=30×4+1030=10×3+0 |
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| 39164. |
x/a^2 + y/b^2 = 2 (solve for x and y) |
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| 39165. |
Show that only one of the number n,n+2 and n+4 is divisible by 3 |
| Answer» Please give me the answer | |
| 39166. |
What is the relation between mean, mode and median |
| Answer» Mean - Mode — 3(Mean - Meadian) | |
| 39167. |
What is Euclid\'s Lemma |
| Answer» Hfh | |
| 39168. |
What is the formula of Lemma |
| Answer» A=bq+r | |
| 39169. |
All chapters formulas in simple way |
| Answer» Check formulae in revision notes :\xa0https://mycbseguide.com/cbse-revision-notes.html | |
| 39170. |
What is mean of root3 |
| Answer» 1.732 | |
| 39171. |
5+√3 it is irrational number prove that |
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Answer» Your Welcome ? Let ✓3 be rational Therefore ✓3 = a/b where a and b are co primes From the above eq ✓3b = a This shows that ✓3 is a. factor of a This contradicts the fact that a is a co prime -1st statement b= a /✓3 This also shows that ✓3 is a factor of b This contradicts the fact that b is a co prime -2nd statement From the statements 1 and 2 we can conclude that ✓ 3 is not a rational number and therefore our assumption is wrong.Hence ✓ 3 is irrational The addition of a rational and an irrational number is always irrational So therefore 5+✓3 is also irrationalHence proved Sssssu |
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| 39172. |
Q. If a right triangle ABC, right-angled at B, if tan A=1 then verify that 2sin A × cos A =1 . |
| Answer» In {tex} \\triangle A B C{/tex},{tex}\\tan A = 1{/tex}{tex}\\Rightarrow \\quad \\frac { B C } { A C } = 1{/tex}{tex}\\Rightarrow {/tex}\xa0BC = x and\xa0AC = xUsing Pythagoras theorem,\xa0{tex}\\Rightarrow A B ^ { 2 } = A C ^ { 2 } + B C ^ { 2 }{/tex}{tex}\\Rightarrow \\quad A B ^ { 2 } = x ^ { 2 } + x ^ { 2 }{/tex}{tex}\\Rightarrow \\quad A B = \\sqrt { 2 } x{/tex}{tex}\\therefore \\quad \\sin A = \\frac { B C } { A B } = \\frac { x } { \\sqrt { 2 } x } = \\frac { 1 } { \\sqrt { 2 } } \\text { and } \\cos A = \\frac { A C } { \\sqrt { 2 } x } = \\frac { x } { \\sqrt { 2 } x } = \\frac { 1 } { \\sqrt { 2 } } {/tex}2 sin A cos A\xa0{tex}= 2 \\times \\frac { 1 } { \\sqrt { 2 } } \\times \\frac { 1 } { \\sqrt { 2 } } = 1{/tex} | |
| 39173. |
(m²+n²)x² -2(mp+np)x + (b² + c²) =0 |
| Answer» We have the following equation,{tex}(m^2+n^2)x^2-2(mp+nq)x+p^2+q^2=0{/tex}where, a =\xa0(m2 + n2), b =\xa0-2(mp + nq), and c = p2 + q2The given equation will have equal roots,if D = 0{tex}\\Rightarrow{/tex}\xa0{tex}b^2-4ac=0{/tex}{tex}\\Rightarrow{/tex}{tex}[-2(mp+nq)]^2-4(m^2+n^2)(p^2+q^2)=0{/tex}{tex}\\Rightarrow{/tex}{tex}4m^2p^2+4n^2q^2+8mnpq-4m^2p^2-4m^2q^2-4n^2q^2-4n^2p^2=0{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}-4m^2q^2-4p^2n^2+8mnpq=0{/tex}\xa0{tex}\\Rightarrow{/tex}{tex}-4(m^2q^2+p^2n^2-2mnpq)=0{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}m^2q^2+p^2n^2-2mnpq=0{/tex}\xa0{tex}\\Rightarrow{/tex}\xa0{tex}(mq-pn)^2=0{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}mq-pn=0{/tex}\xa0{tex}\\Rightarrow{/tex}{tex}mq=pn{/tex} | |
| 39174. |
Class 10 cbse pattern 2018-19 pleas tell |
| Answer» Check exam pattern here :\xa0https://mycbseguide.com/cbse-syllabus.html | |
| 39175. |
How to prove Thale\'s therome |
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Answer» Let be ∆ABC in which D and E are the mid points of AB and AC respectivily. BC parrallel to DE. in triangle ADE draw the two perpendicular on AD and AE. And D point meet with C and point E meet with B . And DM and NE are the two perpendicular on side AE And AD respectively.In ∆ADE, Ar ∆ ADE = 1/2 X AD X NE. (1)AND, AR ∆ ADE= 1/2 X AE X MD (2) And in AR. ∆ BCD = 1/2 X NE X BD (3) IN AR. ∆ DCE = 1/2 X MD X EC. (4) FROM 1 and 3AR ∆ ADE = AD_________________AR. ∆ BCD = BD. FR0M 2and 4AR.∆ ADE = AE________________Ar. ∆ DCE = EC. Ar.∆ BCD = Ar∆ DCE. (Both triangles lie on same base and parallel lines)AR. ∆ ADE. AR.∆ ADE___________ Equals to (=)____________Ar. ∆BCD. AR.∆ BCEHence, AD. AE ____. = ______ BD. CE By BPT theorem. |
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| 39176. |
Addition 22+42 |
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Answer» Sixty four ( 64 )? Are your eally a student of class 10th? 64 |
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| 39177. |
By using completing the square method,solve:a²x²-3ab+b² |
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| 39178. |
Acidity |
| Answer» Joine the group to get a satisfying answer and many material for your 10th preparation like sample paper, notes practise set and many more click the link to join the group "https://chat.whatsapp.com/9FLm03yKyEMLvBXqjE0v5r". Fast or message on number 7258970378 to join in the group.fast | |
| 39179. |
,,22+2 |
| Answer» 22 + 2 = 24 | |
| 39180. |
6x+3y=6xy2x+4y=5xy |
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Answer» 6x+3y=6xy ____ eqn. 12x+4y=5xy_____ eqn. 2 6x+3y/xy=6 (from eqn. 1), 6x/xy+3y/xy=66/y+3/x=62x+4y/xy=5 (from eqn. 2)2x/xy+4y/xy=52/y+4/x=5Let 1/x be p and 1/y be qSo, 6q+3p=6______ eqn. 3 2q+4p=5______eqn. 4By elimination method Multiply eqn. 4 by 3 3*2q+4p=56q+12p=15 _______5 Now subtract eqn. 3 from 5 6q+3p=6 6q+12p=15 - - -______________ 0 +9p= 9 p=9/9 p= 1 Put the value of p in eqn. 42q+4*1=5 2q= 5-4 q=1/2p=1=1/x p=x=1q=1/2=1/yq=y=2 I don\'t no |
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| 39181. |
Definition of consuctive numbers |
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| 39182. |
Find the other zeros of the polynomial 4x³+3x²+7x+1 if one of its zero is √5 |
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| 39183. |
A-b |
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Answer» A-b=A-b Silly question bro? What?? A-b!! Write ur ques properly brthr... |
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| 39184. |
3_5root 2 |
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| 39185. |
Evaluate sin²60° +cos²30°+tan²45°+sec²60°-cosec²30° |
| Answer» Ghjj | |
| 39186. |
The difference of 2 numbers is 4 and difference of their reciprocals is 4/2. Find the numbers |
| Answer» Let the two numbers be x and x – 4.Given that the difference of two numbers is 4.By the given hypothesis, we have\xa0{tex}\\frac{1}{x-4}-\\frac{1}{x}=\\frac{4}{21}{/tex}{tex}\\Rightarrow \\frac{x-x+4}{x(x-4)}=\\frac{4}{21}{/tex}{tex}\\Rightarrow 84 = 4x(x-4){/tex}{tex}\\Rightarrow x^2 - 4x -21=0{/tex}{tex}\\Rightarrow x^2 -7x+3x-21{/tex}{tex}\\Rightarrow (x-7)(x+3) = 0{/tex}{tex}\\Rightarrow x=7 \\ or\\ x=-3{/tex}If x = 7, then\xa0{tex}x-4=7-4=3{/tex}If\xa0x = -3, then\xa0{tex}x-4=-3-4=-7{/tex}Hence, required numbers are 3,7 and -3,-7 | |
| 39187. |
Which sample paper is best for 10 th exam |
| Answer» Google | |
| 39188. |
The sum of two number is 9 the sum of reciprocal is 1/2 |
| Answer» 6 & 3 | |
| 39189. |
Find the zeros of polynomial x3 - 5x2 - 2x+24 if it is given that the product of its two zeros is12 |
| Answer» Let {tex} \\alpha , \\beta , \\gamma{/tex}\xa0be the zeroes of polynomial f(x) = x3 - 5x2 - 2x + 24 such that {tex} \\alpha\\beta{/tex}\xa0= 12.{tex} \\alpha + \\beta + \\gamma = - \\left( - \\frac { 5 } { 1 } \\right) = 5{/tex} ................ (i){tex} \\alpha \\beta + \\beta \\gamma + \\gamma \\alpha = \\frac { - 2 } { 1 } = - 2{/tex}and,\xa0{tex} \\alpha \\beta \\gamma = - \\frac { 24 } { 1 } = - 24{/tex}Putting,\xa0{tex} \\alpha \\beta = 12{/tex}\xa0in\xa0{tex} \\alpha \\beta \\gamma = - 24{/tex}, we get{tex} 12 \\gamma = - 24{/tex}{tex} \\Rightarrow \\gamma = - \\frac { 24 } { 12 } = - 2{/tex}Putting\xa0{tex} \\gamma= -2{/tex} in eq.(i), we get{tex} \\alpha + \\beta - 2 = 5{/tex}{tex} \\Rightarrow \\quad \\alpha + \\beta = 7{/tex}Now,\xa0{tex} ( \\alpha - \\beta ) ^ { 2 } = ( \\alpha + \\beta ) ^ { 2 } - 4 \\alpha \\beta{/tex}{tex} \\Rightarrow \\quad ( \\alpha - \\beta ) ^ { 2 } = 7 ^ { 2 } - 4 \\times 12{/tex}{tex} \\Rightarrow \\quad ( \\alpha - \\beta ) ^ { 2 } = 1{/tex}{tex} \\Rightarrow \\quad \\alpha - \\beta = \\pm 1{/tex}Thus, we have{tex} \\alpha + \\beta= 7{/tex} and {tex} \\alpha - \\beta = 1 {/tex} or, {tex} \\alpha + \\beta= 7{/tex} and {tex} \\alpha - \\beta= -1{/tex}CASE I: When {tex} \\alpha + \\beta= 7{/tex} and {tex} \\alpha - \\beta= 1{/tex}Solving {tex} \\alpha + \\beta= 7{/tex} and {tex} \\alpha - \\beta= 1{/tex} , we get{tex} \\alpha = 4{/tex} and {tex}\\beta= 3{/tex}\xa0CASE II: When {tex} \\alpha + \\beta= 7{/tex} and {tex} \\alpha - \\beta= -1{/tex}Solving {tex} \\alpha + \\beta= 7{/tex} and {tex} \\alpha - \\beta= -1{/tex} , we get{tex} \\alpha = 3{/tex} and {tex}\\beta= 4{/tex}\xa0.Hence, the zeros of the given polynomial are 3, 4 and - 2 or 4,3 , -2. | |
| 39190. |
The nth term of the AP is 6n+2. Find the common difference |
| Answer» 6 | |
| 39191. |
Class 10th cbse me jitne questions syllabus me rahte hai vo usi me se aata hai |
| Answer» Yes, you can check the syllabus here :\xa0https://mycbseguide.com/cbse-syllabus.html | |
| 39192. |
2-7 |
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Answer» Are you just checking does it work or you are a kid. -5 |
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| 39193. |
मैटरिक्स क्या ह |
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| 39194. |
,show that 38 and 45 are prime numbers |
| Answer» 38 is divisible by 2 and 45 is divisible by 5 that\'s why it is not prime no. | |
| 39195. |
What is compodendo and dividendo |
| Answer» If 4 numbers are proportion then it is written as a/b = c/d By componendo , we have (a+b)/b = (c+d)/dBy dividendo , we have (a-b)/b = (c-d)/dAnd by componendo and dividendo , we have (a+b)/(a-b) = (c+d)/(c-d). | |
| 39196. |
To prove that ac sq = ab sq + bc sq |
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| 39197. |
The value of under root3 is1.73 so what is the vslue of under root 5? |
| Answer» The value of √5=2.236!! | |
| 39198. |
If in triangle ABC , EF is parallel to BC Then can we write By BPT,AE/AB=AF/AC? |
| Answer» Yes | |
| 39199. |
To prove that 1uponf =1by u - 1by v |
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| 39200. |
Do char vale raikhik samikrn vugm |
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