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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.
| 4001. |
Find the coordinate of the point on y-axis which is neatest to the point (-2,5) |
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Answer» (0,5) (0,5) (0,5) ( 0, 5) (0,5) |
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| 4002. |
Find the LCM and HCF of 26676 and 337554 |
| Answer» Ispe kese kre?? | |
| 4003. |
The sum of two number is 8. |
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Answer» 6+2=8 4+4=8 |
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| 4004. |
Why has |
| Answer» | |
| 4005. |
Hhe |
| Answer» | |
| 4006. |
Solve x/a-y/b=0 and ax+by=a²+b² |
| Answer» {tex}\\frac { x } { a } - \\frac { y } { b } = 0{/tex}{tex} \\Rightarrow x = \\frac { a y } { b }{/tex}....................(i)ax + by = (a2\xa0+ b2) .......................(ii)Substituting (i) in (ii),we get{tex}a \\left( \\frac { a y } { b } \\right) + b y = \\left( a ^ { 2 } + b ^ { 2 } \\right){/tex}{tex}\\Rightarrow \\frac { a ^ { 2 } y } { b } + b y = \\left( a ^ { 2 } + b ^ { 2 } \\right){/tex}{tex}\\Rightarrow a ^ { 2 } y + b ^ { 2 } y = \\left( a ^ { 2 } b + b ^ { 3 } \\right){/tex}{tex}\\Rightarrow y \\left( a ^ { 2 } + b ^ { 2 } \\right) = b \\left( a ^ { 2 } + b ^ { 2 } \\right){/tex}{tex}\\Rightarrow y = b{/tex}Substituting in (i),we get{tex} x = \\frac { a b } { b }{/tex}\xa0⇒ x = a.So, solution of given equation is x = a and y = b. | |
| 4007. |
12+12 |
| Answer» 24 | |
| 4008. |
How is your xam preprations |
| Answer» Going on | |
| 4009. |
b sinA+ a cosA= c prove that a sinA - b cosA = +- under root a^2+b^2-c^2 |
| Answer» | |
| 4010. |
SecA+TanA=p find SecA-TanA |
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Answer» Whole solution plzzzzzzzz 1/p |
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| 4011. |
Any technique to learn the values of trigonometric ratio (ie ,sinA=0) |
| Answer» | |
| 4012. |
Prove that an=Sn-Sn-1 |
| Answer» consider RHS i.e Sn-Sn-1 | |
| 4013. |
Can I get Blue print of mathematics |
| Answer» U wil gst on this app only... go in mathematics | |
| 4014. |
Polonomial |
| Answer» Polynomial* | |
| 4015. |
In triangle PQR angle Q is right triangle , PR = 7, PQ = 6 determine angle QPR and angle PRQ |
| Answer» In\xa0{tex}\\triangle{/tex}PQR{tex}\\sin ( \\angle \\mathrm { PRQ } ) = \\frac { \\mathrm { PQ } } { \\mathrm { PR } }{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}\\sin ( \\angle \\mathrm { PRQ } ) = \\frac { 6 } { 12 } = \\frac { 1 } { 2 }{/tex}Also, sin 30°\xa0{tex}= \\frac { 1 } { 2 }{/tex}{tex}\\Rightarrow \\angle P R Q = 30 ^ { \\circ }{/tex}In right\xa0{tex}\\triangle{/tex}PQR,\xa0{tex}\\angle Q + \\angle R + \\angle P = 180 ^ { \\circ }{/tex}{tex}90 ^ { \\circ } + 30 ^ { \\circ } + \\angle P = 180 ^ { \\circ }{/tex}{tex}\\Rightarrow \\angle P = 60 ^ { \\circ }{/tex}{tex}\\therefore{/tex}\xa0{tex}\\angle \\mathbf { Q P R } = 60 ^ { \\circ } , \\angle \\mathbf { P R Q } = 30 ^ { \\circ }{/tex} | |
| 4016. |
Geometry means |
| Answer» | |
| 4017. |
Find the roots of 2x-2/x=6 |
| Answer» | |
| 4018. |
The sum of first n terms of AP is given by An=2n²+3n.Find the sixteenth term of the AP. |
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Answer» 65 T16 = a + ( n-1) d = 5 +15×4= 5 +60 = 65 sorry sum nahi term we get 16th term S1= 5 , s2= a1 + a2 = 14 therefore a2 = 14 - 5 = 9 therefore d = a2 - a1 = 4 now a= 5 and d =4 putting the value of 1st term a and common difference d we get sum of 16th term of the ap Plzz explain me How How 65 145 It is 65 |
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| 4019. |
2345+7377 |
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Answer» 9722 I think que me problem hai? |
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| 4020. |
Ratio of this email address and telephone number |
| Answer» | |
| 4021. |
If X-1 is a factor of X²+2X+k then find the value of k. |
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Answer» k=-3 -3 K= -3 |
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| 4022. |
Find the HcF of 31310 and 3100. |
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Answer» Solve by ecluid division algorithm 31310 > 310031310 = 3100 × 10 + 3103100 = 310× 10 + 0So HCF = 310By Euclid division lemme How ? I think 310 310 |
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| 4023. |
Find hCf of 31310 and 3100.And also tell me how you find it. |
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Answer» See ncert chapter 1 HCF =310.....by euclid division lemma.... |
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| 4024. |
Formula of ch 7 |
| Answer» which book | |
| 4025. |
Value of tan 30 |
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Answer» 1 by root 3. 1/√3 |
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| 4026. |
If x/a(cos A)+y/b(sin A)=1 and x/a(sin A)-y/b(cos A)=1 then prove that x^2/a^2+y^2/b^2=2. |
| Answer» On squaring and then adding | |
| 4027. |
If in a circle ABCD is quadrilateral ab is 6cm bc is 7cm and cd is 4cm. Find ad |
| Answer» AD=3(AB+CD=BC+AD) | |
| 4028. |
Does sample released by cbse work as the blueprint of the final paper |
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Answer» Yes No |
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| 4029. |
Basic propasnality theirem |
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Answer» Thales theorem proportionality |
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| 4030. |
A cylinder, a cone and a hemisphere have same base and height . Find the ratio of their volumes. |
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Answer» 1:1:2r 1:1:2r |
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| 4031. |
If the radius of the circle is 6 cm and the length of the arc is 12 cm. Find the area of the sector. |
| Answer» r=6cm l=12cmar(sector)=1%2×12×6 =6×6 =36 | |
| 4032. |
How many mark want to score for A grade in board exam |
| Answer» Above 60 | |
| 4033. |
2+5+7-8+5+97-85 |
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Answer» Sorry 23 25 |
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| 4034. |
G de ff |
| Answer» Sorry wrong typed | |
| 4035. |
in an equilateral triangle ABC, D is a point on side BC such that BD=1/3BC. Prove that 9AD2=7AB2 |
| Answer» We have {tex}BD = \\frac{1}{3}BC{/tex} Draw AP \u200b{tex} \\bot {/tex}\u200b BCIn \u200b{tex}\\triangle {/tex}\u200bAPB,AB2 = AP2 + BP2 = AP2 + (BD + DP)2= AP2 + BD2 + DP2 + 2BD.DP= (AP2 + DP2) + BD2 + 2BD.DP=AD2 + DB2 + 2BD.DP [AP2 + DP2 = AD2]=AD2+{tex}{\\left( {\\frac{1}{3}BC} \\right)^2} + 2\\left( {\\frac{1}{3}BC} \\right)(BP - BD){/tex}={tex}AD^2 + \\frac{1}{9}BC + \\frac{2}{3}BC\\left( {\\frac{1}{2}BC - \\frac{1}{3}BC} \\right){/tex}{tex}\\left[ {BP = \\frac{1}{2}BC,BD = \\frac{1}{3}BC} \\right]{/tex}={tex}A{D^2} + \\frac{1}{9}A{B^2} + \\frac{2}{3}AB\\left( {\\frac{1}{2}AB - \\frac{1}{3}AB} \\right){/tex} [BC = AB, Sides of an equilateral{tex}\\vartriangle {/tex}\u200b ]={tex}A{D^2} + \\frac{1}{9}A{B^2} + \\frac{1}{3}A{B^2} - \\frac{2}{9}A{B^2}{/tex}={tex}A{D^2} + \\frac{1}{9}A{B^2} + \\frac{1}{9}A{B^2}{/tex}\u200b{tex}\\Rightarrow {/tex}\u200b {tex}A{B^2} = A{D^2} + \\frac{2}{9}A{B^2}{/tex}\u200b{tex}\\Rightarrow {/tex}\u200b 9AB2 = 9AD2 + 2AB2\u200b{tex}\\Rightarrow {/tex}\u200b9AB2 = 9AB2 + 2AD2\u200b{tex}\\Rightarrow {/tex}\u200b 7AB2 = 9AD2\u200b{tex}\\Rightarrow {/tex}\u200b 9AD2 = 7AB2 Proved | |
| 4036. |
Write 0.32in fraction form |
| Answer» 8/25 | |
| 4037. |
2+5x+10 find the nature if roots |
| Answer» incomplete eq.... | |
| 4038. |
From where the value of pi came 22/7 |
| Answer» Greek | |
| 4039. |
as we know that now our pattern is cg so all questions come from our ncert book |
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Answer» To......? i think....90% from ncert ...pkka? No |
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| 4040. |
For some integer m, what is the every even integer is of the form |
| Answer» 2m | |
| 4041. |
Qudatirc equation |
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Answer» Have highest degree of 2 Complete it quadratic equation???? |
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| 4042. |
Prove that√8+√8 is irrational |
| Answer» | |
| 4043. |
Prove that√8+√8 |
| Answer» | |
| 4044. |
If n is an odd integer then show that square of n_ 1 is divisible by 8 |
| Answer» Let n = 4q + 1 (an odd integer){tex}\\therefore \\quad n ^ { 2 } - 1 = ( 4 q + 1 ) ^ { 2 } - 1{/tex}{tex}= 16 q ^ { 2 } + 1 + 8 q - 1 \\quad \\text { Using Identity } ( a + b ) ^ { 2 } = a ^ { 2 } + 2 a b + b ^ { 2 }{/tex}{tex}= 16{q^2} + 8q{/tex}{tex}= 8 \\left( 2 q ^ { 2 } + q \\right){/tex}= 8m, which is divisible by 8. | |
| 4045. |
TanA÷1__cotA+cotA÷1_tanA=1+secA×cosecA |
| Answer» We have,{tex}\\mathrm { LHS } = \\frac { \\tan A } { 1 - \\cot A } + \\frac { \\cot A } { 1 - \\tan A }{/tex}{tex}\\Rightarrow \\quad \\mathrm { LHS } = \\frac { \\tan A } { 1 - \\frac { 1 } { \\tan A } } + \\frac { \\frac { 1 } { \\tan A } } { 1 - \\tan A }{/tex}{tex}\\Rightarrow \\quad \\text { LHS } = \\frac { \\tan A } { \\frac { \\tan A -1 } { \\tan A } } + \\frac { 1 } { \\tan A ( 1 - \\tan A ) }{/tex}{tex}\\Rightarrow \\quad \\mathrm { LHS } = \\frac { \\tan ^ { 2 } A } { \\tan A - 1 } + \\frac { 1 } { \\tan A ( 1 - \\tan A ) }{/tex}{tex}\\Rightarrow \\quad \\text { LHS } = \\frac { \\tan ^ { 2 } A } { \\tan A - 1 } - \\frac { 1 } { \\tan A ( \\tan A - 1 ) }{/tex}{tex}\\Rightarrow \\quad \\mathrm { LHS } = \\frac { \\tan ^ { 3 } A - 1 } { \\tan A ( \\tan A - 1 ) }{/tex}\xa0[Taking LCM]{tex}\\Rightarrow \\quad \\mathrm { LHS } = \\frac { ( \\tan A - 1 ) \\left( \\tan ^ { 2 } A + \\tan A + 1 \\right) } { \\tan A ( \\tan A - 1 ) }{/tex}\xa0[{tex}\\because{/tex}\xa0a3\xa0- b3\xa0= ( a - b )(a2\xa0+ ab + b2)]{tex}\\Rightarrow \\quad \\mathrm { LHS } = \\frac { \\tan ^ { 2 } A + \\tan A + 1 } { \\tan A }{/tex}{tex}\\Rightarrow \\quad \\mathrm { LHS } = \\frac { \\tan ^ { 2 } A } { \\tan A } + \\frac { \\tan A } { \\tan A } + \\frac { 1 } { \\tan A }{/tex}{tex}\\Rightarrow{/tex}\xa0LHS = tanA + 1 + cotA [ since (1/tanA) =cotA ].= (1 + tanA + cotA){tex}\\therefore \\quad \\frac { \\tan A } { 1 - \\cot A } + \\frac { \\cot A } { 1 - \\tan A }{/tex}\xa0= 1 + tanA + cotA ...........(1)Now, 1 + tanA + cotA = 1 +\xa0{tex}\\frac { \\sin A } { \\cos A } + \\frac { \\cos A } { \\sin A }{/tex}\xa0= 1 +\xa0{tex}\\frac { \\sin ^ { 2 } A + \\cos ^ { 2 } A } { \\sin A \\cos A }{/tex} = 1 +\xa0{tex}\\frac { 1 } { \\sin A \\cos A }{/tex}\xa0[{tex}\\because{/tex}Sin2A + Cos2 A = 1 ]\xa0= 1 + cosecAsecASo, 1 + tanA + cotA = 1+ cosecAsecA.......(2)From (1) and (2), we obtain{tex}\\frac { \\tan A } { 1 - \\cot A } + \\frac { \\cot A } { 1 - \\tan A }{/tex}\xa0= 1 + tanA + cotA = 1 + cosecAsecA | |
| 4046. |
1/x+1/x-10=78/5 |
| Answer» | |
| 4047. |
If 51 X + 49 Y =150 and 49 X + 51 Y =50 then obtain the value of X_Y : X + Y. |
| Answer» | |
| 4048. |
Last term of AP |
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Answer» But where is AP An or L |
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| 4049. |
is any important question for 2018 for math you know |
| Answer» Buy oswall sample paper book | |
| 4050. |
(1-cos2A) cosec2A=1 |
| Answer» | |