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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.
| 6951. |
If did HCF of 45,27. Find x and y satisfying d=27x + 45y |
| Answer» HCF of 45 and 27 -45 = 27 x 1 + 1827 = 18 x 1 + 918 = 9 x 2 + 0 So, HCF (45, 27) = 9 d = 9 = 27x + 45y Make a linear combination- 9=27-18×1 =27-(45-27×1)×1 (18=45-27×1) =27-45×1+27×1 =2×27-1×45 =27x+45y (x=2,y=-1) | |
| 6952. |
Find the sum of 1st prime number devisible by2 |
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| 6953. |
Explain why 3*5*7+7 is a composite number |
| Answer» We have, (3\xa0× 5\xa0× 7) + 7 = 105 + 7 = 112Prime factors of 112 = 2\xa0× 2\xa0× 2\xa0× 2\xa0× 7 = 24\xa0{tex} \\times {/tex}\xa07So, it is the product of prime factors 2 and 7, i.e. it has factors other than 1 and itself. Hence, it is a composite number. | |
| 6954. |
Explain why3*5*7*+7isacomposite |
| Answer» We have, (3\xa0× 5\xa0× 7) + 7 = 105 + 7 = 112Prime factors of 112 = 2\xa0× 2\xa0× 2\xa0× 2\xa0× 7 = 24\xa0{tex} \\times {/tex}\xa07So, it is the product of prime factors 2 and 7, i.e. it has factors other than 1 and itself. Hence, it is a composite number. | |
| 6955. |
Eonenmbers |
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| 6956. |
What is mean |
| Answer» Ex. Deta hon kudh samaj jaoge bhahii mera ans hai 10 worker hain unki salary alag alag hai malik sirf10000 rupees deta hai magar bich me jo person rupey batta hai bhe kisi ko kam ya kisi jayada deta hai aur bache huye kudh rakh leta hai yahan per mean ye hai ki 10 workers ko barabar kitne rupey milte hai matla per worker ko malik ke hiasab se 1000 rupey ja rahe hai yahi mean hai per person anything me jo cheez barabar batti hai bahi mean hai isse jayada mai nahi samja sakta i hope apko samaj me aagaya go to thanks ki jagah mere questions ke answer dena kaise bhi mere liye ye thanks hoga thank you | |
| 6957. |
Nzivzkvzvdgs |
| Answer» | |
| 6958. |
Prove that the tangents drawn at the end of a diameter of a circle are parallel |
| Answer» Given: PQ is a diameter of a circle with centre O. The lines AB and CD are the tangents at P and Q respectively.\xa0To prove: AB\xa0{tex}\\parallel{/tex} CDProof: AB is a tangent to the circle at P and Op is the radius through the point of contact{tex}\\because{/tex}{tex}\\angle{/tex}OPA = 90o .......(1) [The tangent at any point of a circle is perpendicular to the radius through the point of contact]{tex}\\because{/tex}\xa0CD is a tangent to the circle at Q and OQ is the radius through the point of contact.{tex}\\therefore{/tex}\xa0{tex}\\angle{/tex}OQD = 90o ........(2) [The tangent at any point of a circle is perpendicular to the radius through the point of contact]From (1) and (2),{tex}\\angle{/tex}OPA = {tex}\\angle{/tex}OQDBut these form a pair of equal alternate angles.{tex}\\because{/tex}\xa0AB\xa0{tex}\\parallel{/tex} CD | |
| 6959. |
To proove that (A-B)x^2+(B-C)x+(C-A)=0 |
| Answer» We have (a-b)x2 + (b-c)x + (c-a) = 0Here A\xa0= (a-b), B\xa0= (b-c), C= (c-a){tex}\\therefore D = B^2 - 4AC = (b-c)^2 - 4(a-b) (c-a){/tex}For equal roots, D = 0{tex}\\implies{/tex}(b-c)2 - 4 (a-b) (c-a) = 0b2+c2-2bc -4(ac-a2-bc+ab) =0b2+c2-2bc -4ac+4a2+4bc-4ab=04a2+b2+c2+2bc-4ab-4ac=0(2a-b-c)2=0i.e. 2a-b-c =0Hence, b + c = 2a or 2a = b + c | |
| 6960. |
The tens digit of a number is twice the unit digit |
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| 6961. |
Is all marks of half yearly exam is add to CBSE or not |
| Answer» Best of two exams wiil be added out of pre mid term , mid term, post mid term. Then best of five subjects will be taken and then all will be added and then marks will be given out of twenty. Rest 80 from final board | |
| 6962. |
If cosA-sinA=1 than solve tanA+cotA=? |
| Answer» 1/1+sin^2 | |
| 6963. |
Trigonometry video 10 class |
| Answer» Check videos here :\xa0https://mycbseguide.com/video/ | |
| 6964. |
1/X |
| Answer» X^-1 | |
| 6965. |
find the hcf of 34 |
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Answer» Agar 3 aur 4 ki baat hai to ,hcf is 1. 2 |
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| 6966. |
Geometrically find value of cos 60 |
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Answer» √3/2 1/2 √3/2 |
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| 6967. |
2/7-^$$yfhj |
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| 6968. |
11+11 |
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Answer» The xi very IV yuh to Chuck gain day so disk 22 =22 |
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| 6969. |
find thea mean |
| Answer» mean is average of data\xa0mean ={tex}sum \\ of \\ observations \\over number \\ of \\ observations {/tex} | |
| 6970. |
Find k so that (10k+7),(4k+6) and (2k+3) are in AP? |
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| 6971. |
Can mean be represented graphically? ? |
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| 6972. |
Sin 60 |
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Answer» √3/2 √3/2 31/2/2 1/2 |
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| 6973. |
M = sec theta,show that m^2-1/m^2+1 =1/cosec theta |
| Answer» We\xa0have,(m2\xa0- 1) = (sec{tex}\\theta{/tex}\xa0+ tan{tex}\\theta{/tex})2\xa0- 1= sec2{tex}\\theta{/tex}\xa0+ tan2{tex}\\theta{/tex}\xa0+ 2sec{tex}\\theta{/tex}tan{tex}\\theta{/tex}\xa0- 1= (sec2{tex}\\theta{/tex} - 1\xa0) + tan2{tex}\\theta{/tex}\xa0+ 2sec{tex}\\theta{/tex}tan{tex}\\theta{/tex}= 2tan2{tex}\\theta{/tex}\xa0+ 2sec{tex}\\theta{/tex}tan{tex}\\theta{/tex}\xa0[{tex}\\because{/tex}\xa0sec2{tex}\\theta{/tex}\xa0- 1 = tan2{tex}\\theta{/tex}]= 2tan{tex}\\theta{/tex}(tan{tex}\\theta{/tex}\xa0+ sec{tex}\\theta{/tex}) ..(i)(m2 +1) = (sec{tex}\\theta{/tex}\xa0+ tan{tex}\\theta{/tex})2 + 1= sec2{tex}\\theta{/tex}\xa0+ tan2{tex}\\theta{/tex}\xa0+ 2sec{tex}\\theta{/tex}tan{tex}\\theta{/tex}\xa0+ 1= (1 + tan2{tex}\\theta{/tex}) + sec2{tex}\\theta{/tex}\xa0+ 2sec{tex}\\theta{/tex}tan{tex}\\theta{/tex}= 2sec2{tex}\\theta{/tex}\xa0+ 2sec{tex}\\theta{/tex}tan{tex}\\theta{/tex}\xa0[{tex}\\because{/tex}\xa01 + tan2{tex}\\theta{/tex}\xa0= sec2{tex}\\theta{/tex}]= 2sec{tex}\\theta{/tex}(sec{tex}\\theta{/tex}\xa0+ tan{tex}\\theta{/tex}) ..(ii)From (i) and (ii), we get{tex}\\frac { \\left( m ^ { 2 } - 1 \\right) } { \\left( m ^ { 2 } + 1 \\right) } = \\frac { \\tan \\theta } { \\sec \\theta } = \\left( \\frac { \\sin \\theta } { \\cos \\theta } \\times \\cos \\theta \\right) = \\sin \\theta{/tex}Hence,\xa0{tex}\\frac { \\left( m ^ { 2 } - 1 \\right) } { \\left( m ^ { 2 } + 1 \\right) } = \\sin \\theta{/tex} | |
| 6974. |
The sum of first n terms of an AP is 3n-1 then find nth term |
| Answer» Use the formula n2- n | |
| 6975. |
Solve by using cross multiplication method.ax\\b-by/a ax-by=2ab |
| Answer» The given equations are{tex}\\frac { a x } { b } - \\frac { b y } { a } - ( a + b ) = 0{/tex}{tex}ax - by - 2ab = 0{/tex}By cross multiplication, we have{tex}\\frac { x } { \\left( - \\frac { b } { a } \\right) \\times ( - 2 a b ) - ( - b ) \\times ( - ( a + b ) ) } ={/tex}{tex}\\frac { y } { - ( a + b ) \\times a - ( - 2 a b ) \\times \\frac { a } { b } }{/tex}{tex}= \\frac { 1 } { \\frac { \\mathrm { a } } { \\mathrm { b } } \\times ( - \\mathrm { b } ) - \\mathrm { a } \\times \\left( - \\frac { \\mathrm { b } } { \\mathrm { a } } \\right) } {/tex}{tex}\\Rightarrow \\frac { x } { 2 b ^ { 2 } - b ( a + b ) } = \\frac { y } { - a ( a + b ) + 2 a ^ { 2 } } = \\frac { 1 } { - a + b }{/tex}or,\xa0{tex}\\frac { x } { 2 b ^ { 2 } - a b - b ^ { 2 } } = \\frac { y } { - a ^ { 2 } - a b + 2 a ^ { 2 } } = \\frac { 1 } { - a + b }{/tex}{tex}\\Rightarrow \\frac { x } { b ^ { 2 } - a b } = \\frac { y } { a ^ { 2 } - a b } = \\frac { 1 } { - ( a - b ) }{/tex}{tex}\\Rightarrow \\frac { x } { - b ( a - b ) } = \\frac { y } { a ( a - b ) } = \\frac { 1 } { - ( a - b ) }{/tex}{tex}\\therefore \\frac { x } { - b ( a - b ) } = \\frac { 1 } { - ( a - b ) }{/tex}\xa0and\xa0{tex}\\frac { y } { a ( a - b ) } = \\frac { 1 } { - ( a - b ) }{/tex}{tex}\\therefore x = \\frac { - b ( a - b ) } { - ( a - b ) } \\text { and } y = \\frac { a ( a - b ) } { - ( a - b ) }{/tex}{tex} \\Rightarrow{/tex}\xa0{tex}x = b,\\ and\\ y = -a{/tex}{tex}\\therefore{/tex} the solution is {tex}x = b, y = -a{/tex} | |
| 6976. |
tanA+sinA =mtanA -sinA =n then prove that (m)2-(n)2= 4√mn |
| Answer» Given, {tex}\\tan \\theta + \\sin \\theta = m{/tex} and {tex}\\tan \\theta - \\sin \\theta = n{/tex}L.H.S =\xa0{tex}m^{2}-n^{2}{/tex}{tex}=(tan\\theta+sin\\theta)^{2}-(tan\\theta-sin\\theta)^{2}{/tex}{tex}=tan^{2}\\theta+sin^{2}\\theta+2tan\\theta sin\\theta-[tan^{2}\\theta+sin^{2}\\theta-2tan\\theta sin\\theta]{/tex}{tex}=tan^{2}\\theta+sin^{2}\\theta+2tan\\theta sin\\theta-tan^{2}\\theta-sin^{2}\\theta+2tan\\theta sin\\theta{/tex}{tex}=4tan\\theta sin\\theta{/tex}R.H.S = 4{tex}\\sqrt{mn}{/tex}{tex}=4\\sqrt{(tan\\theta+sin\\theta)(tan\\theta-sin\\theta)}{/tex}{tex}=4\\sqrt{tan^{2}\\theta-sin^{2}\\theta}{/tex}{tex}=4\\sqrt{\\frac{sin^{2}\\theta}{cos^{2}\\theta}-sin^{2}\\theta}{/tex}{tex}=4\\sqrt{\\frac{sin^{2}\\theta-sin^{2}\\theta cos^{2}\\theta}{cos^{2}\\theta}}{/tex}{tex}=\\frac{4}{cos\\theta}\\sqrt{sin^{2}\\theta-sin^{2}\\theta cos^{2}\\theta}{/tex}{tex}=\\frac{4}{cos\\theta}\\sqrt{sin^{2}\\theta (1-cos^{2}\\theta)}{/tex}{tex}=\\frac{4}{cos\\theta} \\times sin\\theta \\times \\sqrt{sin^{2}\\theta}{/tex}{tex}=4tan\\theta sin\\theta{/tex}Hence, L.H.S = R.H.S | |
| 6977. |
A number x is chosen at random from the numbers -3, -2, -1, 0, 1,2,3. What is probability that |x| |
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| 6978. |
6x2+x-12=0 |
| Answer» X= 0 | |
| 6979. |
Tan²+cos²-tan³+cos³= |
| Answer» | |
| 6980. |
A dice is thrown twice.What is probability of getting number greater than three on each dice? |
| Answer» 1by2 | |
| 6981. |
Prove that √2 +3/7 is irrational |
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| 6982. |
√1+sinA/1-sinA = secA +tanA (( root is in whole of 1+sinA/1-sinA. And we have to show LHS =RHS)) |
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| 6983. |
find the quadratic polynomial whose zeroes are |
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| 6984. |
Find the value of k if 2 is a zero of the polynomial 2x^2-5x2^2+5x-k . |
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| 6985. |
Compeleting the squaring |
| Answer» Called as completing square method formulae | |
| 6986. |
3cosA=4sinA,findvalue of 5cosA-2sinA/5cosA+3sinA |
| Answer» 7/13 | |
| 6987. |
if a and b are the roots of the quadratic equation x²+ax-b=0.Then find a and b |
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| 6988. |
what is statics |
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| 6989. |
All theorem of triangle chapter |
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| 6990. |
Hindi me best marks kaise laye |
| Answer» Padh ke | |
| 6991. |
Polynomile |
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| 6992. |
Volume and surface area |
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| 6993. |
Define euclids division lemma |
| Answer» Euclid’s Division Lemma: If we have two positive integers a and b,then there exists unique integers\xa0q\xa0and\xa0r\xa0which satisfies the condition\xa0a = bq + r (where 0\xa0≤ r ≤ b)Example: If we have two integers a=27 b=4Then {tex}27 = 4 × 6 + 3,{/tex}Here q = 6 and r = 3 | |
| 6994. |
If a and b are two positive integers such that a=14b, find the HCF of a and b |
| Answer» If a and b are the two +ve integers then According to Euclid\'s division lemma a=bq+r. where 0<=rAs a=14b then 14b=b×14+0So the HCF is b | |
| 6995. |
1+secA/secA=sin2/1-cosA (prove this) |
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Answer» For more explanations or answer . Whatsaap me 9102084705 RHS- 1+secA/secA changes into 1/secA + secA/SecA . Then SecA/SecA is equal to one and 1/SecA is equal to cosA . Then RHS is equal to 1 + cosA.LHS- sin^2A is equal to 1-cos^2A. 1 -Cos^2 change into (1+cosA)(1-cosA). Then 1-cosA cancel by 1-cosA. Then LHS is equal to 1+cosA.Hence, RHS=LHS. PROVED.... |
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| 6996. |
Can I get sample question paper of B.com? |
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| 6997. |
Harshit |
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Answer» Quality Harshit bo hai jo log nahi samaj paye |
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| 6998. |
Define electrical Current |
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Answer» The rate of flow of charge is known as Electric current The flow of elcetric charge in conductor is called electrical current |
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| 6999. |
Prove that any positibe integer is a form of 6q+2 ,6q+3, for p is some integer |
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| 7000. |
Prove that sin square theta + cos square theta is equal to 1 |
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Answer» p square / h square plus b square / h square then result is p square plus b square /h square we know that p square plus b square is equal to h square and now h square / h square is 1 Take a right angle triangle ABC right angled at B now its measurement are AC=5;BC=4 : AB= 3 . for sinc C hy=5 and per =3 also for cos C hy= 5 and base = 4 now sin 2C+ cos2C= 9/25+16/25=1 ( sin2c mean sin square c)proved |
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