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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.
| 38951. |
SinA=4/5 find perpendicular |
| Answer» 4 | |
| 38952. |
Can the number 4n , n being a natural number , ends with digit 0 ? Give reason |
| Answer» | |
| 38953. |
What is the linear combination of real numbers |
| Answer» Using variables, representing a real no. in the form of an equation. | |
| 38954. |
Find A qaudratic eqauation whose one root is 3 |
| Answer» since 3 is the root of the equation, x = 3 must satisfy the equation.Applying x = 3\xa0in the equation {tex}{x^2} - 5x + 6 = 0{/tex}gives, {tex}{\\left( 3 \\right)^2} - 5 ( 3) + 6 = 0{/tex}{tex} \\Rightarrow {/tex}{tex}9 - 15 + 6 = 0{/tex}{tex} \\Rightarrow {/tex}{tex}15 - 15 = 0{/tex}{tex} \\Rightarrow {/tex}{tex}0 = 0{/tex}{tex} \\Rightarrow {/tex}L.H.S. = R.H.S.Hence, {tex}{x^2} - 5x + 6 = 0{/tex} is a required equation which has 3 as root. | |
| 38955. |
Ifone zero of the polynomial (asquare+9)xsquare+45x+6a is reciprocal of other ,find the value ofa |
| Answer» Please answer quickly | |
| 38956. |
Cos thita equals to |
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Answer» Cos θ = (Adjacent side)/(Hypotenuse side) = √(1-sin^2 θ) = 1/(sec θ) = sin(90-θ) = sin((π/2)-θ) = sin(90+θ) = sin((π/2)+θ) = (sin 2θ)/(2 cos θ) = √(1+cos 2θ)/2 = (tan θ)/(sin θ) = (cos θ)/(cot θ) 1/sin thita 1/sec thita, |
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| 38957. |
What is consecutive positive integer? |
| Answer» The no.s. of the form m and m+1 | |
| 38958. |
Draw a line segment AB=7 cm. Take a point P on AB such that AP : PB =3 : 4 |
| Answer» | |
| 38959. |
Sample paper of board 2018 |
| Answer» | |
| 38960. |
Show that the points A (7,10),(-2,5) and (3,-4) are the vertices of of an isosceles right triangle |
| Answer» Answer please | |
| 38961. |
If alpha and beta are the zeroes of polynomial 6xsquare+x-2.find alpha/beta +beta/alpha. |
| Answer» f(x) = 6x2\xa0+ x - 2a = 6, b = 1, c = -2Let zeroes be {tex}\\alpha{/tex}\xa0and β.ThenSum of zeroes=\xa0{tex}\\alpha{/tex} +\xa0β {tex}=\\;-\\frac ba\\;=-\\frac16{/tex}Product of zeroes {tex}\\alpha{/tex}× β {tex}=\\;\\;\\frac ca\\;=\\;\\frac{-2}6\\;=\\;-\\frac13{/tex}{tex}\\frac { \\alpha } { \\beta } + \\frac { \\beta } { \\alpha } = \\frac { \\alpha ^ { 2 } + \\beta ^ { 2 } } { \\alpha \\beta }{/tex}{tex}= \\frac { ( \\alpha + \\beta ) ^ { 2 } - 2 \\alpha \\beta } { \\alpha \\beta } \\left[ \\because ( \\alpha + \\beta ) ^ { 2 } = \\alpha ^ { 2 } + \\beta ^ { 2 } + 2 \\alpha \\beta \\right]{/tex}{tex}= \\frac { \\left[- \\frac { 1 } { 6 } \\right] ^ { 2 } - 2 \\left[ - \\frac { 1 } { 3 } \\right] } { \\left[ - \\frac { 1 } { 3 } \\right] }{/tex}{tex}= \\frac { \\frac { 1 } { 36 } + \\frac { 2 } { 3 } } { - \\frac { 1 } { 3 } }{/tex}{tex}= \\frac { \\frac { 1 + 24 } { 36 } } { - \\frac { 1 } { 3 } }{/tex}{tex}= \\frac { 25 } { 36 } \\times \\frac { - 3 } { 1 }{/tex}{tex}= \\frac { - 25 } { 12 }{/tex} | |
| 38962. |
________+________+_________=30 fill with odd numbers from 3-15. Only |
| Answer» 3! + 13+ 11 = 30! Means factorialOr(19-1) + 7 + 5 = 30Other than these, there is no valid solution of it. | |
| 38963. |
65%80+12*25%58 |
| Answer» | |
| 38964. |
If 77th term of Ap=1/9 and9th=1/7 find 63th term |
| Answer» Let a be the first term and d be the common difference of the given AP. Then,{tex}T _ { 7 } = \\frac { 1 } { 9 } \\Rightarrow a + 6 d = \\frac { 1 } { 9 }{/tex}\xa0...(i){tex}T _ { 9 } = \\frac { 1 } { 7 } \\Rightarrow a + 8 d = \\frac { 1 } { 7 }{/tex}\xa0...(i)On subtracting (i) from (ii), we get{tex}2 d = \\left( \\frac { 1 } { 7 } - \\frac { 1 } { 9 } \\right) = \\frac { 2 } { 63 } \\Rightarrow d = \\left( \\frac { 1 } { 2 } \\times \\frac { 2 } { 63 } \\right) = \\frac { 1 } { 63 }.{/tex}Putting d =\xa0{tex}\\frac { 1 } { 63 }{/tex}in (i), we get{tex}a + \\left( 6 \\times \\frac { 1 } { 63 } \\right) = \\frac { 1 } { 9 } \\Rightarrow a + \\frac { 2 } { 21 } = \\frac { 1 } { 9 } \\Rightarrow a = \\left( \\frac { 1 } { 9 } - \\frac { 2 } { 21 } \\right) = \\left( \\frac { 7 - 6 } { 63 } \\right) = \\frac { 1 } { 63 }.{/tex}Thus, a =\xa0{tex}\\frac { 1 } { 63 }{/tex}\xa0and d =\xa0{tex}\\frac { 1 } { 63 }.{/tex}{tex}\\therefore{/tex}\xa0T63 = a + (63-1)d = (a + 62d){tex}= \\left( \\frac { 1 } { 63 } + 62 \\times \\frac { 1 } { 63 } \\right) = \\left( \\frac { 1 } { 63 } + \\frac { 62 } { 63 } \\right) = 1.{/tex}Hence, 63rd term of the given AP is 1. | |
| 38965. |
Find the value of Y such that the points A(a,a) B{Y,(3a)1/2} and C{-a,a} |
| Answer» | |
| 38966. |
If a positive integer p is divided by 3,then find the possible remainder |
| Answer» 1,2 | |
| 38967. |
Set of all tallest persons in your classroom is set or not give reason |
| Answer» yes their will be a grp of tallest person | |
| 38968. |
Prove that 6^n can never end with the digits 0 or 2 or 5 |
| Answer» | |
| 38969. |
Polynamials |
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Answer» About polynomial What u need |
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| 38970. |
(3k+1)x^2 + 2(k+1)x +1 find value of k which has equal roots also find roots |
| Answer» The given equation is:(3k + 1)x2 + 2(k + 1)x + 1 = 0This is of the form\xa0ax2\xa0+ bx + c = 0, wherea = 3k + 1, b = 2(k + 1) = 2k + 2 and c = 1As it is given that the given equation has real and equal roots, i.e.,\xa0D = b2\xa0- 4ac = 0.\xa0{tex}\\Rightarrow{/tex}\xa0(2k + 2)2\xa0- 4(3k + 1) (1) = 0{tex}\\Rightarrow{/tex}\xa04k2\xa0+ 8k + 4 - 12k - 4 = 0{tex}\\Rightarrow{/tex}\xa04k2\xa0- 4k = 0{tex}\\Rightarrow{/tex}\xa04k(k - 1) = 0\xa0Therefore, either 4k = 0 or k - 1 = 0{tex}\\Rightarrow{/tex}\xa0k = 0 or k = 1Hence, the roots of given equation are 1 and 0. | |
| 38971. |
Hlw DV |
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Answer» Hm hi I am online now Tum kitne bje online hoge Hii |
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| 38972. |
Thales theorem |
| Answer» Parallel to one side of single a triangle to intersect the other two side in distance. Then the other two sides are divided in the same ratio | |
| 38973. |
10 application problem related to Euclid division lemma |
| Answer» | |
| 38974. |
If 2/3 and 1/5 are the zeros of the polynomial p(x)=3x^4+4x+9kxThen find k |
| Answer» | |
| 38975. |
End term of A.p:17,14,11..-40 |
| Answer» A.P. is 17,14,11,..., - 40We have,l = Last term = -40 , a = 17 and, d = Common difference= 14 - 17 = - 3{tex}\\therefore{/tex}\xa06th term from the end = l - (n -1)d=\xa0l - (6-1) d= -40 - 5 {tex}\\times{/tex} (-3 )= -40 + 15= -25So, 6th term of given A.P. is -25. | |
| 38976. |
Find the value of k for which the quadratic equation 9x2– 3kx + k = 0 has equal roots. |
| Answer» Poda mythandi mone | |
| 38977. |
Easy revise for lesson |
| Answer» Get revision notes from here :\xa0https://mycbseguide.com/cbse-revision-notes.html | |
| 38978. |
Bhai ncert syllabus kab half hoga |
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Answer» 2019 .. next batch of class 10th??? ??? Half nhi hoga whole course aayega final me |
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| 38979. |
Find the coordinates of the point which divides the join of (-1,7)( 4,-3)in the tatio 2:3 |
| Answer» (1,3) | |
| 38980. |
If x=2/3 & x=-3 are the roots of equation ax2 +7x+b=0 find the value of a and b . |
| Answer» a=3,b=-6 | |
| 38981. |
Arithmetic progression exercise no 3 |
| Answer» given a3=15,S10=125,find d and a10 | |
| 38982. |
What is ogive explain in detail |
| Answer» In statistics, an ogive is a graph showing the curve of a cumulative distribution function. The points plotted are the upper class limit and the corresponding cumulative frequency. The ogive for the normal distribution, resembles one side of an Arabesque or ogival arch. | |
| 38983. |
Easy formula. Of. Hcf |
| Answer» Prime Factorisation method-Express each one of the given numbers as the product of prime factors.The product of least powers/index of common prime factors gives H.C.F.Find the H.C.F. of 8 and 14 by Prime Factorisation method.Solution: 8 = 2 x 2 x 2 and 14 = 2 x 7Common factor of 8 and 14 = 2.Thus, Highest Common Factor (H.C.F.) of 8 and 14 = 2. | |
| 38984. |
Are oswall sample papers good for board exams?? |
| Answer» Yes but only for practising.just don\'t carry it too long.it is recommended to read the book☺☺☺☺ | |
| 38985. |
Tutak tutak tutiya |
| Answer» Is a movie by prabhudeva | |
| 38986. |
maximum marks of board exam |
| Answer» 500 | |
| 38987. |
What would be the best guide book for mathematics class10? |
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Answer» Ncert is best but for more practice you take RS AGRAWAL or RD SHRMA ND Rd sharma Monica sharma Rd sharma ,to gether with , Agrawal together with, Ncert exampler book Rs agrawal , rd sharma , |
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| 38988. |
sin A (1+ tan A)+ cos A(1 + cot A) (sec A+ cosec A) |
| Answer» | |
| 38989. |
Hello.friends |
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Answer» hi Is anyone online |
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| 38990. |
Trigonometry ratios |
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Answer» Sinθ=opposite/hypotenuseCosθ=adjacent/hypotenuseTanθ=opposite/adjacentCosecθ=hypotenuse/oppositeSecθ=hypotenuse/adjacentCotθ=adjacent/oppositeTanθ= Sinθ/CosθCosecθ=1/SinθSecθ=1/CosθCotθ=1/Tanθ sinA = p\\hcosA=b\\htanA=p\\bcosecA=h\\psecA=h\\bcotA=b\\p |
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| 38991. |
If the product of the zeros of the polynomial 3x square +5x -4k is -8,then find the value of k |
| Answer» put x = -8\xa0in\xa0{tex}3x^2 + 5x - 4k = 0{/tex}{tex}3(-8)^2 + 5(-8)-4k = 0\xa0{/tex}{tex}3(-8)^2 + 5(-8)\xa0- 4 k = 0{/tex}\xa0192 - 40 - 4k\xa0= 0152 = 4k{tex}\\Rightarrow{/tex}\xa0k =\xa028{tex}\\therefore{/tex}\xa0k = 28 | |
| 38992. |
Whi which of the following cannot be the unit digit of a perfect square number a 6 V 1 C 98 |
| Answer» | |
| 38993. |
if the sum of p term of an Ap is same as the sum of q term show that the sum of q and p is zero |
| Answer» Sp = Sq{tex} \\Rightarrow \\frac{p}{2}\\left[ {2a + (p - 1)d} \\right] = \\frac{q}{2}\\left[ {2a + (q - 1)d} \\right]{/tex}{tex} \\Rightarrow p\\left[ {2a + pd - d} \\right] = q\\left[ {2a + qd - d} \\right]{/tex}{tex} \\Rightarrow {/tex}\xa02ap + p2d - pd = 2aq + q2d - qd{tex} \\Rightarrow {/tex}\xa02a (p - q) + (p2 - q2)d - (p - q)d = 0{tex} \\Rightarrow {/tex}\xa02a (p - q) + (p + q) (p - q)d - (p - q)d = 0{tex} \\Rightarrow {/tex}\xa0(p - q) [2a + (p + q - 1)d] = 0{tex} \\Rightarrow {/tex}\xa0Sp+q = 0 | |
| 38994. |
[A+b]^2 |
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Answer» a²+2ab+b² A^2 +2 AB +^2 A^2+b^2+2Ab |
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| 38995. |
3 × 3 |
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Answer» Go! before classes 1 9 6 |
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| 38996. |
2017 sa1question paper class 10 |
| Answer» Check Question Papers here :\xa0https://mycbseguide.com/cbse-question-papers.html | |
| 38997. |
Divide 123089 by10764 |
| Answer» 11.43524712. | |
| 38998. |
What is tan theta |
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Answer» tan theta = 1÷cot theta also tan theta=sin theta ÷cos theta Ratio of sin theta to cos theta |
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| 38999. |
If (x2+ y2) (a2+ b2) = (ax + by)2. Prove that . |
| Answer» Given,\xa0{tex}\\left( x ^ { 2 } + y ^ { 2 } \\right) \\left( a ^ { 2 } + b ^ { 2 } \\right) = ( a x + b y ) ^ { 2 }{/tex}or,{tex}x ^ { 2 } a ^ { 2 } + x ^ { 2 } b ^ { 2 } + y ^ { 2 } a ^ { 2 } + y ^ { 2 } b ^ { 2 } = a ^ { 2 } x ^ { 2 } + b ^ { 2 } y ^ { 2 } + 2 a b x y{/tex}or,\xa0{tex}x ^ { 2 } b ^ { 2 } + y ^ { 2 } a ^ { 2 } - 2 a b x y = 0{/tex}or,{tex}( x b - y a ) ^ { 2 } = 0{/tex}\xa0{tex}\\left[ \\because ( a - b ) ^ { 2 } = a ^ { 2 } + b ^ { 2 } - 2 a b \\right]{/tex}or,\xa0xb = ya{tex}\\therefore \\quad \\frac { x } { a } = \\frac { y } { b }{/tex}\xa0Hence Proved. | |
| 39000. |
Prove that root 5 minus root 3 is not a rational number |
| Answer» | |