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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.
| 35601. |
Trigonometry kaise seekhen |
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Answer» 1.2.3.P.B.P.H.H.B.Pakistan bhuka pet hindustan hara bhara Pandit bol pandit har har bhole P/H ; B/H; p/B papa bedi pilo; Ha ha beta Practice the solved questions.....Do NCERT properly.....N examplar...... Please fast |
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| 35602. |
2x2+6x+12 factorise |
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Answer» Isme square nahi h to ye factorization possible nahi h This question is wrong Is the question correct ? |
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| 35603. |
Probability of getting 53 saturday in a year |
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Answer» In ordinary year p(e) will be 1/7 and in leap year 2/7 53/365 1/7 |
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| 35604. |
how many solution of the pair of linear equations c+y-10=0 and x+y-7=0 has? |
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| 35605. |
The points (k+1,1) (2k+1,3) and (2k+2,2k) are collinear. Then find k |
| Answer» {tex}{/tex}We know that if three points A, B, C are collinear, then area of\xa0{tex}\\Delta ABC =0{/tex}Given that points A(k + 1, 1), B(2k + 1, 3) and C(2k + 2, 2k) are collinear.Here, x1= k+1 , x2\xa0=2k+1, x3= 2k+2, y1= 1, y2= 3, y3= 2k{tex}\\Rightarrow {/tex}\xa0Area of\xa0{tex}\\Delta ABC =0{/tex}{tex}\\Rightarrow {/tex}\xa0{tex}\\frac{1}{2}|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|=0{/tex}{tex}\\Rightarrow {/tex}{tex}\\frac{1}{2}|(k+1)(3-2k)+(2k+1)(2k-1)+(2k+2)(1-3)|=0{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}|k(3-2k)+1(3-2k)+(2k)^2-(1)^2+(2k+2)(-2)|=0{/tex}{tex}\\Rightarrow{/tex} |3k - 2k2 + 3 - 2k + 4k2 - 1 -\xa04k - 4 |= 0{tex}\\Rightarrow{/tex}\xa0|2k2 - 3k - 2 |= 0{tex}\\Rightarrow{/tex}\xa02k2 - 4k + k - 2 = 0{tex}\\Rightarrow{/tex}\xa02k(k - 2) + 1(k - 2) = 0{tex}\\Rightarrow{/tex}\xa0(2k + 1) (k - 2) = 0{tex}\\Rightarrow{/tex}\xa0k = -\xa0{tex}\\frac { 1 } { 2 }{/tex}, k = 2 | |
| 35606. |
Prove root 3 as irrational |
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Answer» Let √3 be a rational number then it can be written in the form of a/ b, b≠0 and a and b are co prime numbers . √3=a/b , now squaring on both sides gives 3=a²/b²........... 3b²=a² it means a² is divisible by 3 , so a is also divisible by 3 ....... Let a=3c now squaring on both sides gives a²=9c² . putting value of a².... 3b²=9c² , it gives us b²=3c² .. it means b² is divisible by 3 so b is also divisible by 3 ..... Therefore our assumption is wrong . They have a common factor as 3 . So √3 is an irrational number . Hence proved NCERT question.....Maths can\'t type here |
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| 35607. |
What is ShreeBhattacharya rulev |
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| 35608. |
x/4+y/3=5/12 ; x/2+y=1 find the value of x and y in the solition of equation |
| Answer» X=11 and y = - 9/2 | |
| 35609. |
BPT thearem |
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Answer» In a triangle ABC if DE II BC then AD/DB = AE/EC given - DE parallel BCto prove - AD÷DB=AE÷EC it\'s very easiy... |
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| 35610. |
Sn=3n2-n |
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Answer» Or of we put value of n in this equation any than we find the vvalue of S1,S2 etc Ky afind karna h |
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| 35611. |
Find value of angle theta if cos theta /1-sin theta+cos theta/1+sin theta÷4 |
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Answer» Given,\xa0{tex}\\frac { \\cos \\theta } { 1 - \\sin \\theta } + \\frac { \\cos \\theta } { 1 + \\sin \\theta } = 4{/tex}Taking LCM\xa0{tex}\\frac { \\cos \\theta ( 1 + \\sin \\theta ) + \\cos \\theta ( 1 - \\sin \\theta ) } { ( 1 - \\sin \\theta ) ( 1 + \\sin \\theta ) } = 4{/tex}\xa0{tex}\\frac { \\cos \\theta [ 1 + \\sin \\theta + 1 - \\sin \\theta ] } { 1 - \\sin ^ { 2 } \\theta } = 4{/tex}\xa0{tex}\\frac { \\cos \\theta ( 2 ) } { \\cos ^ { 2 } \\theta } = 4{/tex}\xa0{tex}\\frac { 2 } { \\cos \\theta } = 4{/tex}\xa0{tex}\\cos \\theta = \\frac { 2 } { 4 } = \\frac { 1 } { 2 }{/tex}\xa0{tex}\\cos \\theta = \\cos 60 ^ { \\circ }{/tex}{tex}\\therefore \\theta = 60 ^ { \\circ }{/tex} Given,\xa0{tex}\\frac { \\cos \\theta } { 1 - \\sin \\theta } + \\frac { \\cos \\theta } { 1 + \\sin \\theta } = 4{/tex}Taking LCM\xa0{tex}\\frac { \\cos \\theta ( 1 + \\sin \\theta ) + \\cos \\theta ( 1 - \\sin \\theta ) } { ( 1 - \\sin \\theta ) ( 1 + \\sin \\theta ) } = 4{/tex}\xa0{tex}\\frac { \\cos \\theta [ 1 + \\sin \\theta + 1 - \\sin \\theta ] } { 1 - \\sin ^ { 2 } \\theta } = 4{/tex}\xa0{tex}\\frac { \\cos \\theta ( 2 ) } { \\cos ^ { 2 } \\theta } = 4{/tex}\xa0{tex}\\frac { 2 } { \\cos \\theta } = 4{/tex}\xa0{tex}\\cos \\theta = \\frac { 2 } { 4 } = \\frac { 1 } { 2 }{/tex}\xa0{tex}\\cos \\theta = \\cos 60 ^ { \\circ }{/tex}{tex}\\therefore \\theta = 60 ^ { \\circ }{/tex} |
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| 35612. |
If tan²Φ=1-a²,prove that secΦ+tan³ΦcosecΦ=(2-a²)³/² |
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| 35613. |
(3x+1)x+3y-2 |
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Answer» (3x+1)x +3y-23x²+x+3y-2 3xsqare+x+3y-2 |
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| 35614. |
Determime the AP whose third term is 16 and 7 th term exceed the 5th term by 12 |
| Answer» An=a+(n-1)d So, 16=a+2d(equ1). , A7 =a+6d, A5= a+4d Now, according to statement we have, a+6d=a+4d+12 solving this equation we get. d=6 putting this value in( equ 1) we get a=4. Now we have a.=4 ,d= 6 so AP will be 4,10, 16 ,22 ,28............. | |
| 35615. |
Most important chapters for boards are? |
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Answer» Haaaa All are important |
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| 35616. |
Find the sum of the 25 terms of an A.P. whose nth term is given by an=An+B |
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| 35617. |
Foctorastion of 1/abx=1/a+1/b+1/x |
| Answer» Given,{tex}\\frac { 1 } { ( a + b + x ) } = \\frac { 1 } { a } + \\frac { 1 } { b } + \\frac { 1 } { x }{/tex}{tex}\\Rightarrow \\quad \\frac { 1 } { ( a + b + x ) } - \\frac { 1 } { x } = \\frac { 1 } { a } + \\frac { 1 } { b } \\Rightarrow \\frac { x - ( a + b + x ) } { x ( a + b + x ) } = \\frac { b + a } { a b }{/tex}{tex}\\Rightarrow \\quad \\frac { - ( a + b ) } { x ( a + b + x ) } = \\frac { ( a + b ) } { a b }{/tex}On dividing both sides by (a+b){tex}\\Rightarrow \\quad \\frac { - 1 } { x ( a + b + x ) } = \\frac { 1 } { a b }{/tex}Now cross multiply{tex}\\Rightarrow{/tex}\xa0x(a + b + x) = -ab\xa0{tex}\\Rightarrow{/tex}\xa0x2 + ax + bx + ab = 0{tex}\\Rightarrow{/tex}\xa0x(x +a) + b(x +a) = 0{tex}\\Rightarrow{/tex}\xa0(x\xa0+ a) (x + b) = 0{tex}\\Rightarrow{/tex}\xa0x + a = 0 or x + b = 0{tex}\\Rightarrow{/tex}\xa0x = -a or x = -b.Therefore, -a and -b\xa0are the roots of the equation. | |
| 35618. |
If -2 is zero of the quadratic polynomial x² + 3x + K , then find the value of k |
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Answer» The -2 is the zero of p(x).then x=-2 put the value of x you got answer. x²+3x+K=0(-2)²+3*(-2)+K=04+(-6)+K=0-2+K=0K=2 |
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| 35619. |
Find the area of the triangle whose vertices are (2,3),(4,-3),(2,6) |
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Answer» Put the value in triangle formula Use the area of triangle formula by taking any of the values as x1y1,x2y2,x3y3 according to ur wish and u will get the same answer |
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| 35620. |
What is empirical relation between mean,mode and median? |
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Answer» Empirical relation between mean, median and mode:Mode = 3 median - 2 mean 3MEDIAN=MODE+2MEAN |
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| 35621. |
If -5 is a root of (2x^2+px-15)=0 and roots of p(x^2+x)+k=0 are equal,then find p and k. |
| Answer» Since (-5) is a root of given quadratic equation\xa02x2 + px + 15 =0,then,{tex}2 ( - 5 ) ^ { 2 } + p ( - 5 ) - 15 = 0{/tex}{tex}50 - 5 p - 15 = 0{/tex}{tex}5 p = 35 \\Rightarrow p = 7{/tex}Now\xa0{tex}p \\left( x ^ { 2 } + x \\right) + k = 0{/tex}\xa0has equal roots\xa0{tex}p x ^ { 2 } + p x + k = 0{/tex}So\xa0{tex}( b ) ^ { 2 } - 4 a c = 0{/tex}{tex}( p ) ^ { 2 } - 4 p \\times k = 0{/tex}{tex}( 7 ) ^ { 2 } - 4 \\times 7 \\times k = 0{/tex}28k = 49{tex}k = \\frac { 49 } { 28 } = \\frac { 7 } { 4 }{/tex}hence p = 7 and {tex}k = \\frac { 7 } { 4 }{/tex} | |
| 35622. |
Without adding find the sum of 1+3+5+7+9+11+13. |
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Answer» If you want to check take number as me as your choice but all number are in odd or in sequence. in my question you will observe that all numbers are odd and in sequence and i give only 7 number so we square 7 and will get answer 49. How 49 |
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| 35623. |
Find the root of |
| Answer» kiska | |
| 35624. |
How many of you think maths a tough subject |
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Answer» ? U all r right but for me maths is tough and science is too easy Subject is not tough We set our mentality that math is tough subjects are not tuff we decide which one is essay or tuff |
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| 35625. |
Please explain theorem 10.2 |
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| 35626. |
Show that cot theta + tan theta is equal to secant theta cosecant theta |
| Answer» Iss question me cot theta ko cos upon sin kro aur tan theta ko sin upon cos kr do aage apne aap aajaye ga ki kya kr naa h | |
| 35627. |
Write the empirical relationship between the three measures of central tendency?? |
| Answer» Empirical Relationships between the three measure of central tendency2 Mean = 3 Median -Mode\xa0 | |
| 35628. |
prove that median of triangle divides the triangle into two parts of equal areas |
| Answer» You are not eligible to sit un 10 class, better you sit in 9 class and then passed ro 10 | |
| 35629. |
in any triangle abc prove that (a-b)^2 cos^2c/2 + (a+b)^2 sin^c/2=c^2 |
| Answer» L.H.S. =\xa0{tex}{(a - b)^2}{\\cos ^2}\\frac{c}{2} + {(a + b)^2}{\\sin ^2}\\frac{c}{2}{/tex}{tex} = ({a^2} + {b^2} - 2ab){\\cos ^2}\\frac{c}{2}{/tex}{tex} + ({a^2} + {b^2} + 2ab){\\sin ^2}\\frac{c}{2}{/tex}{tex} = {a^2}{\\cos ^2}\\frac{c}{2} + {b^2}{\\cos ^2}\\frac{c}{2} - 2ab\\;{\\cos ^2}\\frac{c}{2}{/tex}{tex} + {a^2}{\\sin ^2}\\frac{c}{2} + {b^2}{\\sin ^2}\\frac{c}{2} + 2ab\\;{\\sin ^2}\\frac{c}{2}{/tex}{tex} = {a^2}\\left( {{{\\cos }^2}\\frac{c}{2} + {{\\sin }^2}\\frac{c}{2}} \\right){/tex}{tex} + {b^2}\\left( {{{\\cos }^2}\\frac{c}{2} + {{\\sin }^2}\\frac{c}{2}} \\right){/tex}{tex} = 2ab\\left( {{{\\cos }^2}\\frac{c}{2} - {{\\sin }^2}\\frac{c}{2}} \\right){/tex}= a2 + b2 - 2 ab cosC{tex} = {a^2} + {b^2} - 2ab\\left( {\\frac{{{a^2} + {b^2} - {c^2}}}{{2ab}}} \\right){/tex}= a2 + b2 - a2 - b2 + c2=c2 R.H.S. | |
| 35630. |
Hey everyone ............ |
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Answer» And ankita Hello shikayna... Hiiiiiiiiiii........ Hiiii Hii...... Hiiii |
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| 35631. |
find the mid point of the following points (-1,0) ,(-5,-1) |
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Answer» (-3,-0.5) (-3,-0.5 |
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| 35632. |
Maths ka paper kaisa Gaya. For govt. School |
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Answer» It was toooooooo good and lengthy Bahut aacha |
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| 35633. |
Syllabus for class 10 2019 |
| Answer» What is | |
| 35634. |
What is probability of choosing M from word MATHEMATICS |
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Answer» 2/11 2/11 https://shrinkearn.com/NtBH 2/11 |
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| 35635. |
find the length of a tangent drawn from a point 13cm away from the centre of a circle of radius 12cm |
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Answer» 5cm 5cm |
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| 35636. |
Root 6 ki value |
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Answer» 36 2.4494897427 |
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| 35637. |
which chapter are removed by ncert in hindi |
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Answer» Tu harsh he kya Haaa animesh r u from barwni... Kshitij me se dev, aatma कथ्य, stri - shiksha ke virodhi kutkro ka khandan and sanskratiThen kratika me se ehi thaiya jhulni herani ho rama and me kyo likhta hu |
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| 35638. |
If 3= y , y = 4 and O =3y , so what is the value of Y+ O . |
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Answer» Y+O = 4+3y = 4+ y × y = 4+2yY+O = 4+2y 4+3y |
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| 35639. |
If x=asin~+bcos~ Y=acos-bsin~ Prove that x2+y2=a2+b2(~ symbolises teta and 2 is square of a;b;x;y) |
| Answer» According to the question,{tex}x^2 + y^2 = (asin\\theta\xa0+ bcos\\theta)^2 + (acos\\theta\xa0- bsin\\theta)^2{/tex}=\xa0{tex}a^2sin^2\\theta+b^2cos^2\\theta+2abcos\\theta sin\\theta+a^2cos^2\\theta+b^2sin^\\theta -2ab sin\\theta cos\\theta{/tex}= a2(sin2{tex}\\theta{/tex}\xa0+ cos2{tex}\\theta{/tex}) + b2(cos2{tex}\\theta{/tex}\xa0+ sin2{tex}\\theta{/tex}){tex}= a^2 + b^2{/tex} [{tex}\\because{/tex}\xa0sin2{tex}\\theta{/tex} + cos2{tex}\\theta{/tex}\xa0= 1].{tex}\\therefore{/tex}{tex}x^2 + y^2 = a^2 + b^2.{/tex} | |
| 35640. |
Using quadratic formula, solve the following equation : abx^2+(b^2-ac)x-bc=0 |
| Answer» We have, abx2 + (b2 -ac) x-bc = 0{tex}\\implies{/tex}abx2 + b2 x - acx - bc = 0{tex}\\implies{/tex}bx ( ax+b) - c (ax + b) = 0{tex}\\implies{/tex}(ax + b) (bx - c) = 0Either ax+b = 0 or bx - c = 0{tex}\\implies x = -{b \\over a},\\, {c \\over b}{/tex}Hence, {tex}x = -{b \\over a},\\, {c \\over b}{/tex} are the required solutions. | |
| 35641. |
Board exams kab se ho rhe hain? |
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Answer» Why pre board held ???? Thnx All the best every one. Pre board from February, board from March. |
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| 35642. |
0+0=0 but 1+1=2 why |
| Answer» Because thenot value of zero | |
| 35643. |
Wat is a polynomial |
| Answer» A polynomial is a mathematical expression that consists of variables and constants combined using addition, subtraction and multiplication. Variables may have non-negative integer exponents. Although division by a constant is allowed, division by a variable is not allowed. | |
| 35644. |
First chapter first exercise 2 questions |
| Answer» Check NCERT solutions here :\xa0https://mycbseguide.com/ncert-solutions.html | |
| 35645. |
( )+( )+( )=30 Taking number 1;3;5;7;9;11;13;25 |
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Answer» What is ! This sign showa If you know then share with us Firstly it is not possible Because if odd numbers are added then even number left .and even numbers cannot give the required result 11+13+3! (We can use this sign) 10 10 10 |
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| 35646. |
12.1 q.2 |
| Answer» Check NCERT solutions here :\xa0https://mycbseguide.com/ncert-solutions.html | |
| 35647. |
196/38220 |
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Answer» 196/38220 = 0.00512 0.00512 |
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| 35648. |
Square of 8 |
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Answer» Its simple 64 64 is the square root of 8 64 64 64 64 |
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| 35649. |
X/a+y/b=a+bX/a2+y/b2=2 Find the value of x and y. |
| Answer» Firstly question is not clear Write in proper sense | |
| 35650. |
If 3=y . Y =4 and o= 3y .so what is the value of y+o. |
| Answer» Y+o Y=3 and o=3y=Y+o =3+ 3y=3(1+y)=3(y+1) | |