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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.
| 37551. |
If 1+4+7+10+______+a=287 find the value of a |
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Answer» 275 Value of a is 274 |
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| 37552. |
Is there any internal choice in 6number sums?????? |
| Answer» Nooo | |
| 37553. |
Why there are only 2 roots of quarditic equation? |
| Answer» A polynomial equation whose degree is 2, is known as quadratic equation. A quadratic equation in its standard form is represented as: ax2 + bx + c = 0 , where a, b and c are real numbers such that a\xa0≠ 0 is a variable.The number of roots of a polynomial equation is equal to its degree. So, a quadratic equation has two roots. | |
| 37554. |
2x+3y=0 find zeros of polynomial |
| Answer» 2x=-3y | |
| 37555. |
A man goes 80 m due to east and then 150 m due to north . How far is he from the starting point |
| Answer» 170m | |
| 37556. |
(√secQ-1/secQ+1)+(√secQ+1/secQ-1) |
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Answer» Sir 0 |
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| 37557. |
Use euclids division lemma to find the HCF of 196 and 38220 |
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Answer» Thx yogita again 38220>196 we always divide greater number with smaller one.Divide 38220 by 196 then we get quotient 195 and no remainder so we can write it as38220 = 196 * 195 + 0As there is no remainder so deviser 196 is our HCF\xa0 |
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| 37558. |
Prove that √n-1 + √n+1 is an irrational number |
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| 37559. |
How to solve under root 1511 |
| Answer» | |
| 37560. |
2x×5x=10 |
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Answer» 1 1 ans =2x×5x=10=10x=10x=10/10= 10 1is the answer Meanns the answer is 1 1 |
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| 37561. |
What is chamical reaction |
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Answer» Goo Those reactions in which new substance with new properties are formed. Chemical reaction, a process in which kne or more substances, the reactants are converted to one or more different substances, the products. Substances are either chemical elements are compounds. A chemical reactions rearranges the constituents atoms of the reactants to create different Substances as products. Type of chamical reaction |
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| 37562. |
What os the general form of ap |
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Answer» a^n = a + ( n- 1 ) d a,a+2d,a+3d.......a+(n-1)d |
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| 37563. |
Solve X + y =14X= 4-y |
| Answer» | |
| 37564. |
Using quadratic formula solve the eq. For xabx+(b-4ac)x-bc |
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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. Okk |
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| 37565. |
Find the middle term of the AP 6,13,20,.....,216 |
| Answer» a = 6 , d = 13-6=7, a^n= 216 , n= ?As we know that,a^ n= a+ (n-1)×d216= 6+(n-1)×7216-6= 7n-7210+7 = 7n217/7=nn=31Thus,the AP contains 31 terms.Therefore,its middle term = 31+1/2= 32/2 =16th term.So, middle term of AP = a + 15 d = 6+15 ×7= 111. | |
| 37566. |
Using quadratic formula3x2+2√5x-5=0 |
| Answer» where is the answer | |
| 37567. |
A/x-b/y =0Ab^2/x+a^2b/y=a^2+b^2 |
| Answer» Taking\xa0{tex}\\frac { 1 } { x } = u{/tex}\xa0and {tex}\\frac { 1 } { y } = v{/tex}, the above system of equation becomes{tex}{/tex}au - vb = 0............ (i){tex}{/tex}\xa0{tex}{/tex}ab2u + a2bv = a2 + b2.......... (ii)By cross-multiplication, using (i) and (ii) we have{tex}\\frac { u } { - b \\times - \\left( a ^ { 2 } + b ^ { 2 } \\right) - a ^ { 2 } b \\times 0 } = \\frac { - v } { a \\times - \\left( a ^ { 2 } + b ^ { 2 } \\right) - a b ^ { 2 } \\times 0 } = \\frac { 1 } { a \\times a ^ { 2 } b - a b ^ { 2 }( - b) }{/tex}{tex}\\Rightarrow \\quad \\frac { u } { b \\left( a ^ { 2 } + b ^ { 2 } \\right) } = \\frac { - v } { - a \\left( a ^ { 2 } + b ^ { 2 } \\right) } = \\frac { 1 } { a ^ { 3 } b + a b ^ { 3 } }{/tex}{tex}\\Rightarrow \\quad \\frac { u } { b \\left( a ^ { 2 } + b ^ { 2 } \\right) } = \\frac { v } { a \\left( a ^ { 2 } + b ^ { 2 } \\right) } = \\frac { 1 } { a b \\left( a ^ { 2 } + b ^ { 2 } \\right) }{/tex}{tex}\\Rightarrow \\quad u = \\frac { b \\left( a ^ { 2 } + b ^ { 2 } \\right) } { a b \\left( a ^ { 2 } + b ^ { 2 } \\right) } = \\frac { 1 } { a } \\text { and } v = \\frac { a \\left( a ^ { 2 } + b ^ { 2 } \\right) } { a b \\left( a ^ { 2 } + b ^ { 2 } \\right) } = \\frac { 1 } { b }{/tex}Now,\xa0{tex}u = \\frac { 1 } { a } {/tex}{tex}\\Rightarrow \\frac { 1 } { x } = \\frac { 1 } { a } {/tex}{tex}\\Rightarrow {/tex} x = aand\xa0{tex}v = \\frac { 1 } { b } {/tex}{tex}\\Rightarrow \\frac { 1 } { y } = \\frac { 1 } { b }{/tex}{tex} \\Rightarrow{/tex} y = bHence, the solution of the given system of equation is x = a, y = b. | |
| 37568. |
Find the zero of quadratic polynomial x square minus x minus 4 |
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Answer» Give me a thanks Use quadratic formula It is (1+√17)/2 and( 1-√17)/2 I\'ve found it I guess there are no zeroes for the above question |
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| 37569. |
Find the value of k(3k+1)x+3y-5=0and2x-3y+5 have infinite solutions |
| Answer» 1 | |
| 37570. |
Find the value of \'k\' for the following equation has a unique solution:4x -5y=k, 2x-3y=12 |
| Answer» The given equations are4x - 5y = k\xa0So, 4x - 5y - k = 0......... (i)And 2x - 3y = 12So, 2x - 3y - 12 = 0 ......... (ii)The system of linear equations is in the form ofa1x + b1y + c1\xa0= 0a2x + b2y + c2\xa0= 0Compare (i) and (ii), we geta1= 4\xa0,b1= -5, c1\xa0= -k,a2=2\xa0,b2= -3\xa0,c2\xa0= -12For a unique solution, we must have{tex} \\frac { a _ { 1 } } { a _ { 2 } } \\neq \\frac { b _ { 1 } } { b _ { 2 } }{/tex}{tex}\\frac { 4 } { 2 } \\neq \\frac { - 5 } { - 3 }{/tex}{tex}2 \\neq \\frac { 5 } { 3 } \\Rightarrow 6 \\neq 5{/tex}Thus, for all real value of k, the given system of equations will have a unique solution. | |
| 37571. |
Check whether the following equation are quadratic or not x\'2-8=0 |
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Answer» Yes yes yes Yes |
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| 37572. |
The angle of quadrilateral are in Ap.THe common difference is 10; find angle |
| Answer» 75 85 95 105 | |
| 37573. |
Prove that root5 is an irrational |
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Answer» Let root5 is rational no. rational no. are p/q, qnot equal to 0 , p&q are co prime number √5=p/q√5q=p Squaring on both side(√5q)square=psquare5qsquare Let, us consider as root 5 is rational,where pandq are co primes √5=p÷q then √5q=p where q divides p ,so in our assumtion the root 5 is rational ia wrong ,so in the contradiction root 5is rational is correct Let us assume that √5 is a rational number.we know that the rational numbers are in the form of p/q form where p,q are intezers.so, √5 = p/q p = √5qwe know that \'p\' is a rational number. so √5 q must be rational since it equals to pbut it doesnt occurs with √5 since its not an intezertherefore, p = √5qthis contradicts the fact that √5 is an irrational numberhence our assumption is wrong and √5 is an irrational number. |
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| 37574. |
Please provide sample papers for 2019 -20 |
| Answer» Net pe mil jaayenge ???? | |
| 37575. |
Prove that SinA+CosA÷SinA-CosA + SinA-CosA÷SinA+CosA =2÷1-2√CosA |
| Answer» Contact me for answer.... | |
| 37576. |
Please U.T ki date sheet de dijiye |
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Answer» Ya different school have different datesheet apne school se pta kro Different schools have different U. T DATE SHEET.... |
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| 37577. |
2×15 |
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Answer» 30Ha ha......... ????? 30... 2nd class ka question hai.... For what reason you had came in class 10 30 But any other do answer....? 30 of course..... Buddy |
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| 37578. |
Show that a1 ,a2, ....... an ........is defined as below:a) an=3+4n |
| Answer» Answer:The given sequences are in A.PStep-by-step explanation:Formula used:\xa0The n th term of A.P a, a+d, a+2d, ........ is\xa01.which is the n th term of A.P 7, 11, 15....... | |
| 37579. |
Find 398-88 |
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Answer» 310 310 310 ??310?? 310 |
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| 37580. |
you have latest blueprint of math |
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Answer» Check marking scheme here :\xa0https://mycbseguide.com/cbse-syllabus.html 1×20=206×2=128×3=246×4=24 Chapter1 -6 marksCh.2,3,4,5-total 20 marksCh.6,10,11-total 15 marksCh.7-6marksCh.8,9- 12marksCh.12,13-10 marksCh.14,15-11 marks Ya |
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| 37581. |
Important quetion of chapter statistics |
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| 37582. |
from where questions come in board in maths |
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Answer» Together with only.... R.D sharma , Ncert ,Prabodh |
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| 37583. |
Which term of AP 5,9,13,17----------------is81 |
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Answer» The 20th term of given ap is 81 a=5 :d=9-5=4 :an=81 : n=..? an=a+(n-1)d81=5+(n-1)481-5=(n-1)476=(n-1)476/4= n-119 =n-119+1 =nn=20 a=5, d=9-5=4. an=81. an=a+(n-1)d. 81=5+(n-1)4. 81-5=(n-1)4. 76÷4=n-1. 19=n-1. 19+1=n. 20=n a = 5 , d = 9-5=4 , Tn=81 , n= ? AP of n th term is given by, Tn= a+ (n-1)d . Then, Tn = 5+(n-1)4 81= 5+4n-4 81+1=4n n= 82/4 n= 2.5 th term An=a+(n-1)d81=5+(n-1)481=5+4n-481=1+4n81-1=4n80=4n80/4=nN=20 |
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| 37584. |
Find the zeroes of P(x)=15x×x+31x-24 |
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| 37585. |
Find the quadratic equation whose zeros are2 and -6 |
| Answer» K(x2-(x+y)x +xy) K[x2-(2+(-6))x +(2*-6)] K[x2-(2-6)x +(-12)] K[x2-(-4)x -12]k[x2+4x-12]put k=1Hence the polynomial is x2+4x-12 | |
| 37586. |
Ek but a + b + X barabar 1 but a + b + 1 by X |
| Answer» ?????? | |
| 37587. |
The first three tremendous an AP 3y-1,3y±5& 5y±1 then find the value of yuan. |
| Answer» Since {tex}(3y-1), (3y+5)\\ and\\ (5y+1){/tex} are in AP, we have{tex}(3y+5)-(3y-1)=(5y+1)-(3y+5){/tex}{tex} \\Rightarrow {/tex}\xa0{tex}3y+5-3y+1=5y+1-3y-5{/tex}{tex} \\Rightarrow {/tex}\xa0{tex}6=2y-4{/tex}{tex} \\Rightarrow {/tex}\xa02y=10{tex} \\Rightarrow {/tex}\xa0y=5 | |
| 37588. |
How to multiply equation?? |
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Answer» Which? By using formula |
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| 37589. |
I have doubt in HCF expansion please explain it in some simplest way |
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Answer» I want to know HCF linear combination in a simplest way For exanple we have to find HCF of 20 and 12 so we first we convert them into its prime factors20= 2×2×512=2×2×6The highest common factor is 2×2. So its HCF is 4 |
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| 37590. |
A circle can hv _______ parallel tangents at most |
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Answer» Niharika sharma....... No ur answer is wrong bro? i m right that is a circle can have two parallel tangents at most. Bro sahi, thanks to you also to confuse me for a moment Sis a circle have infinite point, so infinite parallel line occur as a tangent of a circle Depending on the point given Infinite two |
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| 37591. |
For what value of k the root of the equation x2+4x+k=0 are real |
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Answer» b²-4ac=0. 4²-4×1×k=0. 16-4k=0. 16=4k. 16÷4=k . 4=k. Value of k must be less or equal to 4 |
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| 37592. |
Prove the thales theorem . |
| Answer» | |
| 37593. |
Divide 31 into two parts such that the sum of their squares is 485 |
| Answer» That two parts are. 17 and 14 | |
| 37594. |
Sir please tell me other extra books for practice questions |
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Answer» You can also take help of exam idea R. D sharma. Best book ever for exams preparation |
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| 37595. |
side book questions |
| Answer» Very important question u have to study | |
| 37596. |
Find the number of terms in each of the following finite ap I 3,8,13,..............,78 |
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Answer» A=1,d=3-1=2,an=78,n=?A+(N-1)D=AN1+(N-1)2=781+2n-2=782n=78+2-12n=79n=79/2Ans A=1D=3-1=2AN=78AN= A+(N-1)D78=1+2N-279=2NN=39.5TH TERM (APPROX) OR 40TH TERM |
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| 37597. |
Who likes maths like me comment here |
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Answer» Me tooooo Me too? |
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| 37598. |
For which value of k, the following pair of linear equation has no solution2x+3y=1(k-1)x+(2x+1)y=k-1 |
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Answer» Plz write question properly I Is this ques correct??? According to me here it must be 2k+1 in the bracket instead of 2x+1 and 2x+3y=1 and ( k-1)x +(2x+1)y=k-1 are two different eq not a single one |
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| 37599. |
Hwo proov that root 2 is a irrational number |
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Answer» Let,√2=p/qSquares both side,so we found 2=p2/q2Multiplying q both sides2q=p2/q2×q=2q=p2/qSince p and q are integer so it is rational and his friction is 1 .so it is not rational So it is rational How! and prove! Is the correct spelling |
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| 37600. |
Rationalize 2√2 |
| Answer» It is already rationAlised | |