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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.
| 37051. |
Use euclid division algorithm to find hcf of 10224 and 9648 |
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Answer» 144 ? Easy yarr |
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| 37052. |
(x-2)(x+1)=(x-1)(x+3) |
| Answer» (x-2)(x+1)=(x-1)(x+3)x (x + 1) - 2( x+ 1) = x (x + 3) - 1 ( x + 3)x2 + x - 2x - 2 = x2 + 3x - 1x - 3x2 - x - 2 = x2 + 2 x - 3- x - 2x = -3 + 2- 3x = -1x = 1/3 | |
| 37053. |
6x+3y=20 |
| Answer» {tex}2x + y = 6{/tex}{tex}\\Rightarrow y =-2x + 6{/tex}\tx{tex}2{/tex}{tex}4{/tex}y{tex}2{/tex}{tex}-2{/tex}\t{tex}6x + 3y = 20{/tex}{tex}\\Rightarrow y = \\frac {20-6x} { 3 }{/tex}\tx{tex}0{/tex}{tex}\\frac{10}{3}{/tex}y{tex}\\frac{20}{3}{/tex}{tex}0{/tex}\tThe graph of the system of equations is parallel lines\xa0{tex}\\therefore{/tex} the system has no solution and hence is inconsistent. | |
| 37054. |
If cosec+cos=x Find the value of cosec-cos |
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| 37055. |
describe zero of the polynomial |
| Answer» The zero of the polynomial is defined as any real value of x, for which the value of the polynomial becomes zero. | |
| 37056. |
factorize x²-7x-10 |
| Answer» (x-5) (x-2) | |
| 37057. |
If(9/7)^3×(49/81)^2x-6=(7/9)^9 |
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| 37058. |
Ap=8,3,-2.........find Sn |
| Answer» Here {tex}a = 8,\\ d = 3 - 8 = -5.{/tex}So, Sn\xa0= {tex}\\frac{n}{2}{/tex}[2a + (n\xa0- 1)d]{tex}\\Rightarrow{/tex}\xa0{tex}S_{n}\xa0= \\frac n2(16 - 5n + 15) = \\frac n2(31 - 5n){/tex} | |
| 37059. |
The 17th of an ap exceed it 10th by 7 .find the common difference. |
| Answer» Let the first term and the common difference of the AP be a and d respectively.Given that, a17 = a10+7\xa0{tex} \\Rightarrow {/tex}\xa0a + (17 - 1) d = a + (10 - 1)d + 7\xa0{tex}\\because {/tex}an = a +(n - 1)d{tex} \\Rightarrow {/tex}\xa0a + 16d = a + 9d + 7{tex} \\Rightarrow {/tex}\xa016d - 9d = 7{tex} \\Rightarrow {/tex}\xa07d = 7{tex} \\Rightarrow d = \\frac{7}{7} = 1{/tex}Hence, the common difference is 1. | |
| 37060. |
Solve for X and y if 2 x is equal to 5y + 4 and 3 X - 2y + 6 is equal to zero |
| Answer» 2x = 5y + 42x - 5y = 4..... (i)3x - 2y + 6 = 03x - 2y = -6 ......... (ii)Multiply (i) by 3 and (ii) by 26x - 15y = 12....... (iii)6x - 4y = - 12 .......(iv)Subtract (iv) from (iii)- 11y = 24y = -24/11Put y = -24/11 in (ii) we get3x - 2(-24/11) = -63x + 48/11 = -63x = - 6 - 48/113x = -144/11x = - 38/11 | |
| 37061. |
give equation of a line which is parallel to 2x-3.4y-9=0 |
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| 37062. |
Koi asise channel hai jo 10th ko padatha ho you tube par |
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Answer» Edumantra PuStack and bhai ki padhai |
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| 37063. |
What are the syllables of maths 2019 2020 |
| Answer» Check the syllabus here :\xa0https://mycbseguide.com/cbse-syllabus.html | |
| 37064. |
3+2√5 |
| Answer» 3+2√5 | |
| 37065. |
Find the value of k for which the quadratic equation kx(x-2)+6=0 has two equal roots |
| Answer» We have, kx(x - 2) + 6 = 0{tex}\\Rightarrow k x^{2}-2 k x+6=0{/tex}........(1)We know that quadratic equation {tex}ax^2+bx+c=0{/tex},a≠0, has equal roots if its discriminant D is 0.D = 0\xa0{tex}\\Rightarrow{/tex}{tex}b^2-4ac=0{/tex}{tex}\\Rightarrow{/tex}4k2 - 4(k)6 = 0 [from given quad. equ.(1); a=k≠0, b= -2k, c= 6]{tex}\\Rightarrow{/tex}4k2 - 24k = 0\xa0{tex}\\Rightarrow{/tex}4k(k - 6) = 0 {tex}\\Rightarrow{/tex}k - 6 = 0 [ since, k≠0]{tex}\\Rightarrow{/tex}k = 6\xa0Hence the value of K is 6. | |
| 37066. |
find the number of integers between 50 and 400 which are divisible by 7 |
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| 37067. |
find 3 numbers in an AP such that their sum is 30 and their product is 960 |
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| 37068. |
find the 31st term of an AP whose 11th term is 38 and 16th term is 73 |
| Answer» Here a11 =38 and a16 =73Using formula a n =a +(n−1)d, to find nth term of arithmetic progression,38=a +(11−1) (d) And 73=a +(16−1)(d)⇒ 38=a +10d And 73=a +15dThese are equations consisting of two variables.We have, 38=a +10d ⇒ a =38−10dLet us put value of a in equation (73=a +15d),73=38−10d +15d ⇒ 35=5dTherefore, Common difference =d =7Putting value of d in equation 38=a +10d ,38=a +70 ⇒ a =−32Therefore, common difference = d = 7 and First term = a = -32Using formula a n =a +(n−1)d , to find nth term of arithmetic progression,a31=-32+(31−1)(7) =−32+210=178Therefore, 31st term of AP is 178. | |
| 37069. |
If the zero of the polynomial p(x) (k+4)x2+13x+3k is resipropal of the then the value of k |
| Answer» Let thw zeroes are a and 1/aProduct = c/aa*1/a = 3k/k+41=3k/k+43k=k+42k=4K=2 | |
| 37070. |
3^40-3^39/3^41-3^40 . Solve it if u can. |
| Answer» Taking common 3^39 from numeratorAnd 3^40 from denominator And the ans is 1/3 | |
| 37071. |
Multiply 15=a+2d by 25=2a+9d |
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| 37072. |
If d is the hcf of ( 1155 and 506) then find x,y satisfing d= 1155x + 506y |
| Answer» We need to find the H.C.F. of 506 and 1155 and express it as a linear combination of 506 and 1155.By applying Euclid’s division lemma1155 = 506 {tex}\\times{/tex}\xa02 + 143.506 = 143 {tex}\\times{/tex}\xa03 + 77.143 = 77 {tex}\\times{/tex}\xa01 + 66.77 = 66 {tex}\\times{/tex}\xa01 + 11.66 = 11 {tex}\\times{/tex}\xa06 + 0.Therefore, H.C.F. = 11.Now, 11 = 77 - 66 {tex}\\times{/tex}\xa01 = 77 - [143 - 77 {tex}\\times{/tex}\xa01] {tex}\\times{/tex}\xa01 {∵ 143 = 77 {tex}\\times{/tex}\xa01 + 66}= 77 - 143 {tex}\\times{/tex}\xa01 + 77 {tex}\\times{/tex}\xa01= 77 {tex}\\times{/tex}\xa02 - 143 {tex}\\times{/tex}\xa01= [506 - 143 {tex}\\times{/tex}\xa03] {tex}\\times{/tex}\xa02 - 143 {tex}\\times{/tex}\xa01 {∵\xa0506 = 143 {tex}\\times{/tex}\xa03 + 77 }= 506 {tex}\\times{/tex}\xa02 - 143 {tex}\\times{/tex}\xa06 - 143 {tex}\\times{/tex}\xa01= 506 {tex}\\times{/tex}\xa02 - 143 {tex}\\times{/tex}\xa07= 506 {tex}\\times{/tex}\xa02 - [1155 - 506 {tex}\\times{/tex}\xa02] {tex}\\times{/tex}\xa07 {∵1155 = 506 {tex}\\times{/tex}\xa02 + 143\xa0}= 506 {tex}\\times{/tex}\xa02 - 1155 {tex}\\times{/tex}\xa07 + 506 {tex}\\times{/tex}\xa014= 506 {tex}\\times{/tex}\xa016 - 1155 {tex}\\times{/tex}\xa07Hence obtained. | |
| 37073. |
What is perfect square of 1/2a2 |
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| 37074. |
Find hcf of 420 and 130 |
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Answer» 420=130×3+30130=30×3+1030=10×3+0HCF. 10 30 is right answer Step:1 Since 420 > 130 we apply the division lemma to 420 and 130 to get ,420 = 130 x 3 + 30Step:2 Since 30 ≠ 0 , we apply the division lemma to 130 and 30 to get130 = 30 x 4 + 10Step:3 Since 10 ≠ 0 , we apply the division lemma to 30 and 10 to get30 = 10 x 3 + 0The remainder has now become zero, so our procedure stops. Since the divisor at this Step is 10, the HCF of 420 and 130 is 10. 420=2.2.5.3.7130=2.5.13HCF=5.Means multiply |
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| 37075. |
Sin^2+cos^2 =1 Prove that |
| Answer» I proved. It question very easily | |
| 37076. |
What are composite numbers? With example. |
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Answer» An integer which have more than 2 factors An integer can be generated by multiplying the two smallest positive integers and it contains at least one divisor other than one and itself is called a composite number. Each positive integers is either prime, composite or the unit 1. Composite numbers are the numbers that are neither prime nor unit .Example : 4( factors are 1, 2 and 4) |
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| 37077. |
Find the largest 4 digit number which is exactly divisible by 12, 15,and 24, |
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Answer» Thnx bhai 15 = 3×524 = 2×2×2×312 = 2×2×3Lcm = 120The greatest 4 digit no. Is 9999Dividing lcm by 9999 we get remainder as 399999-39=9960The answer is 9960 Find the largest 4 digit number which is exactly divisible by 12, 15,and 24, |
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| 37078. |
Koi asise channel hai jo 10th class ko padatha ho maths you tube par |
| Answer» Han ho sakta hai .Search karo milega | |
| 37079. |
A3 |
| Answer» | |
| 37080. |
Koi yaha par gayatri institute ka h kya.????????!! |
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Answer» Or center konsa h? 10th mein abhi hi aaye 11 mein Jane h मैं हूं ,,,,,,???? |
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| 37081. |
Split the middle term x^2-5x+8 and 4x^2-8x-6 |
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| 37082. |
given a=5 d=3 an=50 find n and Sn |
| Answer» n=16_-_-_-_-_-_ Sn=248??? | |
| 37083. |
(X-2r)^3 |
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| 37084. |
Who is laks****@***** ...jo b h wo sirf thanks kr rha h ... |
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Answer» Don\'t know Pata ni |
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| 37085. |
Sb gye ke .... |
| Answer» Not | |
| 37086. |
Best of luck for your result guys ??? |
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Answer» Thnx Tq Thq yr.... But result jaldi ajay ab or kitna wait krwayega..... Thanks yrr |
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| 37087. |
If the HCF of 210 and 55 is expressible in the form 210×5+55y then find y.. |
| Answer» 210=55×3+45...........(1)55=45×1+10..............(2)45=10×4+5................(3)10=5×2+0So here HCF (210,55)=5From equation (3), 5 can be written as5=45—10×4..........(4)Now from (2) 10=55—45×1,now this value substitute in equation (4) in place of 10Then we get,5=45—(55—45×1)×4After simplifying we will get5=45×5—55×4.......(5)Now,from (1) 45=210—55×3, this value substitute in equation (5)5=(210—55×3)×5—55×4, after simplifying we will get5=210×(5)+55×(-19)....Now this one compare with given equation that 5=210×5+55×xThen X=–19 is the answer | |
| 37088. |
55x +67y =31167x +55y =299 |
| Answer» For this type of questions. We simple first + the both terms and then substract the both equation. So we get 2 short equations. After that solve the both equation by eliminating method or substitution method. This is only work when coffecient of x is equal to coffecient of y in 2nd equation and coffecient y in 1st equation is equal to coffecient of x in 2nd equation. Hope you like the answer. | |
| 37089. |
6x+5y=7x+3y+1=2(x+6y-1) |
| Answer» The given equations are{tex}6x + 5y = 7x + 3y + 1 = 2(x + 6y -1){/tex}Therefore,we have{tex}6x + 5y = 2(x + 6y -1){/tex}{tex}6x + 5y = 2x + 12y - 2{/tex}{tex}6x - 2x + 5y -12y = -2{/tex}{tex}4x - 7y = -2 {/tex}......(i)Also,\xa0{tex}7x + 3y + 1 = 2(x + 6y - 1){/tex}{tex}7x + 3y + 1= 2x + 12y -2{/tex}{tex}7x - 2x + 3y -12y = -2 -1{/tex}{tex}5x - 9y = -3 {/tex}.........(ii)Multiplying (i) by 9\xa0and (ii) by 7, we get{tex}36x -\xa063y = -18{/tex} ......(iii){tex}35x -\xa063y = -21{/tex} .....(iv)Subtracting (iii) and (iv),we get{tex}x = 3{/tex}Substituting x = 3\xa0in (i),we get{tex}\\Rightarrow4 \\times 3 - 7 y = - 2\\\\ \\Rightarrow - 7 y = - 2 - 12{/tex}{tex}\\Rightarrow-7y = -14{/tex}{tex}\\Rightarrow{/tex}y = 2{tex}\\therefore{/tex}\xa0Solution is x = 3, y = 2 | |
| 37090. |
For how many integers n is √9 - ( n + 2 )^2. |
| Answer» 9 -(n+2)^2=[3^2-(n+2)^2)=using identity a^2-b^2=(a+b)(a-b)=(3-n+2)(3+n-2)=n=-5,1 | |
| 37091. |
Hii..m bapas aa gyi ... |
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Answer» Ok amu bestie... Mast Hii but bye bestie.... din mai baat krege.... Hi..aastha bestie kaisi ho Hlo |
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| 37092. |
Hii..gm....sb kaise ho.. |
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Answer» Hi ...amu bestie nd aastha bestie Tension me Hlo... gm bestie.....?? |
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| 37093. |
When result come |
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Answer» Arre aj nhi aega Think so after 12 00 clock |
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| 37094. |
If one zero of polynomial (k-1)x^2+kx+1 is -3 , find the value of k. |
| Answer» X=-3 p(x)=(k-1)x^2+kx+1=0Putting the value of x in the p(x)P(x)=(k-1)(-3)^2+k(-3)+1=0 =(k-1)(9) - 3k +1=0 =9k-9-3k+1=0 =6k-8=0 =6k=8 =k=8/6 =k=4/3Hence, value of k is 4/3 | |
| 37095. |
Aastha bestie ..kl to rcb jeet gyi ..... |
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Answer» Hmmmm ???? |
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| 37096. |
Doraemon.....❤️❤️ |
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Answer» Ki saayad hota to result badal dete ?? Doraemon ki yaad aa rhi h kya ?? |
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| 37097. |
2+2=5 prove it |
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Answer» Tu pagal hai You plz give me answer to my question Given 2+ 2 To prove 2+ 2 = 5Proof : - 2 + 2 = 4 (Common sense)Adding 1 to both sides2 + 2 + 1= 4 + 15 = 5LHS = RHS Hence proved??? |
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| 37098. |
X + Y is equal to 14 |
| Answer» x + y\xa0= 14 ....(1)x - y = 4 ....(2)x\xa0= 4 + y from equation (2)Putting this in equation (1), we get4+y +y =14⇒ 2y =10⇒ y = 5Putting value of y in equation (1), we getx + 5 = 14⇒ x = 14 - 5 = 9Therefore, x =9 and y =5 | |
| 37099. |
P(x)=x3_x2_x+1 |
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| 37100. |
what is the last digit of the number u get by multiplying the first 2002 odd prime numbers together |
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