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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.
| 39051. |
Solutions for 6x+10y=4,3x+5y=-11 |
| Answer» 6x+10y-4=0..............eq13x+5y+11=0.............eq2now subtract eq2 from eq1gives 3x+5y=-14.............eq 3hat is not possible so i recommend u to recheck the question | |
| 39052. |
what is the standard form in linear equations in two variables |
| Answer» ax+by+c=0 | |
| 39053. |
√45 +7√45 |
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Answer» 24√5 6√5+7 |
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| 39054. |
The hcf of 441,567,693by using Euclid\'s division algorithm |
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| 39055. |
if cosA=(√4-x^2)/2 then find the value secA×tanA |
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| 39056. |
Examine whether (3-√2)(3+√2) is a rational or irrational |
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Answer» Rational Number Rational no. Rational |
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| 39057. |
Types of irrational numbers |
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| 39058. |
If m and n are the zeroes of the polynomial ax2+5x+c find the value of a and c if m+n+mn=10 |
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| 39059. |
Consistency |
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| 39060. |
Write coordinates of origion |
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| 39061. |
Factorization of 11x^2 -33X-368 ? |
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| 39062. |
2x+5=0 x= |
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Answer» -5/2 -5/2 -5/2 -2.5 -5/2 -5/2 |
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| 39063. |
Find the number to be added to the polynomial xsqure -5x+4, so that 3is the zero of the polynomial |
| Answer» If 2 is added to the polynomial.....x²-5x+4+2=x²-5x+6x²-3x-2x+6=x(x-3)-2(x-3)(x-3)(x-2)=x=3,2 | |
| 39064. |
Cosec2(90-theta)-tan2 theta/2(cos2 48+cos2 42)-2tan2 30 sec2 52 sin 2 38 /cosec2 70 -tan2 20 |
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| 39065. |
If cotA + tanA =xAnd secA - cosA = yProve that-(x²y)⅔ -(xy²)⅔=1 |
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| 39066. |
68 is thewater term ofto AP 7,10,13....... |
| Answer» 68 is not the term of given AP | |
| 39067. |
Mean find 37 |
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| 39068. |
Find the composite two digit number and coprime two divit number ? |
| Answer» composite two digit number = 99and\xa0coprime two digit number are 12 and 13 | |
| 39069. |
Write the zeroes of the polynomial x2-x+6 |
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Answer» x2-(3-2)x+6=0x2-3x+2x+6=0-x(x+3)2(x+3)=0(-x+2)(x+3)=0-x+2=0-x=-2x=2x+3=0x=-3Hence the zeroes of polynomial are -3 & 2 U write wrong question. |
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| 39070. |
The HCF of two numbers is 16 and thier product is 3072 find the lcm |
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Answer» HCF = 16,Product of Numbers = 3072According to formula, if a and b are two given numbers, then{tex}HCF(a, b) \\times LCM(a, b) = a \\times b{/tex}{tex}16 \\times LCM(a,b)=3072{/tex}{tex}LCM(a,b)=\\cfrac {3072}{16}{/tex}{tex}LCM(a,b)=192{/tex} 192 |
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| 39071. |
265+564 |
| Answer» 829 | |
| 39072. |
Show that 12to the power n cannot end with digits 0 or 5 any natural minumber n |
| Answer» The factors of 12 are 1,2,3,4,6,12 So to end with 0 the 5 is also required as a factor Since 12 does not have 5 as a factor Do there is no natural no.that end with 0 | |
| 39073. |
Proof of BPT theoram |
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| 39074. |
Hcf of 144 |
| Answer» There must be another number too | |
| 39075. |
If 3cot A = 4, find the value of Cos A- Sin A. |
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Answer» Cos A=b/h=4/3 3cotA=4So,cotA=3/4=b/pH=?H=5CosA=b/h=3/5SinA=p/h=4/5Value=cosA-sinA =3/5-4/5 =-1/5Where b is base ,p is perpendicular and h is hypotenuse CotA=4/3, then h=5 so cos a -sin a=1/5 |
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| 39076. |
Small prime number |
| Answer» 2 | |
| 39077. |
If the HCF of 65 and 117 is expressible in the form of 65m-117 then find the value of m? |
| Answer» 5600 | |
| 39078. |
Hcf of 184230 and 276 by euclide division algorithms |
| Answer» Given numbers are 184, 230, and 276.Applying Euclid\'s division lemma to 184 and 230, we get230 = 184\xa0{tex}\\times{/tex}\xa01 + 46184 = 46\xa0{tex}\\times{/tex}\xa04 + 0The remainder at this stage is zero.So, the divisor at this or the remainder at the previous stage i.e., 46 is the HCF of 184 and 230.Also,276 = 46\xa0{tex}\\times{/tex}\xa06 + 0∴ HCF of 276 and 46 is 46HCF (184, 230, 276) = 46Hence, the required HCF of 184, 230 and 276 is 46. | |
| 39079. |
400÷200 |
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Answer» You are right. Is it 10 class question 400/200=2 |
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| 39080. |
Example 10 chapter 1 class 10 ncert cbse |
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Answer» Show that 5 minus under root 3 is irrational Are you telling or not I want before night Tell fast |
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| 39081. |
Prove (1+cosA)/sinA=sinA/(1-cosA) |
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| 39082. |
Write a rational number between under root3 and under root2. |
| Answer» {tex}\\text {Value of } \\sqrt3 = 1.732... \\text { and value of } \\sqrt2 = 1.414...{/tex}{tex}\\text {Rational number between } \\sqrt 3 \\text { and } \\sqrt 2 \\text { is } \\cfrac {3}{2} \\text { or } 1.5{/tex} | |
| 39083. |
Problems on leap year |
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| 39084. |
can you prove that sin 30 degree =1/2 |
| Answer» Yes | |
| 39085. |
5448 /555 |
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| 39086. |
When was class 10 math compartment paper ... |
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Answer» What is current Exercise 4.3 question 11 |
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| 39087. |
Factorise of x2 -55x+ 750 |
| Answer» x2-30x-25x+750x(x-30)-26(x-30)(x-26)(x-30)x-26=0x=26x-30=0x=30 | |
| 39088. |
express the number 0.3178 In the form of in the form of ractional number a&b |
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| 39089. |
( sin theta + cos theta ) ( tan theta + cot theta ) = sec theta + csc theta prove it |
| Answer» L.H.S.\xa0{tex}= (\\sin \\theta + \\cos \\theta )(\\tan \\theta + \\cot \\theta ){/tex}{tex}= (\\sin \\theta + \\cos \\theta )\\left( {\\frac{{\\sin \\theta }}{{\\cos \\theta }} + \\frac{{\\cos \\theta }}{{\\sin \\theta }}} \\right){/tex}{tex}= (\\sin \\theta + \\cos \\theta )\\left( {\\frac{{{{\\sin }^2}\\theta + {{\\cos }^2}\\theta }}{{\\sin \\theta \\cos \\theta }}} \\right){/tex}{tex} = (\\sin \\theta + \\cos \\theta )\\left( {\\frac{1}{{\\sin \\theta \\cos \\theta }}} \\right){/tex}{tex} = \\frac{{\\sin \\theta + \\cos \\theta }}{{\\sin \\theta \\cos \\theta }}{/tex}{tex} = \\frac{{\\sin \\theta }}{{\\sin \\theta \\cos \\theta }} + \\frac{{\\cos \\theta }}{{\\sin \\theta \\cos \\theta }}{/tex}{tex}= \\frac{1}{{\\cos \\theta }} + \\frac{1}{{\\sin \\theta }}{/tex}{tex}= \\sec \\theta + \\cos ec\\theta {/tex}= R.H.S. | |
| 39090. |
find the sum og firdt 8 multiples if 3 |
| Answer» S = 3 + 6 + 9 + 12 + .... + 24= 3(1 + 2 + 3 + ... + 8)= 3\xa0{tex}\\times{/tex}\xa0{tex}\\frac{8 \\times 9}{2}{/tex} = 108 | |
| 39091. |
If 1&-2 are two zeroes of the pollynomial xcube-4xsquare-7x+10. Find its third zero. |
| Answer» let alpha and beta be two 0\'sThen by app. Formula.. Find third 0ro | |
| 39092. |
If Sn= 3n^2 + 5n and an= 164, find n? |
| Answer» the value of n=27 | |
| 39093. |
If a_n = 3-4n , find AP and sum of 20 terms |
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| 39094. |
Chapter 10 theoram |
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| 39095. |
36x2_12a2x+(a2_b2)=0 |
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| 39096. |
use euclid division algorithum hcf of 216 and 297 |
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| 39097. |
Do optional exercises will come in cbse board |
| Answer» Yes . It may come but in a different language and data . | |
| 39098. |
If f(x)=ax2+bx+chas no real zeros and a+b+c |
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| 39099. |
What is quadratic formula |
| Answer» (X= -b±√b²-4ac/2a) | |
| 39100. |
5x+2y=2k2(k+1)x+ky=(3k+4) |
| Answer» {tex}5x + 2y = 2k{/tex}{tex}2(k + 1)x + ky = (3k + 4){/tex}These are of the form{tex}a_1x + b_1y + c_1\xa0= 0 , a_2x + b_2y + c_2\xa0= 0{/tex}where,{tex}a_1= 5\xa0,b_1= 2, c_1\xa0= -2k,{/tex}{tex}a_2= 2(k +1)\xa0,b_2= k\xa0,c_2\xa0= -(3k + 4){/tex}For infinitely many solutions, we must\xa0have{tex}\\frac { a _ { 1 } } { a _ { 2 } } = \\frac { b _ { 1 } } { b _ { 2 } } = \\frac { c _ { 1 } } { c _ { 2 } }{/tex}This holds only when{tex}\\frac { 5 } { 2 ( k + 1 ) } = \\frac { 2 } { k } = \\frac { - 2 k } { - ( 3 k + 4 ) }{/tex}{tex}\\Rightarrow \\frac { 5 } { 2 ( k + 1 ) } = \\frac { 2 } { k } = \\frac { 2 k } { ( 3 k + 4 ) }{/tex}Now, the following cases arises:Case 1:{tex}\\frac { 5 } { 2 ( k + 1 ) } = \\frac { 2 } { k }{/tex}[Taking I and II]{tex}\\Rightarrow 5 k = 4 ( k + 1 ) \\Rightarrow 5 k = 4 k + 4{/tex}k = 4Case 2:{tex}\\frac { 2 } { k } = \\frac { 2 k } { ( 3 k + 4 ) }{/tex}[Taking II and III]{tex}\\Rightarrow{/tex}2(3k +\xa04) = 2k2\xa0{tex}\\Rightarrow{/tex}6k + 8 = 2k2{tex}\\Rightarrow{/tex}{tex}2k^2\xa0-\xa06k + 8 =\xa00{/tex}{tex}\\Rightarrow{/tex}{tex}2(k^2\xa0- 3k + 4)=\xa00{/tex}{tex}\\Rightarrow{/tex}{tex}k^2\xa0- 3k - 4 = 0{/tex}{tex}\\Rightarrow{/tex}{tex}k^2\xa0- 4k + k - 4 = 0{/tex}{tex}\\Rightarrow{/tex}k(k - 4) + 1(k - 4) = 0{tex}\\Rightarrow{/tex}(k - 4)(k +\xa01) = 0(k - 4) = 0 or k + 1 = 0{tex}\\Rightarrow{/tex}\xa0k = 4 or k = -1Case 3:{tex}\\frac { 5 } { 2 ( k + 1 ) } = \\frac { 2 k } { ( 3 k + 4 ) }{/tex}[Taking I and II]{tex}\\Rightarrow 15 k + 20 = 4 k ^ { 2 } = 4 k{/tex}{tex}\\Rightarrow{/tex}4k2\xa0+ 4k - 15k - 20 = 0{tex}\\Rightarrow{/tex}\xa04k2\xa0- 11k - 20 = 0{tex}\\Rightarrow{/tex}4k2\xa0- 16k + 5k - 20 =0{tex}\\Rightarrow{/tex}4k(k - 4) + 5(k - 4) = 0{tex}\\Rightarrow{/tex}(k - 4)(4k + 5) = 0{tex}\\Rightarrow k = 4 \\text { or } k = \\frac { - 5 } { 4 }{/tex}Thus, k = 4, is the common value of which there are infinitely many solutions. | |