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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.
| 16851. |
Sbb kahan hoooo!!!!!????? |
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Answer» Koi nhii...? Oh mujhe lga..hai??? Hlo ...kaha h aastha bestie??? Hii diya and aastha bestie..hlo aashu Hloo Hii? Ghar me Aur sunao..Kya kr rhe ho yash? Ho hoi...? |
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| 16852. |
Find the equation of :3x-5y=4 |
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| 16853. |
Find the solutions of the equation:2x+7=9y |
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| 16854. |
Find the mean :Class intervals- 0-6,6-12,12-18,18,24,24-30,30-36,36-42Fi: 10,11,7,4,4,3,1 |
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| 16855. |
Hlo koi h????? |
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Answer» Hii Hmm..hlo Yes...hii Hlo hello Hello... |
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| 16856. |
Can two numers have10 as their HCF and 105 as their LCM |
| Answer» {tex}\\frac{105}{10}=10.5{/tex}Hence 105\xa0is not divisible by 10But LCM of two numbers should be divisible by their HCF.{tex}\\therefore{/tex}\xa0Two numbers cannot have their HCF as 10\xa0and LCM as 105. | |
| 16857. |
If the HCF and LCM of two number are 9 and 90 respectively. If one number is 18 . findthe other. |
| Answer» Using the result,{tex}HCF{\\text{ }} \\times {\\text{ }}LCM{/tex} = Product of two natural numbers{tex} \\Rightarrow {/tex} the other number = {tex}\\frac{{9 \\times 90}}{{18}}{/tex} = 45 | |
| 16858. |
Find sum of all n odd natural number |
| Answer» The given n odd natural numbers are,\xa01, 3, 5, .... nClearly,it is an AP with ,\xa0First term (a) = 1Common difference (d) = 3 - 1 = 2Number of terms = {tex}{/tex}nSum of n terms {tex} = \\frac{n}{2}\\left[ {2a + (n - 1)d} \\right]{/tex}{tex} = \\frac{n}{2}\\left[ {2 \\times 1 + (n - 1) \\times 2} \\right]{/tex}{tex} = \\frac{n}{2}\\left[ {2 + 2n - 2} \\right]{/tex}{tex} = \\frac{n}{2} \\times 2n{/tex}= n2 | |
| 16859. |
Graph of quadratic polynomial is a 1 circle 2 straight line 3 porable 4 none of the above |
| Answer» Parabola | |
| 16860. |
3cot tith =2find the value of 4sin t-3cos t/2sin t+6cos t |
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| 16861. |
For what value of k the system of equation kx ➕3y=1 and 12x ➕ky=2 has no solution |
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Answer» sorry for that is for infinite solution for no solution a1/a2 -- b1/b2 not equal to c1/c2k/12 -- 3/kk2 -- 36k -- root36k -- 6 in case of no solutiona1/a2 --- b1/b2 -- c1--c2k/12 -- 3/k -- 1/2k/12 -- 1/2 & 3/k -- 1/2 (solve it by cross multiplication)then2k -- 12k -- 12/2k -- 6&3/k -- 1/2k -- 6so the value is 6 of k for no solution 6 is answer 6 |
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| 16862. |
How many solutions does the pair of equations x+2y =3 and 1/2x+y-3/2 have |
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Answer» infinite solutions What a question ? |
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| 16863. |
Anyone tell me extra nd important question of chapter 1 real number?? |
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| 16864. |
Exirsize, questions,1,3 |
| Answer» Which chapter | |
| 16865. |
Squareroot of 588 |
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Answer» 24.248 24.668 Squareroot of 588 |
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| 16866. |
Can you send me a 20 extra questions and solution of chapter 1 to 4 |
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Answer» This app has all the NCERT solutions of Mathematics as well as extra questions (in cbse important questions) . So, there is no need to send any NCERT solutions. No |
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| 16867. |
How a cubic polynomial can be formed when two zeroes are given |
| Answer» Third zero can be 0 | |
| 16868. |
(x-a)(x-b)(x-c)(x-d) ...........................(x-y)(x-z) |
| Answer» Very easy question\xa0i think we can do it like this,\xa0(X-A)(X-B)(X-C)....(X-Y)(X-Z)= (X-A)(X-B)(X-C)(X-D)....(X-W)(X-X)(X-Y)(X-Z)= (X-A)(X-B)(X-C)(X-D)....(X-W)(0)(X-Y)(X-Z) (since X-X = 0)= 0 [Product of any term multiplied with zero always results in zero].Hence the answer is 0. | |
| 16869. |
If 0.3528 is expressed on the form p/2m 5n, find the smallest value of n,m and p . |
| Answer» Question:\xa0If 0.3528 is expressed on the form p/2m 5n, find the smallest value of n,m and p .Explanation: | |
| 16870. |
What is the probability of 53 fridays in leap year |
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Answer» 53/366 1 |
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| 16871. |
What is the probabilty of 53 fridays on lep yeae |
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Answer» 53/366 1 |
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| 16872. |
Use euclid\'s division algorithm to find HCF of 1651 and 2032.write steps aldo |
| Answer» \xa02032 = 1651 {tex} \\times{/tex} 1 + 381 .1651 = 381 {tex} \\times{/tex} 4 + 127\xa0381 = 127 {tex} \\times{/tex} 3 + 0.\xa0Since the remainder becomes 0 here, so HCF of 1651 and 2032 is 127.{tex} \\therefore{/tex}\xa0HCF (1651, 2032) = 127.Now,{tex} 1651 = 381 \\times 4 + 127{/tex}{tex} \\Rightarrow \\quad 127 = 1651 - 381 \\times 4{/tex}{tex} \\Rightarrow \\quad 127 = 1651 - ( 2032 - 1651 \\times 1 ) \\times 4{/tex}\xa0[from 2032 = 1651 {tex} \\times{/tex} 1 + 381]{tex} \\Rightarrow \\quad 127 = 1651 - 2032 \\times 4 + 1651 \\times 4{/tex}{tex} \\Rightarrow \\quad 127 = 1651 \\times 5 + 2032 \\times ( - 4 ){/tex}Hence, m = 5, n = -4. | |
| 16873. |
If one of the zero polynomial p(x) 4usquare-8ku-9 is negative of the value of \'a\',if 3x+2beta=20 |
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| 16874. |
If x+a is a factor of polynomia X2+Px+g and x2+mx+n, prove that a=n-g/m-p |
| Answer» Since x + a is a factor of x2+ px + qthen (-a)2\xa0- pa + q = 0{tex}\\Rightarrow{/tex}a2 = pa -q ..........(i)also (x + a) is a factor of x2+ mx + n then we get(-a )2\xa0- am + n = 0{tex}\\Rightarrow{/tex} a2 - am + n= 0{tex}\\Rightarrow{/tex}a2\xa0=\xa0am - n........(ii)From eq(i) and (ii), we getam - n = ap - q{tex}\\Rightarrow{/tex}am - ap = n - qHence, a =\xa0{tex}\\left[ \\frac { n - q } { m - p } \\right]{/tex} | |
| 16875. |
State Euclid\'s division lemma |
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Answer» For any two given integers a and b , there exists unique integers p and q such that : a = bq + rDividend = divisor × quotient + remainder Dividend = divisor × quotient + remainder.a=bq+r Where a, b are integer. a=bq+r |
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| 16876. |
If root of ax2+ bx+ c = 0 be equal then the value of c is what |
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Answer» Thanks equation = ax² + bx + c = 0\xa0It has equal roots .So ,\xa0D = 0\xa0b² - 4ac = 0\xa0b² = 4ac\xa0=> 4 a c = b²\xa0=> c = b² / 4a\xa0So ,\xa0the value of c = b² / 4a\xa0 |
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| 16877. |
What is the 2x+2x |
| Answer» Your answer is 4x | |
| 16878. |
If alpha and beta are the zeroes of the polynomial p(x)= x square +x-1 then find 1/alpha and 1/ beta |
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Answer» Since we have the equation of x square minus x minus 1Through the the equation making formula of a x square + bx + cOur a is is 1 our b is minus one and our c is -1 Since we have a equation X square + X - 1Then through the the equation making formula of of a x square + bx + c |
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| 16879. |
Find the sum and product of the polynomial P(x)=ax2+bx+c |
| Answer» B/C;c/a | |
| 16880. |
Divide 27 into a two parts such that the sum of there resiprocal |
| Answer» Incomplete question my friend | |
| 16881. |
X square minus 12x+35 =0 |
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Answer» xsquare-12x+35=035=-7*-5 -12= -7-5xsquare -7x-5x+35=0x(x-7) -5(x-5)=0(x-5) (x-7)=0So x-5=0,x=5x-7=0,x=7 xsquare-12x+35 -- 0x2-(5x+7x)+35 --0x2-7x-5x+35 --0x(x-7)-5(x-7) -- 0(x-5)(x-7) --- 0if x-5 -- 0x -- 5if x-7 -- 0x -- 7 |
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| 16882. |
HCF of 18,36 is |
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Answer» The prime factorise of 18 is 2×3×3 and 36 is2×2×3×3.2×3×3is common in both factorise . So its common factor is 18. The greatest common factor 18 and 36 is 18. |
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| 16883. |
1/3(2 1/2+3 1/3)÷2/9(3 1/8-1 1/12 |
| Answer» Its answer is 1/10 | |
| 16884. |
1+ 2 ? |
| Answer» 3 stupid..?? | |
| 16885. |
Aftab tells his daughter |
| Answer» Let the present ages of Aftab and his daughter be x year and y year respectively. Then the algebraic representation isGiven by the following equations:x - 7 = 7(y - 7){tex}\\Rightarrow{/tex} x - 7y + 42 = 0 ...(1)And x + 3 = 3(y + 3){tex}\\Rightarrow{/tex}\xa0x - 3y - 6 = 0 ...(2)To, represent this equation graphically, well find two solution for each equation, These solution are given below;For Equation (1) x - 7y + 42 = 0{tex}\\Rightarrow{/tex} 7y = x + 42{tex}\\Rightarrow y = \\frac{{x + 42}}{7}{/tex}Table 1 of solutions\tx07y67\tFor Equation (2) x - 3y - 6 = 0{tex}\\Rightarrow{/tex} 3y = x - 6\xa0{tex}\\Rightarrow y = \\frac{{x - 6}}{3}{/tex}Table 2 of solutions\tx06y-20\t We plot the A(0, 6) and B(7, 7)Corresponding to the solutions in table 1 on a graph paper to get the line AB representing the equation (1) and the points C(0, -2) and D(6, 0) corresponding to the solutions in table 2 on the same graph paper to get the line CD representing the equation (2), as shown in the figureWe observe in figure that the two lines representing the two equations are intersecting at the point P(42, 12). | |
| 16886. |
(Cos^3theta-sin^3theta/costheta-sintheta)-(cos^3theta+sin^3theta/costheta+sintheta)=1 |
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Answer» Not able to understand complrtely.. Use the identity (a^3+b^3)= (a+b)(a^2+b^2-ab)sin^3A + cos^3A÷(sinA +cos ). + sinA.cosA(sinA+cosA)(sin^2+cos^2 -sinA.cosA)÷ (sinA + cosA)+ sinA.cosAor( sinA^2+cosA^2 -sinA.cosA) +sinA.cosA1-sinA.cosA+sinA.cosA1 |
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| 16887. |
The 11 term of an ap is zero prove that 25th term of an ap is 3 times its 11 term |
| Answer» We have given that4th\xa0term of an A.P.= a4\xa0= 0∴ a + (4 – 1)d = 0∴ a + 3d = 0∴ a = –3d ….(1)25th\xa0term of an A.P. = a25= a + (25 – 1)d= –3d + 24d ….[From the equation (1)]= 21d3 times 11th\xa0term of an A.P. = 3a11= 3[a + (11 – 1)d]= 3[a + 10d]= 3[–3d + 10d]= 3 × 7d= 21d∴ a25\xa0= 3a11i.e., the 25th\xa0term of the A.P. is three times its 11th\xa0term. | |
| 16888. |
Find the value of :(Cos0°+sin45°+sin30°) (sin90°-cos45°+cos60°) |
| Answer» (cos 0° + sin 45° + sin 30°)(sin 90° − cos 45° + cos 60°) …(i)By trigonometric ratios we have\xa0By substituting above values in (i), we get | |
| 16889. |
If sinA+sin2=1 and find the Value of cos2+cos4=1 |
| Answer» Question is wrong, send it again correctly | |
| 16890. |
If p is a prime number than p is also an |
| Answer» Natural number | |
| 16891. |
Find the hcf of 65 and 117 and express it in the form 65m+117n such that 117765 |
| Answer» First find the HCF of 65 and 117 by Using Euclid\'s division algorithm,117 = 65{tex}\\times{/tex} 1 + 5265 = 52{tex}\\times{/tex} 1 + 1352 = 13{tex}\\times{/tex} 4 + 0So, HCF of 117 and 65 = 13HCF = {tex}65m + 117n{/tex}For, {tex}m= 2{/tex} and {tex}n = -1{/tex},HCF = 65{tex}\\times{/tex} 2 + 117{tex}\\times{/tex} (-1)= 130 - 117= 13Hence, the integral values of m and\xa0n are 2 and -1 respectively and the HCF of 117 and 65 is 13. | |
| 16892. |
What must be subracted from x^3-3x^2+5x-1 to get 2x^3+x^2-4x+2 |
| Answer» Just subtract the second one from the first one.. | |
| 16893. |
HCF of 18 and 36 is |
| Answer» 18 = 2 × 3^236 = 2^2 × 3^2HCF = 2 × 3^2 = 18Thus HCF of 18 and 36 is 18. | |
| 16894. |
1÷2×1÷√2-√3÷2×1√2 |
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| 16895. |
For what value of k the roots of the equation x2 + 4 x+ k = 0 are real |
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Answer» I can\'t understand x2+4x+k=0D=0b2-4ac=0(4)2-4*1*k=016-4k=0-4k=-164k=16k=16/4k=4 |
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| 16896. |
What is tens no or ones no |
| Answer» Digit at tens place or ones place | |
| 16897. |
Shhcn cd:?! Hxbj,v n.m |
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Answer» …….?? Vnzfbxx |
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| 16898. |
The first term of ap is 5 |
| Answer» Here, a = 5l = 45S = 400We know that{tex}S = \\frac{n}{2}(a + l){/tex}{tex} \\Rightarrow 400 = \\frac{n}{2}(5 + 45){/tex}{tex} \\Rightarrow 400 = \\frac{n}{2}(50){/tex}{tex} \\Rightarrow 400 - 25n{/tex}{tex} \\Rightarrow n = \\frac{{400}}{{25}}{/tex}{tex} \\Rightarrow n = 16{/tex}Hence, the number of terms is 16.Again, we know thatl = a + (n - 1)d{tex} \\Rightarrow {/tex}\xa045 = 5 + (16 - 1)d{tex} \\Rightarrow {/tex}\xa045 = 5 + 15d{tex} \\Rightarrow {/tex}\xa040 = 15 d{tex} \\Rightarrow d = \\frac{{40}}{{15}} = \\frac{8}{3}{/tex}Hence, the common difference is\xa0{tex}\\frac{8}{3}{/tex}. | |
| 16899. |
The sum of two numbers is 8. Determine the number if the sum of there reciprocal is 8 by 15 |
| Answer» Let the two\xa0numbers are\xa0x and 8 - x.\xa0According to question,{tex}15\\left( {\\frac{1}{x} + \\frac{1}{{8 - x}}} \\right) = 8{/tex}{tex} \\Rightarrow 15\\left( {\\frac{{8 - x + x}}{{x(8 - x)}}} \\right) = 8{/tex}{tex} \\Rightarrow {/tex}\xa0{tex}15 \\times 8=8x(8-x){/tex}{tex} \\Rightarrow 15 = \\frac{{8x}}{8}(8 - x){/tex}{tex} \\Rightarrow {/tex} 15 = x(8 - x){tex} \\Rightarrow {/tex} 15 = 8x - x2{tex} \\Rightarrow {/tex} x2 - 8x + 15 = 0Factorise the equation,{tex} \\Rightarrow {/tex} x2\xa0- 5x - 3x + 15 = 0{tex} \\Rightarrow {/tex} x(x - 5) - 3(x - 5) = 0{tex} \\Rightarrow {/tex} (x - 5)(x - 3) = 0{tex} \\Rightarrow {/tex} x = 5 or x = 3Hence, required numbers are 3 and 5. | |
| 16900. |
State wheather 5.2323.....+3/4 is rational or not |
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Answer» its rational number Rational |
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