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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.
| 16001. |
For any positive integer n, prove that N cube - n is divisible by 6 |
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| 16002. |
Prove that the product of three consecutive positive integer is divisible 6 |
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Answer» Let the 3 consecutive no.s be X,X+1,X+2 Answer bata do plz???? |
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| 16003. |
Find the naturevof the roots of the following quaddratic equation if the real roots exist find them |
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| 16004. |
4-5×8÷8+67=? |
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Answer» 66 66 |
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| 16005. |
Sintheta +costheta =cosectheta |
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| 16006. |
What is what is volume volume of |
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Answer» 1/3πH(R2 + Rr + r2) Here is the answer Thanks |
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| 16007. |
What is volume of frustum |
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Answer» hello diya sis 1/3πh(r²+R²+rR) cubic units |
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| 16008. |
5×5 |
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Answer» How do u ask such high level questions.....Well to help u in ur pittyful condition the answer is 50/2 10 class m pass kaise h gye 25 Hame itna difficult question nahi aata,,, Koi easy question pucho 25 |
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| 16009. |
Differentiate between alternate current and direct current |
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Answer» ??? Direct current =current do not stop Alternate current =current stop time to time |
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| 16010. |
Value of 2 root 2 |
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Answer» 2 is the right answer 2√2 =√8Or 2√2 = 2.828...... 2√2 ==√8 |
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| 16011. |
Solve the quadratic equation 9xsquare-3/4x-√2=0 by method of completing the square |
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| 16012. |
Factorise this 120-33+260=0 |
| Answer» 347 | |
| 16013. |
(tan*3÷1+tan*2)+ (cot*3÷1+cot*2)= (1-2sin*2×cos*2)÷sin×cos |
| Answer» ??? | |
| 16014. |
If l want to choose commerce in 11 then which math l should take |
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Answer» ??? You can check it on official website or ask to your teachers......Deepika ji? Commerce keep sath standard lena hai, basic nahi,,, aadya g I\'m right, we can choose basic maths if we wanna study commerce. No aadya u r wrong Tell me your stream, which u would like to choose in 11th You can take basic maths if you want to choose commerce . Basic math is better for you. No i don\'t think.... For that u can confirm more with your teacher... Mejhe btao maths kw paper k bare me mujhe nahi pta....gourav Can l take basic maths U can take standard maths.... |
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| 16015. |
p, +90,p+180,p+270,........, where p=(999)999 |
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| 16016. |
Deep thank u sorry accept karne ke liye |
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Answer» Mtlb wo ab chat nhi kr skti Matlab Wo report ho chuki hai |
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| 16017. |
if p( -1, 1) is the midpoint of the line segment joining a(-3,b) and b(1,b+4) then find b |
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| 16018. |
if 7×11×13 +5 it is composite number |
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Answer» 7 × 11 × 13 + 13Take 13 common there we get\xa0= 13(7 x 11 +1 )= 13(77 + 1 )= 13(78)It is the product of two numbers and both numbers are more than 1. So, it is a composite number. 1106 I thought aapne question me kuch likhna chod diya hh Kartik |
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| 16019. |
Solve the following pair of linear equation:Px+qy=p-qqx-py=p+q |
| Answer» The given pair of equations ispx + qy = p - q .....(1)qx - py = p + q ....(2)Multiplying equation (1) by p and equation (2) by q, we getp2x + pqy = p2 - pq....(3)q2x - pqy = pq + q2.....(4)Adding equation (3) and equation (4), we get(p2 + q2)x = p2 + q2{tex}\\Rightarrow \\;x = \\frac{{{p^2} + {q^2}}}{{{p^2} + {q^2}}} = 1{/tex}Substituting this value of x in equation (1), we getp(1) + qy = p - q{tex}\\Rightarrow{/tex} qy = -q{tex}\\Rightarrow \\;y = \\frac{{ - q}}{q} = - 1{/tex}So, the solution of the given pair of linear equations is x = +1, y = -1.Verification, Substituting x = 1, y = -1,We find that both the equations (1) and (2) are satisfied as shown below:px + qy = p(1) + q(-1) = p - qqx - py = q(1) - p(-1) = q + p = p + qThis verifies the solution. | |
| 16020. |
Find the areaof the triangle whoose vertices are i) (2,3),(-1,0),(2,4) ii)(-5,-1),(3,-5),(5,2) |
| Answer» I solved this question but how i can show the solution of this question ? | |
| 16021. |
Write first four term of ap when the first term a and the common difference d are given as follow |
| Answer» 1. a2. a+d 3. a+2d4. a+3d. It\'s your first four term of an AP... | |
| 16022. |
Tan 60 |
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Answer» √3 √3 √3 √3 ✓3=tan60 √3 |
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| 16023. |
If the mean of y,1/y is M then what is the mean of y^3+1/y^3 |
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| 16024. |
Agar ham 10 commerce le with out maths to future me Acountant ban sakte hai....? |
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Answer» Ok Without maths bhi accountant ban sakte hai Aaliya But my teacher say to me I can choose commerce without maths to mai Acountant ban sakti hu. No aaliya,Bcoz in Accountancy maths is important......Take maths ,it will help u in future !! Which stream are you taking? Thanks. Yes |
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| 16025. |
Are there any question paper or samples paper for half yearly Examination? |
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Answer» ???????? U haveto buy the packkage !!! Not yet |
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| 16026. |
(a+b) (a-b) = ? |
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Answer» a-2_b-2 a^2 - b^2 a^2 - b^2 a ka sqare -b ka square "a" ka Square -"b" ka square a^2-b^2 It is a formula (a+b)(a-b) = a^2-b^2Or else u can also solve the equation in a general manner . The same will be as follows(a+b)(a-b) = a*a + a*-b + a*b + b*-b=a^2-ab+ab-b^2=a^2-b^2 a ka square plus b ka square |
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| 16027. |
X cube + 10 X square + 21 X + 5 Where x is equal to 0 2 3 7 8 and 6 |
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| 16028. |
How to prove root 5 |
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Answer» Thank you sir Question:\xa0Prove that root 5 is an irrational number.Answer:Let us assume that\xa0√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,\xa0√5 = p/q p =\xa0√5qwe know that \'p\' is a rational number. so\xa0√5 q must be rational since it equals to pbut it doesnt occurs with\xa0√5 since its not an intezertherefore, p =/=\xa0√5qthis contradicts the fact that\xa0√5 is an irrational numberhence our assumption is wrong and\xa0√5 is an irrational number. |
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| 16029. |
Exercise 5.3 question no. 3 (IX) solve it |
| Answer» As we know that, S= n/2(2a+(n-1)d). Then, 192 = 8/2 ( 2×3 + ( 8-1 ) d) , 192 = 24 + 56/2× d, 168 = 56/2 × d, 168× 2 = 56 d, 336/56 = d , 6 = d, So, the value of d is 6. | |
| 16030. |
If the third term and the 9th term of an ap are 4 and -8 respectively which term of this ap is zero |
| Answer» a(3) = a + 2d4 = a\xa0+ 2da(9) = a + 8d-8 = a + 8d(elimination method)a + 2d = 4(-) a + (-)8d =(+) -8 (opposite sign)0\xa0- 6d = 12d = -2a + 2(-2) = 4 (substitution)a = 8a(n) = a + (n-1)d0 = 8 + (n-1)(-2)= 8 - 2n + 2= 10 - 2n2n = 10\xa0n = 5 | |
| 16031. |
Find the sum of first 40 positive divisible by 6 |
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Answer» Arithmetic progression Real number hai kya Ab ye kaun se chapter ka question hai |
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| 16032. |
How can i acquire masterresy in trigonometry ? |
| Answer» By dear sir on youtube | |
| 16033. |
what is the maximum value of 1/secQ |
| Answer» May Be, One, \'cause\xa01/sec(theta)=1/(1/cos(theta))=cos(theta),\xa0so the maximum value is 1. | |
| 16034. |
2+2×2÷2=36 |
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Answer» In Bachpan Classes, we have learnt how to do calculation,may be you not but i have learnt it, so it was "DMAS",i.e., 2 + 2 x 2 divided by 2 = 36=> 2 + 2 x 1 = 36 => 2 + 2 = 36 => 36 - 4 => 32 9 32 4 9 |
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| 16035. |
How alpha+beta =1 |
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Answer» Its not 1 its -b/c x^2-{α+β}x+αβ=1 |
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| 16036. |
find a cubi polynomiL whose zero are 3,1/2 and _1 |
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| 16037. |
Integrate 1/x+4-x squaredx with limits 0 and 2 |
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| 16038. |
If sin Q = √3 by 2 , find the value of all T_ ratios of Q |
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| 16039. |
Sin0-cos0+1/sin0+cos0-1=1/sec0-tan0 |
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Answer» Thanks bhai Divide both numerator and denominator of LHS by cos0 and solve both lhs and rhs |
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| 16040. |
Class 10 rs aggarwal chapter coordinate Gemortery exercise 16 B |
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| 16041. |
Find the sumof all multiples of 9 lying between 400 and 800 |
| Answer» The numbers divisible by 9 between 400 and 800 are:405, 414, 423,.................792Here a=414, d=9 and l=792Let the number of these terms be n, then{tex}\\mathrm{We}\\;\\mathrm{know}\\;\\mathrm{that}\\;{\\mathrm a}_{\\mathrm n}=\\mathrm a+(\\mathrm n-1)\\mathrm d{/tex}an=792Or, a + (n-1)d=792{tex}\\Rightarrow{/tex}\xa0405 + (n-1){tex}\\times{/tex}9=792{tex}\\Rightarrow{/tex}\xa09(n - 1) = 387{tex}\\Rightarrow{/tex}\xa0(n - 1) = 43{tex}\\Rightarrow{/tex}\xa0n = 44So, S44\xa0=\xa0{tex}\\frac{n}{2}{/tex}(a+l)\xa0=\xa0{tex}\\frac{{44}}{2}{/tex}(405\xa0+ 792)\xa0= 22\xa0{tex}\\times{/tex}\xa01197 = 26334Hence, Sn=26334 | |
| 16042. |
Paper style |
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| 16043. |
All chapters are difficult |
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Answer» No yar each chapter of our books is very easy but main thing is that we have to focus each chapter very clearly and we have to clear our each doubt. Any chapter are not difficult Seek into each chapter and you shall find enjoyment No all chapter are not difficult |
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| 16044. |
Solve using cross multiplication method is x+y=1 2x-3y=11 |
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| 16045. |
Tan 2A=cot(A-18°) where 2A is an acute angle ,find the value of A |
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Answer» We have, tan 2A = cot (A - 18°){tex}\\Rightarrow{/tex}\xa0tan 2A = tan {90° -(A -18°)}{tex}\\Rightarrow{/tex}\xa0tan 2A = tan (108° - A){tex}\\Rightarrow{/tex}\xa02A = 108° - A{tex}\\Rightarrow{/tex}\xa03A = 108°{tex}\\Rightarrow{/tex}\xa0A = 36° Thank you so much |
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| 16046. |
Prove that root 2 + root 3 is irrationl |
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| 16047. |
X2-2ab=5 |
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| 16048. |
Has the pattern for math has been changed I mean basic maths and standarised |
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Answer» The existing Mathematics examination is the Standard Level Examination. Standard-Level and Basic-level Question papers shall be based on the same syllabus. However the Standard-Level Mathematics assesses higher Mathematical abilities compared to Basic-Level. Accordingly, the difficulty level of the Mathematics – ‘Basic’ is less than that of Mathematics-‘Standard’.for more info : Click the link;http://cbse.nic.in/newsite/circulars/2019/03_CircularFAQ_2019.pdf Yes |
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| 16049. |
What is the best way to solve the math qoestion |
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Answer» But bro time ttoh sabhi mai dena padta hai Give more time in thinking about it |
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| 16050. |
Sin square theta + cos power 4 theta equal to cos squared theta + sin power 4 theta |
| Answer» Sin2+cos4=cos2+sin41-cos2 +(1-sin2)2=1-cos2 +1^2 +sin2^2- 2sin2=1- cos2 +1+sin4- 4sin2=1-cos2 +1 +sun4 - 4(1- cos2)= | |