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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.
| 33801. |
Find the sum of all multiples of 7 lying between 500 to 900 |
| Answer» All multiples of 7 lying between 500 and 900 are504,511,518,...,896This is an AP in which a = 504, d=7 and l = 896.Let the given AP contain n terms. Then,Tn = 896\xa0{tex}\\Rightarrow{/tex} a + (n\xa0- 1)d = 896 {tex}\\Rightarrow{/tex}504 + (n -1) {tex}\\times{/tex}\xa07 = 896 {tex}\\Rightarrow{/tex}497 + 7n = 896{tex}\\Rightarrow{/tex}7n = 399 {tex}\\Rightarrow{/tex}n = 57.{tex}\\therefore{/tex}required sum = {tex}\\frac{n}{2}{/tex}(a + l)={tex}\\frac{{57}}{2}{/tex}{tex}\\cdot{/tex}(504 + 896) = ({tex}\\frac{{57}}{2}{/tex}{tex}\\times{/tex}1400) = 39900.Hence, the required sum is 39900. | |
| 33802. |
If p^th ,qth and rth terms of an ap are a,b,c respectively then show that (q-r)+b(r-p)+c(p-q)=0 |
| Answer» Given that the pth, qth and rth terms of an AP be a, b, c respectively.Suppose, x be the first term and d be the common difference of the given arithmetic progression. Therefore,Tp = x+(p-1)d, Tq =x+(q-1)d and Tr =x+(r-1)dNow, Tp\xa0= a\xa0{tex}\\Rightarrow{/tex}\xa0x+(p-1)d = a ....(i)Tq\xa0= b\xa0{tex}\\Rightarrow{/tex}\xa0x+(q-1)d = b....(ii)Tr\xa0= c\xa0{tex}\\Rightarrow{/tex} x+(r-1)d = c....(iii)On multiplying (i) by (q-r), (ii) by (r-p) and (iii) by (p-q), we get,(q-r)[x+(p-1)d]=a(q-r)....(iv)(r-p)[x+(q-1)d]=b(r-p)....(v)(p-q)[x+(r-1)d]=c(p-q)....(vi)Now adding we get,a(q - r) + b(r - p) + c(p - q) = x{(q - r) + (r - p) + (p - q)} + d{(p - 1)(q - r)+ (q - 1)(r - p)+ (r - 1)(p - q)}= (x\xa0{tex}\\times{/tex}\xa00) + (d {tex}\\times{/tex}\xa00) = 0Therefore, a(q - r) + b(r - p) + c(p - q) = 0.Hence\xa0proved. | |
| 33803. |
The nth term of AP is 6n+2.find its common difference |
| Answer» an=6n+2a1=6(1)+2a1=8a2=6(2)+2a2=14d=a2-a1 =14-8 =6 | |
| 33804. |
What is the date of cbse class 10 results of session 2017-2018 |
| Answer» 30 may | |
| 33805. |
If alfa and bita are the zeroes of polynomial f(x)=xsquare-5x+k |
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Answer» View full question PtA nahi |
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| 33806. |
What is the probability of 53 sundays in a leap year ? |
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Answer» 2/7 is right answer , 53/1465 53/366 2/7 |
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| 33807. |
1234.68÷457 simple equation |
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Answer» Ya Hlo is anyone online? ?? |
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| 33808. |
Trigonometry ex-8.4question5 solution nerct |
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Answer» 5 number ki (iv) galat hai 5 number ki kon si bit |
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| 33809. |
-4(1)(-1452) multiply it |
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Answer» -4(-1452)=+5808 5929 |
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| 33810. |
What is the formula of perimeter of triangle |
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Answer» Sum of all sides Sum of all sides of triangle Adding all sides. all sides are add |
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| 33811. |
The differencebetween two numbers is 26 and one number is three times the other. Find them\xa0 |
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Answer» 13 and 39 The no.s are 13 and 39 Let the numbers be x and 3xDifferences between the two numbers is 26 So, 3x-x =26 =2x=26 =x=13 other number is 39Hence, the numbers are 13 and 39 |
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| 33812. |
If one of the quadratic polynomial f(x)=4 square-8k-9 is negative of the other. Find the value of k |
| Answer» Let, the two zeroes of the polynomial f(x) = 4x2 - 8kx - 9 be α and -α.Comparing f(x) = 4x2\xa0- 8kx - 9 with ax2+bx+c we geta = 4; b = -8k and c = -9.Sum of the zeroes = α + (-α) ={tex}-\\frac ba=\\;-\\frac{-8k}4 {/tex}0 = 2kk = 0 | |
| 33813. |
Cos 45°/sec 30° +cosec30° |
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| 33814. |
Solution of question no. 5 of exercise 1.1 of class 10th |
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| 33815. |
2x+y=122x-y=4 |
| Answer» Y= 4 and X = 4 . Solve it by eliminating 2x? | |
| 33816. |
The product of two numbers is 768 and their L.C.M. is 96 find their H.C.F. |
| Answer» HCF= 768÷96=8 | |
| 33817. |
Prove (tan A+cosecB) square - (cotB-secA)square is equal to 2tanA cotB ( cosecA + secB) |
| Answer» We have,LHS = (tanA + cosec B)2 - (cotB - sec A)2{tex}\\Rightarrow{/tex}\xa0LHS = (tan2A + cosec2B + 2tanA cosecB ) -\xa0(cot2B + sec2A\xa0- 2cotB secA){tex}\\Rightarrow{/tex}\xa0LHS = (tan2A - sec2A)+(cosec2B - cot2B)+2tanA cosecB +2cotB secABut, Sec2A - tan2A =1 & cosec2A - cot2\xa0A = 1{tex}\\therefore{/tex}\xa0LHS = -1 + 1 + 2 tanA cosecB + 2cotB secA{tex}\\Rightarrow{/tex}\xa0LHS = 2 (tanA cosecB + cotB secA){tex}\\Rightarrow{/tex}\xa0LHS = 2 tanA cotB{tex}\\left( \\frac { cosec\\: B } { \\cot B } + \\frac { \\sec A } { \\tan A } \\right){/tex}\xa0[Dividing and multiplying by tanA cotB]{tex}\\Rightarrow{/tex}\xa0LHS = 2tan A cotB{tex}\\left\\{ \\frac { \\frac { 1 } { \\sin B } } { \\frac { \\cos B } { \\sin B } } + \\frac { \\frac { 1 } { \\cos A } } { \\frac { \\sin A } { \\cos A } } \\right\\}{/tex}\xa0[Since, CosecA.SinA =1 , SecA.cosA =1, (sinA/cosA)= tanA & (cosA/SinA) =cotA ]{tex}\\Rightarrow{/tex}\xa0LHS = 2 tanA cotB{tex}\\left( \\frac { 1 } { \\cos B } + \\frac { 1 } { \\sin A } \\right){/tex}\xa0= 2tanA cotB ( secB + cosecA ) = RHS. Hence, proved. | |
| 33818. |
Find the no. Of divisors of 7056 |
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| 33819. |
Ex 1.4all sums |
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| 33820. |
x + 6/y = 63x - 8/y =5 |
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Answer» Let 1/y = aThen, x + 6a = 6-------------------(1) 3x - 8a = 5--------------------(2)solving the above we get a = 1/2 ; but a= 1/y . therefore y = 2x= 3 What we have to do |
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| 33821. |
2+2=4why? |
| Answer» 2 + X = 24 therefore X is equals to 2 | |
| 33822. |
Explain why 7*6*5*4*3*2*1+5 is a composite number |
| Answer» Because when we solve we see that in last we have 1009 and we know that the factor of 1009 is 1 so it is a composite no. | |
| 33823. |
What is the history of pie |
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| 33824. |
The sum of two number is 137 and their difference is 43 find the numbers. |
| Answer» Let one no. Be xIf the difference of the numbers is 43 then second no. Will be...x-43...soAcc. To questionX+(x-43) = 1372x -43 = 1372x = 180X=90->X-43=90-43=47;NO\'s are 90 and 47 | |
| 33825. |
In an ap the sum of first ten terms is -80 and the sum of its next ten terms is (-280 ) find the ap |
| Answer» The series will be 1, -1, -3, -5, -7,........How to get it ?The sum of n terms of an AP is given by , S = n/2 {2a+(n-1)d}According to problem , S (10) = sum of first ten terms = -80 = 10/2 {2a + (10-1)d} = 5 (2a + 9d)------------(1)Again sum of next 10 terms is = -280Therefore we can write , S (20) = sum of first twenty terms = -80-280 = 20/2{20+(19-1)d} = 10(2a+19d)---------(2)solving (1) and (2) we get a= first term = 1 and d= common difference = -2. | |
| 33826. |
Prove that is an irrational number. |
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| 33827. |
Find the roots of the quadratic equation by the ethod of completing the square : 16x square -8x+1=0 |
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| 33828. |
Multiply 1/2x + 1/3y = 2 and 1/3x + 1/2y = 13/6 with 6 |
| Answer» {tex}{\\frac{1}{{2x}} + \\frac{1}{{3y}} = 2}{/tex} ... (1){tex}\\frac{1}{{3x}} + \\frac{1}{{2y}} = \\frac{{13}}{6}{/tex} ...(2)Let {tex}\\frac{1}{x}{/tex}= p and {tex}\\frac{1}{y}{/tex}= qPutting this in equation (1) and (2), we get{tex}\\frac{p}{2} + \\frac{q}{3} = 2{/tex}\xa0and\xa0{tex}\\;\\frac{p}{3} + \\frac{q}{2} = \\frac{{13}}{6}{/tex}{tex}\\Rightarrow{/tex}\xa03p + 2q = 12 and 6(2p +3q) = 13 (6){tex}\\Rightarrow{/tex}\xa03p + 2q = 12 and 2p + 3q = 13{tex}\\Rightarrow{/tex}\xa03p + 2q - 12 = 0 .................. (3) and2p + 3q - 13 = 0 ................. (4){tex}\\frac{p}{{2( - 13) - 3( - 12)}} = \\frac{q}{{( - 12)2 - ( - 13)3}}{/tex}{tex} = \\frac{1}{{3 \\times 3 - 2 \\times 2}}{/tex}{tex}\\Rightarrow \\frac{p}{{ - 26 + 36}} = \\frac{q}{{ - 24 + 39}} = \\frac{1}{{9 - 4}}{/tex}{tex} \\Rightarrow \\frac{p}{{10}} = \\frac{q}{{15}} = \\frac{1}{5} \\Rightarrow \\frac{p}{{10}} = \\frac{1}{5}\\,{\\text{and}}\\,\\frac{q}{{15}} = \\frac{1}{5}{/tex}{tex}\\Rightarrow{/tex}\xa0p = 2 and q = 3But {tex}\\frac{1}{x}{/tex}\xa0= p and {tex}\\frac{1}{y}{/tex}\xa0= qPutting value of p and q in this we get,x = {tex}\\frac{1}{2}{/tex}and y = {tex}\\frac{1}{2}{/tex} | |
| 33829. |
If x=3 is one root of the quadratic equation x×x -2kx+6=0 then find the value of k |
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Answer» The quadratic equation f (x) = x*x -2kx+6=0if one of the root is 3 then , f(3) = 3*3 - 2k*3+6 =0 (replacing x by 3 in the equation i.e., k = 15/6 Ans is . 9 How Answer is 9 |
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| 33830. |
Find indicated terms in each of following sequence whose n th term are A(n)=5n-4,A12 and A15 |
| Answer» Tell | |
| 33831. |
Find the list no.which when divided by 20,25,35,40 leaves reminder 14,19,29,34 respectively |
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| 33832. |
If the HCF of 408 and 1032 is expressible in the form 1032m - 408*5 , find m. |
| Answer» Given integers are 408 and 1032 where 408 < 1032By applying Euclid’s division lemma, we get 1032 = 408 {tex}\\times{/tex}\xa02 + 216.Since the remainder ≠ 0, so apply division lemma again on divisor 408 and remainder 216, we get the relation as408 = 216 {tex}\\times{/tex}\xa01 + 192.Since the remainder ≠ 0, so apply division lemma again on divisor 216 and remainder 192216 = 192 {tex}\\times{/tex}\xa01 + 24.Since the remainder ≠ 0, so apply division lemma again on divisor 192 and remainder 24\xa0192 = 24 × 8 + 0.Now the remainder has become 0. Therefore, the H.C.F of 408 and 1032 = 24.Therefore,24 = 1032m - 408 {tex}\\times{/tex}\xa051032m = 24 + 408 {tex}\\times{/tex}\xa051032m = 24 + 20401032m = 2064 {tex}m = \\frac{{2064}}{{1032}}{/tex}Therefore, m = 2. | |
| 33833. |
If d is the Hcf of 56 and 72 , find x and y satisfy x and y are not unique . D=56x + 72y |
| Answer» HCF of 56 and 72{tex}\\begin{array}{l}56=8\\times7=2^3\\times7\\\\72=8\\times9=2^3\\times3^2\\\\So\\;HCF(56,72)=2^3=8\\end{array}{/tex}d = 56x + 72y⇒ 8 = 56x + 72yDividing by 8 both sides1= 7x + 9yPut x = 4 and y = –31 = 7 × 4 + 9(–3)= 28 – 271 = 1L.H.S = R.H.S.Put x = –5 and y = 41 = 7(–5) + 9 {tex} \\times {/tex}\xa04= –35 + 361 = 1L.H.S = R.H.S.x = 4, and y = –3x = –5 and y = 4Satisfy the equations\xa0∴ x and y are not unique. | |
| 33834. |
\'Right hand thumb rule\' explain. |
| Answer» According to the right hand thumb rule when we take a conductor in our hand and wrap it with our fingers then the wrapped fingers show the direction of magnetic field lines and thumb show the direction of current | |
| 33835. |
1/2x-1/y=-11/x+1/2y=8Find value of x and y |
| Answer» Let 1/x = a and 1/y = beqn 1/2x-1/y=-1 is now a/2 - b = -1 or a - 2b = -2 ----------(1)eqn 1/x+1/2y=8 is now a + b/2 = 8 or 2a + b = 16 ---------(2)solving (1) and (2) we get , a= 6. but a = 1/x . Therefore x= 1/6similarly b = 4 . But b = 1/y. Therefore y = 1/4 | |
| 33836. |
If Tn =Sin n A +Cos n A Then Prove That - 6T10- 15 T 8+ 10 T 6 -1= 0 |
| Answer» 0 | |
| 33837. |
Tell me imp ques from exercize 6.2 and 6.3 ncert class 10 wich could come in board exam |
| Answer» In ex.6.2, ques.3,4,5,6,10In ex.6.3, ques.2,3,4,7,9,10,11,15,16These ques.are important and come inboard exam | |
| 33838. |
Which book is best for maths |
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Answer» RD Sharma is the best....... not sure ........but it all types of questions......so it is the best according to me All books are good but RD Sharma is one of the best book according to me Rd sharma All the books are best Ncert and ncert exampler All in one |
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| 33839. |
Find condition that the zeroes of the polynomial f(x)=x\'3+3px\'2+3qx+r may be in A.P. |
| Answer» The given quadratic polynomial is:f(x) = x3 + 3px2 + 3qx + rwe have to show that the zeroes of given polynomial are in the form of AP.Let, a - d, a, a + d be the zeroes of the polynomial, thenThe sum of zeroes = {tex}\\frac{{ - b}}{a}{/tex}a + a - d + a + d = -3p3a = - 3pa = - pSince, a is the zero of the polynomial f(x),Therefore, f(a) = 0f(a) = a3 + 3pa2 + 3qa + r = 0{tex}\\Rightarrow{/tex}\xa0a3 + 3pa2 + 3qa + r = 0{tex}\\Rightarrow{/tex}\xa0(-p)3 + 3p(-p)2 + 3q(-p) + r = 0{tex}\\Rightarrow{/tex}\xa0-p3 + 3p3 - 3pq + r = 0{tex}\\Rightarrow{/tex} 2p3 - 3pq + r = 0Which is the required condition. | |
| 33840. |
Which are considered as consistent and inconsistent solution |
| Answer» The equation having unique or infinite solutions are consistent solutions.the equation which does not have any solution is inconsistent. | |
| 33841. |
Find the discriminant of quadratic equation X2-4x+1=0 |
| Answer» X2 - 4x + 1 = 0 A= 1 B = -4 c = 1 D = b2 - 4ac D = (-4)2 - 4 × 1 × 1 D = 16 - 4 D = 12 | |
| 33842. |
What is the nature of the roots of quadratic equation 3x^ -x+1/3=0 |
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| 33843. |
Ffxff |
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| 33844. |
Sin60 cos30 + sin30 cos60 |
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Answer» Xyz (√3/2)(√3/2) + (1/2)(1/2)= (3/4) + (1/4)= 4/4= 1 |
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| 33845. |
2-2×6÷10×5+3 |
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Answer» 0 according to BODMAS 0 = 2 - 2 x 0.6 x 5 + 3= 2 - 6 + 3= 0 3 |
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| 33846. |
Find the LCM of 30240 and 29700 |
| Answer» 1663200 | |
| 33847. |
Solve for X and Y : 2x_3/y=9,3x+7/y=2 ( y not equal 0 |
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| 33848. |
cosec A + sec A= cosec A * sec A |
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| 33849. |
85*89 |
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| 33850. |
New sullabus of cbse 10 board exam |
| Answer» CBSE Class 10 SyllabusCBSE Syllabus for Class 10 MathsCBSE Syllabus for Class 10 ScienceCBSE Syllabus for Class 10 EnglishCBSE Syllabus for Class 10 HindiCBSE Syllabus for Class 10 Social Science | |