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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.
| 23801. |
If two positive integers pand q are written as p=asqare b |
| Answer» LCM(p,q) = a3b3and HCF(p,q) = a2bLCM(p,q)\xa0{tex}\\times{/tex}\xa0HCF(p,q) = a3b3\xa0{tex}\\times{/tex}\xa0a2b{tex}\\begin{array}{l}\\mathrm{LCM}(\\mathrm p,\\mathrm q)\\;\\times\\mathrm{HCF}(\\mathrm p,\\mathrm q)\\;=\\mathrm a^5\\mathrm b^4\\;-----(1)\\\\\\mathrm{and}\\;\\mathrm{pq}=\\mathrm a^2\\mathrm b^3\\;\\times\\mathrm a^3\\mathrm b=\\mathrm a^5\\mathrm b^4\\;------(2)\\\\\\mathrm{from}\\;\\mathrm{eq}^\\mathrm n\\;(1)\\;\\mathrm{and}\\;(2)\\\\\\mathrm{LCM}(\\mathrm p,\\mathrm q)\\;\\times\\mathrm{HCF}(\\mathrm p,\\mathrm q)\\;=\\mathrm{pq}\\\\\\\\\\end{array}{/tex}\xa0 | |
| 23802. |
Find first fourth terms of all |
| Answer» a+3d | |
| 23803. |
Question papers of last10years |
| Answer» Bought an together with.Which is of Orange colour and it contains 10 yrs.questions chaptervise. | |
| 23804. |
Which is the best book for board exam? |
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Answer» R. S AGRAWAL Vidya exam book R.D. Sharma Yes Arya is r8 Ncert. |
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| 23805. |
^\'s |
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| 23806. |
Constructions |
| Answer» Sach mai kya construction karna hai | |
| 23807. |
Here is ananya Singh I have a question |
| Answer» | |
| 23808. |
How to find altitude of sun |
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| 23809. |
CSA of cone |
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Answer» πrl pie r l |
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| 23810. |
if the common difference of A. P.is 3.a20-a15 |
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Answer» \xa020 th term a20\xa0= a+(20-1)d = a+19d15 th term a15\xa0= a+(15-1)d = a+14da20 - a15\xa0= a+19d-(a +14d) = 19d-14d = 5d = 5*3=15 17 |
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| 23811. |
How can we learn trignometric ratio values easily |
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Answer» By a formula Pandit Badri prasadHar Har BoleHere ratio pandit/har = perpendicular/hypotenuse=sinA Ratio Badri /har =base/hypotenuse=cosA Ratio Prasad/bole=perpendicular/base=tanAThen after these 1/sinA=cosecA=hypotenuse/perpendicular 1/cosA= secA =hypotenuse/base 1/tanA=cotA = base/perpendicular Friends you can use this formula to learn trigonometry ratios easily. Thank you By trick that you can watch on YouTube By learning |
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| 23812. |
Awsome ♥ ♥ as always |
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| 23813. |
Applying quadratic formula solve ques no 3of ex 4.3 |
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| 23814. |
Cosec^263°+tan^24°/cot^266°+sec^227° |
| Answer» Cosec^263°+tan^24°/cot^266°+sec^227°= sec^2 27+ cot^2 66/ cot^2 66°+sec^2 27°=1[cosecA=Sec(90-A), tanA=cot(90-A)] | |
| 23815. |
Perimeter of right triangle is 60cm.hypotenuse is 25cm.find the area of the triangle. |
| Answer» Given: a + b + 25 = 60\xa0{tex}\\Rightarrow{/tex}\xa0a + b = 35a2 + b2 = 252ab\xa0{tex}=\\frac{1}{2}\\left[(a+b)^{2}-\\left(a^{2}+b^{2}\\right)\\right]{/tex}{tex}=\\frac{1}{2}[1225-625]{/tex}{tex}=\\frac{1}{2}[600]{/tex}= 300{tex}\\therefore{/tex}\xa0area of\xa0{tex}\\triangle{/tex}\xa0= 150cm2 | |
| 23816. |
How to score good marks |
| Answer» Study well | |
| 23817. |
if the sum of n term of an A.P is 3(n)2 +5n, then which of its terms is 164 |
| Answer» Given:Sn = 3n2 + 5n\xa0and am = 164Put n = 1 and n = 2 in Sn we get, S1 = 3 + 5 = 8\xa0{tex}\\Rightarrow {/tex}\xa0a = 8 ..........(i)and S2 = 3(4) + 5(2) = 12 + 10 = 22We have an = Sn - Sn-1{tex}\\therefore{/tex}\xa0a2 = S2 - S1{tex}\\Rightarrow{/tex}\xa0a2 = 22 - 8 = 14Now\xa0d = a2\xa0- a1\xa0= 14 - 8 = 6{tex}\\therefore{/tex}\xa0am = 164\xa0{tex}\\Rightarrow {/tex}\xa0a + (m - 1)d = 164\xa0{tex}\\Rightarrow {/tex}\xa08 + (m - 1)6 = 164{tex}\\Rightarrow {/tex}\xa08 + 6m - 6 = 164{tex}\\Rightarrow {/tex}\xa06m = 162{tex}\\Rightarrow m = {{162} \\over 6} = 27{/tex} | |
| 23818. |
if 6/3,a,4 are A.P find the value of a |
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| 23819. |
If sec theta + tan theta =p .Prove that =p×p-1÷p×p+1 |
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| 23820. |
If Sn =2n2+2n+1 then find the value of a10 |
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| 23821. |
Rational number class 10 |
| Answer» Rational numbers are in the form of p/q and q does not equaĺ to zero | |
| 23822. |
Ncert question |
| Answer» Check NCERT Solutions here :\xa0https://mycbseguide.com/ncert-solutions.html | |
| 23823. |
EF//DC FG//CB GH//BA prove that HE//AD |
| Answer» | |
| 23824. |
How to do study of chapterwise |
| Answer» Read and learn one chapter everyday | |
| 23825. |
How to make our full focus on studies? ? |
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Answer» Or other useless activities By leaving use of mobile and tv |
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| 23826. |
Which triangle have equal sides |
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Answer» Equilateral triangle have same sides Equilateral triangle |
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| 23827. |
Prove that tan²A+cot²A+2=sec²A cosec²A |
| Answer» LHS {tex}= tan^2\xa0A + cot^2\xa0A + 2{/tex}{tex}= sec^2\xa0A - 1 + cosec^2\xa0A - 1 + 2{/tex}= sec2\xa0A + cosec2\xa0A{tex}=\\frac{1}{\\cos ^{2} A}+\\frac{1}{\\sin ^{2} A}{/tex}{tex}=\\frac{\\sin ^{2} A+\\cos ^{2} A}{\\cos ^{2} A \\cdot \\sin ^{2} A}{/tex}{tex}=\\frac{1}{\\cos ^{2} A \\cdot \\sin ^{2} A}{/tex}{tex}=\\frac{1}{\\cos ^{2} A} \\times \\frac{1}{\\sin ^{2} A}{/tex}= sec2 A. cosec2\xa0A = RHS | |
| 23828. |
Value of sin30 |
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| 23829. |
CBSE allowded question paper |
| Answer» Check last year question papers here :\xa0https://mycbseguide.com/cbse-question-papers.html | |
| 23830. |
What is perimeter of semi circle |
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Answer» Diameter + half of circumference of circle R(π+2).. |
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| 23831. |
Tan2A=cot(A-18°) where 2A is acute angle |
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Answer» Its similiar to ex 8.2 question 5 of trigonometry I think what I have written its difficult to understand Cot (90- 2A) = cot (A- 18) = 90-2A = A - 18 = 90 +18 = A + 2A =. 108 = 3A = A = 36 |
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| 23832. |
If sinA=½ find the value of 2secA/(1+tan²A) |
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| 23833. |
What is the HCF of n+7 and n+13? |
| Answer» 1 | |
| 23834. |
Find roots of equation x + ^x-2 = 4 |
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| 23835. |
14 important question |
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| 23836. |
How many chapters are coming in class 10 social science |
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| 23837. |
The base of an isosceles triangle is 16 cm if it\'s perimeter is 36 cm find the area |
| Answer» Base=16cmPermiter=36cmLength of two side= 20cmSince it is a isosceles triangle the length of other side is equal.Length of othe side=10cm eachSo by Heron\'s formulaArea of triangle= 48sq cm | |
| 23838. |
Prove that ( a square +b square ) |
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| 23839. |
Find the value(s) of k, if the quadratic equation 3x^2 k±3x4 =0has equal roots. |
| Answer» \xa0If Discriminant of quadratic equation is equal to zero, or more than zero then roots are real.3x2 -k{tex}\\sqrt3{/tex}x + 4 = 0Compare with ax2 + bx + c = 0then a = 3, -k{tex}\\sqrt3{/tex}\xa0and c = 4D = b2- 4acFor real roots, b2- 4ac > 0{tex}( - k \\sqrt { 3 } ) ^ { 2 } - 4 \\times 3 \\times 4 \\geq 0{/tex}3k2- 48 {tex}\\geq{/tex}0k2\xa0- 16\xa0{tex}\\geq{/tex}0(k - 4)(k + 4) {tex}\\geq{/tex}\xa00{tex}\\therefore{/tex}\xa0k {tex}\\leq{/tex}\xa0- 4 and k{tex}\\geq{/tex}\xa04 | |
| 23840. |
Find the values of k when the quadratic equation is 3x^2- |
| Answer» \xa0If Discriminant of quadratic equation is equal to zero, or more than zero then roots are real.3x2 -k{tex}\\sqrt3{/tex}x + 4 = 0Compare with ax2 + bx + c = 0then a = 3, -k{tex}\\sqrt3{/tex}\xa0and c = 4D = b2- 4acFor real roots, b2- 4ac > 0{tex}( - k \\sqrt { 3 } ) ^ { 2 } - 4 \\times 3 \\times 4 \\geq 0{/tex}3k2- 48 {tex}\\geq{/tex}0k2\xa0- 16\xa0{tex}\\geq{/tex}0(k - 4)(k + 4) {tex}\\geq{/tex}\xa00{tex}\\therefore{/tex}\xa0k {tex}\\leq{/tex}\xa0- 4 and k{tex}\\geq{/tex}\xa04 | |
| 23841. |
3 is rational |
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| 23842. |
Check whether (20,3),(19,8),(2,-9) are the vertices of a triangle |
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| 23843. |
If tanA=ntanB, sinA=msinB then prove that cos2A=m2 - 1/n2-1 |
| Answer» SinA = m Sin B , Sin A/ Sin B = m and\xa0tan A= n tan B, tan A/tan B= n or Sin A/ cos A/Sin B/cos B= n , SinA Cos B / Cos A. Sin B= nm2\xa0-1/ n2\xa0-1 = (Sin A / Sin B )2\xa0-1/(Sin ACos B /Cos A Sin B)2\xa0-1 = Sin2A/ Sin2B -1/ Sin2ACos2B/Cos2ASin2B -1 = Sin2A-Sin2B/ Sin2B/Sin2ACos2B - Cos2A Sin2B/ Cos2A Sin2B = Sin2A - Sin2B x Cos2A Sin2B/Sin2B (Sin2A Cos2B - Cos2A Sin2B) = Sin2A - Sin2B x Cos2A/ Sin2A(1- sin2B) - (1-Sin2A) Sin2B = (Sin2A - Sin\xa02B ) x Cos2A /Sin2A - Sin2A Sin2B - Sin2B +Sin2A Sin2B = (Sin2A - Sin2B) x Cos2A/ ( Sin2A - Sin2B) = Cos\xa02A | |
| 23844. |
How to find |
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| 23845. |
Prove that:3cos68°cosec22°-1/2 tan47° tan60° tan78°=6-√3/2 |
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| 23846. |
How to solve trigonometry identity ouestion |
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| 23847. |
What is frustum of cone. |
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| 23848. |
5÷0 |
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| 23849. |
If 1kg us of 64 rupee so 270 kg will be of how many rupee.? |
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Answer» 17280 rupee 270 ×64 |
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| 23850. |
R=5 t=2p=1000 si=? |
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