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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.
| 26801. |
Hello eva |
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| 26802. |
Sahil me jau please |
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| 26803. |
Who live in gujarat |
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| 26804. |
Koi nhi Ji I\'d to possible nhi |
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| 26805. |
Hmm sahil bolo |
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| 26806. |
How many hours do you all study in a day |
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Answer» I don\'t study at all ???? Don\'t you have any idea also ayush Only three hours Haven\'t count yet |
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| 26807. |
What is the timetable of 2018 class 10th bird exams |
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| 26808. |
Anyone missing me |
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| 26809. |
Helo...❤❤Any one |
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| 26810. |
Sahil kaha kaha dil deke rakhe ho |
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| 26811. |
how to learn maths formulas in short times???? |
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| 26812. |
RIMSHA if u r online then plzz reply |
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| 26813. |
Prove Thales theorem |
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| 26814. |
Solve ax+by=1 |
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| 26815. |
Sahil kaha ho jii kya Kar rahe ho |
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| 26816. |
Hey Pragya kya bol rhi ho main to shaam ko online bhi nhi tha |
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| 26817. |
Is derivation of theorem will come on exam or not |
| Answer» Yes 100% | |
| 26818. |
How to prepare for maths exam |
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| 26819. |
Find the greatest number of the six digit excatly divisible by 18 , 24 and 36? |
| Answer» LCM of 18,24 and 3618 = 2 {tex}\\times{/tex}\xa03224 = 23{tex}\\times{/tex}\xa0336 = 22{tex}\\times{/tex}\xa032LCM (18,24,36) = 23\xa0{tex}\\times{/tex}\xa032 = 72Hence if any number is divisible by 72 that will be divisible by 18,24 and 36 also.Now the largest 6 digit number is 999999On dividing 999999 by 72 we get{tex}\\;999999=13888\\times72+\\;63=999936+63{/tex}Hence 999936 is the greatest 6 digit number divisible by 18,24 and 36. | |
| 26820. |
When your pre board start |
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| 26821. |
Who is the best student of my CBSE guide 2017 |
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| 26822. |
What is kreb cycle |
| Answer» This is the cycle through which plant get glucose and carbondioxide which it has used . it mainly consists of three process :- carboxilation , reduction & regeneration | |
| 26823. |
Prove that sinA - cosA+1/sinA + cosA - 1 = 1+ secAcosecA |
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| 26824. |
Any method solve and bring ans 10 question 999999 |
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| 26825. |
If 15 oranges cost Rs 70 what is the cost of 20 oranges |
| Answer» Approx Rs.93 or RS 93.33333 | |
| 26826. |
I love crush |
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| 26827. |
What is the number of solutions of pair of linear equation x+2y-8=0 2x+4y=16 |
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Answer» Xmas nhi x Sorry Linear equations in 2 variables ke 2 hi solutions hote hai one for Xmas and other for y |
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| 26828. |
Fine the nature of roots |
| Answer» Let an equation be ax2 + bx +c =0Then D= b2\xa0-4ac here D is called Discriment.If D= b2\xa0-4ac > 0 , Roots are equalIf D = b2\xa0-4ac = 0, roots are equal.If D= b2\xa0-4ac < 0, roots are imagenary. | |
| 26829. |
if cosA=2/5,find the valueof 4+4tan2A\xa0 |
| Answer» Given that Cos A = 2/5, it means that base = 2 units and Hypotenuse = 5 units. Using Pythagorus therem perpendicular = √ hypotenuse2- base\xa02\xa0= √ 52\xa0-22\xa0= √25-4 = √21Tan A = Perpendicular / base = √21/2. Tan 2 A = 2 x√21/2 = √214+4 Tan 2A = 4+4*√21 = 4(1+√21) | |
| 26830. |
What are your aim in life |
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| 26831. |
How do I find the value of graph ??How to take the values ?? |
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| 26832. |
2×(2x+3)/x-3 -25 (x-3)/2x+3 =5 |
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| 26833. |
Various aspects of Pythagoras theorem. |
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| 26834. |
Important questions of maths class 10 for 2018 examination |
| Answer» Check last year questions :\xa0https://mycbseguide.com/cbse-question-papers.html | |
| 26835. |
The ratio of the sum of two APs is 7n+1:4n+27Then finď the ratio of mth term. |
| Answer» Let a, and A be the first terms and d and D be the common difference of two A.PsThen, according to the question,{tex}\\frac { S _ { n } } { S _ { n } ^ { \\prime } } = \\frac { \\frac { n } { 2 } [ 2 a + ( n - 1 ) d ] } { \\frac { n } { 2 } [ 2 A + ( n - 1 ) D ] } = \\frac { 7 n + 1 } { 4 n + 27 }{/tex}or,\xa0{tex}\\frac { 2 a + ( n - 1 ) d } { 2 A + ( n - 1 ) D } = \\frac { 7 n + 1 } { 4 n + 27 }{/tex}or,{tex}\\frac { a + \\left( \\frac { n - 1 } { 2 } \\right) d } { A + \\left( \\frac { n - 1 } { 2 } \\right) D } = \\frac { 7 n + 1 } { 4 n + 27 }{/tex}Putting,\xa0{tex}\\frac { n - 1 } { 2 } = m - 1{/tex}{tex}n-1 = 2m - 2{/tex}{tex}n= 2m - 2 + 1{/tex}or, {tex}n = 2m - 1{/tex}{tex}\\frac { a + ( m - 1 ) d } { A + ( m - 1 ) D } = \\frac { 7 ( 2 m - 1 ) + 1 } { 4 ( 2 m - 1 ) + 27 }{/tex}{tex}\\frac { a + ( m - 1 ) d } { A + ( m - 1 ) D } = \\frac { 14 m - 7 + 1 } { 8 m - 4 + 27 }{/tex}{tex}\\frac { a + ( m - 1 ) d } { A + ( m - 1 ) D } = \\frac { 14 m - 6 } { 8 m + 23 }{/tex}Hence,\xa0{tex}\\frac { a _ { m } } { A _ { m } } = \\frac { 14 m - 6 } { 8 m + 23 }{/tex} | |
| 26836. |
If sin + cos = √2 then prove tan + cot = 2 |
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| 26837. |
Happy new year frnds????❤️❤️❤️ |
| Answer» Happy new year...??? | |
| 26838. |
Sahil u are not my sahil |
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| 26839. |
5+3+4+5+78+0++6+8×0 |
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Answer» 0 0+0×3÷6×0 0 |
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| 26840. |
Area related to circle ch ke kuch bi note samaj ni aye????? |
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| 26841. |
In a certain A.P the 24term is twice the 10term. prove that the 72 term is twice the 34term |
| Answer» Given,a24\xa0= 2{tex}\\times{/tex}\xa0a10{tex}\\Rightarrow{/tex}\xa0a + (24\xa0- 1)d = 2{tex}\\times{/tex}[a + (10 - 1)d]{tex}\\Rightarrow{/tex}a + 23d = 2[a + 9d]{tex}\\Rightarrow{/tex}a + 23d = 2a + 18d{tex}\\Rightarrow{/tex}a - 2a = 18d - 23d{tex}\\Rightarrow{/tex}-a = -5d{tex}\\Rightarrow{/tex}a = 5d...........(i)To prove: a72\xa0= 2a34Proof:LHS = a72= a + (72\xa0- 1)d= 5d\xa0+ 71d [From (i)]= 76dRHS = 2a34= 2[a + (34\xa0- 1)d]= 2[5d + 33d]= 2\xa0{tex}\\times{/tex}38d= 76d{tex}\\therefore{/tex} LHS = RHS | |
| 26842. |
What is easterrification |
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Answer» The reaction between carboxylic acid and ethanol or alcohol is called esterifixation formation of ester\xa0 |
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| 26843. |
Sahil don\'t worry I completely trust u |
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| 26844. |
If (n-1),(4n-1)and (5n+2) are in ap find n |
| Answer» N=3/2 | |
| 26845. |
How to solve substituting method |
| Answer» By substituting the values given in the questions | |
| 26846. |
Sahil aapko happy hona ab mujha jaise problem se holidays |
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| 26847. |
Prove that the points A(7,10);B (-2,5);and C(3,-4) are the vertices of isosceles Right triangle |
| Answer» Let A(7,10),B(-2,5) and C(3,-4) be the vertices of given isosceles triangle.Then,\xa0{tex}AB=\\sqrt{(7+2)^2+(10-5)^2}=\\sqrt{106}{/tex}{tex}AC=\\sqrt{(7-3)^2+(10+4)^2}=\\sqrt{212}{/tex}{tex}BC=\\sqrt{(3+2)^2+(-4-5)^2}=\\sqrt{106}{/tex}Thus,AB=ACHence, given vertices are coordinates of an isosceles triangle. | |
| 26848. |
Express as a decimal representation (77/8+7/80) |
| Answer» {tex}\\frac { 77 } { 8 } + \\frac { 7 } { 80 } = \\frac {770 + 7 } { 80 } = \\frac { 777 } { 80 }{/tex}=\xa097.125 | |
| 26849. |
Prove sin 60 geometrically |
| Answer» Root3/2 | |
| 26850. |
2√2-√2 |
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Answer» Root 2 √2 |
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