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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.
| 23851. |
What type of question will come |
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Answer» Must be tough.Focus more on maths.equations .Attend all questions you will get 1/2 marks for attending.study texts for social.write all equations of all subjects in a chart and stick on your walls.everytime you sees it your memory cathes it.Practise all sample papers. To the SSLC |
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| 23852. |
Progressions 2.3 |
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| 23853. |
If sin(A+B)=1 and tan(A-B)=1\\1. 73, find the value of (i)tan A+cot B (ii) sec A-cosec B |
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| 23854. |
If cosA =2/5, find the value of 4+4 tan² A |
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| 23855. |
Find the sum of first 100 natural numbers which are neither divisible by 3 nor 6 . |
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| 23856. |
Can you please derive the formula of sine 11A |
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| 23857. |
Determine if the points (1,5),(2,3)and (_2,_11) are collinear. |
| Answer» Let A {tex}\\rightarrow{/tex} (1, 5)B {tex}\\rightarrow{/tex} (2, 3)C {tex}\\rightarrow{/tex} (-2, -11)Then {tex}AB = \\sqrt {{{(2 - 1)}^2} + {{(3 - 5)}^2}} = \\sqrt {1 + 4} = \\sqrt 5{/tex}{tex}BC = \\sqrt {{{( - 2 - 2)}^2} + {{( - 11 - 3)}^2}}{/tex}{tex}= \\sqrt {{{( - 4)}^2} + {{( - 14)}^2}}{/tex}{tex}= \\sqrt {16 + 196} = \\sqrt {212}{/tex}{tex}CA = \\sqrt {{{[1 - (-2)]}^2} + [5 - {{( - 11)]}^2}}{/tex}{tex}= \\sqrt {{{(3)}^2} + {{(16)}^2}}{/tex}{tex}= \\sqrt {9 + 256} = \\sqrt {265}{/tex}We see thatAB + BC\xa0≠ CABC + CA ≠ ABand CA + AB ≠ BCHence, the given points are not collinear. | |
| 23858. |
Square root of 4 |
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Answer» 44 2 2 |
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| 23859. |
why be cannot use segment formula inboard exam |
| Answer» Because it\'s not illustrated in ncert | |
| 23860. |
The largest number which divides 70 and 125 leaving reminder 5 and 8 is |
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| 23861. |
a quadrilateral ABCD circumscribes a circle AB=6cm BC=8cm CD=15cm find DA |
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| 23862. |
If a and b are the zeroes of the polynomial such that a+b=6 |
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| 23863. |
Kunlo pikono che |
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| 23864. |
How many tangent can circle have |
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Answer» zero tagents\xa0inside a circle andinfinte tangents\xa0outside a circle Infinite. |
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| 23865. |
The two sides in a right angled triangle are15cm and20cm each.find its hypotenuse |
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Answer» as per Pythagorus theoremhp2\xa0=side2\xa0+ side2 = 15^2\xa0+ 20^2 = 225 + 400 = 625hp = 25 cm Given that:two sides of triangle are Perpendicular(P)=20\xa0cmBase(B)=15cmwe know that:H2=P2+B2H2= 202+152H=25 25 |
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| 23866. |
If 1 and 3 are zeroes of of polynomial x^3-ax^2-13x+b find value pf a and b |
| Answer» A- 1/5 and B= -69/5 | |
| 23867. |
If sin alpha=n sin beta,tan alpha=m tan beta find cos aplha=m²-1 n²-1 |
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| 23868. |
In this year board paper MCQs are include or not |
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Answer» There will be no any MCQ question... Instead of them one word questions are included... No First 2 wiil MCQ No |
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| 23869. |
If (x-2)&(x-1/2) are the factors of a polynomials qx² +5x+r prove that q = r |
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| 23870. |
Proobe that the tangents drawn from the external points are equal |
| Answer» Given: P is an external point to the circle C(O,r).PQ and PR are two tangents from P to the circle.To Prove: PQ = PRConstruction: Join OPProof:{tex}\\because{/tex}\xa0A tangent to a circle is perpendicular to the radius through the point of contact{tex}\\therefore{/tex}{tex}\\angle{/tex}OQP = 90o = {tex}\\angle{/tex}ORPNow in right triangles POQ and POR,OQ = OR [Each radius r]Hypotenuse. OP = Hypotenuse. OP [common]{tex}\\therefore{/tex}{tex}\\triangle{/tex}POQ\xa0{tex}\\cong{/tex}{tex}\\triangle{/tex}POR\xa0[By RHS rule]{tex}\\therefore{/tex} PQ = PR | |
| 23871. |
What is sin 120° |
| Answer» I think so it is √3/2 | |
| 23872. |
2÷3 |
| Answer» 0.66666...... | |
| 23873. |
What is a parabora |
| Answer» Parabola is used to represent quadratic polynomial or the zeroes of quadratic polynomial in graph. It is curve like structure | |
| 23874. |
The mean of 20 numbers is 18 . If 2 is added to each number , what is the new mean ? |
| Answer» I think it the new mean is 36 | |
| 23875. |
If two of the zeros of a cubic polynomial are zero then does it have linear and constant terms |
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| 23876. |
Prove that 1+tan^2x/1+cot^2x=(1-tanx/1-cotx)^2 |
| Answer» To prove :\xa0{tex} \\frac{{1 + {{\\tan }^2}\\theta }}{{1 + {{\\cot }^2}\\theta }} = {\\left( {\\frac{{1 - \\tan \\theta }}{{1 - \\cot \\theta }}} \\right)^2}{/tex}Consider : {tex} \\frac{{1 + {{\\tan }^2}\\theta }}{{1 + {{\\cot }^2}\\theta }} = \\frac{{1 + \\frac{{{{\\sin }^2}\\theta }}{{{{\\cos }^2}\\theta }}}}{{1 + \\frac{{{{\\cos }^2}\\theta }}{{{{\\sin }^2}\\theta }}}}=\\frac{{\\frac{{{{\\cos }^2}\\theta + {{\\sin }^2}\\theta }}{{{{\\cos }^2}\\theta }}}}{{\\frac{{{{\\sin }^2}\\theta + {{\\cos }^2}\\theta }}{{{{\\sin }^2}\\theta }}}}{/tex}{tex}= \\frac{{\\frac{1}{{{{\\cos }^2}\\theta }}}}{{\\frac{1}{{{{\\sin }^2}\\theta }}}} = \\frac{{{{\\sin }^2}\\theta }}{{{{\\cos }^2}\\theta }}{/tex}\xa0{tex} \\left[ {\\because {{\\sin }^2}\\theta + {{\\cos }^2}\\theta = 1} \\right]{/tex}{tex}= {\\tan ^2}\\theta {/tex}Consider\xa0{tex} {\\left( {\\frac{{1 - \\tan \\theta }}{{1 - \\cot \\theta }}} \\right)^2} = \\frac{{1 + {{\\tan }^2}\\theta - 2\\tan \\theta }}{{1 + {{\\cot }^2}\\theta - 2\\cot \\theta }}{/tex}{tex}= \\frac{{{{\\sec }^2}\\theta - 2\\tan \\theta }}{{\\cos e{c^2}\\theta - 2\\cot \\theta }}{/tex}\xa0{tex} \\left[ {\\because 1 + {{\\tan }^2}\\theta = {{\\sec }^2}\\theta } \\right]{/tex}{tex}= \\frac{{\\frac{1}{{{{\\cos }^2}\\theta }} - \\frac{{2\\sin \\theta }}{{\\cos \\theta }}}}{{\\frac{1}{{{{\\sin }^2}\\theta }} - \\frac{{2\\cos \\theta }}{{\\sin \\theta }}}} = \\frac{{\\frac{{1 - 2\\sin \\theta \\cos \\theta }}{{{{\\cos }^2}\\theta }}}}{{\\frac{{1 - 2\\sin \\theta \\cdot \\cos \\theta }}{{{{\\sin }^2}\\theta }}}}{/tex}{tex} = \\frac{{{{\\sin }^2}\\theta }}{{{{\\cos }^2}\\theta }} = {\\tan ^2}\\theta {/tex} | |
| 23877. |
A coin is tossed two times find the probability of getting not more than one head |
| Answer» 1/2 | |
| 23878. |
Sina+cosa=√2 than evaluate: Tana+cota |
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| 23879. |
Find the sum of all two digit numbers which are either multiple of 2 or 3 |
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| 23880. |
Show that 1/7-4root3 is not a rational number |
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| 23881. |
a + a 5 +a 10 + a 15 + a20 +a24=225 Then find out the sum of first 24 terms |
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| 23882. |
Sin a_ cos a=1/2 find 1/sin a + cos a |
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| 23883. |
Use euclid\'s divison (2) 196and38220 |
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| 23884. |
Difference between and and or in probability |
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| 23885. |
Can two numbers have 18 as their HCF and 380 is the LCM |
| Answer» We know that HCF of two numbers is a divisor of their LCM. Here, 18 is not a divisor of 380.But 380 = 18×21+2Here 2 is remainder so 380 is not divisible by 18.So, 18 and 380 cannot be respectively HCF and LCM of two numbers. | |
| 23886. |
Which chapter have more marks in CBSE board exam |
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Answer» Trigonometry is important and in statistics more than less then and ogiv is imp Number system 6 marksAlgebra 20 ma...Coordinate geom... 6 m..Geom.. 15 ma..Trigonometry 12 marksMensuration 10 marksStati..& probability 11 marks Algebra |
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| 23887. |
What does co-ordinate mean ? Explain with examples |
| Answer» You can locate any point on the coordinate plane by an ordered pair of numbers (x,y), called the coordinates.Example :\xa0coordinates\xa0of the origin are (0, 0). | |
| 23888. |
Weightage of chapters in maths X in board exam |
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| 23889. |
If two vertices of of an equi. triangle are (0,0);(3,0) fimd the third vertex |
| Answer» (0,3) | |
| 23890. |
Construct an angle aob =75°. Draw a circle of radius 2.7cm touching the arms of and ob |
| Answer» So easy | |
| 23891. |
What\'s about 10 date sheet |
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| 23892. |
What is the common difference of an AP in which T18-T14 ?. |
| Answer» 4 | |
| 23893. |
Find the centre of a circle passing through the points(6,-6) , (3,-7) and(3,3) |
| Answer» Let\xa0A → (6, –6), B\xa0→ (3, –7) and C\xa0→ (3, 3).Let the centre of the circle be I(x, y)Then, IA = IB = IC [By definition of a circle]{tex}\\Rightarrow{/tex} IA2 = IB2 = IC2{tex}\\Rightarrow{/tex} (x - 6)2 + (y + 6)2 = (x - 3)2 + (y + 7)2 = (x - 3)2 + (y - 3)2Taking first two, we get(x - 6)2 + (y + 6)2 = (x - 3)2 + (y + 7)2{tex}\\Rightarrow{/tex} x2 - 12x + 36 + y2 + 12y + 36 = x2 - 6x + 9 + y2 + 14y + 49{tex}\\Rightarrow{/tex} 6x + 2y = 14{tex}\\Rightarrow{/tex} 3x + y = 7 ......(1) ....[Dividing throughout by 2]Taking last two, we get(x - 3)2 + (y + 7)2 = (x - 3)2 + (y - 3)2{tex}\\Rightarrow{/tex} (y + 7)2 = (y - 3)2{tex}\\Rightarrow{/tex} (y + 7) = {tex}\\pm{/tex}(y-3)taking +e sign, we gety + 7 = y - 3{tex}\\Rightarrow{/tex} 7 = -3which is impossibleTaking -ve sign, we gety + 7 = -(y - 3){tex}\\Rightarrow{/tex} y + 7 = -y + 3{tex}\\Rightarrow{/tex} 2y = -4{tex}\\Rightarrow y = \\frac{{ - 4}}{2} = - 2{/tex}Putting y = -2 in equation (1), we get{tex}\\Rightarrow{/tex} 3x - 2 = 7{tex}\\Rightarrow{/tex} 3x = 9{tex}\\Rightarrow{/tex} x = 3Thus, I {tex}\\rightarrow{/tex} (3, -2)Hence, the centre of the circle is (3, -2). | |
| 23894. |
What number should be added to the polynomial x2+7x-35 so that 3 is the zero of the polynomial? |
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Answer» And will be 5. -5 5 |
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| 23895. |
What is best way to attempt sections in maths board paper ? |
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Answer» From last Start in the opp direction- D,C,B then A |
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| 23896. |
How can we draw three equation in same graph. Please reply |
| Answer» By adjusting them all | |
| 23897. |
Draw the graph of the following 2x-3y = -1 and 4x-6y=11Only points and at least three point please |
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| 23898. |
If the sum of n, 2n and 3n terms of an AP be S1, S2 and S3 respectively then prove that S3=3(S2-S1). |
| Answer» Let a be the first term and d be the common difference of the given AP. Then,S1 = sum of first n terms of the given AP,S2 = sum of first 2n terms of the given AP,S3 = sum of first 3n terms of the given AP.S1 ={tex}\\frac{n}{2}{/tex}{tex}\\cdot{/tex}{2a+(n-1)d}, S2={tex}\\frac{{2n}}{2}{/tex}{tex}\\cdot{/tex}{2a+(2n-1)d}, and S3= {tex}\\frac{{3n}}{2}{/tex}{tex}\\cdot{/tex}{2a+(3n-1)d}{tex}\\Rightarrow{/tex}3(S2-S1) = 3{tex}\\cdot{/tex}[{2na+n(2n - 1)d}\xa0- {na+{tex}\\frac{1}{2}{/tex}n(n-1)d}]= 3{tex}\\cdot{/tex}[na + {tex}\\frac{3}{2}{/tex}n2d-{tex}\\frac{1}{2}{/tex}nd] = {tex}\\frac{{3n}}{2}{/tex}{tex}\\cdot{/tex}[2a+3nd - d]= {tex}\\frac{{3n}}{2}{/tex}{tex}\\cdot{/tex}[2a+(3n-1)d}=S3.Hence, S3=3(S2-S1). | |
| 23899. |
Can you please tell the marking scheme |
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| 23900. |
Zeros of x2- kx + 6 are in the ratio 3 ratio 2, find k |
| Answer» K=3 | |