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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.
| 23951. |
What is the level of board examinations? Is it too tough this time? |
| Answer» | |
| 23952. |
In board exam quetions are easy or difficult |
| Answer» Normal | |
| 23953. |
The 24th rerm of an A.P. is twice its 10th term. Show that its 72nd term is 4 time its 15th term |
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Answer» a+23d=2a+18da=5d(23d-18d=5d) (i)a15=5d +14d (from i ) a15=19d (ii)now,a72=5d+71d (from i ) a72=76da72=4(19d)a72=4a15H.P. a24\xa0= 2a10a+23 d= 2(a+9d)a+23d = 2a+18d23d-18 d= 2a-a5d= aa72 =\xa04a15a+71d = 4(a+14d)5d+71 d= 4a+56d76d= 4(5d)+56d76 d= 20d+56d76d=76dhence proved |
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| 23954. |
After taking maths in 11th class what to do after 12th class what are the options |
| Answer» Are yaar pahle board ka tension lo.. | |
| 23955. |
In which class logarithm came |
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Answer» Pdna chaho to 10th m bhi teacher se kh k pd skte ho it cn be possible 11and 12 th |
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| 23956. |
Verify the quadratic equation x^3-6x^2+3x+10 |
| Answer» Use the zeroes alpha,beta and gamma. Add them one by one. Then add them two at a time. Then multiply all of them. Equate this to -b/a ,c/a and -d/a. | |
| 23957. |
If two zeros of polynomial x⁴_6x³_26x²+138x-35 are 2±√3, find other zeros |
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Answer» 5;-7 -5,7 5,7 |
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| 23958. |
Prove that the point (0,0) (5,5) (-5,5) are the vertices of right vertices triangle. |
| Answer» Let A= (0,0) , B = (5,5) and C = (-5,5)AB = √(5-0)2\xa0+(5-0)2\xa0= √25+25 = √50BC = √(-5-5)2\xa0+ (5-5)2\xa0= √(-10)2\xa0+0 = √100AC = √(-5-0)2\xa0+(5-0)2\xa0= √(-5)2\xa0+(5)2\xa0= √25+25 = √50BC2\xa0= AB2+AC2(√100)2\xa0= (√50)2\xa0+(√50)2100= 50+50100=100So the given points are vertices of right angled triangle. | |
| 23959. |
Sin\'6 theta_cos\'6 theta |
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| 23960. |
Nth term |
| Answer» In AP the Nth term is the number of units in a series or more like the number of numbers in a series. For example 1, 2, 3, 4, 5, 6. Here the Nth term is 6 as there are 6 units in this A.P. | |
| 23961. |
If circumference and the area of circle are numerically equal, find the diameter of the circle |
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Answer» 4 2πr=πr×r =>r=2. =>. D=4 |
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| 23962. |
What is the value of cos a |
| Answer» Sin(90-a)=cos a | |
| 23963. |
Sin2a+cos2a=3/2 find sin4a |
| Answer» sin2a+cos2a=3/2 to sin4a= | |
| 23964. |
34+56-+45×π45=? |
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Answer» 6451.7251235 Question aache se likho. |
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| 23965. |
What is the value of tan teetha? |
| Answer» Sin teetha/cos teetha | |
| 23966. |
When will 2018 board exams |
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| 23967. |
Draw a line segment of 6cm |
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| 23968. |
Convert 60m to 120cm |
| Answer» (120×50)cm | |
| 23969. |
how 1=2 |
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| 23970. |
How to find the 11th term from the last term of the ap27,23,19,............ - 65. |
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Answer» WRite the above A.P. in reverse order, we get -65 ,-61, -57,..........19,23,27Here First term a = -65 and common difference d = 4So 11 th term a11\xa0= a+(n-1)da11\xa0= -65+(11-1) x4a11\xa0= -65+10x4a11 = -65+40= -25 a=-65d=23-27=-5a11=a+10da11=(-65)+(10)-5 |
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| 23971. |
Find the nature of the roots 5 xsquare -3 x+3=0 |
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Answer» Given eq is 5 x2-3x +3 = 0Here a= 5, b=-3 and c= 3So D= b2\xa0-4acD= (-3)2\xa0-4 x5x3D= 9-60= -51D< 0 . So roots are imaginary or not real -51 , Two no real Roots |
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| 23972. |
A cylinder , hemisphere and cone have same base and height . Find the ratio of their volume. |
| Answer» The height of a hemisphere is its radius. So volume of hemisphere = 2π r3/ 3 = 2π r2\xa0h/3\xa0volume of cylinder : volume of hemisphere : volume of coneπ r2h : 2π r2h/3 : π r2h /31:2/3:1/3Multiply the above ratio by 3, we get\xa03:2:1 | |
| 23973. |
99,81,44 ________? |
| Answer» 31 | |
| 23974. |
If cos a = 2/5 find the value of 4+4 tan A |
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| 23975. |
(x-2)3 |
| Answer» 3x-6 | |
| 23976. |
The 4th term of ap is zero prove 25th term of ap is three times it\'s 11th term |
| Answer» We have,a4\xa0= 0{tex}a + 3d = 0{/tex}{tex}3d = -a{/tex}or {tex}-3d = a{/tex}..........(i)Now,a25\xa0= {tex}a + 24d{/tex}= {tex}-3d + 24d{/tex} [Putting value of a from eq(i)]= {tex}21d{/tex}...........(ii)a11\xa0= {tex}a + 10d{/tex}= {tex}-3d + 10d{/tex}= {tex}7d{/tex}.........(iii)From eq(ii) and (iii), we geta25 = 21 da25 = 3(7d)a25\xa0= 3a11Hence Proved | |
| 23977. |
What is limitations of herons formula |
| Answer» These formulas can only be used only when an angle is given, as well as either the length of the side that is opposite of the angle, or the length of the hypotenuse. | |
| 23978. |
Prove that root2 is a irrational number |
| Answer» Refer maths ncert | |
| 23979. |
What is hardy-ramajun numbers |
| Answer» 1729 is the natural number following 1728 and preceding 1730. It is known as the Hardy–Ramanujan number, after an anecdote of the British mathematician G. H. Hardy when he visited Indian mathematician Srinivasa Ramanujan in hospital. | |
| 23980. |
What is locus of circles |
| Answer» A locus is a set of points that meet a given condition. | |
| 23981. |
1+2+3+6+4+5+x+3=8 |
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Answer» right answer is -16 -16 16 |
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| 23982. |
for what value of k will the linear equation have no solution 3x+y= 1 and ( 2k-1)x+( k-1)y=2k+1. |
| Answer» The given pair of linear equation is3x + y = 1, (2k - 1) x + (k - 1) y = 2k + 1{tex}\\Rightarrow{/tex} 3x + y - 1 = 0Here, (2k - 1)x + (k - 1) y - (2k + 1) = 0a1 = 3, b1 = 1, c1 = -1a2 = 2k - 1, b2 = k - 1, c2 = -(2k + 1)for having no solution, we must have {tex}\\frac{{{a_1}}}{{{a_2}}} = \\frac{{{b_1}}}{{{b_2}}} \\ne \\frac{{{c_1}}}{{{c_2}}}{/tex}{tex}\\Rightarrow \\frac{3}{{2k - 1}} = \\frac{1}{{k - 1}} \\ne \\frac{{ - 1}}{{ - (2k + 1)}}{/tex}From above we have\xa0{tex}\\frac{3}{{2k - 1}} = \\frac{1}{{k - 1}}{/tex}{tex}\\Rightarrow{/tex} 3(k - 1) = 2k - 1{tex}\\Rightarrow{/tex} 3k - 3 = 2k - 1{tex}\\Rightarrow{/tex} 3k - 2k = 3 - 1{tex}\\Rightarrow{/tex} k = 2Hence, the required value of k is 2. | |
| 23983. |
If secA + tanA= p find secA and tanA |
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| 23984. |
Polynomial is 3x^2-x^2+9x^2-6 |
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| 23985. |
Find the eleventh term from the last term of the AP |
| Answer» a = – 62, d = 3a11 = a + 10d= – 62 + 10(3)= – 32 | |
| 23986. |
If one of the zeroes of the polynomial 3x^2-8x+2k+1 is 7 times the other, find the value of k |
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Answer» How is it\'s solution 69 |
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| 23987. |
Find lcm of a^2b^3 , ab^4 , a^4b^2 |
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| 23988. |
How to get good marks when I am a losser in maths? |
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Answer» you have study mathmatics in the form of game No chance..m |
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| 23989. |
What is terminating number ? check 7/75 is terminating or not ? |
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Answer» It is non-terminating decimal. A terminating decimal is a decimal that ends. It\'s a decimal with a finite number of digits. |
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| 23990. |
When will our exam start????? |
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Answer» Time table will Declared in january but exam will start in month of March March 2nd itseems Not declared yet. It may be declared in January. |
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| 23991. |
What is real number?????? |
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| 23992. |
What is real no |
| Answer» | |
| 23993. |
Kx34-3 |
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| 23994. |
Prove that the tangent drawn from an external point to a circle are equal in length |
| 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 | |
| 23995. |
Solve the following quadratic equation by factorization method:9x*x-6b*bx-(a*a*a*a-b*b*b*b) |
| Answer» We have,9x2 - 6b2x - (a4 - b4) = 0Here, Constant term = a4 - b4 = (a2 - b2)(a2\xa0+ b2)Also, the coefficient of middle term is - 6b2 = [3(a2 + b2) - 3(a2\xa0- b2)]Now using the above two values in given equation, 9x2 - 6b2x - (a4 - b4) = 0{tex}{/tex}\xa0we have, 9x2 - [3(a2 + b2) - 3(a2 - b2)]x - (a2 - b2)(a2\xa0+ b2) = 0{tex}\\Rightarrow{/tex} 9x2 - 3(a2\xa0+ b2)x + 3(a2 - b2)x - (a2 - b2)(a2 + b2) = 0{tex}\\Rightarrow{/tex} 3x[3x - (a2\xa0+ b2)] + (a2 - b2)[3x - (a2 + b2)] = 0{tex}\\Rightarrow{/tex} [3x + (a2\xa0- b2)][3x - (a2 + b2)] = 0{tex}\\Rightarrow{/tex}either [3x + (a2\xa0 - b2)] = 0 or, [3x - (a2 + b2)] = 0{tex}\\Rightarrow{/tex} 3x = -(a2 - b2) or 3x = a2 + b2{tex}\\Rightarrow x = -(\\frac{{{a^2} - {b^2}}}{3}){/tex} or {tex}x = \\frac{{{a^2} + {b^2}}}{3}{/tex}{tex}\\Rightarrow x = \\frac{{{b^2} - {a^2}}}{3}{/tex} or {tex}x = \\frac{{{a^2} + {b^2}}}{3}{/tex}Hence, the roots of given quadratic equation are\xa0{tex}\\frac{{b^2 - a^2}}{3}{/tex}and\xa0{tex}\\frac{{a^2 + b^2}}{3}{/tex} | |
| 23996. |
PHYTHAGORUSTheorm |
| Answer» H2=B2+p2 | |
| 23997. |
If the quadractic equation(k+1)x-2(k-1)x+1=0 have real root then, find the value of k |
| Answer» The question is wrong because the equation doesn\'t has power on it | |
| 23998. |
If Sn denotes the sum of the first n terms of an A.P prove that S12=3(S8-S4) |
| Answer» Let a be the first term and the common difference be d.Sn\xa0=\xa0{tex}\\frac n2{/tex}[2a + (n - 1)d]S12 = {tex}\\frac {12}2{/tex}[2a + (12 - 1)d]= 6[2a + 11d]= 12a + 66dS8 =\xa0{tex}\\frac 82{/tex}[2a + (8 - 1)d]= 4[2a + 7d]= 8a + 28dS4 =\xa0{tex}\\frac 42{/tex}[2a + (4- 1)d]= 2[2a + 3d]= 4a + 6d3(S8 - S4) = 3[(8a + 28d) - (4a + 6d)]= 3[8a + 28d - 4a - 6d]= 3[4a + 22d]= 12a + 66d= S12 | |
| 23999. |
Solve for x√3xsquare-2√2x-2√3=0 |
| Answer» {tex}\\sqrt { 3 } x ^ { 2 } - 2 \\sqrt { 2 } x - 2 \\sqrt { 3 } = 0{/tex}{tex}\\sqrt { 3 } x ^ { 2 } - 3 \\sqrt { 2 } x + \\sqrt { 2 } x - 2 \\sqrt { 3 } = 0{/tex}{tex}\\sqrt { 3 } x [ x - \\sqrt { 6 } ] + \\sqrt { 2 } [ x - \\sqrt { 6 } ] = 0{/tex}or,\xa0{tex}( x - \\sqrt { 6 } ) ( \\sqrt { 3 } x + \\sqrt { 2 } ) = 0{/tex}or,\xa0{tex}x = \\sqrt { 6 } , - \\sqrt { \\frac { 2 } { 3 } }{/tex} | |
| 24000. |
Find lcm of (x²-4)and(x⁴-16) |
| Answer» x²-4 x2- 16 | |