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InterviewSolution
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.
| 51. |
Find the sum of roots of a quadratic equation 2x²+ 214 x – 49=0Your answerBackNext |
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| 52. |
If the sum of three terms of an AP is 42, than middle term isYour answer |
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Answer» a1=14,a2=28,a3=42 The middle term is 28 Let the three consecutive terms be (a-d) (a) & (a+d) resp.(a - d) + (a) + (a + d) =423a = 42a = 14 |
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| 53. |
The sum of first n odd natural numbers is\xa0* |
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| 54. |
The roots of the quadratic equation 6x² – x – 2 = 0 are\xa0* |
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Answer» 6x^2 - x - 2 = 06x^2 -4x +3x - 2= 02x(3x-2)+1(3x-2) =0 (3x-2) (2x+1) =03x-2=0,2x+1=03x=2,2x=-1Therefore X=2/3 , x=-1/2 \xa06x² – x – 2 = 0\xa06x² – 4x+ 3x – 2 = 02x(3x - 2) + 1 (3x - 2) = 0(3x -2) ( 2x + 1) = 03x - 2 = 0 and 2x + 1 = 03x = 2 and 2x = -1x = 2/3 and x = -1/2 |
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| 55. |
If p(x) = ax² + bx + c, then c/a is equal to:\xa0* |
| Answer» Product of the roots | |
| 56. |
The sum of first 5 multiples of 3 is:\xa0* |
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Answer» 45 Sum of first 5 multiples of 3:Sum of an AP is:\xa0n/2 \u200b×(2a+(n−1)d)a=3,d=3,n=5Then,\xa0Sum =\xa05/2 \u200b×(2×3+(5−1)×(3))=\xa05/2\u200b×(6+12)=\xa045 |
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| 57. |
Nature of roots of the quadratic equation x² – 4x + 4 = 0 is\xa0* |
| Answer» Real and equal roots | |
| 58. |
The 10th term from the end of the A.P. -5, -10, -15,…, -1000 is\xa0* |
| Answer» The 10th term from last term is -955 | |
| 59. |
The polynomial equation x (x + 1) + 8 = (x + 2) (x – 2) is\xa0* |
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Answer» It is linear polynomial Not quadratic polynomial |
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| 60. |
In an, A.P if d = 2 , n = 5 and last term is 0 , the value of a is\xa0* |
| Answer» a=-8 | |
| 61. |
(1+ cot² A)/(1+ tan² A) is equal to\xa0* |
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| 62. |
sec A =\xa0* |
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| 63. |
The distance of the point (α, β) from the origin is\xa0* |
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| 64. |
The distance of the point P(2, 3) from the x-axis is\xa0* |
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| 65. |
Theorem 6.1 (triangles) |
| Answer» (Ans) if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are dividing in the same ratio. | |
| 66. |
USE Euclids division algorithm to find the HCF of 867 and 255 |
| Answer» By using Euclids division algorithma = bq+rwhere a is > bso a = 867 and b=255867=255 × 3+102here r≠0 so a=255 and b=102255=102×2+51here r≠0 so a=102 and b=51102=51×2+0here r=0so, Hcf of (867,255) is =51 | |
| 67. |
The sum of given A.P.1, 4 , 7 , 10 , . . . upto 20 terms is____________ . |
| Answer» a = 1 d = 4-1 = 3an = a + (n-1)da20 = 1 + (20-1)3a20 = 1 + (19)3a20= 1 + 57a20= 58 | |
| 68. |
If the sum of three terms of an AP is 42, than middle term is Your answer |
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| 69. |
The sum of given A.P.1, 4 , 7 , 10 , . . . upto 20 terms is____________ .Your answer |
| Answer» Given AP is 1,4,7,10,.....\xa0In given AP first term a=1, common difference d=3\xa0Since, Sn= n/2[2a+(n−1)d] ....... sum of n terms\xa0Put n=20 for sum of first 20 terms\xa0S20=20/2 [2(1)+(20−1)3]\xa0S20=10[2+57]\xa0S20=590 | |
| 70. |
If (b²-4ac) = 0 , then roots will be equal and real.\xa0*TRUEFALSE |
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Answer» True If b2\xa0- 4ac = 0 then the roots will be\xa0a real number because b and a are real. |
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| 71. |
what is the product of (cosecA-cotA), (cosecA+cotA) |
| Answer» (cosecA-cotA) (cosecA+cotA)= cosec2 A - Cot2A=\xa0cot2A + 1\xa0- Cot2A (cot2A + 1 =\xa0cosec2A)= 1 | |
| 72. |
find the distance between A (7,-4) and B (-5,1) |
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| 73. |
if the sin o = 3/5, what is the value of 3 cot o |
| Answer» If\xa0sino = 3/5Then, opposite side of the right angled triangle containing angle\xa0o = 3 and hypotenuse\xa0=5Using pythagoras theorem, we get the adjacent side\xa0= 4\xa0Now we have, coso =Adjacent\u200b/Hypotenuse=4/5 \u200bAnd cot o =Adjacent/Opposite\u200b= 3/43 cot o =3( 3/4) = 9/4 | |
| 74. |
find the sum of roots of quadratic equations 2x^2+214x-49=0 |
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| 75. |
Question 1exercise 8.4 exemplar |
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| 76. |
x+15=8/5then what is X value? |
| Answer» X = (8/5) - 15X = 1.6 - 15X = -13.4 | |
| 77. |
-5,-1,3,7 find the next four terms of ap |
| Answer» 11 | |
| 78. |
Examples of irrational numbers |
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Answer» You can see the page number. the page number is (12) √2,√3,√15, π, _√2/√3, 0.10110111011110...., etc |
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| 79. |
Q.30 If tan (A+10) = 1 then value of A is |
| Answer» tan (A+10) = 1we know that tan 45 = 1So, tan (A+ 10) = tan45A + 10 = 45A = 45 - 10A = 35 | |
| 80. |
Given a=5,d=3,an=50,find n and sn |
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Answer» Given,\xa0a=5,d=3,an\u200b=50⇒a+(n−1)d=50⇒5+(n−1)3=50⇒5+3n−3=50⇒3n=48⇒n=16∴S16\u200b=16/2\u200b[2a+(16−1)d]=8[2×5+15×3]=440Hence,\xa0n=16,S16\u200b=440 añ=a+(n-1)d50=5+(n-1) 350=5+3n-350=2+3n50-2=3n48=3n48/3=n16=nSn=n/2(a+añ)Sn=16/2(5+50)Sn=8(55)Sn=440 |
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| 81. |
Plz tell converse of Basic Proportional Theorum |
| Answer» Converse of Basic Proportionality Theorem is if a line divides the any two sides of a triangle in the same ratio, then the line must be parallel to the third side... hope it would be helpful?✌ | |
| 82. |
Solve x and y : x + y = 15 and x - y = 5 |
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Answer» x+y=15........(i)x-y=5.........(ii),From equation (ii), we getx=5+y Put the value of x in equation (i),we get5+y+y=152y=15-5y=10/2=5Now, put the value of y in equation (ii), we getx-5=5x=5+5=10:.x=10 and y=5 Adding by eliminTion method=2x=20X=10Putting in 1 eq x+y=15Y=15-10 = 5 |
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| 83. |
(Cos4 theta - sin2 theta) is equal to |
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| 84. |
Worksheet 37 |
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| 85. |
Ifα and β are the zeroes of the polynomial x²+5x+6,then the value of 1/α+1/β+αβ is\xa0* |
| Answer» Here, a=1,b=5,c=6 alpha+beta=-b/a =-5/1Alpha+beta=-5,.......(i) And alpha×beta=c/a=6/1 Alpha×beta=6.......(ii)Now, 1/alpha+1beta+alpha×beta= Beta+alpha/alpha×beta+alpha×beta=-5/6+6 (from equation i and ii)=31/6.......It may helpful to you | |
| 86. |
16/9-7/9 how to solve explain |
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Answer» 16/9-7/99/9= 1 Right answer..... 16/9-7/9=9/9=1 hi hoga ?✌ 16/9-7/9=9/9=1 hai 16/9-7/9=9/9=1 16/9-7/9=9/9=1 |
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| 87. |
How to solve ex.1.3 question answer in short trick |
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Answer» Yes leave it Leave it . yes |
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| 88. |
Point(4 , -3) is at perpendicular distance of _____ from the x-axis |
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| 89. |
If sin theta is = 4/5 find the value of 4 tan theta -5cos theta/ sec theta + 4cot theta |
| Answer» sin theta=p/h=4/5 so we should find b from this equationb=✓h²-p²✓5²-4²✓25-16✓9=3\xa0\xa0\xa04tan theta - 5 cos theta by sec theta + 4 cot theta4×p/b - 5×b/h divide by h/b + 4×b/p4×4/3 - 5×3/5 divide by 5/3 + 4×3/416/3-3 by 5/3+ 37/3 by 14/33_3cancel7/14=1/2 | |
| 90. |
Solve (2x-3)=25 |
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Answer» 2x - 3 = 25 ,, 2x = 25 + 3 ,,x = 28 / 2 ,,x = 14 ...... 2x-3=25 =》2x=25+3 =》x=28/2 =》x=14 2x=25+3=》2x=28=》x=28/2=》x=14 (2x-3)=252x=25+3=x=28/2=x=14 2x-3=25 2x=25+3 x=28/2 x=14 |
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| 91. |
what is square of34 |
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Answer» 34² = 1156 . (34)^2=1156 1156 1156 1156 |
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| 92. |
Point (4,_3) is at a perpendicular distance ot from the x_ axis |
| Answer» 3 unit | |
| 93. |
For the given quardratic equcation 5x+3y=5 state the root. 1,0 2,1 _1,_2 0,1 |
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| 94. |
Embacment question |
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| 95. |
Sin2B=2SinB |
| Answer» 0 | |
| 96. |
Is there anybody who know that when cbse will release sample paper of board 2020-21? |
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Answer» Yes I know CBSE will release sample in the end of year ..Probably November... There is no such information given by CBSE . Don\'t know Sorry but I don\'t know ?? No |
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| 97. |
Given sin(A+B)=SinA cosB +cosA sinB then find the value of sin 75° |
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Answer» Let A=45°,B=30°(so that sum of A and B =75° and we can get Sin74°)(put the value of A and B in equation) - Sin75°=Sin45°*Cos30° + Cos 45°*Sin30°- Sin75° = 1/√2*√3/2 + 1/√2*1/2- Sin75° = √3/2√2 + 1/2√2- Sin75° = (√3+1)/2√2Sin75° = (√6+√2)/4 sin ( A+ B) = sin75°Let A = 45° and B = 30°Thensin75° = sin45° cos30° + cos45° sin30°= 1/√2 * √3/2 + 1/√2 * 1/2= √3 / 2√2 + 1/2√2= √3 + 1/2√2You can rationalise the denominator. |
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| 98. |
Find the nature of the equation 2X^2-5X-7=0. |
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Answer» 2x^2-5x-7x=02x^2-5x-7(2x^2-7x)+(2x-7)x(2x-7)+1(2x-7)(x+1)(2x-7) Answer:the roots are not realStep-by-step explanation:comparing the equation in ax^2+bx+c=0a=2b=-5c=7Delta =b^2-4ac =25-4 x 2 x 7 =25-56 = -31It describes that the roots are imaginary or do not exist . Answer:the roots are real and equalStep-by-step explanation:comparing the equation in ax^2+bx+c=0a=2b=-5c=7Delta =b^2-4ac =25-4 x 2 x 7 =25-56 = -31It describes that the roots are imaginary or do not exist . |
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| 99. |
Sin 60 = ???? |
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Answer» Sorry for wrong answer the answer is. √3/2 Sin60° = √3/3 Sin 60°=√3/2 |
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| 100. |
Find the zeroes of the polynomial 4X^2+3X+7 ????? |
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Answer» 4x^2+3x+7=4x^2+7x-4x+7=4x(x+7/4)-4(x+7/4)=(4x-4) or (x+7/4)=0=x=1 or x=-7/4 1,-7/4 4x^2 +3x+7(4x^2+4x)+(7x+7)4x(x+1)+7(x+1)(4x+7)(x+1) |
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