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This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
(iii)pxt q |
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Answer» 1 |
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| 2. |
, where p and q are integers and q0.Express the following in the formD 0.6(m) 0.47(iii) 0.001? With |
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| 3. |
- Express each of the following in the form(a) (i) 0.47.(ii) 0.235(b) (i) 0.001(ii) 0.076923where p and q are(iii) 2.4178(iii) 0.06 |
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Answer» and (i) 0.47 bar (means repeating term)let the no.x=o.47...eq1100x= 47.4747... eq2 sub 100x-x=47.4747..-(0.4747..99x=47x=47/99 |
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| 4. |
Abox contains 135 balls. How many balls are contained in 24bo |
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Answer» If one box contains 135 ballsthenin 84 balls 84*135=11340balls |
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| 5. |
) 1014 3)613 4) 2104 4, g (a)-2 then32. IE(a)-2,/(a)-1,g(a)--f(x),f(a)-g(a)lim =x-a1) 1/53) -1/5 4) -5 |
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Answer» if we put x=a numerator and denominator both will be 0. so 0/0 form using L'Hospital Rulelim (f'(x)f(a)+f(x)f'(a)-g'(a))/1 x->a =f'(a)f(a)+f(a)f'(a)-g'(a)=1*2+2*1-(-1)=2+2+1=4+1=5 |
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| 6. |
6 How many times is the HCF of 48, 36, 72 and 24 contained in their LCM ? |
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Answer» As per the question, 48=2×2×2×2×3=2^4×336=2×2×3×3=2^2×3^272=2×2×2×3×3=2^3×3^224=2×2×2×3=2^3×3 HCF=2×2×3=12LCM=2×2×2×2×3×3=144 144/12=12. So, 12 times is the HCF of 48,36,72 and 24 contained in their LCM. |
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| 7. |
\left. \begin{array} { l } { 1 ^ { 2 } , 3 ^ { 2 } , 5 ^ { 2 } , 7 ^ { 2 } , 9 ^ { 2 } , \dots } \\ { 0 , - 4 , - 8 , - 12 , \dots } \\ { 16,18 \frac { 1 } { 2 } , 20 \frac { 1 } { 2 } , 23 , \dots } \end{array} \right. |
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Answer» option d is in APas the common difference is same for all the convective numbersthat is -4 how |
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| 8. |
рди.L: ren(eAd |
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Answer» sinA=cos(90-A) sin18=cos(90-18) =cos(72) So sin18/cos72 = sin18/sin18 =1 |
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| 9. |
Quadrilateral EFGH is a rectangle in which J is the point of intersection of the diagonals. find the value of x, if JF = 8x +4 and EG =24x - 8 . |
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| 10. |
AB is diameter of a circle and AC is it chord such that |
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Answer» Let O be the centre of the circle. By Tangent Chord Theorem ∠CAB =∠BCD=30° …(1) Also ∠OCD=90° [∵Angle between Tangent and radius is 90° ] ∴∠BCO=90°-30°=60°. Also ∠CBO=60° because, Angle opposite to equal sides (radii) are equal. And hence ΔOBC must be a Equilateral Triangle. ∴∠COB=60°. Also in Right Triangle OCD, ∠CDO=90°-60°=30° …(2) From (1) and (2) BC=BD because Sides opposite to equal angles are equal in a triangle |
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| 11. |
ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and AD respectively. Show that the quadrilateral PQRS is a rhombus. |
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| 12. |
EXAMPLE 16 By using the concept of slope, prove that the diagonals of a rhombus are at right angles |
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| 13. |
ABCD is a quadrilateral in which AD=BC. P,Q,R,S are the mid points of sides AB,BC,CD,DA. Then PQRS is a rhombus prove it. |
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| 14. |
D, E, F are mid points of sides BC, CA, AB ofAABC. Find the ratio of areas of ADEFand ΔABC. |
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| 15. |
"tan'të¸).tan'"()-2-then.findthe vauveofx.if tan (x -4x+4) 4 |
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Answer» Like my answer if you find it useful! |
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| 16. |
Example 3. ABCD is a rectangle and P, Q, Rand S are mid-points of the sides AB, BC, CD and DArhombus.[CBSE 2012] |
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| 17. |
52. In the given figure, ABCD is a kite in whichAB AD and O is the mid-point of EG and FH. Thequadrilateral EFGH is always a(1) Parallelogram(3) Square(2) Rectangle(4) Rhombus |
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Answer» 1 Secondary School Math 8 points ABCD is a kite having AB = AD and BC= CD. prove that the figure formed by joining the mid - points of the sides , in order , is a rectangle. Advertisement Ask for details Follow Report byKenypiyupriyaagarw30.10.2016 Answers  prabhjyot Ambitious ABCD is a kite and AB=AD and BC=CD.EFGH are mid points of AB ,BC,CD,AD resp.To prove that EFGH is a rectangle. Construction: join AC and BD.In ΔABD , E and F are mid points.so we can write ,EF ║BD andEF = 1/2 BD ---------(1) (using mid point theorem) Now , in Δ BCD , G and H are mid points.So, GH║ BD andGH=1/2 BD ------------>(2) (by mid point th.)From (1) and (2) EF║ GH and EF = GH ( opposite sides of quad. EFGH are parallel and equal so EFGH is a parallelogram) Now since ABCD is a kite ,diagonal intersect at 90.So ∠AOd= 90. |
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| 18. |
\frac { \operatorname { tan } ( \frac { \pi } { 4 } + x ) } { \operatorname { tan } ( \frac { \pi } { 4 } - x ) } = ( \frac { 1 + \operatorname { tan } x } { 1 - \operatorname { tan } x } ) ^ { 2 } |
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| 19. |
10. Show that the plane 8X6Y+Z+50 touchesthe paraboloid- z, and find the pointof contact |
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Answer» yes this answer is correct |
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| 20. |
New Wordshonoursutherexperience pain, umess or inrycourage/confidencelange posts that help to hold somethingExerciseOral1 Recite the poem "Nation's Strength2. Describe the given picture in your own words.Written1. Answer the following questions:1. What makes a nation great? Brave r MOKOWhat type of people are referred to as 'Brave menthe most 23. List down some 'Brave men' of India. Who inspires you the most? |
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Answer» 2. the people who are least bothered about the consequences of their act are termed as brave men eg. armed men3.Rani Lakshmi bai, Netaji Subash Chandra Bose , Bhagat Singh, and all the soldiers of India. |
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| 21. |
â Given logo 4 = 0.6021, calculate approximately logo 404. |
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Answer» log10 4 = 0.6021 log10 404 = log10(101*4) = log10 101 + log10 4 = 2 + 0.6021 = 2.6021(approx) May you give answer by using any series like maclaurin's or taylor 's series |
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| 22. |
9. In AABC, the bisector of ZA is perpendicular to side BC. Prove that AB-AC. |
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Answer» Like if you find it useful |
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| 23. |
Given log10 4 = 0.6021, calculate approximately logo 404. |
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| 24. |
The value of 4 x logo.52, is(a) -2 (b) 4 (c) 2 (d) none of these |
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Answer» Thank you ..but the correct answer is option c sorry correct answer is option b |
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| 25. |
12 AB and AC are two tangents of a circle. O is the centerof the circle. MH is a tangent on the circle oleintersects AB & AC at M&N. The tangent MN does nottouch the circle at the point where the line onintersect the circle . |
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| 26. |
A, B, C and D are centres of equal circles which touch extemally inD is a square of side 7 cm. Find the area of the shaded region. |
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| 27. |
D)tauu tant /cot -I |
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Answer» Write tanx as sinx/cosx and cotx as cosx/sinx I= √((sin^2x/sinxcosx) + √(cos^2x/sinxcosx))dx I= (sinx+cosx dx)/√(sinxcosx) sinx-cosx=t (sinx+cosx)dx=dt (sinx-cosx)^2=t^2 1–2sinxcosx=t^2 sinxcosx=(1-t^2)/2 Therefore, I=dt/√((1-t^2)/2) I=(√2)sin^-1(t) + C I=(√2)sin^-1(sinx-cosx) + C |
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| 28. |
rence, ADEXAMPLE 10rhombusIf all the side of a parallelogram touch a circle, show that the parallelogram is aINCERT, CBSE 2000C, 2002,2008, 2012, 2013, 20141OR |
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| 29. |
m pey tantFind Her of the nunbeb |
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Answer» K=Kx1 2K=Kx2 3K=3xk 5K=5xK since K is common among all so K will be the hcf of question. Like my answer if you find it useful! |
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| 30. |
tant21+ tan x1-tan x4tan4 |
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| 31. |
prove thattemt 2n + tant (4)Con-243Iton (A41242) |
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Answer» check outmy solution |
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| 32. |
10. D isany point on side AC ofAABC with AB = AC . Show that CD < BD |
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| 33. |
3.1If a +b =lä-bl, a.b are non - zero vectorsthen the angle between a and bis :(a) 30°(b) 60°(c) 90°(d) 120° |
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Answer» c) 90°..................... (c) 90°is a correct answer |
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| 34. |
tan(90+θ)-sin(180-9)sin (270-8)+cosec(360-0)52. Iftan θ--and Bis not in the fourth Quadrant then12 |
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| 35. |
10.D is any point on side AC ofA AABC with AB =AC. Show that CD < BD |
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Answer» Given - AB = AC To prove -- BD > CD Proof -- Since AB = AC∠ABC = ∠ACB (By Isosceles Triangle property) ----(i) Here clearly,∠ABC > ∠CBD ∠ACB > ∠CBD ---from (i)∠DCB > ∠CBDBD > CD (Angle opposite to greater side is greater in a triangle) Hence Proved. |
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| 36. |
.| âXâX64s 7| |
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| 37. |
Express 7 g 6 dg 2 cg into cg. |
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Answer» 1 g = 100 cg and 1 dg = 10 cg 7 g 6 dg 2 cg = 700+60+2 = 762 cg |
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| 38. |
lim _________.___..*/â 3+x -5-xxâ1 X i 1 |
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| 39. |
âŹl =471 + x¢b=4+x (g |
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Answer» thanks |
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| 40. |
The focii of the curve represented by x 3 (cost+ sin t), y=4(cos t-sint) is(A)(0,-+root7)(B) (-+root7,0)(C)(-+3/4root7,0)(D)(0,-+3/4root7)(D) (0,Âą4.7(A) (0,tv7(B) (V7,0(C) |7,0 |
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Answer» we have x/3 = cos(t) + sin(t) and y/4 = cos(t) - sin(t) ii) Squaring them and adding, x²/9 + y²/16 = 1This is of the form x²/b² + y²/a² = 1, with a > b.This ellipse center is (0, 0) and major axis along y-axis. b = 3 and a = 4; eccentricity e = √7/4Parametric form of any point on this is (3cos(t), 4sin(t))hence foci is (+-ae,0)hence (+-3/4√7,0) |
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| 41. |
$. x=sint,y = cos 2t |
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| 42. |
sint - “2 5111 6: tantMoo < "32¢co0s” 0 cosl |
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| 43. |
12- If cost and then find Sint and tant13 |
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| 44. |
"०06 > 6 > o0 0 = € — §;500 कु : 6 गण [0S () °I e |
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| 45. |
3BisSin (Sint ICestal = 1, then find the value of x.telu |
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| 46. |
four reional oumbers betweenand |
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Answer» hit like if you find it useful |
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| 47. |
(a) logo (1012)(b) logo ()(c) 2log1yG+31°gioan. Simplify: |
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Answer» does log0 the base 10 exist answer is 0 |
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| 48. |
If x is a positive real number. then what is the value of |
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| 49. |
If x is a positive real number then calculate the value of √⁴√³√x² |
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Answer» x^((2/2)*(1/3)*(1/4)) =x^(1/12) |
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| 50. |
12. Ifx be a positive real number such that x2+_2=50, then evaluate x+-. |
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Answer» (x+1/x)^2=x^2 +1/x^2 +2x(1/x) (x+1/x)^2=50/7+2=64/7x+1/x=(64/7)^(1/2) |
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