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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. |
: /\% Find the roots of the following equations: |
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| 2. |
ueW1real and equal roots? Solve the equations |
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| 3. |
1. Findthe roots of the quadratic equations 2x2 + x+4 |
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| 4. |
3 Find the roots of the following equations:(i)x-_, = 3,170 |
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| 5. |
2 AB is a line segment. AX and BY are two equal linesegments drawn on the opposite sides of AB such thatAXI BY. If AB and XY intersect each other at P, prove Abat(f) AAPXミABPY(i) AB and XY bisect each other.3 ) In the given figure, AD and BC are perpendicularto line segment AB and AD=BC. If CD intersectsAB at P, show that P is the mid-point of AB as Awell as that of CD |
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| 6. |
otfienIn Δ ABC, LM is a line segment parallel to side BC ofBC where L and M divide AB and AC respectivelyAAin the ratio 1:3. If BC 8.4 cm, find LM |
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Answer» Given AL/LB= AM/MC= 1/3 or LB= 3AL and MC= 3 AM So, AL= 1/4 AB and AM= 1/4 AC or AL= 1/2 (1/2AB) and AM= 1/2 (1/2 AC) or L is the midpoint of 1/2 AB and M is the midpoint of 1/2 AC, So, LM= 1/2 (1/2BC) [Midpoint theorem] or LM= 1/4* 8.4 or LM= 2.1cm |
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| 7. |
find the slope of the line x_y+3=0 |
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Answer» Given line : x - y = 3 So, in standard form y = x - 3 So, slope is 1 |
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| 8. |
\left| \begin{array} { l l l } { 1 } & { 1 } & { 1 } \\ { a } & { b } & { c } \\ { b c } & { c a } & { a b } \end{array} \right| = ( a - b ) ( b - c ) ( c - a ) |
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| 9. |
5. In figure, the vertices of AABC are A(4, 6), B(1, 5) and C(7, 2). A line-segment DE is drawn tointersect the sides AB and AC at D and E respectively such thatAD AE 1AB AC 314M]Calculate the area of AADE and compare it with area of AABCA(4,6)B(1C(7,2) |
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| 10. |
10. In the given figure ABCD is a parallelogram andline segment AE and CF bisect ZA andrespectively. Prove that AE || CF.DE |
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| 11. |
७ छहिक 8५० 5 217० का] %८ DI पट कु € Lk 000'8 2.el |
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| 12. |
s in the adjoining figure, ABCD is aparallelogram in which BAO 35,105°DO = 40" and ..coD = 105°.Calculate (i) 'ABO, (ii) <ODC, (iii)and (iv) 2CBD40°ACB |
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Answer» Thanks |
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| 13. |
8. If A and B are (-2 -2) and (2, -4), respectively find the coordinates of P such tAPラABand P lies on the line segment AB.nta uthich divide the line segment joining A(-2, 2) and |
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| 14. |
5. In figure, CP and CQ are tangents to a circle with centre O. ARB is another tangenttouching the circle at R. If CP 11 cm , BC 7 em, then find the length of BR.0.2 x2 |
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| 15. |
The wave front due to a point source at a finite distancefrom the sounce is- |
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Answer» Wave front and wave normal Wavefront and Wave NormalConcept of wavefront :Consider a point source ‘S’ of light placed at a point in air or vacuum. The point source of light will emit the light waves in all directions with same velocity in the same medium. Let ‘v’ be the velocity of light in air medium. In time ‘t’ all the waves covers a distance equal to ‘vt’. In the other words in time ‘t’ all the waves will reach on a spherical surface of radius vt and centre S. The spherical surface so formed is called wavefront. It should be noted that all points on the spherical surface should be in the same state of vibration.Definition of wavefront :“The locus of all the points of the medium to which all the waves will reach simultaneously, so that all the points are in the same state of vibration is called wavefront” Types of wavefronts: There arethreetypes of wavefronts.1. Spherical wavefront.2. Plane wavefront.3. Cylindrical wavefront. 1.Spherical wavefront: A wavefront originating from the point source of light at finite distance is called spherical wavefront. 2.Plane wavefront: As time passes the spherical surface of spherical wavefront is so large that a small part of it can be considered as a plane wavefront.A wavefront originating from the point source of light at infinite distance i.e. at very large distance is called as plane wavefront. The plane wavefront can be obtained by keeping point source at the focus of the convex lens.The light from the sun reaches the earth in the form of plane wavefronts. 3.Cylindrical wavefront: A wavefront originating from a linear source of light is called cylindrical wavefront.Wave normal :A normal or perpendicular drawn to the surface of a wavefront at any point, in the direction of propagation of light, is called a wavenormal.The wavenormal gives the direction of propagation of light. |
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| 16. |
20 men complete one third of a work in 20days. How many more men should beemployed to finish the rest of the work in25 more days? |
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Answer» but what are M1 D1 W1. M2 D2 W2 |
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| 17. |
cos A -sin A + 1= cosec A + cot A, using the identicos A + sinA -1 |
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Answer» CosA-sinA+1/cosA+sinA-1 =(cosA-sinA+1)(cosA+sinA+1)/(cosA+sinA-1)(cosA+sinA+1) =(cos²A-cosAsinA+cosA+cosAsinA-sin²A+sinA+cosA-sinA+1)/{(cosA+sinA)²-(1)²} =(cos²A-sin²A+2cosA+1)/(cos²A+2cosAsinA+sin²A-1) ={cos²A+2cosA+(1-sin²A)}/(1+2cosAsinA-1) [∵, sin²A+cos²A=1] =(cos²A+2cosA+cos²A)/2cosAsinA =(2cos²A+2cosA)/2cosAsinA =2cosA(cosA+1)/2cosAsinA =(cosA+1)/sinA =cosA/sinA+1/sinA =cotA+cosecA =cosecA+cotA (Proved) Hit like if you find it useful! Thanks |
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| 18. |
e4-4—(s( L XL__flu) L ठप." दो 9- स-6s 1\o ( X .;.)_L “15 | = न हि + २2-98 ७ |
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| 19. |
Prove each of the following identi\frac{\csc A+1}{\csc A-1}=\frac{1+\sin A}{1-\sin A} |
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Answer» cosec A= 1/sinA (CosecA+1)/(cosecA - 1) = (1/sinA +1)/(1/sinA - 1) = (1+sinA)/(1-sinA) proved |
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| 20. |
43.If x = 1, then the solution ofx_y31sb)(l,-15a)(1, 2)c) (3, 1)d) (3, 2) |
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| 21. |
Find x if distance between points L(x, 7) and M(1, 15) is 10 |
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Answer» Distance = √(1-x)^2+64= 10(1-x)^2= 100-641-x= 6x= -5 |
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| 22. |
Find x if distance between points L(x, 7) and M(1, 15) is 10. |
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Answer» Say (x1,y1) = (x,7)(x2,y2) = (1,15)Distance between two points = √[(x2-x1)^2+(y2-y1)^2]10 = √[(1-x)^2+(15-7)^210 = √(1+x^2-2x+64)10 = √(x^2-2x+65)Squring to each side.100 = x^2-2x+65Bring all terms to one side.x^2-2x+65-100 = 0x^2-2x-35 = 0x^2-7x+5x-35 = 0x(x-7)+5(x-7) = 0(x+5)+ (x-7) = 0x+5 = 0 and x-7 = 0x = -5 and x = 7 |
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| 23. |
13m long tremany men should be employed for digging a 27 mI men can dig 6nch in one day. How4long trench in one day? |
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Answer» 11 men can dig 27/4 m long trench in one dayTo dig 27m we need (27/(27/4))x11 = 44 men |
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| 24. |
11 men can dig 6metre-long trench in one day. How many men should be employed fordigging 27-metre-long trench of the same type in one day?34 |
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| 25. |
10. The number of people employed in a firm decreased from 240 to 210. Find the percentage decrease. |
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Answer» percentage decrease=(240-210)/240*100=300/24=12.5% |
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| 26. |
10. How many men need to beemployed to complete a job in5 days, if 15 men can completeof the job in 7 days?(1) 20 (2) 21 (3) 45 (4) 63 |
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Answer» 63 is right answer of this question |
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| 27. |
23.2 men and 7 boys can do a piece of work in 4 days. The same work is done in 3 days by4 men and 4 boys. How long would it take one man and one boy to do it? |
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| 28. |
2 men and 7 boys can do a piece of work in 4 days. It is done by 4 men and 4 boys in 3 days. Howlong would it take for one man or one boy4- |
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| 29. |
In the below figure, ABCDE is any pentagon. BP drawn parallel to AC mects DC produced at Pand EQ drawn parallel to AD meets CD produced at Q. Prove that ar (ABCDE) ar (APQ) |
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Answer» Like my answer if you find it useful! |
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| 30. |
AD is a median of ∆ABC, E is the mid point of AD, BE produced meets AC in F. Prove that AF=1/3AC. |
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Answer» thanks a lot |
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| 31. |
In △ABC, AD is a median and E is the midpoint of AD. If BE isproduced, it meets AC in F. Show that AF 3AC ICBSE 2006c1 |
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| 32. |
23.in a parallelogram ABCD, ABAE and BC produced meet at F. Find the length of CF10 cm and AD-6 cm. The bisector of LA meets DC İnE. |
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| 33. |
LICEJUICELEducation+Class- 10th1. If x = 2/3 and x = -3 are the roots of the equation ax2 + 7x + b = 0, find the values of a and b.2. Solve the following quadratic equations by factiruzatuib nethod:. (1, +*+1 = 34, x = 0,x = -1x+1 X 151.-L. Clarina undratic equations by factorization |
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Answer» It is for 2nd question 1)ax^2+7x+b=0x=2/3a(2/3)^2+7(2/3)+b=04a/9+14/3+b=04a+9b+42/3=04a+9b+42=0-----(1)ax^2+7x+b=0x= -3a(-3)^2+7(-3)+b=09a+b=21----(2)×94a+9b= -42----(1)81a+9b= 1894a+9b= -42by substrates 77a= 231a=231/77=3putting the value of a in eq24a+b=214×3+b=21b= 21-12=9 |
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| 34. |
InAABC, the perpendicular AD from A to the bisector BD of the LABC meets the baseIn A16.(produced if necessary) at E. Prove that AD = DE. |
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Answer» thanks |
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| 35. |
Inf Fig.9.ABCD is a square. Mis the mid-point of AB andPQ I CM meets AD at P and C'B produced at QProve thata) CP-CQ |
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| 36. |
Example 6: ABCD is a parallelogram. The circle through A, B and Cintersects CD (producedifnecessary) at E. Prove that AE AD. |
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Answer» thanks! |
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| 37. |
the sum of two opposite angles of a parallelogram is 130°.find the measure of each of its angles |
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Answer» 115 is the correct answer 115 is correct answer |
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| 38. |
classmateDateand.dhu.noasuru.trmoach-the-wearingind15o |
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| 39. |
leasuie l each of its angles5. The sum of two opposite angles of a parallelogram is 130°. Find the measure of each ofits angles |
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| 40. |
135. In a rectangle ABCD, AB = 25 cm and BC = 15. In what ratio doesthe bisector of ZC divide AB? |
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Answer» Answer: 2 : 3 Step-by-step explanation: in a rectangle ABCD AB=25 cm BC=15 cm in what ratio bisector of angle C divide AB Let say CE is bisector of angle C ∠C = 90° Bisector of ∠C = 45° in Δ CEB ∠BCE = 45° & ∠B = 90° => Tan 45° = BE /BC => 1 = BE/15 => BE = 15 cm AB = 25 cm AB = AE + BE => 25 = AE + 15 => AE = 10 cm AE : BE :: 10 : 15 => AE : BE :: 2 : 3 bisector of angle C divide AB in 2:3 ratio |
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| 41. |
29. Two opposite angles of a parallelogram are (8x -2) and (63 -2x). Find all the angles of parallelogram |
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| 42. |
6. The sum of two opposite angles of a parallelogram is 130. Find themeasure of each oiits angles. |
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Answer» opposite angles of parallelogram are equal so, 2x = 130 => x = 130/2 = 65° now the other angle is y = 180-65 = 115° |
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| 43. |
2) Solve the following quadratic equations. |
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| 44. |
) Solve the following quadratic equationsfactorizationi) x2 15x +54 0(2 mar |
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Answer» X =10. or X = 5 X = 9. or X = 6 |
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| 45. |
Find the roots of the following quadratic equations by factorisation:2x2+7x+5v2 |
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| 46. |
Ex. Solve the following quadratic equations by factorisation.(1) m - 14 130 (2) 3x2 x10 0 |
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| 47. |
sample 13 : Find the roots of the following quadratic equations, if they exist, usingthe quadratic formula:dr +522-2.5.10 |
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| 48. |
If two circles intersect at two points, prove that their centres lie on the perpendiculatbisector of the common chord.3. |
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| 49. |
2. Find the values of the discriminant for thefollowing quadratic equations :(1) x2 + 7x-1=0. |
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Answer» For quadratic equation ax^2 + bx + c = 0.Discriminant is b^2 - 4ac Therefore,For equation x^2 + 7x - 1 = 0Discriminant = (7)^2 - 4*1*-1= 49 - (-4)= 49 + 4 = 53 |
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| 50. |
Check whether the following are quadratic equations: |
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Answer» (X+1)^2=x^2+1+2xx^2+2x+1=2x-6x^2+7=0hence it's a quadratic equation as highest power is 2 |
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