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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. |
is 27 years older than her daughter Rekha. After 8 years she will be twice asa. Find their present agespresent agesGoel |
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Answer» thanks a billion |
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| 2. |
Alka is 27 years older than her daughter Rekha. After 8 years she will be twice as old as Rekha. Find theirresent ages. |
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| 3. |
Mrs Goel is 27 years older than her daughter Rekha. After 8 years she will be twice as old asRekha. Find their present ages |
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| 4. |
. Mrs Goel is 27 years older than her daughter Rekha. After 8 years she will be twice as old asRekha. Find their present ages |
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| 5. |
Mrs goel is 27 years older than her daughter Rekha . After 8 years she will be twice as old as Rekha .find their present ages. |
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| 6. |
A=\left[ \begin{array}{ll}{1} & {1} \\ {2} & {2}\end{array}\right], B=\left[ \begin{array}{ll}{1} & {1} \\ {3} & {4}\end{array}\right] \text { then find }|A B| |
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| 7. |
A=\left[ \begin{array}{lll}{2} & {-1} & {1} \\ {2} & {-3} & {5}\end{array}\right], B=\left[ \begin{array}{rr}{3} & {-3} \\ {4} & {-1} \\ {5} & {0}\end{array}\right], \text { find } A B |
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| 8. |
Theorem 10.3 : The perpendicular from the centre of a circle to a chord bisectsthe chord |
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Answer» Thanks😊😊😊 |
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| 9. |
THEOREM 3The perpendicular from the centre of a circle to a chord bisects the chord. INCERT |
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Answer» Angles OMA and OMB are both right-angles. OA is thehypotenuseof triangle OAM. OB is the hypotenuse of triangle OBM. OA = OB as both areradiiof the circle. OM is common to both triangles. Therefore, triangles OAM and OMB arecongruent(RHS – right-angle, hypotenuse, side). Therefore, the remaining sides of the triangles are equal, AM = MB. So, M must be the mid-point of AB, and the chord has been bisected. |
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| 10. |
Theorem 10.6: Equal chords of a circle (or of congruent circles) are equidistantfrom the centre (or centres). |
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Answer» Given a circle with centre O and chords AB = CDDraw OP⊥ AB and OQ⊥ CDHence AP = BP = (1/2)AB and CQ = QD = (1/2)CDAlso ∠OPA = 90° and ∠OQC = 90°Since AB = CD⇒ (1/2) AB = (1/2) CD⇒AP = CQIn Δ’s OPA and OQC,∠OPA = ∠OQC = 90°AP = CQ (proved)OA = OC (Radii)∴ ΔOPA ≅ ΔOQC (By RHS congruence criterion)Hence OP = OQ (CPCT) |
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| 11. |
Theorem 10.6 Equal chords of a circle (or of congruent circles) are equidistantfrom the centre (or centres) |
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Answer» Thanksyou |
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| 12. |
ollowingTheorem 10.6 Equal chords of a circle (or of congruent circles) are equidistantfrom the centre (or centres). |
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| 13. |
result:Theorem 10.3 : The perpendicular from the centre of a circle to a chord bisectsthe chord.o olear what is |
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Answer» Thank you |
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| 14. |
5 {HD},12. यदि =यदि 4=तो दिखायें कि[1-2-4 |- 2 || |
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| 15. |
A = \left[ \begin{array} { c c c } { 2 } & { 1 } & { 3 } \end{array} \right] B = \left[ \begin{array} { l } { 2 } \\ { 2 } \\ { 1 } \end{array} \right] |
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| 16. |
Prind thevalue of a4 HD |
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Answer» 2x+3x+10=180; 5x=170; x=170/5=34 |
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| 17. |
urroundod by a path 4 m wide. The diamA cicular flower bad is shd ls 66m what is the area of this path?(π=3.14)1 |
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Answer» thanks |
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| 18. |
Evaluate l Cos'kdsĎ/2 |
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| 19. |
If 4 tan θ= 3, evaluate l4sina-eesen19, |
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Answer» Given,4 tan = 3∴ tan = 3/4 As we know,Tan = Perpendicular / baseTan = 3/4Now,Hypotenuse = √3² + 4² = √9+16 = √25 = 5 Using this, we can find:Sin = Perpendicular / Hypotenuse = 3/5Cos = Base / Hypotenuse = 4/5 ATQ,(4sin - cos + 1)/(4sin + cos - 1)= (4 × 3/5 - 4/5 + 1)/(4 × 3/5 + 4/5 - 1) [putting values]= (12 - 4 + 5)/(12 + 4 - 5)= 13/11 hit like if you find it useful |
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| 20. |
Evaluate l dxx +1 |
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Answer» put in the limits[1-tan^-1(1)]-[0-tan^-1(0)]1-π/4-0-01-π/4 |
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| 21. |
lfa = í and b ,s, then evaluate l (1) + (111 ęłJ(a+b |
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| 22. |
1+2-3&TaT3)Find the remainder when the polynomial p(y)-y4-3уг +7y-10īs divided by (y-2). |
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Answer» By using remainder theroem, when p(y) = y⁴ - 3y² + 7y - 10 us divided by (y-2) remainder will be p(2) p(2) = 2⁴ - 3 × 2² + 7 × 2 - 10 p(2) = 16 - 12 + 14 - 10 p(2) = 4 + 4 = 8 Remainder is 8 tysm... |
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| 23. |
Pove tat the points (3, 0) (6,4) and (1,3)ore the veshtes 2 a nght d įso sceles △anal e |
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| 24. |
\frac{x^{2}-2 x}{x^{2}+2 x} \times \frac{3 x+6}{x-2}= |
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Answer» (x*x-2x)/(x*x+2x) * (3x+6)/(x-2) = x(x-2)/x(x+2) * 3(x+2)/(x-2) = 3 |
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| 25. |
x = 2 + 2 ^ { 2 / 3 } + 2 ^ { 1 / 3 } , \text { then the value of } x ^ { 3 } - 6 x ^ { 2 } + 6 x |
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| 26. |
x-2/x-2+6(x-2/x-6)=1 |
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| 27. |
Angles subtended by opposite sides of anyquadrilateral at the centre of incircle are(a) Complementary Jy Supplementary(d) ObtuseAc) 90° |
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Answer» Let ABCD be a quadrilateral circumscribing a circle with centre O.Now join AO, BO, CO, DO.From the figure, ∠DAO = ∠BAO [Since, AB and AD are tangents] Let ∠DAO = ∠BAO = 1 Also ∠ABO = ∠CBO [Since, BA and BC are tangents] Let ∠ABO = ∠CBO = 2 Similarly we take the same way for vertices C and DSum of the angles at the centre is 360° Recall that sum of the angles in quad. ABCD = 360° ⇒2(1 + 2 + 3 + 4) = 360° ⇒1 + 2 + 3 + 4 = 180° In ΔAOB, ∠BOA= 180 – (a + b)In ΔCOD, ∠COD = 180 – (c + d)Angle BOA + angle COD = 360 – (a + b + c + d)= 360° – 180°= 180°Hence AB and CD subtend supplementary angles at OThus, opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle. |
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| 28. |
: If the angles subtended by the chords of a circle at the CentreTheoremre equal then the chords are equal. |
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Answer» Let AB and CD be the chords of a circle with centre 0where angle AOB= angle CODTo prove:AB equal toCDProof: in triangle AOB and triangle COD,we haveOA = OC (radii of a circle) angle AOB = angleCOD OB = OD(radii of a circle)therefore triangle AOB congruent to triangleCOD(by SAS rule)therefore AB = CD (by cpct) hit like if you find it useful |
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| 29. |
A circle touches the sides of a quadrilateralABCD at PQRS respectively. Show thatthe angles subtended at the centre by a pairof opposite sides are supplementary |
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Answer» Step-by-step explanation: Let PQRS be a quadrilateral circumscribing a circle with centre O. Join OA, OB, OC and OD. Since the two tangents drawn from an external point to a circle subtend equal angles at the centre. ∴ ∠1 = ∠8 Similarly, ∠2 = ∠3, ∠4 = ∠5 and ∠6 = ∠7 ........ (1) Now, Sum of the angles at the centre = 360° ∠1 + ∠2 + ∠3 + ∠4 + ∠5 + ∠6 + ∠7 + ∠8 = 360° ⇒ 2(∠1 + ∠2 + ∠5 + ∠6) = 360° [From equation (1)] and 2(∠7 + ∠8 + ∠3 + ∠4) = 360° ⇒ ∠1 + ∠2 + ∠5 + ∠6 = 180° and ∠3 + ∠4 + ∠7 + ∠8) = 180° Thus, opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles of the centre of the circle. Here is your answer: ∠AOD + ∠BOC = 180° ∠AOB + ∠COD = 180° Now the area OSDR = ΔODS x ΔODR ∡1 + ∡4 + ∡5 + ∡8 = 180° ∡2 + ∡3 + ∡6 + ∡7 = 180° ∠AOD + ∠BOC = 180° (∡1 + ∡4 + ∡5 + ∡8 = 180°) ∠AOB + ∠COD = 180° (∡2 + ∡3 + ∡6 + ∡7 = 180°) |
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| 30. |
Theequationlog-log (1+x)=0maybewrittena)x-+-1-0,b)x,+x+1=0,c)as- |
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| 31. |
23. State Rolle's theorem and verify the theorem forf(x) = x2 +2x8, x€[-4,2] |
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| 32. |
The equationlog-log (1+x) = 0be written as - a)r+x-1-0,b)x2+x+1=0,mayc) |
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| 33. |
II, (4.β theequation cre-bx+ a= Q[Ans.-6]·β are the roots of equation ax 2 +bx+c=0 then what are the roots of2[Ans.-_ ,--] |
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| 34. |
9.Solve the followirg quadratic equation for x: abx2 + (b2 |
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| 35. |
[Ans. [2x + log(1 +3,2-c]] 5(b) Find the equation of the tangent to the curve y=4-5 at the point x =-2.x+1[Ans. 8x + y + 9=015Evaluate l-- |
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| 36. |
Solve the following equation:2(x+4) = 12 |
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| 37. |
Ans.Q. 5. Solve the given equation and verify youranswer:1.5x+0.3 3103 x |
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Answer» ( 1.5x + 0.3 )/3x = 3/10 1.5x + 0.3 = 3x × 0.3 1.5x + 0.3 = 0.9x 1.5x - 0.9x = - 0.3 0.6x = -0.3 x = -0.3/0.6 x = -1/2 = -0.5 |
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| 38. |
f(a)Solve the differential equation-\frac{d y}{d x}=-\left(\frac{1+y^{2}}{y}\right)[Ans. [2x + log(1 + y2) c]l |
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| 39. |
4. Multiply by repeated addition:4 x 3-4 x6-6 x 2 |
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Answer» 4×2= 4+4=87×3=7+7+7=213×6=3+3+3+3+3+3=182×5=2+2+2+2+24×3=4+4+4=125×5=5+5+5+5+5=253×9=3+3+3+3+3+3+3+3+3=274×6=4+4+4+4+4+4=246×2=6+6=12 |
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| 40. |
find the equation of the tangento the cuore x=h t-sent).y = r(1-Cost) at tat2(Ans > 2y = 2x - 4 (7-4) |
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Answer» Dy/dx=2 |
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| 41. |
13.Solve the following pair of linear equation-23x+4y = 10, 2x-2y = 2 |
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| 42. |
solve the following equation:- (1) 2Y + 5/2 = 37/2 |
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| 43. |
00. The cm 10.2 : If the angles subtended by the chords of a circle at the centrem equal, then the chords are equal. |
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| 44. |
1.Solve the following equation for z.42 + 2 |
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Answer» hit like if you find it useful thank you Bhai |
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| 45. |
abx? + (b* - ac)x âbc =0 |
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Answer» hit like if you find it useful plz tap the like button your answer is ready |
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| 46. |
3)2Solve the following equation:(x+2)(x+6x +1 |
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Answer» hit like if you find it useful |
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| 47. |
4. Solve the following equation:(a)-2(mi) = 12-(d)-4(2 + x) = 8 |
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| 48. |
abx^2+(b^2-ac)x-bc=0 |
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| 49. |
23. The quadratic equation abx^2+ acx +b(bx+c) =0has non-zero equal and rational roots. The values ofa and c respectively cannot be equal to (ab # 0)(1) 4 & 49(3) 4 & 64(2) 49 & 16(4) 8 & 49 |
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| 50. |
x^2 + axy + abx + by |
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