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. |
1, Find the coordinates of a12. 3hand Bis(1.4)point A, where AB is the diameter of a circke whose centre is |
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and its coetfi cients32 Divide (-20) by o+3) and verify division algorithmgifα and β are the zeroes ofx2 + 7x + 12, then find the value of-+-2apr For what value of k, is -2 a zero of the polynomial 3+4x+2k |
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Answer» 2. 4. |
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| 3. |
SECTION-AWhich one of the following species of honeybee is an Italian species?a. Apisdorsatab. Apis floraec. Apisceranaindicad. Apismellife |
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Answer» option d) Apiss mellifera is the correct answer. வ் |
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2. State any two Euclid's axioms.3/ In the given figure, AP LI, PR > PQ and PS = QPProve that AR AQ.S R |
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Answer» Given: PR > PQ Apply Pythagoras theorem in triangle APQ, AP² = AQ² - PQ² ... (1) Apply Pythagoras theorem in triangle APR, AP² = AR² - PR² ... (2) From equation (1) and (2), AR² - PR² = AQ² - PQ² Given PR > PQ, So, AR² - PQ² > AQ² - PQ² AR² > AQ² AR > AQ Hence Proved |
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SECTION A1. liia and n'aie aros ofthe poiyuonial as"-6x . e, find tire values of a and c, ifm+n-10-mnAP İs langent to the circle with centro O. OA16, AP30 QĂ,x, |
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| 6. |
2. In the given figure, AP is tangent to the circle with centre O. OA- 16, AP 30 QPxfind x3016 |
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| 7. |
(İİİ)xLt-1[1 + x + x2 ++ x10]lit |
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Answer» please like my answer if you find it useful |
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| 8. |
If angle between two tangents drawn from. a point P to a circle of radius 'a, and centre 0 is 60°then prove that AP av3 |
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| 9. |
the sum of all terms of this APTwo tangents TP and TQ are drawn to a circle with centre O from an external point t6. |
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24. Which tern of the AP: 3, 15, 27, 39,...will bewill be 132 TlOTo25. Two tangents TP and TQ are drawn to a circle with centre O from an external point T. Prove that4d in n hox and mixed thorougltly. One card is drawn |
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Answer» We know that, the lengths of tangents drawn from an external point to a circle are equal.∴ TP = TQIn ΔTPQ,TP = TQ⇒ ∠TQP = ∠TPQ ...(1) (In a triangle, equal sides have equal angles opposite to them)∠TQP + ∠TPQ + ∠PTQ = 180º (Angle sum property)∴ 2 ∠TPQ + ∠PTQ = 180º (Using(1))⇒ ∠PTQ = 180º – 2 ∠TPQ ...(1)We know that, a tangent to a circle is perpendicular to the radius through the point of contact.OP ⊥ PT,∴ ∠OPT = 90º⇒ ∠OPQ + ∠TPQ = 90º⇒ ∠OPQ = 90º – ∠TPQ⇒ 2∠OPQ = 2(90º – ∠TPQ) = 180º – 2 ∠TPQ ...(2)From (1) and (2), we get∠PTQ = 2∠OPQ |
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| 11. |
SELF-ASSESSMENT TESTminutesMaximum marks aig. 8.86, the pair of tangents AP and AQ drawn from an external point. A to a drdetre O are perpendicular to each other and length of each tangent is 5 cm. Fiad the radcircle. |
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| 12. |
ShortAISWU2 & 4 MarksQ. 1. Show that the tangents drawn at theend points of a chord of the circle make equaln angles with the chord.(U. P. 2007, 11)tSolution. Given : AB is a chord of circleeiniC(0, r) and AP and BQ are the tangents. |
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Answer» Let AB be a chord of a circle with center O and let AP and BP be the tangents at A and B.Let the tangent meet at P.Join OP Suppose OP meets AB at C.To prove : ∠PAC = ∠PBCProof : In ΔPAC and ΔPBCPA = PB [Tangents from an external point to a circle are equal]∠APC = ∠BPC [PA and PB are equally inclined to OP]PC = PC [Common]ΔPAC ≅ ΔPBC [SAS Congruence]∠PAC = ∠PBC [C.P.C.T] |
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| 13. |
TA and TB are tangents to a circle withO from an external point T. OTcircle at point P. Prove that AP bisects the angleangle TABcentre |
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Answer» GivenTA and TB are tangent to a circle with centre O from point TConsider, ΔOAT and ΔOBTHere, OA = OB [Radii of circle]OT = OT [Common side]TA = TB [Tangents from an external point to a circle are equal in length]∴ ΔOAT ≅ΔOBT [By SSS congruence criterion]Recall that exterior angle of a triangle is equal to sum of interior opposite angles.Hence in ΔAPT, ∠APO = ∠a + ∠bIn ΔAPO, OA = OP [Radii of circle]∠OAP = ∠OPA = ∠a + ∠b ... (1) [Angles opposite to equal sides are equal]From the figure, ∠OAP=90°- ∠b ... (2) [Since radius is perpendicular to tangent]From equations (1) and (2), we get∠a + ∠b = 90° - ∠b⇒ ∠a = 90° - 2∠b … (3)In ΔOAT, ÐOAT + ÐAOT + ÐATO = 180° [Angle sum property of triangle]90° + ∠a + ∠b+∠c+=180°∠a+∠b+∠c=90°90°- 2∠b + ∠b + ∠c=90°- ∠b + ∠c=0∠b = ∠cHence, AP bisects ∠TAB |
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| 14. |
Show that two tangents can be drawn from the point (9, 0) to the circle xy16; also find theequation of the pair of tangents and the angle between them. |
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| 15. |
In the givem figure, P is a point on the side BC of AABC. Prove that (AB + BC + AC)> 2AP |
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| 16. |
1. Determine the limits between which n must lie in order that theequation2ax(ax +nc)+(nt -2)0 |
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| 17. |
(s)=If f (x) = |cos x-sin x| , then f'1)VE2)-v3)04) does not exists |
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Answer» As mod is always positiveso answer will be√2 |
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If nC,-"C , find nC2?2* |
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Answer» if ⁿCₓ = ⁿCᵤ then, x + u = n ] Here, ⁿC₈ = ⁿC₂ so, according to results , n = 8 + 2 = 10 hence, n = 10 |
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| 19. |
Subtract the second term from first term.(i) 8x, 5x(ii) 5p, 11p(iii) 13㎡, 2㎡ |
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Find the value of(i) 25% of 720(j), 12% of 18.6 kg(ii) 125% of 80 |
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| 21. |
frequency10233137. The ratio of 7th to the 3rd term of an APis 12:5. Find the ratio of 13th to the 4th term.QR |
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Answer» Like my answer if you find it useful! |
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| 22. |
The ratio of 7th to the 3rd term of an APis 12:5. Find the ratio of 13th to the 4th term.ORind five numbers in Arithmetic progression whose sum is 25 and the sum of whose squares4x4 16 |
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Answer» Formula of n th term in AP = a+(n-1)d Where, a is the first term n is the term number d is the common ratio Now when n = 7So, a+(7-1)d=a+6d When n = 3 So, a+(3-1)d=a+2d Since we are given that the ratio of 7th to 3rd term of an AP is 12:5 ⇒(a + 6d)/(a + 2d) = 12/5⇒5a+30d=12a+24d⇒6d = 7a⇒a=6d/7 Now 13th term =a+(13-1)d =a+12d = 4th term =a+(4-1)d =a+3d = Now ratio of 13th and 4th term= (a+ 12d)/(a + 3d)= (6d/7 + 12d) / (6d/7 + 3d)= (90d/7) / (27d/7)= 90/27= 10/3 Thus the ratio of the 13th and 4th term is 10:3 |
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| 23. |
The sides AB and CD of a cyclicquadrilateral ABCD are producedto meet at P. The sides AD and BCare produced to meet at Q. If<ADC = 85° and <BPC = 40°, then<BAD is equal to |
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Answer» in triangle 🔺 ADP, A+D+P=180THUS, A=180-(85+40) A=180-125 A=55 |
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| 24. |
ABCD is a quadrilateral in which the bisectors ofmeet DC produced at Y and BA produced at X respectively8,A and < C |
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Answer» In the ΔABY: ∠Y + ∠A /2 + ∠B = 180°In the ΔDCX: ∠D +∠C /2 + X = 180° Add them to get: (∠A +∠C)/2 +∠B +∠D +∠X +∠Y = 360° we know∠A + ∠B +∠C + ∠D = 360° From these two equations , we get: angle X+ ∠Y = (∠A+∠C) /2 |
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19)ABCD is a cyclic quadrilateral. Sides AB andDC produced meet at point E; whereas sidesBC and AD produced meet at point F.If LDCF : LF: LE3 5 4, find theangles of the cyclic quadrilateral ABCD |
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| 26. |
. In the given figure, PQRS is a parallelogram. The bisectors of ZPQR andzQPS meet PS produced and QR produced at X, Y respectively. Showthat PQ = XY |
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| 27. |
11Na(IIT JEE 2007) |
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Answer» please specify the question in english |
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| 28. |
The function f : [0, 3]-> [1, 29], defined by f(x)-2x3-15x2 + 36x + 1, is(A) one-one and onto(C) one-one but not onto(B) onto but not one-one(D) neither one-one nor onto[IIT-JEE 20 |
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| 29. |
what will be the number of real zeroes of the polynomial x^2+1 |
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Answer» Draw a rough diagram of a solid showing the combination of a cone and cylinder,whose base radii are same. |
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| 30. |
Theorem 6.2: If a transversal intersects two parallel lines, then each pair ofalternate interior angles is equal. |
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Answer» thanx.... 😊 It was really helpful... ☺ |
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| 31. |
Theorem 6.2 : If a transversal intersects two parallel lines, then each pair ofalternate interior angles is equal |
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| 32. |
10.Prove that if a transversal intersect two parallel lines, then each pair ofalternate interior angles is equal. |
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| 33. |
1. Find any two irrational numbers between 0.1Board Term I, 2014]and 0.12. |
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| 34. |
Prove that thè diagonals of a square are equal andperpendicular to each other [Board Term 11 2013] |
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| 35. |
8. If the quadratic equation, (1 b' 2abcx(c -m2) 0, in x has equal roots, prove that[Board Term-2, 2014]c2m2(1 + a) |
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| 36. |
Find the number of plates. 1.5 cm in diameterand 0.2 cm thick, that can be fitted completelyinside a right circular cylinder of height 10 cmand diameter 4.5 cm.[Board Term-2, 2014] |
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Answer» Vol. of cylinder = πr^2h= 22/7*(4.5/2)^2*10= 22/7*5.0625*10= 159.107143 cm^3vol.of plate = π*r^2*h= 22/7*(1.5/2)^2*0.2= 22/7*0.5625*0.2=0.35357143 cm^3No.of plates = vol.of cylinder/vol.of plates= 159.107143/0.35357143= 450 plates. nhi samajh mai aaya |
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| 37. |
12. Show that x is rational, if:(i) x2 = 16 (ii) x2 = 0.0004 (iii) x2 = 10 |
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Answer» a)If x²=16 then x=±4b)If x²=0.0004 then x=±0.02c)If x²=1(7/9) that is x=16/9 then x=±(4/3).Hence all the values of X are rational. |
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The 24th term of an A.P. is twice its 10th term. Show that 72nd term is 4 times its15th term.Tind the number of natural numbers between 101 and 999 which are divisible22 |
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| 39. |
7.Show that the polynomials given below have no real zeroes:(ii) x2 +6x+ 10(2) x2- 2x +5 |
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Answer» 1) x^2-2x+5It is of form Ax^2+Bx+c where A= 1, B=-2, C=5B^2-4AC= (-2)^2-(4*5)=4-20=-16the value is negative , hence it has no zeroes2) x^2+6x+10It is of form Ax^2+Bx+c where A= 1, B=6, C=10B^-4AC= 6^2-(4*1*10)=36-40=-4the value is negative , hence it has no zeroesLike if you find it useful |
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| 40. |
6. In a parallelogram PQRS, PQ 12 cm and PS 9 cm. The bisector of Pmeets SR in M. PM and QR both when produced meet at T. Find thelength of RT |
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| 41. |
Find the difference between simple interest and compound interest on 20000t 8 р.с.р.а. |
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Answer» For 1 year both simple interest and compound interest are same.So the difference will be 0 |
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| 42. |
i. The elevator shown in figure (5-Es) is descending withan acceleration of 2 nmis'. The mass of the block A is05 kg. What foree is ekerted by the block A on theblock B:TFigure &-8S |
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| 43. |
an Ap consist of 50 terms of which 3rd term is 12 and the last term is 106 find the 29th tearm |
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| 44. |
Calculate the distance between two electrons if the force one exerts upon the other equals the5.06 m)weight of the electron.1.6 x 1019 C, m = 9 x 1028 g |
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| 45. |
मुकेश ने एक बाइक 20000 रुपये के अंकितमूल्य पर क्रमशः 10% और 15% बढ़ती छूट केबाद खरीदी। 700 रुपये बीमा और मरम्मत मेंव्यय किये। फिर उसने यह बाइक 20000 रुपयेमें बेच दी। लाभः प्रतिशत ज्ञात कीजिए। |
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Answer» 10000 is the right answer 10000 is the correct answer 10000 is the right answer 10,000 is the correct answer 10000................... |
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| 46. |
If a transversal intersects two parallel lines, then which of the pairs of angles is equal? |
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Answer» ∠1 = ∠4 ( vertically opposite angle)∠1 = ∠5 ( corresponding angles)∠4 = ∠5 ( alternative angles) Similarly∠2 = ∠3 ( vertically opposite angle)∠2 = ∠6 ( corresponding angles)∠3 = ∠6 ( alternative angles) |
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| 47. |
1000 %2B 20000 |
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Answer» 1000 add 90000 we will get 91000 the answer of this question is 1000+90000=91000 91000 is the correct answer 91000 is the correct answer 91000 is the correct answer of the given question 1000+90000=91,000 very easy to do it When we 1000+90000 we get 91000 |
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| 48. |
afrifdl20000 for 2 years and 2 months at 3% pa. compounded annually |
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| 49. |
Find the compound interest onRs. 20000 at 10% per annum for 3 years. |
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| 50. |
A transversal intersects two parallel lines. If the measure of one of the angles is40°, then find the measure of its corresponding angle. |
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