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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. |
(a)700(b)70, (c)7ă(d)ziarandl |
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Answer» 4900 ko hum 70 × 70 lekh sakte hai to √(4900) = 70 hoga. b) option sahi hai |
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| 2. |
5) Find the continued product of x + y, x-y, x2 + xy + y2, X2-xy + y2 |
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| 3. |
12. In the adjoining figure, ABC is a triangle inwhich AB AC If D and E are points on ABand AC nspe tively such that AD" АЕ, showthat the points B, C. E and D are concyclic |
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| 4. |
Prectice set 3.3In figure 3.37, points G, D, E, Fre concyclic points of a circle with1.centre CECF = 70°, mt(arc DGF) = 200。find m(are DE) and m(are DEF)Fig. 3. |
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Answer» sum of angles around the point should be 360° => m(arc DE) +70 + 200° = 360°=> m(arc DE ) = 360-200°-70°=> m(arc DE) = 90° so m(arc DEF) = 90°+70° = 160° |
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| 5. |
15.Two circles intersect at points A and B. From a pointP on the common chord BA produced, secants PCDand PEH are drawn one to each circle. Prove that thepoints C, D, H, E are concyclic. |
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Answer» 1 The quadrilateral CDHE is required to be proved to be a cyclic quadrilateral. CDH+CEH=180=DCE+EHD. Join CB and AD, and join AH and BE. From this construction we get two pairs of similar triangles: APD and BPC, and APH and BPE, because of the common angle at P, and equal angles PCB=DAP, PAH=BEP (angles in the same segment). We can therefore write: PD/PB=PA/PC=DA/BC (triangles PCB and DAP) and PB/PH=PE/PA=BE/HA (triangles PAH and BEP). From this we get: PB.PA=PC.PD=PE.PH. So PC.PD=PE.PH PC/PE=PH/PD. Therefore, triangles PCE and PHD are also similar because P is theincluded common angle. PDH=PEC. But CDH=180-PDH (supplementary angles on a straight line), so CDH=180-PEC (PEC is the same angle as CEH). These are opposite angles of the quadrilateral CDHE, and this is a definitive property of cyclic quadrilaterals, so CDHE is cyclic. The other two angles must also be supplementary because the angles of a quadrilateral add up to 360 degrees. |
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| 6. |
2) In figure, points G,D,E,Fare concyclic point of acircle with center CECF 700 M ( arc DGF ) 2000Find m (arc DE) and m (arc DEF) |
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Answer» given by:- 》∠ECF = 70°,》m(arc DGF) = 200°》m(arc DE) = 360°-[m(arc DGF) + ∠ECF]》= > 360° - (200+70)》=> 360 - 270》m(arc DE) =90° ans》m(arc DEF) = m(arc DE) + ∠ECF》=> 90+70》m(arc DEF) = 160°ans》hence,》m(arc DEF) = 160°》m(arc DE) =90° |
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| 7. |
If the line y- V3x +3=0 cuts the curve y2 = x+2 at A and B and point on the line P is (V3,0) then |PA.PB| is |
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Answer» We may write y-√3x+3=0 as y-0/√3/2=x-√3/1/2=r....(1) Substituting x = √3+r/2, y=√3r/2 in y^2=x+2, we get 3r^2/4=r/2+√3+23r^2-2r-(4√3+8) =0AP.AQ=|r1.r2|=(4√3+8)/3=4(√3+2)/3. |
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| 8. |
Find three rational number between /2 and 3. |
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| 9. |
e 8 If yz-1, show that:-1 |
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Answer» (1 + x + y^-1)-1+(1 + y + z-1)-1+ (1 + z + x-1)-1= (1 + x + 1/y)-1+(1 + y + 1/z)-1+ (1 + z + 1/x)-1= (1 + x + xz)-1+(1 + y + 1/z)-1+ (1 + z + 1/x)-1[xyz = 1, or xz = 1/y]= (1 + x + xz)-1+{(z + yz + 1)/z}-1+ {(x + xz + 1)/x}-1= (1 + x + xz)-1+{(z + 1/x + 1)/z}-1+ {(x + xz + 1)/x}-1[xyz = 1, or yz = 1/x]= (1 + x + xz)-1+{(xz + 1 + x)/xz}-1+ {(x + xz + 1)/x}-1= 1 / (1 + x + xz) + xz / (1 + x + xz) + x / (x + xz + 1)= (1 + xz + x) / (1 + x + xz)= 1 |
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| 10. |
के.o R Photatririr—1f 36 Gyey S Te R — |
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Answer» Total outcomes = 6Favourable outcome = 3Probability of getting a composite number = 3/6 = 1/2 |
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| 11. |
11. a) By how much is a -2ab- b more than 2a+2ab - 2b?b) By how much is 2(a+b) less than a? +2ab +2 Multinhr (17 |
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Answer» thank you |
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| 12. |
Three normals are drawn from the points (14,7) to the curve y2-16x-8y = 0 find the coordinatesof the feet of the normals. |
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Answer» Let the coordinates of foot of perprndicular be (x1,y1) Differentiating the equation to find the slope of the normal, => dy/dx = 8/(y-4) slope = -(dx/dy) hence we can find the equation of normal in terms of x1 and y1. then passing through the (x1,y1) thus we find a cubic equation on solving which we get, y1= 0, 16 , -4 thus corresponding x1 = 0, 8 , 3 |
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| 13. |
Find the equation of normals to the curvey-X3+2x +6 which are parallel to the line x +14y +4-0 |
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| 14. |
find three rational number between-8/5and8/5 |
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Answer» sorry the photo is clicked by mistake 80/50 80/49 80/48 this is a great answer -8/5, -7/5, -6/5, -5/5, -4/5........8/5 |
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| 15. |
Nita starts with the number 27 andkeeps subtracing 6 from it. Findthe number she will arrive atcarries out the subtactionshe6 time |
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Answer» 6 subtracted 6 times from 27 = 27-(6*6) = 27-36 = -9 |
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| 16. |
28. If x +yz 0 show that, |
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Answer» We know x³ + y³ + z³ - 3xyz= (x+y+z) (x²+y²+z²-xy-yz-zx)Given x + y + z = 0So,x³ + y³ + z³ - 3xyz = 0x³ + y³ + z³ = 3xyz thank you |
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| 17. |
25x by x85x^2 yz by 5xyz |
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| 18. |
34. The foot of the perpendicular of the point (2, 3) in x-axis1) (2,0) 2) (0,- 5) 3) (7,-4) 4) (-2,-3)is |
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Answer» what is your question option (1) that is (2, 0) is correct answer Mark as best option 1 is the correct answer of the given question |
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| 19. |
the given fig. BAC is a line. Find 'a(3a 20)3aăBAC is a line. Find y'8 |
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| 20. |
Find the equation of the perpendicular dropped from the point (2,-3) on the line forjoining the points (3, 4) and (-4, 5). |
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Answer» Slope of the line formed by joining (3,4) and (-4,5) is (5-4)/(-4-3) =1/(-7)=-1/7So, slope of the perpendicular will be 7Let the equation be y=mx+bput m=7y=7x+bIt passes through (2,-3)-3=7(2)+b-3-14=bb=-17So, equation is y=7x-17 |
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| 21. |
15. In fig. BAC is a line Pind y(8y + 30)316. In the given fig, BAC is a line. Find '. Hence find ZCAB and BAD55°(3x - 5)x+20 |
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| 22. |
5. In the given figure, BD DC andCBD30°, find <BAC. |
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| 23. |
3./fn Fig. 6.41, ifABI DE, < BAC # 35° and < CDE53°, find < DCE. |
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Answer» please post figure 6.41 |
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| 24. |
3. In Fig. 3.41, ifAB I DE, BAC 35° and ZCDE-53°, find DCE |
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| 25. |
Find a point on y-axis from where graph of linear equation x - 5 y = 3 will pass |
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| 26. |
In Fig. 6.41, ifAB II DE, 2 BAC = 35° and < CDE = 53°, findDCE. |
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| 27. |
8 Three normals to y' 4x pass through the points (15, 12). Show that one of the normals is givenby y #x-3. Also find the equation of other normals. |
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| 28. |
13. In the given fig. BAC is a line. Find y(2y-20)(2y + 40*3 |
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| 29. |
Evaluate:\frac{\cot 54^{\circ}}{\tan 36^{\circ}}+\frac{\tan 20^{\circ}}{\cot 70^{\circ}}-2 |
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Answer» cot (90-54) = tan 36tan(90-20)= cot70so tan 36/tan36+tan20/cot20-22-2=0 |
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| 30. |
9. In which of the following quadrants does the points P(3,6) lie?a.b. 11c. IIId. IV |
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Answer» a) first quandrantbecause x=3 y=6 both are positive a) first quandrant because x and y are both positive a.)First quadrant because (x=3)and(y=6)are both positive (a). First quadrant because (x=3) and (y=6) are both positive Option (A)In first quadrants does the points P(3,6) lie. |
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| 31. |
Find the value of:\frac{-\tan \theta \cdot \cot \left(90^{\circ}-\theta\right)+\sec \theta \cdot \csc \left(90^{\circ}-\theta\right)+\sin ^{2} 35^{\circ}+\sin ^{2} 55^{\circ}}{\tan 10^{\circ} \tan 20^{\circ} \tan 30^{\circ} \tan 70^{\circ} \tan 80^{\circ}} |
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| 32. |
5.In which quadrants y co-ordinates are negative? |
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Answer» The y coordinates are negative in 3rd and 4th quadrants. thank you😊 y co-ordinates are negative in 2 nd and 3 rd quadrants |
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| 33. |
The sign of the trigonometric ratios in different quadrantsare as under |
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| 34. |
11. What is ordinate ? |
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Answer» Theordinateis the y-coordinate of a pointon the coordinate plane. |
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| 35. |
convert 2dm 5cm to dm |
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Answer» 2.5 dm |
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| 36. |
cast of134I0. Find the breadth ot the heama issn 18m lang witho |
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| 37. |
CATEHoH.27x4- 13x625-5X2 |
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Answer» using BODMAS= (108 -- 78 ) / ( 5 × 2)= 30 / 10= 3 (27×4-13×6)/(25÷5×2)=(108-78)/(5×2)=30/10=3.....(Ans) using BODMAS = (108 - 78) / 10= 30/10= 3 answer of this question is 3 (108 - 78) / (5×2)30 /103 answer (27*4-13*6)/25÷5*2(108-78)/25÷1030/2.5=12 ans 3 is right answer of this question =(108- 78) /(5X2)=30/10=3 यहाँ BODMAS का नियम लगाने पर -108-78/5*2=30/10=3 is answering 30_10 answer correct 30/10 =3 is answer cheak my answer 108_78/10=30/10=3 iska answ 3 hoga Good question and I like this question 3 is a right ans for your question using BODMAS= ( 27×4-13×6) / (5×2)= (108 - 78 ) /10= 30/10= 3 (Ans.) 3 is the correct answer 27*4-13*6=18250/10=3 answer=(27×4)-(13×5) /5×2=108-78 / 5×2=30/10=3 answer correct answer of this question is 3 (27×4-13×6)/(25÷5×2 ) =(108-78) /(5×2) =30/10=3 (answer) 27×4-13×6/25÷5×2 108-87/25÷10 21/5/242/5 108-78 / 5x2 =30/10=3BODMOS formula 108-78/25÷10108-78/2.530/2.512 (27×4-13×6) /(25÷5×2) =(108-78)/(5×2)= 30/10=. 3 =(108-78)/5×2=30/10=3 3 is the right answer of this question 4 is the right answer 3 is the right answer |
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| 38. |
on 1) oRe T 3R S Itan 20 tan 20 tan 54 tan 55 |
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Answer» tan(π/20) tan(3π/20) tan(5π/20) tan(7π/20) tan(9π/20) = 1. To this end:tan(π/20) tan(3π/20) tan(5π/20) tan(7π/20) tan(9π/20)= tan(π/20) tan(3π/20) tan(5π/20) cot(π/2 - 7π/20) cot(π/2 - 9π/20)via cofunction identity = tan(π/20) tan(3π/20) tan(5π/20) cot(3π/20) cot(π/20)= tan(π/20) tan(3π/20) tan(5π/20) * 1/tan(3π/20) * 1/tan(π/20)= tan(5π/20)= tan(π/4)= 1. |
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| 39. |
The points (other than origin) for which abscissa is equal to the ordinate will lie in(A) I quadrant only (B) I and II quadrants(C) I and III quadrants (D) II and IV quadrants |
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Answer» B is ans Please mark best 👈🙏 b is the correct answer of this question b) is the right answer of the following B is the correct answer |
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| 40. |
ВСindl theAABC? |
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Answer» thnx for sending me this answer |
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| 41. |
indl the asauita tional force hctueen thfnas s 2x1022 kg Qnd.-5kg--and he cl is andbelwcen thenis 38X10 5 mams 38 X10m |
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Answer» wrong |
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| 42. |
5६6७ 2 (0)$४6 2-फठिप056 2 (8)SL6 2 (V)ehglp1Bp Ah Dhid Ry (4 20 312] के 228RIS 1db %S | | ६8 1-19 1फक %0ट [8 Ikl128 20 Dhid [ 0TL 2 blo 2k कफ 1%फ 9५०IJN\ 00६ 0)08८ (3)हद पद.छठ ४)1:91 हि।2ि | हि 5६1 31019 %५५४ 19% | |
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Answer» 45 percentage = 1351percentage = 135/45°1percent = 3°100 percentage होगा 3*100=300 जब 80 परसेंट =7201 परसेंट=720/809105 परसेंट=105*9=945 rupee |
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| 43. |
The area of a circle of radius ris 10 sq.cm. Then the area of a circle of radius2ris (20/30/40) sq.cm.2 |
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Answer» area=πr^2hencer=√10/πcmhence2r=2√10/πcmArea=πr^2=π*4*10/π=40cm^2 |
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| 44. |
Sub: 18m 2dm 5cm from 43m 9cm |
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| 45. |
Sub tract:-ar y foom 70 - 11 from & |
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Answer» -7/9 - 4/9 = (-)11/98/3 - (-)11/6 = 27/6 -11/9 &. 27/6 is correct answer -11/927/6. is the right answer |
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| 46. |
\begin { equation } 2 y+\frac{5}{3}=\frac{26}{3}-y \end { equation } |
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Answer» 2y+5/3= 26/3 -y2y+y=26/3-5/33y=21/3=7y=7/3 |
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| 47. |
ofIn Fig. 9.24, ABC and ABD are two triangles on »xthe same base AB. Ifline- segment CD isby AB at O, show that ar(ABC)-ar (ABD). |
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Answer» thnx |
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| 48. |
In Fig. 9.24, ABC and ABD are two triangles onthe same base AB. If line-segment CD is bisectedby AB at O, show that artABC) ar (ABD) |
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| 49. |
Two triangle having the same base and equal area lie between the same parallels. |
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Answer» ABC And ∆ADC Both Lie On The Same Base In A Way That ar( ∆ABC ) = ar( ∆ABD) to prove:-∆ABC And ∆ADB Lie Between The Same Parallel.. ie:- CD || AB construction:- Altitude CE And DF Of ∆ACB & ∆ADB.. On AB .Now According To Question It's Said That ∆ABC And ∆ABD Both lie On The Same Base ... And Both Have Equal Area. Now By Construction We Have ... CE Perpendicular To AB And DF Perpendicular To AB Now We Know That Lines Perpendicular To Same Line Are Parallel To Each Other Therefore ... CE || DF --- EQ (1) So Now It's Given That ar ( ABC ) = ar ( ABD ) Now We Know That .. Area of triangle=1/2*base*area Therefore ar(ABC) = 1/2 × AB × CE And ar(ABD) = 1/2 × AB × DF Now Since Area Of ABC = Area OF ABD Therefore We Have ... CE=DF ---- EQ. (2) Now In Quadrilateral CDEF CE = DF And CE || DF So We Know That If In A Quadrilateral If One Pair Of Opposite Sides Are Equal And Parallel Then The Quadrilateral Is Parallelogram. Therefore CDEF Is A Parallelogram ... Therefore CD || EF ( opposite Sides Of Parallelogram ) That Is .. CD||ABHence proved |
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| 50. |
a) (a2-22a +117) (a-1 |
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