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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. |
Q.1 In the given figure, OP and OQ are opposite rays, fnd x.rh(x+20 |
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Answer» according to the figure (x+10) + x + (x+20) = 180=> 3x +30 = 180=> 3x = 150=> x = 150/3 = 50° thx I have a question may I ask u ask in community |
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| 2. |
1a) The radius of a circle is 10 cm. Its one chord is 16 cm long, Find theperpendicular distance of this chord from the centre |
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Answer» Given that, Radius of circle (OA) = 10cm Chord (AB) = 16cm DrawOC⊥AB We know that The perpendicular from centre to chord bisects the chord The perpendicular from centre to chord bisects the chord ∴AC=BC=16/2=8cm Now inΔOCA, by Pythagoras theorem AC^2+ OC^2= OA^2 =>8^2+ OC^2= 10^2 =>64 + OC^2= 100 =>OC^2= 36 => OC^2= 36 =>OC =6cm |
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| 3. |
DOMS Page No.Date /Om log (or ty) at Clog s t legy]thenprove thatez tys-txy |
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| 4. |
9+69+23 |
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Answer» 9+69+23 = 78+23 = 101 |
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| 5. |
the required distance is 4 V2 units.LEs Find the distance of the line 4x +7y+5 Ofrom the point (1, 2) along the line 2x-y oINCERT |
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| 6. |
जे 532 : 6 : : 556 : 4 : 923 : 8 : 492 :(1) 17 2 N1(3) 16 (4) 13 ¥ |
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Answer» it will be 11 thnks |
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| 7. |
Find the number of terms in this polynomial.1096 + 492 |
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Answer» The number of terms in this polynomial is 2. the no of terms in polynomial are 2 |
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| 8. |
0451492-21 gt-12 |
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| 9. |
a milkman has 35.500l milk.he gives 9.250 l to one customer and 13.300l to a third customer. how much milk does he have now |
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Answer» Please check your question once! |
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| 10. |
c)lzd) T217Find the value for x2x 3x-10)°(a) 43(c) 18.5(b) 23°(d) 37o4x-17) |
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| 11. |
A car is priced at $7200. The car dealer allows a customer toPay a one-third deposit and 12 payments of $420 per monthHowMuch extra does it cost the customer? |
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| 12. |
5,70% of students in a school are boys. If the number of girls is 240, how many boys areschool? |
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| 13. |
12. Ranjeeta scored 492 marks out of 600 whereas her brother scored 435 marks out of 500. Whoperformed better ? |
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Answer» Percentage got by Ranjeeta = 492/600 x 100 = 82% Percentage got by Ranjeeta's brother = 435/500 x 100 = 87% Therefore, Ranjeeta's brother performed better. |
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| 14. |
Solve: sin7x + sin 4x + sinx = 0 and 0 < x < |
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| 15. |
००7... एड 295 |
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Answer» sin30=1/2 Tan45=1 cosec60=2/√3 sec30=2/√3 cos60=1/2 cot 45=1hence 1/2+1-2/√3/2/√3+1/2+1=3/2-2/√3/4+√3/2√3+1=3√3-4/2√3/4+3√3/2√3=3/√3-4/4+3√3 |
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| 16. |
41ABCD is parallelogram. Δ DEC is drawn such that BE--AE. Sum of areas ofan DE and ABEC is3F Ea) ar (ABCD)b)ar (ABCD) ar(ADEC)ar (ADEC)ARC 33 then ZCED isal to AR I |
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| 17. |
29.89222What is the value of x in the circle?39826922889399233833938958838633183333333333393395913029525 |
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Answer» Angle = sector/radius65= x/rx= 65r |
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| 18. |
the original ne of bye and is te les 5 s |
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Answer» After 6 boys were admitted and 6 girls left, the total number of students remained the same. Let X be the original number of girls and y be the original number of girls. Then, [(X + 6)/(X+ y) - (X / X+Y) ] * 100 = 75-60=> 600/(X+y) = 15=> X+y = 40 So, original number of boys = 60% of 40= (60/100)*40= 24. original number of girls = 40-24 = 16. Please hit the like button if this helped you. |
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| 19. |
To log on ty) at Clog et lesthen prove that p2tys try |
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| 20. |
29. The radius of a circle is v2 cm. Thisdivided into two parts by 2 cm long chord. Thenprove that an angle of 45 is subtends atpoint of major segment by this arc. |
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| 21. |
Full form of www |
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Answer» www full form of wwwis world wide web wrong answer |
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| 22. |
APTITUEMPLYSIMPLIFICATIONDirection: In the question given below, if the givenam, Phathematical symbols are changed from '+' to ':', '-o'x', '', to '- and from 'x' to '+', then choose yourinswers from the following options.544.67 x 119 + 17 - 27 - 259 = ?[DMRC-Customer Relationship Assistant (CRA) Exam, 2016](a) -13(6) -3(c) 4(d) 7xam, 2 |
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Answer» the answer will be -3 |
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| 23. |
Differentiate tan i() |
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| 24. |
1- tan x1+ tan xDifferentiatew.r.t. x. |
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| 25. |
28·P, Q, R, S are respectively the midpoints of thesides AB, BC, CD and DA of lgm ABCD. Showthat PQRS is a parallelogram and also show thatAPar(lgm PQRS) ar(lgm ABCD). |
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| 26. |
(1 + cos x)Differentiate tan-1with respect to x.sin x ) |
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| 27. |
(aris the centre ofthecircle. Given osep-130cmmProve that 22QLR + ZQMR 295130(aon eeds a tloo |
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| 28. |
Limit)x-Wsin 4xsin7x43.Evaluate ( |
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| 29. |
avream dignuter lesmillinga sphere ofGerm. What will bethe weetpreparedradusthe denne(3M)o |
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Answer» Volume of sphere Vs= 4/3*pi*r^3 = 4/3*22/7*6*6*6 Let length of wire = lRadius of wire = 1 cm Assuming wire in the form of cylinder Volume of wire Vw= pi*r1^2*l= 22/7*(1/2)^2*l Now, Vw = Vs22/7*1/4*l = 4/3*22/7*6*6*6l = 4*4*6*6*6/3l = 16*36*2l = 32*36l = 1152 cm Therefore, length of wire= 1152 cm |
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| 30. |
CIssdISPUSfourth quadrant.5. If the height and radius of a cone of volume V are doubled, find the volume of the cone.Solution. If V. rand h denote the volume, radius and height of the cone then |
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| 31. |
wwwwwwww* ** |
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| 32. |
5. If the height and radius of a cone of volume V are doubled, find the volunSolution. If V, r and h denote the volume, radius and height of the cone the |
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| 33. |
04If the height of a cone of volume V is doubled, find the volume of the cone. |
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| 34. |
17.Differentiate x cos-1. |
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| 35. |
Differentiate r cos x from the first principle. |
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| 36. |
Differentiate x sin r with respect to .. |
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| 37. |
Differentiate tan(2x +3) w. r. t. x. |
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| 38. |
Differentiate sin (2r1-2 with respect to r if |
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| 39. |
A192 शत न uboicl Covdathg Some et v,F of W«. W w B2 wedor, The levd = W बैडदि - 9८ दिन नि 6८ ८ ४. सी g नट्ेट्ैस्टालदनिक+ e wlhit o है |
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Answer» Volume of water displaced = Volume of sphere=> Area* Height increase = (4/3)*(22/7)*r*r*r=> (297/7)*2 = (4/3)*(22/7)*r*r*r=> r^3 = 20.25=> r = 2.725 cm |
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| 40. |
Differentiate (cos x)sin xw. r. t. (sin x)cos x |
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Answer» please sir give me like |
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| 41. |
3√16+3√54+3√192+3√372 |
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| 42. |
Differentiate w. r. t. x=x to the power x to the power x |
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| 43. |
3√16+3√54+3√192+3√375 |
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| 44. |
1-nta nu.sec.. |
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Answer» it is very simpple |
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| 45. |
9. In the adjoining figure,ABCD is a trapezium inwhich ABIl DC. If ZA = 55°and <B = 70°, find Cand ZDCl |
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| 46. |
70°In the given figure, O is centre of circle. QPR 70° andm(arc PYR) 160°, then find the value of each of the following:(a) m(arc QXR)(b) ZQOR(c) PQR |
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Answer» (a) m(arcQXR)=140°----(By inscribed angle theorem) (b) angle QOR=140°----(By definition of a minor arc) (c) anglePQR=80°____(By inscribed angle theorem) |
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| 47. |
ABCD is a kite and LA-a. Ite and <A = <C. If LCAD-70°, <CBD65°, find : (a) 〈BCD(b) LADC. |
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| 48. |
E. which number in each of thefollowing pairs is to the rightof the other on the number fine:(1)0,4 |
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Answer» what is your question 4 is right to 0 on number line 4 is right to 0 on number line 4 is right to 0 on number line 4 is right to 0 on number line 4 is right on zero number line 4 is right to the 0 on number line 4 is on the right to 0 |
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| 49. |
Factorise 49 a^{2}+70 a b+25 b^{2} |
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| 50. |
vowels. In the given igure, O isthe centre of the cirde The area of sector OAPBtimes of the area of the circde. Find18the value ofx |
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