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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. |
75-25= |
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Answer» 75-2550is the right answer 75-25=50is correct answer 50 is the correct answer 50 is the right answer 50 is the right answer 50 is a correct answer 50 is a correct answer 50 is the right answer 50 is the right answer 50 is the right answer 50 is the right answer 50 is the right answer of this questions correct answere is 50 only 50 is the right answer |
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| 2. |
24. In the given figure, AB|| CD. If ZCAB80°and LEFC = 25°, thenCEF = ?(a) 65。(c) 45°(b) 55°(d) 75°25°80° |
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| 3. |
75+25 |
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Answer» adding 75 and 25 we get 100 adding 75 plus 25 and her answer is 100 |
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| 4. |
-240 - 25 %2B 2*75 %2B 4*75 |
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Answer» It will be 2.45according to bodmas |
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| 5. |
2*75^circ - 25 %2B 4*75 - 240 |
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Answer» 2.75 - 1.25 + 4.75 - 3.80 = (2.75 + 4.75) + (-1.25 - 3.80) = 7.5 - 5.05 = 2.45 2.75-1.25+4.75-3.80=(2.75+4.75)+(-1.25-3.80)=7.5-5.05=2.45 |
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| 6. |
3. If atte=5, find the value of cs Zote Ois 2+to |
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| 7. |
wr) o Y7”Df TTOA IVb T T4०] A 5 = dq?g, [ ~ Y IAg h‘g“)?f\ oo ig?/ozmg) OIS ‘8}'Xorve—foox o ruq«/} YopOpr Uro%? |
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| 8. |
Ifu is root154egual toat-- ther~ then -value -of P and Q.2eguation |
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| 9. |
4D and BC are egual perpendiculars to a linesegment AlB (see Fig. 7.18) Show that CD bisectsA.Fig. 7.18 |
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Answer» In ∆AOD and ∆BOC AD = BCOAD = OBC (90°)angle BOC = angle AOD (vertically opposite angle)By AAS rule ∆AOD and ∆BOC are congurent. so by CPCT OB = OA |
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| 10. |
n is to check thering for variousA walks 30m towards east direction and then turns to thenorth and walk 40m. how far is he from his originalposition.Ex.1South, East and |
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Answer» The resultant of the walk will be √(30)^2+(40)^2=50mhence he is 50m away from his original position. |
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| 11. |
2) For the network shown in the figure 2 pointsassume that the port impedance isZab= R_ L - tan-1 WRC.V(1+w2R²C2)Find the average power consumed by thenetwork when R= 10k1, C = 200nF,and i = 33 sin(377t + 22°)mA.+OLinearnetworkOT0.8 W1.7 w3.4 W6.8 W |
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| 12. |
ind the Velume ?-9,2 cmcm |
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Answer» volume of cuboid =l×b×hL=9.2cmb=6.1cmh=8cm9.2×6.1×8cm=448.96cm |
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| 13. |
Evaluate the following :- \sin ^{2} 25^{\circ}+\sin ^{2} 65^{\circ}+\sqrt{3}\left(\tan 5^{\circ} \tan 15^{\circ} \tan 30^{\circ} \tan 75^{\circ} \tan 85^{\circ}\right) |
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Answer» Sin²25°+sin²65°+√3(tan5°tan15°tan30°tan75°tan85°)=sin²25°+sin²(90°-25°)+√3{tan5°tan15°×(1/√3)×tan(90°-15°)tan(90°-5°)}=sin²25°+cos²25°+√3×1/√3(tan5°tan15°cot15°cot5°)=1+1=2 Ans. |
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| 14. |
A rectangular tank is 9m by Hm hlode tous into t threagha pipe whese cross section is scmx 3om al the rate of 1om/sln how much time ollwater be Im odeep?oatey Hows into |
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| 15. |
EXERCISESe correct alternative:stands for(b) Star To(d) InterneLocal Area NetworkWide Area Networktch processing jobs are served on the basis of:FIFORequesttrue and'F for false statement:ork makes a system fast.(b) LIFO(d) |
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Answer» LAN stands for Local Area Network Please hit the like button if this helped you. |
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| 16. |
\sqrt { \frac { 1 + \operatorname { cos } 30 ^ { \circ } } { 1 - \operatorname { cos } 30 ^ { \circ } } } = \operatorname { sec } 60 ^ { \circ } + \operatorname { tan } 60 ^ { \circ } |
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Answer» thank you very much |
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| 17. |
\frac \operatorname tan 60 \operatorname sin 60 %2B \operatorname cos 60 |
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| 18. |
If sin θcos2 θ.cos θ, then find the value of 2tan θ + |
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| 19. |
If sin θ-, cos θ, then find the value of 2tan θ + cos2 θ.4. If nth term of an A Ps (2 t 1hat e t |
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Answer» Like if you find it useful |
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| 20. |
4 ( \operatorname { sin } ^ { 4 } 60 ^ { \circ } + \operatorname { cos } ^ { 4 } 30 ^ { \circ } ) - 3 ( \operatorname { tan } ^ { 2 } 60 ^ { \circ } - \operatorname { tan } ^ { 2 } 45 ^ { \circ } ) + 5 \operatorname { cos } ^ { 2 } 45 ^ { \circ } |
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| 21. |
()(1) 27a" by-9a3 |
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Answer» -3a is the correct answer. |
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| 22. |
कि - tan 30| + tan 60° tan 30tan 30°. |
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Answer» Substitute the values √3 - 1 /√3= ------------------- 1 +√3x 1 /√3 = 2 /√3/ 2 = 2 /√3 x 1 / 2 = 1 /√3tan 30°please like the solution 👍 ✔️ |
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| 23. |
a rectangular pillar (cuboid), 10 m in height. Its base is athe cost of painting its four surfaces at the rate 50 permle of a garden of the following shape is 20 m. Find thegarden at the rate of 50 per m2 |
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Answer» 1. |
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| 24. |
(i) ff2 tan 30 cos60° tan 45o60° cot 601+ tan 301- tan2 30° |
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Answer» tan30°=1/√3tan60°=√3tan45°=1cos60°=1/2cot60°=1/√3 substitute all{2(1/√3)(1/2)}/{1+(1/√3)^2} - {1 -1/2(1/2)(1/√3)}/{1-(1/√3)^2}=√3/4 - (4√3-1)√3/2=√3(1-8√3+2)/2=(3√3 - 24)/2 |
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| 25. |
\frac{\tan 60^{\circ}-\tan 30^{\circ}}{1+\tan 60^{\circ} \tan 30^{\circ}}+\cos 60^{\circ} \cos 30^{\circ}+\sin 60^{\circ} \sin 30^{\circ} |
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Answer» tan60=√3tan30=1/√3cos60=1/2cos30=√3/2sin60=√3/2sin30=1/2put in the values√3-1/√3/1+√3*1/√3+1/2*√3/2+√3/2*1/23-1/2√3+√3/4+√3/41/√3+√3/4+√3/41/√3+2√3/44+6/4√310/4√310√3/125√3/6 rong |
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| 26. |
Show that tan 75-tan 30 Ian 75 tan 30 |
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| 27. |
2tan x15, -बनाsin 2x = 2 sinx cos x = |
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Answer» We know, one important identity , sin( A + B) =sinA.cosB + cosA. sinB let A = x B = x then, sin( x + x ) = sinx.cosx + cosx .sinx sin2x = 2sinx.cosx. Now, take RHS =>2tanx / (1+tan^2x) = 2tanx / sec^2x = (2sinx/cosx) / (1/cos^2x) = (2sinx/cosx)(cos^2x/1) = 2sinxcosx = sin(2x) = LHS |
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| 28. |
\frac \operatorname tan 60 ^ \circ - \operatorname tan 30 ^ \circ 1 %2B \operatorname tan 60 ^ \circ \cdot \operatorname tan 30 ^ \circ %2B \operatorname cos 60 ^ \circ \cdot \operatorname cos 30 ^ \circ %2B \operatorname sin 60 ^ \circ \cdot \operatorname sin 30 ^ \circ |
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Answer» tan60=√3 tan30=1/√3cos60=1/2 cos30=√3/2sin60=√3/2 sin30=1/2√3-1/√3/1+1+√3/2*1/2+√3/2*1/23-1/2√3+√3/4+√3/4=1/√3+2√3/4=1/√3+√3/22+3/2√3=5/2√3 bujhina |
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| 29. |
X.(1) Find the value of 2tan 45° + cos 30 |
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Answer» =2tan 45+ cos30=2(1) +√3/4=2+√3/4=(8+√3) /4 |
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| 30. |
(2*tan(30^circ))/(-tan(30^circ)^2 %2B 1) |
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Answer» 2Tan30°/1-Tan^30° =2×1/√3/1-(1/√3)^ =2/√3/1-1/3 =2/√3/3-1/3 =2/√3/2/3 =2/√3×3/2 =√3 =Tan60° tan 60 |
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| 31. |
Prove that sec^4(1-sin^2)-2tan^2=1 |
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| 32. |
N A e S SR T T09 xC — Zx (A[)09+ X¢ — रॉ 0)i |
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Answer» x² - 5x + 6 = 0 x² - 2x - 3x + 6 = 0 x ( x - 2) - 3 ( x - 2) = 0 ( x - 3) ( x - 2) = 0 |
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| 33. |
Expand(i) (5a + 6b) |
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Answer» (5a+6b)²=(5a)²+(6b)²+2(5b)(6b) =25a²+36b²+60ab |
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| 34. |
factorise_27a^3+8b^3-9a-6b |
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| 35. |
OIf n(A) 110, n(B) 300, n(A-B) 50, then lindgn(A UB)。af7 n(A) = 110, n(B)-300 IH.S. '09 1 |
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Answer» Thnq |
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| 36. |
हा +lo=0\7,.09)(,"3 =0- 0 N |
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Answer» Multiply the equation number 2 by 7hence3x-7y=-10-14x+7x=21now add the equationshence-11x=11x=-1put in equation 2y-2*1-3=0y-2-3=0y=5 |
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| 37. |
15. Find the median of the following distributionx 1012 14 16 18 20344 |
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Answer» please like my answer if you find it useful |
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| 38. |
2*cos(e*(c*theta))=tan(theta/2) %2B cot(theta/2) |
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| 39. |
10. Pranali and Prasad started walking to theEast and to the North respectively, from thesame point and at the same speed. After2hours distance between them was 15 2 km.Find their speed per hour.Solution:15v2 Km |
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| 40. |
Mayank travelled a distance of S0m towards North. Then he turns right and travels 65 m, then travem Northwars He further travels by turning 45° clockwise. In which direction was he travelling tirection was he travelling fi) East2) South - West3) North - East4) South - East[ |
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| 41. |
Cectric currentd. of these5. An alpha particle is moving towards west is deflected towards north by a field. The field istowards downward d.towards upwardmagnetic. What will be direction of field |
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Answer» A positively charged particle projected towards west is deflected towards north by a magnetic field. Here the direction of positively charged alpha particle is towards west this implies the direction of current is also towards west and the direction of magnetic force is towards north. |
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| 42. |
. A train of length 150 meters,moving towards north directionat a speed of 144 km/hr, cancross a 250 meters long bridgein |
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| 43. |
\cot \theta=\sqrt{3} \text { pevaluate } \frac{\csc ^{2} \theta+\cot ^{2} \theta}{\csc ^{2} \theta-\sec ^{2} \theta} |
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| 44. |
N EQUALAteRAL TRIA NGWETHAT O IS A POINTON BC HE SHOW THATBc'e ë (AC.AO) |
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| 45. |
If θ be an acute angle such thatSolve the quadratic equation (x)--5x)+6 0.4 5em Now construct another tria |
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Answer» (x*x - 5x)(x*x - 5x) - 7(x*x - 5x) + 6 = 0 (x*x - 5x)(x*x - 5x) - 6(x*x - 5x) - (x*x - 5x) + 6 = 0 (x*x - 5x)(x*x - 5x - 6) - 1(x*x - 5x - 6) = 0 (x*x - 5x - 1)(x*x - 5x - 6) = 0 x*x - 5x - 6 = 0 (x-6)(x+1) = 0 x = -1,6 x*x - 5x - 1 = 0 x = (5+√(25+4))/2 or (5-√(25+4))/2 = (5+√29)/2 or (5-√29)/2 |
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| 46. |
\cot \theta=\sqrt{3}, \text { evaluate } \frac{\csc ^{2} \theta+\cot ^{2} \theta}{\csc ^{2} \theta-\sec ^{2} \theta} |
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Answer» sorry in my last attempt I forgot to mutiply 2 with 3I deeply regret for my mistake Its,okay I'll do it right |
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| 47. |
te the hexagonal and star shaped Rangolies [see Fig. 5.53 () and (i)] by fillingthemtriawith as mauy equilateral triangles of side l cm as you can. Count the number ofnglesin each case. Which has more triangles?S cmcm5 cm/h5 cm |
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Answer» Tq |
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| 48. |
ig. 7.31|earlieABC is a triangle in which altitudes BE and CF tosides AC and AB are equal (see Fig. 7.32). Showthat4.(seetria(ii) AB AC, i.e., ABC is an isosceles triangle. |
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Answer» given BE =CF are altitude AC=AB Are equle to prove : ∆ABE=~ ∆ACF ∆ABC is isosceles Proof : |
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| 49. |
\csc ^{6} \theta-\cot ^{6} \theta=1+3 \csc ^{2} \theta \cot ^{2} \theta |
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| 50. |
\frac{\cos ^{2} \theta}{\cot ^{2} \theta-\cos ^{2} \theta}=3 |
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