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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. |
the cost of fencing a square field at rupees 14 per metre is rupees 28000. find the area of the field |
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| 2. |
A rectangular grass field is 75 m * 55 m, it has a path of 2.5 m wide all round it on the outside. Find the area of the path and the cost of constructing it at Rs.2 per sq m? |
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| 3. |
lities and represent the solution in the number line:2(x-1 ) < x+5,3 (х + 2) > 2-х. |
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| 4. |
1 The radius of a circular garden is 56 m. What would it cost to put a 4-round fencearound this garden at a rate of 40 rupees per metre? |
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Answer» Radius=56mPerimeter=2πr=2(22/7)56==352m 4 rounds=4*352=1408m 1m =Rs 401408m=40*1408=56320Rs |
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| 5. |
a room of 9.5 M long and 6M wide is surrounded by a verandah 1.25 m wide calculate the cost of cementing the flow of this veranda at rupees 6.80 per metre square |
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| 6. |
ХNIUthen x =V1-X |
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| 7. |
3. What is the value of a rectangular plot of land 75.5 m long and 30.5 m broad at therate of 1000 rupees per square metre? |
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Answer» the cost would be $23015 |
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| 8. |
Identify the like terms in the following:(ii) -2xy, xy. 5yx, x'z(iv) abc. ab'c, acb. c'ab, bac. a'bc. cab一2.xyx. у, 5U"Х. х z |
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| 9. |
The following observations have been arranged in ascending order. If the median ofthe data is 63, find the value of x.50,29,32,48,х+2,72,78,84,95х, |
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Answer» write the quotients 1.(_124)division by4 |
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| 10. |
Solve:"Oх1-X,* # 0 and x #1V1-X |
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| 11. |
3.The following observations have been arranged in ascending order. If the median ofthe data is 63, find the value of x.50,29,32,48,х,х+2,72,78,84,95 |
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| 12. |
l The following observations have been arranged in ascending order. If the median ofthe data is 63, find the value of x29,32,48,50,х,х+2,72,78,84,95 nnan |
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| 13. |
A door-frame of dimensions 3 m × 2 m is fixed on the wall of dimension10 mx 10 m. Find the total labour charges for painting the wallifthelabour charges for painting I m*of the wall is 2.50 |
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| 14. |
A room 8m x 6m is to be carpeted by a carpet 2 m wide. Find the length of the carpet, required. |
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Answer» area of carpet = area of room = 8*6 = 48 m² width of carpet = 2m length of carpet = 48/2 = 24 m can u give in more steps |
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| 15. |
A door-frame of dimensions 3 mx 2 m is fixed on the wall of dimension10 m x 10 m. Find the total labour charges for painting the wall if thelabour charges for painting 1 m2of the wall is 2.50 |
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| 16. |
14. A laboratory is 9 m long, 8 m broad and 6 m high. It has 2 doors cach of size 3 m x 1.5 m and fourwindows each of size 1.5 m x 1 m. Find the cost of whitewashing the walls of the laboratory at the rateof 1.75 per sq. m |
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Answer» Area of four walls of laboratoryAr= 2 (length*height + breadth*height)= 2(9*6 + 8*6) = 2(54 + 48)= 2(102) = 204 m^2 Area of window Aw = 1.5*1 = 1.5 m^2 Area of door Ad = 3*1.5 = 4.5 m^2 Area for white washing = Ar - 2*Ad - 4*Aw= 204 - 2*4.5 - 4*1.5= 204 - 9 - 6= 204 - 15= 189 m^2 Cost of white washing = 189*1.75 = Rs 330.75 |
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| 17. |
the area of a rectangular field is 786 m2 and its length is 32 m . find:1) the breath of the field2) the cost of fencing it at rupees 20 per metre |
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| 18. |
аQX2ІІ0х8 | 8XyXY |
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| 19. |
A hall 20 m long and 15 m broad is surrounded by a verandah of uniform width of 2.5 m. The cast of flooring the verandah at the rate of Rs 3.50 per sq. Meter is |
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| 20. |
2. Zeroes of polynomial a2-11 are:(A) VII, VII11,-411none of these(D) |
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Answer» x²-11 =0=> x²= 11=> X = ±√11=> X = √11, -√11. Please hit the like button if this helped you. |
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| 21. |
(vii) If L[F()] f(s), then prove that:VIILeat F(t)] = f(s-a) |
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| 22. |
-1/2x×4/2x |
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Answer» -1/x is right answer for your question |
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| 23. |
A verandah 2 m wide is constructed all around a room of dimensions 8 mx 5 m. Find thearea of the verandah |
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| 24. |
A verandah 2 m wide is constructed all around a room of dimensions 8 m 5 m. Find thearea of the verandah |
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Answer» Area of the room= length×breadth =8×5=40 sq.mLength of the room including verandah=8+2+2=12mBreadth of the room including verandah=5+2+2=9mHence,Area of the room including verandah=12×9=108 sq.m∴,the area of the verandah is=(area of the room including verandah-area of the room)=108-40 =68sq.m |
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| 25. |
A verandahof the verandah.2 m wide is constructed all around the outside of a room of size 8 m by 5 m. Find the area |
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Answer» Area of the room= length×breadth =8×5=40 sq.mLength of the room including verandah=8+2+2=12mBreadth of the room including verandah=5+2+2=9mHence,Area of the room including verandah=12×9=108 sq.m∴,the area of the verandah is=(area of the room including verandah-area of the room)=108-40 =68sq.m |
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| 26. |
1 A verandah of width 2.25m is constructed all along outside a room which is 5m long and 3m wide. Find1.the area of verandah and2.the cost of cementing the floor of verandah at the rate of rupees 275 per m2 |
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Answer» Area of room Ar = length*breadth= 5*3 = 15 m^2 Area of outer of verandah Ao= length*breadth= (5+4.5)*(3+4.5)= 9.5*7.5 = 71.25 1)Area of verandah = Ao - Ar = 71.25 - 15 = 56.25 m^2 2)cost of cementing floor= 56.25*275 = Rs 15468.75 |
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| 27. |
A verandah 2 m wide is constructed all around a room of dimensions 8m x Sarea of the verandah |
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Answer» Area of the room= length×breadth =8×5=40 sq.mLength of the room including verandah=8+2+2=12mBreadth of the room including verandah=5+2+2=9mHence,Area of the room including verandah=12×9=108 sq.m∴,the area of the verandah is=(area of the room including verandah-area of the room)=108-40 =68sq.m |
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| 28. |
3. A verandah 1.25 m wide is constructed all along the outside of a room 5.5 m long and 4 m wide. Findthe cost of cementing the floor of this verandah at the rate of15 per sq m. |
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Answer» Area of the room is given by 4 * 5.5 = 22m^2 Area of the verandah with room is (5.5 + 1.25 + 1.25) * (4 + 1.25 + 1.25) = 8 * 6.5 = 52 Area of only verandah = 52 - 22 = 30m2 Cost of cementing the floor of the verandah = 15 Rs/m^2 Total cost of cementing 30m2 verandah = 30 * 15 = 450 Rs |
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| 29. |
Q22. In an isosceles triangle ABC, AB-AC D and E are points on BC such that BE-CD Show that AD-AE |
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Answer» InΔabe and inΔacdab =ac ∠B=∠C BE=CDBY SAS RULEΔABE≡ΔACDBY CPCTAE=AD |
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| 30. |
4 ^ { 2 x - 1 } - 16 ^ { x - 1 } = 384 \text { find } x |
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| 31. |
(4)^2x-1 - (16)^x-1=384 |
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| 32. |
4 ^ { 2 x - 1 } - 16 ^ { x - 1 } = 384 |
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Answer» wrong answer should be 11/4 |
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| 33. |
( 4 ) ^ { 2 x - 1 } - ( 16 ) ^ { x - 1 } = 384 |
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| 34. |
\frac { 1 } { 1 + \sqrt { 2 } } + \frac { 1 } { \sqrt { 2 } + \sqrt { 3 } } + \frac { 1 } { \sqrt { 3 } + \sqrt { 4 } } + \ldots + \frac { 1 } { \sqrt { 8 } + \sqrt { 9 } } |
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| 35. |
2. Find an antiderivative of (ax + b) by the method of inspection. |
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| 36. |
Find:\frac{1}{3+\sqrt{7}}+\frac{1}{\sqrt{7}+\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{3}}+\frac{1}{\sqrt{3}+1} |
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| 37. |
parallel to PQFind an antiderivative of (ar + b)? by the method of inspection.2. |
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| 38. |
Prove that\frac{1}{3+\sqrt{7}}+\frac{1}{\sqrt{7}+\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{3}}+\frac{1}{\sqrt{3}+1}=1 |
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| 39. |
Show that $ (3 x+7)^{2}-84 x=(3 x-7)^{2} $ |
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| 40. |
(3 x+7)^{2}-84 x=(3 x-7)^{2} |
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| 41. |
Show that.(3 x+7)^{2}-84 x=(3 x-7)^{2} |
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| 42. |
Write antiderivative of tan2 x |
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Answer» sec^2x-tan^2x=1so tan^2x=sec^2x-1so antiderivative(tan^2x)=antiderivative(sec^2x-1)=tanx-x+cwhere c is a constant |
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| 43. |
28, cos2A = Istana1+ tan2 A |
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Answer» 1 - tan^2(A) / 1 + tan^2(A) = cos2A Solving from LHS is easy I guess So LHS = (1 - tan^2 A)/ (1+tan^2 A) = {( 1 - sin^2 A / cos^2 A) } / Sec^2 A = {( Cos^2 A - Sin^2 A)/Cos^2 A } / (Sec^2 A) = { ( Cos^2 A - Sin^2 A)/Cos^2 A } * Cos^2 A = Cos^2 A - Sin^2 A = Cos 2A = RHS ... Hence the proof... Important formulas used : Cos2A = Cos^2 A - Sin^2 A tan^2 A +1 = Sec^2 A Sec^2 A = 1/ Cos^2 A Tan^2 A = Sin^2 A / Cos^2 A |
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| 44. |
2*acot(8) %2B acot(7) %2B 2*acot(5) |
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| 45. |
acot(1/(sqrt(3))) |
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Answer» π/3 is the correct answer 60 degree answer hi right h |
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| 46. |
sec(acot(-5/12)) |
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Answer» sec [cot^-1(-5/12)] = sec [pi - cot^-1(5/12)] = - sec(cot^-1(5/12)) = - sec(tan^-1(12/5)) = - sec(cos^-1(1/sqrt(1 + (12/5)^2) = - sec(cos^-1(1/sqrt((25 +144)/25)) = - sec(cos^-1(5/13)) = - sec(sec^-1(13/5)) = - 13/5 |
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| 47. |
a*(sqrt(3)/n) - acot(-sqrt(3)) |
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Answer» We know that tan¯¹ √3 is at the angle π/3.Now, ⇒ tan¯¹ √3 - cot-1 (-√3)⇒ π/3 - (π-π/3)⇒ π/3 - 5π/6⇒ -3π/6⇒ -π/2 |
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| 48. |
Derivative(-acot(x), x)=1/(x^2 %2B 1) |
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Answer» property:cot^-1=-1/1+x^2then d/dx of -cot^-1 will be -(-1/1+x^2)=1/1+x^2 cot^-1=-1/1+x^2d/dx of-cot^-1will be-(1/1+x^2)=1/1+x^2 |
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| 49. |
products.() (3x + 7) (3x + 7) |
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Answer» 3x(3x+7)+7(3x+7)9xsquare+42x+49 |
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| 50. |
24. Show that: (3x + 7,--84 x = (3x-7)2 |
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