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. |
Find the area of square of side 4*1/3 m. |
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Answer» side = 13/3so area of a square is 13/3*13/3169/9 m^2 thanks |
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| 2. |
12. Find the area of square of side 4 metre.3 |
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| 3. |
4)1. x varies directly as y and x varies inversely as the square of z. When y=75 and x =6, thenz=5. Find the value of x when y=24 and z=4.(2) 3(4) 61)13)4 |
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Answer» Option (2) is correct. |
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| 4. |
3 \sqrt{75}+5 \sqrt{48}-\sqrt{243} |
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Answer» √243 = √81.3 = 9√3 √48 = √16.3 = 4√3 √75 = √25.3 = 5√3 15√3+20√3-9√326√3 |
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| 5. |
25+√75+√27 |
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Answer» 25+√75+√27=25+√5×5×3+√3×3×3=25+5√3+3√3=25+8√3 25+ 8√ 3 is the correct answer of the given question 25+V75+V27=25+V5x5x3+V3x3x3=25+V5x5x3+V3x3x3=25+5V3+3V3=25+8V3 25+8√3 is right answer of this question. 25+V75+V27=25+V5x5x3+V3x3x3=25+5V3+3V3=25+8V3 |
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| 6. |
10. A capsule of medicine is in the shape of a sphere of diameter 3.5 mm. How muchmedicine (in mm2) is needed to fill this capsule? |
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| 7. |
A solid cylinder of brass 8 m high and 4 m in diameter is melted and recast into a cone ofdiameter 3 m. Find the height of the cone.4. |
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| 8. |
Simplify.(i) 7147-,(iii) V216-5/6tyl294--s4719292-./75(ii) 5A+2V27+-To/3675-7238-75(1V)4V12-75-7V48 |
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Answer» can you tell 4 and 5 |
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| 9. |
(ii) A cubical block of side 10 cm is surmounted by a hemisphereWhat is the largest diameter that the hemisphere can have? Findthe cost of painting the total surface area of the solid so formed, at[CBSE 2015the rate of5 per 100 sq cm. [Use π3.14.] |
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| 10. |
A cubical block of side 10 cm is surmounted by a hemisphere. What is the greatest diameter thehemisphere can have? Also find the cost of painting the total surface area of the solid so formed, at therate of t 5 per 100 sq. cm.16.[AICBSE 2015] |
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| 11. |
0Page NoDate:Evaluate162 |
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| 12. |
. A cubical block of side 10 cm is surmounted by a hemisphere. What isthe largest diameter that the hemisphere can have? Find the cost ofpainting the total surface area of the solid so formed, at the rate of 5CBSE 2015per 100 sq cm. Use π3.14.] |
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Answer» isme area of circle minus kyu karenge |
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| 13. |
(ii) A cubical block of side 10 cm is surmounted by a hemisphere.What is the largest diameter that the hemisphere can have? Findthe cost of painting the total surface area of the solid so formed, at[CBSE 2015the rate of ? 5 per 100 sq cm. [Use Tt 3.14.] |
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| 14. |
6. sin 75+cos 75N3/23)4) 2) |
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Answer» √3/2= √6/2 |
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| 15. |
(3*75)/1000 |
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Answer» 3.75 ÷ 1000 = (375/100) ÷ 1000 = 375 / 100000 = .00375 |
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| 16. |
The centre of circle 4 x^{2}+4 y^{2}-24 x-8 y-24 = 0 is |
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Answer» centre is (-g, -f)=(3,1) |
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| 17. |
4B) A circle is inscribed in a square of side 4 m. IfAyou drop a dice at random on the square regionwhat is the probability that dice will landbetween square and circle.44.4 |
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Answer» Area of square = 4² = 16Area of circle = 3.14×2² = 12.56Probability = 12.56/16 = 1.27 |
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| 18. |
2Ifthe equationrepresents a circle4then λ-(a) 14(c) 04 |
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Answer» The denominator of both the variables should be same for the equation to represent circle => L/3 = 1/4=> L = 3/4 option B |
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| 19. |
(27/((-3)))*(75*(-5)) |
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| 20. |
3 - 75 z ^ { 2 } |
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Answer» sonakshi acharya can you tell me more question |
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| 21. |
0. A capsule of medicine is in the shape of a sphere of diameter 3.5 mm. How muchmedicine (in mm3) is needed to fill this capsule? |
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| 22. |
10. A capsule of medicine is in the shape of a sphere of diameter 3.5 mm. How muchmedicine (in mm3) is needed to fill this capsule? |
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| 23. |
Moh date: 0 |
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Answer» x(x+4)=0 xx(x+4)=0x x+4=0 x+4−4=0−4 ∗x=−4∗. x(x+4)=0 x(x+4x+4)=(0x+4). ∗x =0∗... |
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| 24. |
PAGE NoDATE0 |
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Answer» 1 2 thank you for giving me answer |
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| 25. |
0)/ Page No.Date / |
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| 26. |
e diameter of a cylinder is 28 cm and its height is 40 cm. Find thecurved surface area, total surface area and the volume of the cylinder1. Th |
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| 27. |
JTJ.JUinl.lIndthetotalsurfacearea of the toy12. A cone and a hemisphere have equal baseand equal volume. Find the ratio of their heights.13 Find th |
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Answer» Given : a cone and hemisphere have equal bases and equal volume.volume of cone is equal to the volume of hemisphere, let height of cone is a and hemisphere is b1/3 π r² a =2/3π r² bπr² had cancleda/b =2/13 and 3 had cancled it meansratio of height is 2:1 |
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| 28. |
22. A cubical block of side 10 cm is surmounted by a hemisphere. What is the largest diameter that thhemisphere can have ? Find the cost of painting the total surface area of the solid so formed, at the ratof ?.5 per 100 sq. cm. [Use 3.14] |
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| 29. |
(81/16)^(3/4)*((9/25)^(3/2)/(5/2)^(-3)) |
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| 30. |
25^(3*x - 4) %2B 5^(x %2B 3) |
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Answer» (5)^x+3 = (25)^3x-4 (5)^x+3 = (5)^6x-8 As base are same so power should be equalx + 3 = 6x - 86x - x = 3 + 85x = 11x = 11/5 |
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| 31. |
8x3. Find x. if ()*x(25) - 3IRIMC, 2009] |
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Answer» try to do it by your self |
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| 32. |
Exercise 4.5I. Evaluate(b) (02)3 -(0.3)3+(01)(d) (-12)3+7+53(a) (25)3 -(75) (50) |
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Answer» Please specify the question to be answered as we are not allowed to answer multiple questions in a single query B,c, d |
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| 33. |
(-25)*(-3) |
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Answer» So(-25)*(-32)= 800now(-32)*(-25)= 800 800 is the correct answer of the given question (-25)×(-32)=800your answer is 800. |
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| 34. |
25/3 |
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Answer» 8.33 is the answer of the question 8.33 is the answer of the following 8.33 is the correct answer of the given question 8 is the answer of the following 8.33 is the right answer. 25 ÷ 3 = 8.33 is the answer. 8.33 . 8.33 is the right answer 8.33 is the right answer 8.33 is right answer oooo 8.33 is the answer of the following questions 8.33 is the right answer 8.33 is the right answer 8.33 is the best answer 8.33 is correct answer 8.33 is the right answers 8.33 is the correct answer 8.33 is the best answer |
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| 35. |
206. The horizontal angle between true meridianand magnetic meridian, is known |
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Answer» The horizontal angle between true meridian and magnetic meridian, is known magnetic declination |
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| 36. |
0The true value of dip at a place is 45If the vertical plane carrying the needleis turned through 60 from themagnetic meridian, the apparent dip is(I) tan 1(1/2) ( )(2) tan(2(4) tan (V3(3) tan (3) |
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Answer» If the plane of the scale of the dip circle is not in the magnetic meridian, then the needle will not indicate the correct direction of earth's magnetic field. The angle made by the needle with the horizontal is known as themagnetic dip. |
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| 37. |
The ages of Ram and Sushma are in the ratio 9: 8. If the ratio between their ages becomes 11:10 after 10 years find their present ages |
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Answer» thanks |
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| 38. |
A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. Thdiameter of the hemisphere is 14 cm and the total height of the vessel is 13 cm. Find theinner surface area of the vessel. |
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| 39. |
. 8in30cos® 0 + cos 38sin’ 0 =3B de की 2cotas1) :—dn«a 2 “meo £ N3 |
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| 40. |
ORIf cosec θ _ sin θ-m 3 and sec θ-cos θn3 then prove that m4n2 + m2n4 = 1 |
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Answer» 1/sinθ - sinθ = a^3(1 - sin²θ)/sinθ = a^3cos²θ / sinθ = a^3 similarlysin²θ / cosθ = b^3 thena^4 b^2 + a^2 b^4= [cos²θ/sinθ]^(4/3) [sin²θ / cosθ]^(2/3) + [cos²θ/sinθ]^(2/3) [sin²θ / cosθ]^(4/3) = (cosθ)^(8/3) / (cosθ)^(2/3) + (sinθ)^(8/3) / (sinθ)^(2/3) = cos²θ + sin²θ= 1 |
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| 41. |
37. The ratio of 7th to the 3rd term of an AP is 12:5. Find the ratio of 13th to the 4th term.OR484 |
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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 = 7 So,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 = 6d/7 + 12d 4th term =a+(4-1)d =a+3d = 6d/7 + 3d Now ratio of 13th and 4th term (6d/7 + 12d)/(6d/7 + 3d)= (6d + 84)/(6d + 21d)= 90d/27d= 10/3= 10:3 Thus the ratio of the 13th and 4th term is 10:3 |
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| 42. |
37. The ratio of 7th to the 3rd term of an AP is 12:5. Find the ratio of 13th to the 4th term.OR4x4=16 |
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Answer» For AP nth term isTn = a + (n-1)d Where,a = first termd = common difference T7 = a + 6dT3 = a + 2d T7/T3 = 12/5a + 6d/a + 2d = 12/55(a + 6d) = 12(a + 2d)5a + 30d = 12a + 24d12a - 5a = 30d - 24d7a = 6da = (6/7)d Ratio of 13th to 4th termT13/T4 = a + 12d/a + 3d Put value of a = (6/7)dT13/T4 = (6/7)d + 12d/(6/7)d + 3dT13/T4 = 90d/27d = 10/3 |
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| 43. |
The ages of 2 men A and B are in the ratio 5:7.6years agotheir ages were in the ratio 2:3.Assuming their presentages to be x and y form equation and solve for x and y. |
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| 44. |
2-6yy+2 36. The ages of Hari and Harry are in the ratio 5:7. Four years from now the ratio oftheir ages will be 3:4. Find their present agesr hu 8 If the |
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| 45. |
2 Tarun's father is 4 times as old as he is now. After 11 years, the ratio of theirages will be 2:5. Find their present ages. |
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| 46. |
ORrom a solid cylinder whosheight and same diameter is hollowed out. Find thethe nearest cm2a solid cylinder whose height is 2.4 cm and diameter 1.4 cm, a conical cavity of thtotal surface area of the remainsolid to |
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Answer» Ans :- The total surface area of the remaining solid will be 18 cm² Step-by-step explanation: The outer surface area of the cylinder = 2πrh = 2 x 3.14 x 0.7 x 2.4 = 10.55 cm² slant height of the cone, L = √(h² + r²) =√(2.4² + 0.7²) = √6.25 = 2.5 cm hence outer surface area of the cone = πrL = 3.14 x 0.7 x 2.5 = 5.5 cm² This outer surface area of the cone is equal to the inner surface area of the hollow portion of the cylinder left. surface area of the cylindrical base = πr² = 3.14 x 0.7² = 1.54 cm² Hence total surface area of the remaining solid = The outer surface area of the cylinder + inner surface area of the hollow portion of the cylinder left + surface area of the cylindrical base = 10.55 + 5.5 + 1.54 = 17.59 cm² = 18 cm² (rounded off to nearest cm²) Hence the total surface area of the remaining solid will be 18 cm² |
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| 47. |
.From a solid cylinder whose height is 2.4 cm and diameter 1.4 cm, a conical cavity of thesame height and same diameter is hollowed out, Find the total surface area of theremaining solid to the nearest cm'. |
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| 48. |
Describe the e buile meth r(1 2)+1() N3(32 x6)31l/22-3/2 |
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Answer» = {3,4,9,17,33} is the answer can you plz solve it further plz |
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| 49. |
99. If by selling an article for P390, ashopkeeper gains 20%, then the costprice of article is(1) P324(3) P323(2) P321(4) P325 |
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Answer» SP=390gain=20%so CP=390/1.2=325 |
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| 50. |
How much per cent above the cost price should a shopkeeper mark his goods sothat after allowing a discount of 20% on the marked price, he gains 12967pt of 5% for |
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Answer» let the CP be Rs. 100 and MP be Rs. xgain % after discount = 12%gain = 12% of 100 = Rs. 12⇒SP of the article = CP + Gain = Rs. (100 + 12) = Rs. 112Discount % = 20%Discount = 20% of Marked Price= Rs.20/100 * x = Rs. 20x/100 = Rs. x/5as da question saysmarked Price - discount = SP=> x - x/5 = 112=> (5x - x)/5 = 112=> 4x/5 = 112=> x = 112 * 5/4 = Rs. 140marked Price = Rs. 140amount marked above the CP = MP - CP = Rs. (140 - 100) = Rs. 40 ∴% amount marked above the CP= Amount increased/CP * 100= 40/100 * 100= 40% |
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