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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. |
sin 25° cos 65° + cos 25° sin 65o |
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| 2. |
।।८।।त्रिकोणमितीय सारणियों के प्रयोग बिना निम्नलिखित का मान निर्धारित कीजिए.1. sin 25°cos 65°. |
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Answer» Sin25 Cos65=Sin(90-25) Cos65= Cos65 Cos65=1/2+1/2=4/2=2 |
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| 3. |
-5. The dimensions of a field are 12 m x10 m.A pit 5 m long, 4 m wide and 2 m deep isdug in one corner of the field and theearth removed has been evenly spread overthe remaining area of the field. The levelof the field is raised by |
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| 4. |
sin 25 cos 65 |
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Answer» Sin 25° cos 65° = sin 25° cos (90-25) = sin 25° sin 25°= sin²25° = 0.17 |
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| 5. |
The dimensions of a field are 15 m x 12 m. A pit 7-5 m x 6 m x 1-5 m is dug in one comerof the field and the earth removed from it, is evenly spread over the remaining area of thefield. Calculate, by how much the level of the field is raised. |
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| 6. |
19. A field is 30 m long and 18 m broad. A pit 6Tmlong, 4 m wide and 3 m deep, is dug out frothe middle of the field and the earth removedevenly spread over the remaining area to the field.Find the rise in the level of the remaining partof the field in centimetres correct to one decimalplace.is(ICSE) |
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Answer» Volume = 6 * 4 *3 m^3= 72m^3Area of Pit = 6*4=24m^2Area of field=30*18=540m^2Remaining area = (540-24)=516m^2 Rise in level = 72*100/516= 13.95cm |
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| 7. |
a circular well is dug . it's diameter is 3 1|2 m and it's depth is 6m.what is the lateral surface area of the well ? what is the volume of the earth removed to make the well ? |
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Answer» tq |
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| 8. |
What istheA circular well is dug. Its diameter is 3 1 m and its depth is 6m.lateral surface area of the well? What is the volume ofmake the well?ithe earth removed to |
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| 9. |
A cylindrical tank of diameter 2m is dug and the earth removed from it is spread uniformly all around the tank to form an embankment 2m in width and 1.6m in height. Find the depth of circular tank. |
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Answer» Ans :- For circular tank external radius R = 3 minternal radius r = 1 m thus volume of embankment V = π× (R^2 - r^2) × h = π× (9-1) × 1.6 = 40.192 m^2 now volume of tank V' = 12 × r^2 ×h = 3.14 × 1 × h since both are equal V = V'thus 3.14 h = 40.192 h = 12.8 m thus height of tank is 12.8 m |
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| 10. |
A cylindrical tank of diameter 2m is dug and the earth removed from it is spread uniformly all around the tank to form an embankment 2m in width and 1.6m in height. find the depth of cylindrical tank. |
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Answer» thnx |
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| 11. |
FiricT DT e diameter of the moon is approximately one fourth the diameter of the earth. Find theratio of the volume of the moon to the volume of the earth.SECTION D |
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Answer» Given that, the diameter of Moon is approximately one-fourth of the diameter of Earth. Let radius on Moon = r Then, radius of Earth = 4r Therefore,VolumeofMoon/VolumeofEarth=4/3πr^3/4/3π(4r)^3=r^3/64r^3=1/64=1 : 64 |
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| 12. |
छट प्रेबन्धक (18885)6. लेखांकन सूचनाओं के बाहा उपयोगकर्ता (D) 3=l (hesearchers}# (External users of accounting information are)R RdaNe L oy |
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Answer» External usersare people outside the business entity (organization) who useaccounting information. Examples ofexternal usersare suppliers, banks, customers, investors, potential investors, and tax authorities. please like the solution 👍 ✔️ |
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| 13. |
कक कक दी पर N e bt"’ 2 G 2 ० ४-6 तथा. oy AT R से हल करें, एवं +जक्त से है'ायांकित करें और छम्याकित s ज्ञात करी । |
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| 14. |
(a) 340(b)60(e) 80(d) 170 |
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| 15. |
13.A vehicle has 1000 kg mass and co-efficient of friction between tyres and road is 0.4what is the maximum breaking force (retarding force) which can stop the vehicle onroad-23924 Nb) 9180 Nc) 1000Nd) 400 N |
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Answer» Force due to friction is mgu where m is the mass, g is gravity, and I is the coefficient of friction. That gives you 1000*9.8*0.4 which equals 3920. hence , the correct option is (a) If you want the acceleration as well( I know you didn’t ask but I’m bored) you use F=ma. Solve for acceleration gives you a=F/m. 3920/1000=3.92. This means that the maximum acceleration that car can experience is 3.92m/s^2. This applies to speeding up and slowing down. |
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| 16. |
a) - 65 + 87 + (-43) +- 55 ) + 65 |
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Answer» -65+87+(-43)+(-55)+65=-11 |
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| 17. |
|. (i) sin 25९ ८08 65% + cos 25° sin 65° |
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Answer» sin(A + B ) = sinA cosB + sinB cosA Here A = 25° and B = 65° sin(25° + 65°) = sin(90) = 1 |
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| 18. |
|. है () sin 25° cos 65° + cos 257 sin 65 |
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Answer» sin 25 cos 65 + cos 25 sin 65[using sin (90 - x) = cos x and cos(90 - x) = sin x] = sin 25 sin(90 - 65) + cos 25 cos (90 - 65) = sin 25 sin 25 + cos 25 cos 25 = sin^2 25 + cos^2 25[using sin^2 x + cos^2 x = 1] = 1 |
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| 19. |
sin 25 cos 65 + cos 25 sin 65 |
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Answer» sin(25+65) sin 90 answer 1. answer is correct like me 1 is the correct answer me to 2 is the correct answer for me 1 is the correct answer 1 is the correct answerof above question according to trignometric identities ans as we know Sin90 =1 hence ,1 is the absolutely correct answer 1 is the correct answer sin 25+sin65sin (25+65)sin 901 1 is the correct answer 1 is the correct answer |
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| 20. |
(i) sin 25° cos 65° + cos 25° sin 65% |
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| 21. |
\sin 65^{\circ}+\cos 65^{\circ}=\sqrt{2} \cos 20^{\circ} |
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Answer» For RHS√2 cos20 = √2 cos (65 - 45) =√2cos65 cos45 +√2 sin65 sin45 =√2 cos65 (1/√2) + √2 sin65 (1/√2) => cos65 + sin65 = √2 cos20° => sin65 + cos65 = √2 cos20 Hence, proved. We used the identity rule - cos (A - B) = cosA cosB + sinA sinB |
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| 22. |
(i) \sin 65^{\circ}+\cos 65^{\circ}=\sqrt{2} \cos 20^{\circ} |
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| 23. |
प्रश्न 14.dr– का मान ज्ञात कीजिए।1 + sin x |
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| 24. |
: ला 2 \/—) Ify= Lt g & gin-1% 2y a® - x® + = sin! - prove that dxy a? - 2%, |
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| 25. |
MICROSOFT POWERPOINT |
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Answer» Microsoft PowerPoint is a presentation program, created by Robert Gaskins and Dennis Austin at a software company named Forethought, Inc. It was released on April 20, 1987, initially for Macintosh computers only. Microsoft acquired PowerPoint for $14 million three months after it appeared. |
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| 26. |
dr = tan(2+a)+b find the value of a and b..10.1f11+sin x |
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| 27. |
it (८५. फेस्टि) ७ |
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Answer» (6x + 8y)² - 192xy 36x² + 64y² + 96xy - 192xy 36x² + 64y² + 96xy (6x + 8y)² |
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| 28. |
"5 &|-, Tब्लड ८५elG U |
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Answer» d1 = (x+2)-(x-1)=3d2 = (x-7)-(x+2)=-9since the sequence had no common differenceso it is not AP |
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| 29. |
a man sold product by 20% profit at price of 660. what is the actual price ? |
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Answer» 20 percentage of profit= 120 percentage cost price= 120 percentage cost price = 6601 percentage will be660/120now 100 percentage or actual price will be660/120*100550 ruppes |
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| 30. |
५०५८० /८५/०० हु? 9१0० हम केदा का «277. BRI NoGwmas छत]्यृसव ठल ! |
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| 31. |
सब ८५Sपन्न करने ।|525% की 95ष क) 5.या । |
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| 32. |
| ८५ कि । |
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Answer» I don't know number hai 24 Ko Hum Log 12 se multiply Karenge Aap Jo answer aayega uska |
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| 33. |
Evaluate : sin 60° . cos 30° + sin 30° , cos 60*S e o |
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Answer» If you find this solution helpful, Please give it a 👍 |
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| 34. |
Evaluate 2 sin2 630 +1+2sin2273 cos2 17 -2+3cos2 73 |
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Answer» 2sin^2 63+2sin^2 27+1 / 3cos^2 17+3cos^2 73-2 2(sin^2 63+sin^2 27)+1 / 3(cos^2 17 +cos^2 73 -22(sin^2 63+cos^2 63)+1 / 3(cos^2 17+sin^2 17-22(1)+1 / 3(1)-22+1 / 3-23/1=3 |
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| 35. |
3. Evaluate:sin2 630+ sin2 27°(1) tOS2 17" + COS2 730 |
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| 36. |
Ifcot evaluate: 0)cs)-cose)(1 + sin 0)l -sin 6), (i) cot e(1 + cos θ) (1-cos0)(ii) cot |
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Answer» Please like the pic and like other solution 👍 ✔️ as well |
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| 37. |
he sales tax on a refrigerator is 9%. Sales tax is t i 170 Find the actual sale price.s Tgol ß discount of 10% on the suit he bought The marked price wast 5000 for 0 e su trefolnysales tax of 10% on the price at which he bought it, how much did he pay?0suid. ithto pay |
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Answer» please give answer with statement |
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| 38. |
Evaluate the following limits:\operatorname{lit}_{x \rightarrow 0} \frac{e^{2 \sin x}-1}{x} |
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Answer» Using L hospital ruleDifferentiate both Numerator and DenominatorHence2cosxe^2sinx/1put in the limitshence2cos0e^2sin0hence2*1=2 |
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| 39. |
Evaluate \begin { equation } \int e^{\sin x} \cos x d x \end { equation } |
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Answer» Letu=sinx, thendu=cosxdxand ∫e^sinxcosxdx = ∫e^udu = e^u = e^sinx |
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| 40. |
(ii) B (d,e, f, g(iv) D (f,g, h, a=(a, b, c(iii)C={a, c, e, g} |
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Answer» purak of A=d,e,f,g,hpurak of B=a,b,c,hpurak of C=b,e,f,hpurak of D=b,c,d,e |
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| 41. |
Given that U ={a,b,c,d,e,f,g, h},A={a, b, f, g), and B={a, b, c), verifDe Morgan's laws of complementation. |
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| 42. |
(iii) the actual percentage profit madWhen a discount of 15% is allowed on the marked price of an article, it is sold forCalculate its marked price. Given that(i) its cost price(ii) the profit in made by the sale of the article.e by the shopkeepel2975.the marked price is 40% above the cost price of the article, calculate |
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| 43. |
Question - 7:(a)It' sin θ + 2 cos θ1, prove that cos θ-2 sino-2 |
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Answer» sinθ+2cosθ=1 Squaring both sides, sin²θ + 2sinθ2cosθ + 4 cos²θ = 1 (1-cos²θ)+4sinθcosθ+4(1-sin²θ)=1 [∵sin²θ + cos²θ = 1] 1- cos²θ + 4sinθcosθ + 4 - 4sin²θ=1 -4sin²θ + 4sinθcosθ - cos²θ = 1-1-4 -(4sin²θ-4sinθcosθ+cos²θ)=-4 (2sinθ)²-2.2sinθcosθ+(cosθ)²=4 (2sinθ-cosθ)²=2² 2sinθ-cosθ=2 |
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| 44. |
In a dress store, a dress is marked at price 600. Due to a clearane sale, the shop offere45% discount and still makes a profit-of 10%. What is lhe actual cost of the dress? |
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Answer» After 45 percentage discount = 600*45/100270rupee330 ,in this he got 10 percentage gain or can say this is the value of 100 percentageso 1 percentage will be 330/110= 3100 percentage = 3*100= 300 ruppes |
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| 45. |
After allowing a discount of 10%, the sale price of anoven is 900. Find the cost price of the oven. |
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| 46. |
BO D010. The dingonals of a qundrilateral ABCD intersect each aher at the point O such thatAO COto Do Bhow that Alen is a apezium |
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Answer» Construction: Through O, draw line EO, where EO || AB, which meets AD at E. Proof: In ΔDAB, we have EO || AB ∴ DE/EA = DO/OB ...(i)[By using Basic Proportionality Theorem] Also, AO/BO = CO/DO (Given) ⇒ AO/CO = BO/DO ⇒ CO/AO = BO/DO ⇒ DO/OB = CO/AO ...(ii) From equation(i)and(ii), we get DE/EA = CO/AO Therefore, By using converse of Basic Proportionality Theorem, EO || DC also EO || AB ⇒ AB || DC. Hence, quadrilateral ABCD is a trapezium with AB || CD. |
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| 47. |
Question 7(a) Prove that (sinA + Cosec A)* + |
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| 48. |
rosed2100 from his friend Mr Basu at the rate of 18- Mr Sharma bho% per annum simple interest forHe lent this mowey to Atr Giupta at the same rate but compounded annually Find his gain aher2 years |
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| 49. |
Write the base and power for each of the following: ,(iii) [(— 2) x 413 (W) (— C)‘Ti) 5’ (ii) (— 7)‘A44 ८५.८८. |
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Answer» 1.base=5 power=22. base=-7 power=43.base=-8 power=24.base=-c power=1 (i) 5³=125(ii) (-7)⁴=2401(iii) [(-2) ×4]^5=32768(iv) (-C)¹=-C 1 base =5power=22 base =-7 power =4 3 base = -8 power =2 4 base = -c power = 1 |
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| 50. |
TI.सरलीकरणSagir -7. (३+३)-(१( मान ।) को सरल करने पर मान = |
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Answer» 49/11 |
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