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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. |
7. Identify the empty set the singleton and the pair set.A-(0)SHB -xx-9 .0.x Z)c-x** - 25. * EN)W D {x15x - 4 -0.x W) |
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Answer» I think it is i) A{0} |
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| 2. |
Of all the closed right circular cylindricalcans ofvolume 128 Ď cm, find the dimensionsofthe can which has minimum surface area |
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Answer» Given pi r^2h = 128 piSo h = 128 / r^2Total surface area S = 2 pi r^2+ 2 pi r hPlug hSo we get S = 2 pi r^2+ 2 pi * 128 / rDifferentiating w r t r we get dS/dr = 4 pi r - 256 pi / r^2As S is to be minimum then dS/dr has to be 0Plugging 0 we have r^3= 64So r = 16 cm |
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| 3. |
y) Draw the graph of the sameIn countries like USA and Canada, temperature is measured in Fahrenheit, whereas incountries like India, it is measured in Celsius. Here is a linear equation that convertsFahrenheit to Celsius:FC+32ve using Celsius for x-axis and Fahrenh Draw the graph of the linear equation abo(ii) If the temperature is 30°C, what is the temperature in Fahrenheit?Gii) If the temperature is 95°F, what is the temperature in Celsius?(iv) If the temperature is 0°C, what is the temperature in Fahrenheit andfor y-axis.temperature is 0°F, what is the temperature in Celsius?(v) Is there a temperature which is numerically the same in both FahrenCelsius? If yes, find it. |
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| 4. |
6.What is the area of a square with perimeter 128 cm? |
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Answer» Perimeter = 4 × side 128cm = 4 × side side = 32 cmArea = (side)^2 = (32)^2 =1024 cm^2 hit like if you find it useful |
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| 5. |
cos (90° + θ) sec (-0) tan (1800-9)1.ve that: sec (360° -0) sin (180 + 0) cot (90-) |
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Answer» cos(90+theeta)= sin thetasec theta= 1/cos thetacot theta= cos theta / sin thetatan theta= sin theta/ cos theta(sin theta/cos theta)tan theta--- numeratornow denominator = 1/cos(360- theta)(-sin theta) ( tan theta) 1/(-1)= -1as cos(-theta) = cos thetacot (90- theta) = tan theta |
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| 6. |
the day time temperature as moon is 100°c. at night the temperature drops -173° c. how much more is the maximum temperature then the minimum temperature? |
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Answer» 273°and 73 °is the given question is the correct answer 273 degree Celsius as in the given Questions. |
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| 7. |
much does Javed's father earn?Ifa 15metre road is widened by 20%, what is the new width of the road?8, |
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Answer» 15 metre road is widened by 20% so the widened part is 15×20/100 = 3m so new width = 15+3 =18m thanks |
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| 8. |
4)If the temperature is oc, what is the temperatureAnsfahrenheit and if the temperature is O°F, what istemperature in celsius?Stat5 Io tho |
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Answer» (0°C× 9/5) + 32 = 32°F 0°F− 32) × 5/9 = -17.78°C F=(9/5)C+32° is the formula of finding fahrenheitby putting 0 F=9/5*0+32F=32° |
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| 9. |
riExpress 3.25 in the form of652134013D) 20x13B)C) |
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Answer» 3.25 can be written as 325/100 in fraction.Now we will reduce the fraction 325/100 Cancel using 5, we get 65/20 Again cancel using 5, we get 13/4 (A) option is correct |
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| 10. |
Express 3.25 in the form of13465B)1340A)C) |
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Answer» 3.25 = 325/100Cancel using 5325/100 65/20 Again cancel using 5 13/4 (A) option is correct |
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| 11. |
हे. _'........०६9 IS ;67 500 + ०५9 इ0० ८50 =६८ द०००+ ०८ हा. हैं17 OIS +o9 - Te B |
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Answer» sin^2(90-27)+sin^227/cos^2(90-17)+cos^273=cos^227+sin^227/sin^273+cos^273=1/1=1 sin25sin(90-25)+cos25cos(90-25)=sin25sin25+cos25cos25=(sin25)^2+(cos25)^2 =as sin square theta+cos square theta =1=hence (sin25)^2+(cos25)^2=1 answer=1. |
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| 12. |
Wi s the cost of uiling a rectangulár plot of land 500m lang und 230 m wide at theODrate of 8 per hundred sqm? |
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Answer» I don't know if you are su area of rectangle=l×b =500×200=1,00,000 sq msince rate of tilling is ₹8 per hundred sq mhence total cost=1,00,000×8/100 =₹8000 8000 is the right answer area of rectangle =l×b=500×200=1,00,000sq msince rate of tilling is Rs 8 per hundreds square m hence total cost = 1 ,00,000×8/100=rs8000 Rs= 8000 is correct answer of this question area of rectangle is length ×breadth=500×200=100000 sq mbut rate of tilling is rs8 per hundred sq mhence total cost=100000×8÷100=RS=8000
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| 13. |
3. If Drakash sowed, jower on 75% of the19500 sqm of his land. on howmรกny. so n dia the actuallyplant sower?A |
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| 14. |
A garden in the form of a right-angledtriangle has an area 128 sqm. If the twosides comprising the right angles areequal, what could be the length of thesesides |
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| 15. |
35. The area of a sector, whose radius is 7 cm and the angle is 120 is(March-2018)51.5 sq.cm. D) 51.6 sqmA) 51.3 sq.cm. B) 51.4 sq.em. C |
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Answer» Area of sector = /360° × pi r² = 120°/360° × 22/7 × 7² = 1/3 × 22 × 7 = 154/3 = 51. 3 cm² A) option is correct |
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| 16. |
A well is dug whose diameter is 4 m and the depthis 10 m. Find the quantity of earth taken out?Calculate the cost of plastering the curved (inner)surface at R 5.00 per sqm. |
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Answer» Diameter=4mRadius=d/2=4/2=2mDepth=10m Volume of earth taken out =πr²h=22/7×2×2×10=8800/7 m³ Curved surface area=2πrh=2×22/7×2×10=8800/ 7m²=88000000/7 cm² Cost of plastering inner curved surface area:-For 1 cm²=Rs. 5For 88000000/7 cm²=62857142.8 rupees. |
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| 17. |
18. A well is dug whose diameter is 4 m and the depthis 10 m. Find the quantity of earth taken out?Calculate the cost of plastering the curved (inner)surface at 5.00 per sqm. |
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| 18. |
There is a road ofequal width around the outside of a rectanthe length 45 m. and breadth 40 m. and the area of the road is und havinge area of the road is 450 sqm. |
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| 19. |
4.The altitude and base of a triangle of area 600 sq. cm are in the ratio 25: 3. Find the altitude and base |
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Answer» The ratio of altitude and base of the triangle = 25 : 3 Let the common multiple of 25 and 3 be 25x and 3x Then, Altitude = 25x cm and Base = 3x cm Area of the given triangle = 600 sq cm Area of triangle = 1/2*Base*Height ⇒ 600 = 1/2*25x*3x⇒ 75x² = 2*600⇒ 75x² = 1200⇒ x² = 1200/75⇒ x² = 16⇒ x√16x = 4 So, Altitude = 25*4 = 100 cm and Base = 3*4 = 12 cm |
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| 20. |
D4Area of a square having perimeter equal to that of a rectangle having sides 10 cm and 6 cm will beA 0.64 sq.mB 6400 sq.mm| 0 90 9 CHD 600 sq. cm46 How many minutes in 3/5 of an hour? |
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Answer» perimeter of square=perimeter of rectangle4(side)=2(length+breadth)4(side)=2(10+6)4(side)=2(16)4(side)=32side=32/4=8cmtherefore area of square=side²=8²=64cm²=(64/100)m²=o.64m²so correct option is A 0.64 sq.m perimeter of sq= perimeter of rectangle4×side= 2(10+6)side of sq= 2×16/4= 8 cmarea of square= (side)^2 = (8)^2 = 64 sq cm |
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| 21. |
of a triangle ares-2d r- 5. What is its perimeser |
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Answer» please like my answer if you find it useful |
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| 22. |
Find the altitude of the triangle if its area is 25 ares and base is 20 m. |
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Answer» Area of triangle = 25 ares We know that 1 are = 100 sq m So, 25 ares = 25*100 = 2500 sq m So, the area of the given triangle is 2500 sq m Base = 20 m Area of triangle = 1/2*Base*Height ⇒ 2500 = 1/2*20*H ⇒ 2500 = 10H ⇒ H = 2500/10 ⇒ H = 250 So, the height or altitude of the given triangle is 250 m Thank you Area of triangle = 1/2 * Base (b) * Altitude (h) Both height and altitude are same in the case of triangle. 1 acre = 4046.85 metre sq.25 acres = 25 * 4046.85 = 10117.25 metre sq. 10117.25 = 1/2 * 20 * h 10117.25 = 10 *hh = 10117.25/10 = 1011.725metres Therefore the altitude of the triangle is 1011.725 metres. |
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| 23. |
Which of the following numbers are in proportion?(i) 5,9,10, 181.(ii) 15, 18, 12, 16 |
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Answer» i.5/9 = 10/18 ii.15/18 is not equal to 12/16 So, i is in proportion |
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| 24. |
+्शी कुल UL O O N e elle BIR S [Pl 12 हि dbabe Gke lkb 9 L I el 2 i4K BYIRLE 2h lnle de o=l | € |
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| 25. |
If length, breadth and total surface area of a cuboid are 15 cm, 10cm and 1300 cmr, then itsa) 15 cmb) 20 cmc) 25 cmd) 30 cm |
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Answer» Surface area = 2(lb + bh + hl)1300 = 2(15 × 10 + 10h + 15h)1300 ÷ 2 = (150 + 25h)650 = 150 + 25h25h = 650 – 15025h = 500h = 500/25h = 20cm |
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| 26. |
dleaisincreasingatrateof1200rsq em/sec. Ans.Example 7. The volume of a cube is increasingat a rate of 9 cubic centimetre per second. Howtast is the surface area increasing when the lengthof an edge is 10 centimetre.NCERT) |
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| 27. |
28 cm10 m long, 45 CH will und27. The decorative solid trophy ismade of two solids of shiningglass as shown in the adjoiningfigure. Find the cost of theglass for making the trophy, ifone cubic centimetre of glasscosts 4.28 cmTuren of its |
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| 28. |
Harsh scored 20% out of 50 marksin hisfirst test. In the next test, there was anincrease of 20% in his marks. In a later test,there was a decrease of 20%. What were hismarks in these three tests? |
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| 29. |
luce ared Ul a Cube is 324cm2.Find its volume and total surt10. The diameter of a sphere is 14 cm. Find the volume of the sphere. |
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Answer» Radius = Diameter/2= 14/2= 7 cm.Volume = (4/3)*(22/7)*7*7*7= 1437.333 cubic metres. |
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| 30. |
Question-2Find the A.P. whose 10th term is 5 and 18th term is 77 |
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| 31. |
dus 4 ies lle uge 8l Savita. Fınd their present ages.(5) In a factory the ratio of salary of skilled and unskilled workers is 5:3. Total salary of one day of both of them is720. Find daily wages of skilled andunskilled workers |
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Answer» Ratio of skilled and unskilled salary is 5 : 3 Let it be 5x and 3x Total of these salary = 720 So 5x + 3x = 720 8x = 720 x = 720 / 8 x = 90 5x = 90 × 5 = 450 3x = 3 × 90 = 270 So the salary of skilled labour is Rs 450 And the salry of unskilled labour is Rs 270 |
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| 32. |
PV DU <V 4065 6 4 Vs 0 [‘.L नव पी एस कम ६ g yiomit 5 |
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| 33. |
Ul d lbe is 216 cm2. what will be its volume?the probabikry of winning a game is 0.995 then wha will be the probability of losing ame?nat is the ordinata of n |
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Answer» Given:- Probability of winning a game is 0. 995. ∴Probability of losing the game=1-0.995=0.005 this is right |
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| 34. |
2. Duis. Examine the following functions for continuity.(b) f(x) =- (a) f(x) = x-5 |
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Answer» w. k. t every polynomial function is continous for all Real numbers, hence the given function is a polynomial function in x and it is continous for all x belonging to set of real numbers. |
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| 35. |
A plastic box 1.5 m long, 1.25 m wide and 65 cm deep is to be made. It is opened at thetop. Ignoring the thickness of the plastic sheet, determine: The area of the sheet required for making the box.(i) The cost of sheet for it, if a sheet me1. |
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| 36. |
ta) 15 cmhallenge!uncovered. If the area orectangular field is to be fenced on three sides, leaving a side of 20 cmthe field ise field is 680 m2, how many metre of fencing is required?m2 |
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Answer» I will compute for the answer to the problem but I will change the unit cm in 20 cm to m because I was thinking 20 cm is a very small side of a rectangular field. It is a very unusual field having one side as 20 cm. Given:Area = 680 m^2; unfenced side = 20 m Formula for perimeter of a rectangle: P = 2W + 2L; one of W was not fenced. The formula can also be written as P = 2(W + L) ; Area =LW. LW = 680 m^2 Since W = 20 m then 20L = 680 m^2 L = 680 m^2 / 20 m L = 34 m So the length of the fencing material is L + L + W. It is 34 m + 34 m + 20 m = 88 m. |
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| 37. |
Find(a) 35 - (20)(b(c) (-15)-(-18)(e) 23-(- 12)Fill in the blanks with >, < or = |
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Answer» a)15b)3c)35 are the right answer answer 15answer 3answer 35 a.15 is the right answer (a) 15(b) 3(e) 35 is the correct answer me 15,335 are you right answer. a- 15b- 3c-35 are the right ans |
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| 38. |
84is 12 kg. what25 per suonofweight8. Theis theorghtweightanofoflove alicaidentical booksbooka35 such books?dual |
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Answer» the weight of 84 boxes 12 KG show the weight of 35 such books is 5 kg the weight of 84 books = 12 kg.1kg = 1000g12kg = 12000gthe weight of 1 book = 12000÷84 =142.86gthe weight of 35 such books = 142.86 × 35 = 5000gso 5000g= 5000÷1000=5kg |
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| 39. |
Ex 1:Find the point on the curve y = 7x - 3x where tangent makesS-03. S-05angle of 45° |
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| 40. |
5. If the price of 5 metres length of wire is 80, then what is the price of 1 decimetrelength of wire? |
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Answer» wrong |
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| 41. |
EXERCISE 3DFill in the blanks to make each of the following a true statement:(i) 246 x1 = ......(ii) 1369 x 0 = .......(iii) 593 x 188 = 188 x .......(iv) 286 x 753 = ...... 286(v) 38 x (91 x 37) = ...... (38 x 37)(vi) 13 x 100 x...... = 1300000MW SOHW MI MOISIVO(vii) 59 x 66 +59 x 34 = 59 x (...... + ......)(vii) 68 x 95 = 68 x 100 - 68 x ....... |
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Answer» Pilleg 24659391100066+34950753 (i) 246× 1 = 246(ii) 1369× 0 = 0(iii) 593× 188 = 188× 593(iv) 286× 753 = 753× 286(v) 38× (91× 37) = 91× (38× 37)(vi) 13× 100× 1000 = 1300000(vii) 59× 66 + 59 × 34 = 59× ( 66 + 34)(viii) 68× 95 = 68× 100 − 68× 5 |
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| 42. |
15. A rectangular field is to be fenced on three sidesleaving a side of 20 cm uncovered. If the area ofthe field is 680 m^2, how many meter of fencingis required ? |
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| 43. |
Fig. 6716. In Fig. 68, the values of xand yare(a) x = 120, y150 (b) : 110, y = 160 (c)x = 150, y = 120(d)x = 110, y160110°409xoFig. 68 |
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Answer» it help u to get answer of same problem. Thanks |
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| 44. |
6. In a circular table cover of radius 32 cm, adesign is formed leaving an equilateraltriangle ABC in the middle as shown inFig. 12.24. Find the area of the design.Fig. 12.24 |
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| 45. |
REAS RELATED TO CIRCLESA rounfind the cost of making the designs at the rate of Rs. 0.35 per e12.53, round table cover has six equal designis as showa in hg, 6. If the mdius of thequal designs as shown in fig. 5. If the |
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| 46. |
10. The interior angles of a pentagon are a°, (a +20), (a +40), (a +60)° and (a + 80) . Find thesmallest angle of the pentagon |
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Answer» Sum of all angles in a Pentagon is 540. a+a+20+a+40+a+60+a+80 =540 5a+200=540 5a=340 a=68 So the smallest angle is 68. |
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| 47. |
Example 7 : A girl of height 90 cm iswalking away from the base of alamp-post at a speed of 1.2 m/s. If the lampis 3.6 m above the ground, find the lengthof her shadow after 4 seconds. |
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| 48. |
angioThe interior angles of a pentagon are in the ratio4:5:6:7:5. Find each angle of the pentagon. |
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| 49. |
Example 7 : A girl of height 90 cm is Awalking away from the base of alamp-post at a speed of 1.2 m/s. Ifthe lampis 3.6 m above the ground, find the lengthof her shadow after 4 seconds.tion |
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| 50. |
66. The angles of a pentagon are x(x 20)(x+40 x 60 and(x+ 80. The smallest angle of the pentagon is |
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Answer» The sum of the interior angles of a pentagon is 540Therefore ,x+x+20+x+40+x+60+x+80=5405x+200=5405x=340x=68Therefore the smallest angle of the pentagon = x = 68x is the smallest angle of the pentagon is x because all the other angles are greater than x. hit like if you find it useful |
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