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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. |
iven figure, O is thecchord and the tangent PT at P makesof 50 with PQ. Then, ZPOQ-?an angle(a) 130°(c) 90(b) 100°(d) 75O4etween two radu of a circle is 120。 |
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| 2. |
156 %2B 217 |
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Answer» √47089= 216.99,√24336=256=256+216 99=472.99 √47089= 216.88√24336= 256√47089+√24336=216.99+256=472.99 472.99 is correct answer of this question. |
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| 3. |
217*f %2B 156 |
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Answer» √ 47089 + √ 24336 217 + 156 373 |
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| 4. |
\tan ^{4} \theta+\tan ^{2} \theta=\sec ^{4} \theta-\sec ^{2} \theta |
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| 5. |
\operatorname { sec } ^ { 4 } \theta - \operatorname { sec } ^ { 2 } \theta = \operatorname { tan } ^ { 4 } \theta + \operatorname { tan } ^ { 2 } \theta |
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| 6. |
\operatorname { sec } ( \frac { \pi } { 4 } + \theta ) \operatorname { sec } ( \frac { \pi } { 4 } - \theta ) = 2 \operatorname { sec } 2 \theta |
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Answer» sec(π/4 + A) sec(π/4 - A) = 1 / [ cos(π/4 + A) cos(π/4 - A)] = 1/ [ {cos(π/4)cos A - sin(π/4) sin A}{cos(π/4)cos A + sin(π/4) sin A}] = 1/ [ (1/√2)cos A - sin A}(1/√2){cos A + sin A}] = 2 / [ (cos A - sin A)(cos A + sin A)] = 2 / [ cos^2(A) - sin^2(A)] = 2 / cos(2A) = 2 sec(2A) |
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| 7. |
ViyayattenthaltBay | Sobre nos20051 5Salve-solve2-3 -1 dere it0-2 5 debatted that a norr1 mult u 16. Edhe number0-9 _Lind the number which wohen miscreated by 72:e d bil6.10theThe differeol Loween two rumberballes place the7 SL brine0:1lange47 th the numberBuches útiled and the multi inseated bywe get 50' find the number20:12. Mrs: JainAfteris 27 year older than her daughton Midi-nfl8 years, she will be twice as old asfind the preseglage |
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Answer» the paper is crushed |
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| 8. |
3) Find in radians and degrees the anglesubtended at the centre of a circle by an arcwhose length is 15 cms, if the radius of thecircle is 25 cms. |
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| 9. |
[cl 0.25 cfl594 cubic metres of earth was dug out tomake a well. If the diameter of the wellthus formed was 6 m, the depth of thewell is |
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| 10. |
OPQ is a sector of a circle with centre O and radiusis 15 cms. If m ZPOQ 30, find the area enclosedby arc PQ and chord PQ. |
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| 11. |
2. Find the cost of flooring a room 6.5 m by 5 m with square tiles of side 25 cm at the rate of1880 per hundred tiles. |
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| 12. |
406-250156 sq. cmAns.5. A well is 8 m long, 6 m broad and 9 m deep.It has water filled up to the height of 6 m. Find thecapacity of the well and the present volume ofwater filled in the well 2 |
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Answer» Volume of the tank= 8 x 6 x 9= 432 m^3 volume of water in the tank= 8 x 6 x 6= 288 m^3 extra volume of water it can hold= 432 - 288= 114 m^3 |
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| 13. |
circular well with a diameter of 2 meters, has a volume 44 m2. What will be the depth of thecircular well? |
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| 14. |
Exercise 1.2Find the length of an arc of circle whichsubtends an angle of 108 at the centre, ifthe radius of the circle is 15 cms.: Here, r 15 cms. andθ 1080 |
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| 15. |
03. Prove that,coseca + CoreCAcoseca-i 'cosecAtI= 2 sec 4. |
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| 16. |
Find the length of an arc of circle whichsubtends an angle of 108 at the centre, ifthe radius of the circle is 15 cms. |
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Answer» answer is length of arc= 9×3.14 cms |
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| 17. |
14. How many rectangular tiles 10cm by 7cm will be required to cover a floor 14m by12- m and what will be the cost at R. 5 per hundred tiles2 |
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Answer» In case of floor Lenght=14m=1400 cm breadth = 25/2m=1250cm In case of tile length = 10cm breadth = 7cm no.of tiles required = area of floor/area of tile =1400×1250/10×7 =25000 |
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| 18. |
Find the length of an arc of circle which subtendsan angle of 108° at the centre, if the radius of thecircle is 15 cms. |
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Answer» but it's Answer is 9π |
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| 19. |
19. Find the area of the shaded region in the following figure. If the ABCD is a square of Sinde 14cm.14cms14 cms8 |
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Answer» Shaded area= area of circle- area of squareArea of circle= πr^2Radius = diagonal of square = 14√2= 19.79Area of circle = 3.14*(19.79)^2/2= 615.7cm^2Area of square= 14*14=196 cm^2Shades area = 615.7-196)= 419.7cm^2Please like the solution 👍 ✔️👍 |
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| 20. |
2)Find the equation of tangent andvalute Sin) Evaluatedx |
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| 21. |
Q. 2. Evaluate (elog(in a)dx |
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Answer» For a functionf(x),f(x), elog(f(x))=f(x)elog(f(x))=f(x) Therefore elog(sinx)=sinxelog(sinx)=sinx ∫sinxdx=−cosx+c∫sinxdx=−cosx+c where c is the integration constant. |
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| 22. |
Evaluate.2dx |
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| 23. |
Calculate the angle between a 2N force and a3N force if the resultant force is 5N.hanked at an angle of 15°. If the coefficient of f |
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Answer» Angle is 90°thanksplease like the solution 👍 ✔️ |
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| 24. |
m rectangular tiles without breaking tiles to Siai8. Find the area in square metre of a piece of cloth 1 m 50 cm wide aand 3 m longniece of land 6 m long and 3 m w |
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Answer» Length=3 mbreadth=1 m 50cm⇒1 m = 100 cm ∴ 50cm=50/100⇒0.50 marea = l x b = 3 x 0.5 m = 1.5m∧2 |
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| 25. |
he perimeter of a parallelogram is 50 cm. One side is 1 times the other. The sides of thearallelogram are: |
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Answer» Let one side be x Then other side will be (3/2)x Perimeter=50 2(x+3/2x)=50 5x=50 x=10 Hence the sides are 10 And 15 |
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| 26. |
If cosecA3then find the values of cot.A, sin A and cos A.fcosecA = _, then find the values of cotA, sin A and cos A |
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Answer» cosec A = 5/4 Then, sin A = 1/cosec A = 1/(5/4) = 4/5 cos A = sqrt(1 - sin^2 A) = sqrt(1 - 16/25) = sqrt(9/25) = 3/5 cot A = cos A/sin A = (3/5)/(4/5) = 3/4 |
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| 27. |
Four Square flower beds of side 1 metre 50 cm are dug on a rectangular piece of land 8m long and 6m 50cm wide. what is the area of the remaining part of land |
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| 28. |
22Evaluatedx |
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Answer» please like my answer if you find it useful |
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| 29. |
2. Evaluate: |
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Answer» hii 5/62 is the correct a nswer |
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| 30. |
a capsule of medicine is in the shape of sphere of diameter 3.5mm. how much medicine (in 3mm)is needed to fill this capsule. |
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| 31. |
solid cuboidal slab of iron of dimensions66 cmpipe. If the outer diameter of the pipe is 10 cmand thickness is 1 cm, then calculate the length of20 cm 27 cm is used to cast an iron.the pipe |
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| 32. |
psule of medicine is in the shape of a sphere of diameter 3.5 mm. How muhmedicine (in imn3) is needed to fill this capsule |
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| 33. |
A capsule of medicine is in the shape of a sphere of diameter 3.5 mm. How Toemedicine (in mm) is needed to fill this capsule? |
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| 34. |
8) IfM 75 an8) IM+Z- 75 and M-Z 1.4, then findvalues. 1 33.2its mean, median and mode |
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Answer» M + Z = 75M - Z = 1.4____________2M = 76.4M = 76.4/2M = 38.2Median = 38.2 M + Z = 75Z = 75 - M= 75 - 38.2= 36.8Z = 36.8Mode = 36.8 Z = 3M - 2X36.8 = 3(38.2) - 2X36.8 = 114.6 - 2X2X = 114.6 - 36.82X = 77.8X = 77.8/2X = 38.9 |
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| 35. |
10. A capsule of medicine is in the sbape of a sphere of diameter 3.5 mm. How muchmedicine (in mm) is needed to fill this capsule'? |
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| 36. |
13. In the given tigtre, AB ll CD. Find the values of x, y and z.75*35*) (Find the values of x, y and z. |
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| 37. |
side IS dema0le、thantheOther by 4m lind tuo allsdlales |
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| 38. |
Each of the two equal sides of a triangle are 4m less than three times the third side. Find thedimensions of the triangle, if its perimeter is 55m.. |
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| 39. |
Find the area of square of side 4 cm. If the side of square is doubled, find the area of resulting new square. |
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Answer» thnx bhaiya |
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| 40. |
Expand (3a-2b)^2Evaluate (16/81) ^1/4" |
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Answer» 9a^2-12ab+4b^2 (2^4/3^4)^1/4=2/3 4th line is 3a(3a-2b)+(3a-2b) 4th line is 3a(3a-2b)+(3a-2b) |
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| 41. |
5) 1000 75%= ? |
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| 42. |
The sum of the radius of the base and the heightof a solid cylinder is 37 m. If the total surfacearea of the solid cylinder is 1628 m, then whatis the volume of the cylinder? |
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| 43. |
Expand (3a -2b)^2Evaluate (16/ 81) ^1/4 |
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Answer» (3a-2b)² = (3a)² + (2b)² - 2×4a×2b =9a² + 4b² - 12ab 3a -2 3a-2b=3a+2b-2×4a×2b=9a+4b-12ab |
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| 44. |
12. A hollow copper pipe of inner diameter 6 cm andouter diameter 10 cm is melted and changed intoa solid circular cylinder of the same height as thatof the pipe. Find the diameter of the solid cylinder |
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| 45. |
5. यदि cos e=N3 तो सिद्ध कीजिए कि tant = = तथा cosece =2V3 |
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Answer» cos= √3/2meansthera= 30so tan 30= 1/√3anr cosec 30= 1/2 cos=root3/2because theta =30so tan 30=1/root3and cosec30=1/2 |
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| 46. |
Q.5) Expand (3a -2b)^26) Evaluate (16/81) ^1/4 |
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Answer» (a-b)^2=a^2+b^2-2abhence (3a-2b)^2=9a^2+4b^2-12ab(16/81)^1/4=(2^4)/(3^4))^1/4=2/3 |
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| 47. |
16. Find the product 24x2 (1-2x) and evaluate it forx 2. |
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Answer» wrong no it's correct sorry for saying 'wrong' |
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| 48. |
20.What is the value of tan 45o + sin 60°- cos 30° as per Trigonometrical ratio?b)仮N32c)d)42 |
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Answer» Correct option: aReason: tan45 + sin60 - cos30= 1 + √3/2 - √3/2= 1. Please hit the like button if this helped you. |
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| 49. |
6) Evaluate (16/ 81) |
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| 50. |
\cos 75 ^ { \circ } + \cos 15 ^ { \circ } |
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