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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. |
37 एक शंकु मैं = 10 सेमी एवं h = 8 रोगी है। इसके व(A) 80 7 सेमी (B) 100 1 सेमी (C) 60 71 से’.का क्षेत्रफल है -(D) 36 , सेमी |
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Answer» 100πसेंटीमीटर |
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| 2. |
the given figure PQ Il ST, -PQR 110°, and RST-130 find <ORS |
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Answer» Interior angles on the same side of thetransversal: The pair of interior angles on the same side ofthe transversal are called consecutive interior angles or allied angles or cointerior angles. If a transversal intersects two Parallel Linesthen each pair of interior angles on the same side of the transversal is supplementary. If a transversal intersects two lines such thata pair of alternate interior angles is equal then the two lines are parallel. PQ || ST, ∠PQR = 110° and ∠RST = 130° Construction, A line XY parallel to PQ and ST is drawn. ∠PQR+∠QRX= 180° (Angles on the same side of transversal.)⇒110° + ∠QRX= 180°⇒∠QRX= 70° Also,∠RST+ ∠SRY= 180° (Angles on the same side of transversal.)⇒130° + ∠SRY= 180°⇒∠SRY= 50° Now,∠QRX+∠SRY+∠QRS = 180°⇒70°+ 50°+ ∠QRS= 180°⇒∠QRS= 60° |
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| 3. |
Hâ˘~â˘~o~0 |
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| 4. |
7. A tent consists of a frustum of a cone, surmounted by a cone. If thediameters of the upper and lower circular ends of the frustum be 14 mand 26 m respectively, the height of the frustum be 8 m and the slantheight of the surmounted conical portion be 12 m, find the area of thecanvas required to make the tent. (Assume that the radii of the uppercircular end of the frustum and the base of the surmounted conicalCBSE 2008portion are equal.) |
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| 5. |
प्रथम पाँच सम प्राकृत संख्\frac{2+4+6+8+10}{5}= |
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Answer» 2+4+6+8+10/56 will be the answer |
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| 6. |
if A3then A'3 -24 -2If Aand I-, find k so that A- kA-20 -1 |
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| 7. |
Spid3 e':â"'.â. |
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| 8. |
23 I \[ââe *~1dx |
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| 9. |
Wirie the following in decimal form and say whatbasIn the prereal numbus to reprhowimal form and say what kind of decL.Supp36100(viiv)13 047811So. Ier |
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Answer» 36/100100 ÷ 360.36 |
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| 10. |
et Us Practisxpress as a percentage:(i) 450 ml out of 5v) 16 hours out of a dayo spent 3500 on jeansent onjeans?nya was given R300 as0. Whaya scored 32 out of 40er nercentage?o saved the great |
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Answer» 5l is 5000ml450÷5000×100%9% |
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| 11. |
us forwill buys 7 bats and 6 balls for 3800. Later, she buys 31750. Find the cost of each bat and each ball.er by 18 degrees. FindBOSSalve the pain of anDate:XTRAPageug uahons by Substitutionmethod12x + 34 20Bx toy a eFerom de me getTo |
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Answer» x=0y=0 is the correct answer X=0&Y=0 it is correct |
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| 12. |
| o4057 u9aZ 1 84qbt 36 b" er-ap |
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Answer» (7a+6b)^2= 49a^2+36b^2+ 84ab 2) 4p(4p^2-1) 4p(2p+1)(2p-1) |
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| 13. |
** VERY SHORT ANSWER1. Multipy : **** |
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Answer» -3/4 * 5/7 * - 8/11 * 7/10 = (-3/4 * - 8/11) * (5/7 * 7/10) = 6/11 * 1/2 = 3/11 840/3060 is right answer |
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| 14. |
10 ka 2% |
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Answer» 10 ka 2%10 × 2/1002/20 = 0.2% |
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| 15. |
7.A 1.8 kg block is moved at constant speed over a surface forwhich H 0.25. The displacement is 2 m. It is pulled by aforce directed at 45° to the horizontal as shown in figure. Findthe work done on the block by(a) the force F(b) friction(c) gravity |
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Answer» M = 1.8 kgμ = 0.25 F acts at 45⁰ with horizontal.s = displacement horizontally = 2 meters.Frictional force along the horizontal in the direction opposite to the displacement = Ff = μ * N = μ m g = 0.25 * 1.8 kg * 10 m/s² = 4.5 NewtonsWork done by the Frictional force = Wf = Ff . s = 4.5 * (-2 meters) = - 9 Joules (it is negative as displacement is in the opposite direction to friction). The component of force F, along the horizontal in the direction of s : F Sin 45⁰ = F/√2Since, the block moves with a uniform speed, there is no net force on the block. Hence F/√2 = Ff = 4.5 Newtons F = 4.5 √2 Newtons Work done by F on the block = F . s = F/√2 * 2 = 9 Joules. This is positive as the force does positive work on the block. Gravitational force, ie., the weight and the normal force N are perpendicular to the displacement vector. Hence, the dot product is 0, for the work done. |
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| 16. |
In the given figure, D and E trisect BC. Prove that:AEVERGREEN 10%SUCCESS-NM THEMA Ka-o |
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Answer» thank you |
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| 17. |
14) Find <1, 2.2, <3 and 1.4AA7060°120°△2415) Find L1, 2.2, <3, <4and 1.5 |
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Answer» 60+70+ang1=180Ang3=50°Ang1=180-120=60°60+60+ang2=180ang2=120° |
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| 18. |
Example 5. Prove that:(i) (2n)!2nn. 1. 3. 5....(2n 1)].. (2n 1)!l12"1.3.5 2 1) (2n 1)].rl |
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| 19. |
tion 4. IIU l1 2 3,4, 5, 6, 7, 8,9)12 4,6, 8 and B- 12, 3, 5, 7), verify that(ii) (ABYA LE |
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| 20. |
L11-Draw a line segment of length 7.6 cm and divide it in the ratio 5:8. Measure the wo part12: Find the area of the sector of a circle with radius 4 cm and of angle 30. Also, find the area of the correspondingmajor sector (Use Î= 3.14) |
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| 21. |
() 120(TT) 40°(G) 600(T) 300 |
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Answer» 2*angle in segment=angle at the centre by same chord Hence 2*angle BAC=angle BOC angle BOC=120 degrees Kase |
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| 22. |
" T e s —————————————AW 2890 = 78G5 ™ = (10 Dewa |
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Answer» 75 rupees 25 paise = 75.25 rupees.1 rupee = 100 paise. Please hit the like button |
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| 23. |
(10) The total cost of 3 chairs and 2 tables is Rs. 4500. The total cost of2 chairs and 3 tables is Rs. 5500. Find the total cost of a chair anda table. |
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Answer» let the cost of chair be x and table be y3x + 2y = 4500......(1)2x + 3y = 5500......(2)(1)×3;9x + 6y = 13500......(3)(2)×2;4x + 6y = 11000......(4)(3)-(4);5x = 2500x = Rs 500then y = Rs 1500 |
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| 24. |
Example 6) Solve22 28Solution:Multiply by 88, the least commmon multiple of the denominatoors352-11(x-9) = 4x-4→4x-44removing brackets, 352-11x + 99--11x-4x =-44-352-99transposing,collecting terms and changing signs, 15x -495.. x-33x-4 2x-3 5x-32+9335(ii) Solve928Solution: |
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| 25. |
A vertical pole of length 6 m casts a shadow 4ong on the ground and at the same time a towasts a shadow 28 m long. Find the height of theower. |
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Answer» intersting question. please give anssr fast |
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| 26. |
perpendicular sides of a right angled triangle are 10cm and 24cm.This triangle is revolved about its hypotenuse.Find the volume of the object formed |
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| 27. |
3 units from this point.17. Т.е hypotenuse of a right angled triangle has its ends at the points (1,3)(-4, 1). Find the equation of the legs (perpendicular sides) of the triangle.and |
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Answer» thanks |
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| 28. |
On both sidesDewa bine and a perpendicular to it as below: |
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Answer» It is a right angle. right angle is the right answer right angle is the right thing |
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| 29. |
If a perpendicular is drawn fromthe vertex of the right angle of a right triangle tothe hypotenuse then triangles on both sides ofthe perpendicular are similar to the whole triangleand to each other |
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| 30. |
Single correct answer.1. The curved surface area ofone cone is twice that of the other cone. The slantis twiceheight of the latter is twice that of the former.Find the ratio of their radii.inder is 'r' and its height |
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| 31. |
In a certain A.P. 31th term is twice the 12th term. Prove that 70th term is twice the 31th term. |
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Answer» T₃₂=a+(32-1)d=a+31dwhere a=1st term and d=common difference of the A.P.t₁₂=a+(12-1)d=a+11dNow by the given condition,a+31d=2(a+11d)or, a+31d=2a+22dor, a-2a=22d-31dor, -a=-9dor, a=9dNow,t₃₁=a+(31-1)d=a+30d=9d+30d=39d∴, t₇₀=a+(70-1)d=a+69d=9d+69d=78d=2×39d=2t₃₁ (Proved) |
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| 32. |
(i){2, 3, 4}{1, 2. 3. 4. 5} |
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| 33. |
Add the product of 2 p^{2} q^{2} \text { and }\left(\frac{-3}{4} p^{3} q\right)to the product of \left(-5 p^{4} q\right) \text { and }\left(p q^{2}\right) |
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| 34. |
Prove that the median of an equilatertriangle is perpendicular to the correspondineside. |
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| 35. |
3)If the diagonal of square is 12 N2 cm. then the perimeterof square is(a) 24cm(b)24 12cm ()48em (d)48 y2em、)24V2 |
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| 36. |
Find the area of a right-angled triangle if its perpendicular sides are of length2(2x +3) and (3x + 2) units.5.r - 2 and |
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| 37. |
the diagonal of a rhombus are 75cm and 12cm . find its area |
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Answer» Area is rhombus is product of diagonals/2=75*12/2=75*6=450cm^2 |
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| 38. |
find the area of isosceles right triangle the length of whose diagonal is12cm(ii)20 cm |
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| 39. |
qual onarea. (c) The surface area and the diagonal of a cuboid are 288sq.cm and 12cmrespectively. Show that the cuboid is cube. |
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| 40. |
4. The diagonal of a quadrilateral shaped field is 24 mand the perpendiculars dropped on it from theremaining opposite vertices are 8 m and 13 m. Find12cm Findis perpeit13 m24 mthe area of the field.8 m |
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| 41. |
HRI P aQ. A(2,-2)阿B(-7, 4)动fire 付可 ταταて前研师R.HTTA f丽Let P and Q be the points of trisection of the line segment joining the pointsA(2, -2) and B(-7, 4) such that P is nearer to A. Find the coordinates ofP and Q. |
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| 42. |
तहत 09-1५ व या] el P )L -9 . 8N -- HON4 ] ) फ़िर,ut <HA Q#/7 (ot ह t %g cad Q'] + (0७०3 4IR od 2 |
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| 43. |
\left(\frac{3}{7} p^{2}+4 q^{2}\right) \text { by } 7\left(p^{2}-\frac{3}{4} q^{2}\right) |
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Answer» +3p power 4 +103/4,p power 2 - 21q power 4 Thats the answer |
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| 44. |
22. AD is an altitude of an equilateral triangle ABC. On AD as base, another equilatertriangle ADE is constructed. Prove that Area ( AADE ) : Area ( Δ ABC)-3 : 4. [CBSE 201 |
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| 45. |
If the mean of 1,3,4,5,7,4 is m and mean of 3,2,2,4,3,3, p is (m-1) and medianis q then p +2=(1) 4(2) 5(3) 6(4) 7यदि 1,3,4,5,7,4 का माध्य ! है तथा 3,2,2,4,3,3, 2 का माध्य ? । एवम् माध्यिका १ है।a p + q =(1) 4(3) 6(2) 5(4) 7 |
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| 46. |
âĄPQRS is an isosceles trapezium1(P7Q) = 7 cm. seg PMIL seg SR,!(SM) ,Distance between two parallelsides is 4 cm, find the[-] PQRS=3cm4area of S3 Mr Let's learn. |
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Answer» Area of parallelogram=(1/2)(sum of parallel sides)*height=(1/2)(7+7+3+3)*4 =(1/2)(20)*4=40cm² |
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| 47. |
0.13. AD is an altitude of an equilateral triangle ABC. On AD as base, another equilatertriangle ADE is constructed. Prove that Area (AADE) :ABC)-3 : 4 |
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| 48. |
of the ptheir tops.D and E are points on the sides CA and CBrespectively of a triangle ABC right angled at CProve that AE+BD-AB+ DEThe perpendicular from A on side BC of aA ABC intersects BC at D such that DB 3 CD(see Fig. 6.55). Prove that2AB2-2AC2+BC. %.In an equilateral triangle ABC, D is a point on side EIn an equilateral triangle, provesthat three times thits altitudes |
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| 49. |
18.. AD is an altitude ofan equilateral triangle ABC. On AD as base another equilateral triangle ADEis constructed. Prove that ar(AADE) : ar(AABC) = 3 ; 4. |
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| 50. |
30. AD is an altitude of an equilateral triangle ABC. On AD as base, another equilateral triangleADE is constructed. Prove thatArea (AADE): Area (AABC) 3:4 |
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