Explore topic-wise InterviewSolutions in Current Affairs.

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.

=e : i= T =A-x9= . e ‘ . टनg=1-s MW - pI=4+x ()मधु 192 & सु नानक हि सर्वर तप SR bk T€€ 2210-06.

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2.

u) नमसका गहा।06.x2-x-(2k+2) का एक शून्यक-4है तोk का मान निम्न मेंसे कौन होगा?

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3.

Illustration 1.60 Write the equivalent (piecewise) definitionof f(x) = sgn(sin x).

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4.

Illustration 1.71 If the function f: Rf(x) is surjection, then find AA given byr2 +1

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5.

There are x students tyring out for a solo in a chorus concert. Only 6 will bechoosen how many students will not be chosen?06.

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Total number of students applying xNumber of students chosen 6Number of students left x - 6

6.

: Gl’ve 0:06 20— oo JJD({[ Ao कर 110 व दे

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7.

28.A particke of mass m is observed from an inertialframe of reference and is found to move in acircle of radius r with a unifrom speed v. Thecentrifugal force on it is)-towards the centreaway from the centremy2(3)along the tangent through the particle

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8.

Illustration 2. 2:8::3:?(a) 20(b) 21(c) 24(d) 27

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2 : 8 = 2 : 2×2×2 3 : ? = 3 : 3×3×3 = 3 : 27 ? = 27

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9.

Illustration 2.(a) 9(b) 5(c) 10 (d) 21uttorn is as follows

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Here B is rightas5+4= 9 then 9+4= 13

10.

Illustration 2. What is the sum of+22 +32+. + 20?(a) 2800(b) 2820(e) 2870(d) 2900

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Sum will ben(n+1)(2n+1)/6here n = 2020*21(41)/62870 is the answer

11.

ILLUSTRATION(i) 5 15 166 18 8

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12.

(K) form bober(2) Give a suitable illustration of Dobereiner'slaw of triads.ndium and notassium form

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Dobereiner’s law of triads states that, the atomic mass of the middle element of a triad is the arithmetic mean of the atomic masses of the other two elements.Example:In the triad of lithium, sodium and potassium. The atomic mass of lithium is 7 and the atomic mass of potassium is 39. The average of masses of lithium and potassium gives atomic mass of sodium 23.

13.

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500rupey vjfdhxdrhbfhhfdgghjvcffiggf

14.

9- 24(3-x)-06(2x-3)=0

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2.4(3-x)-0.6(2x-3)=07.2-2.4x-1.2x+1.8=09-3.6x=03.6x=9x=90/36=10/4=5/2=2.5

15.

21.A cone of radius 8 cm and height 12 cm is divided into two parts by a plane through themid-point of its axis parallel to its base. Find the ratio of the volumes of two parts.

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16.

न न डिनडान ला ल िनि v b % :सिर. |® bbb BE 13 1k 1 lkok| [0 चर अफसर सु bk 8T RS Le ERBAy

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17.

4. In a circle of radius 5 cm, AB and CD are two parallel chords of lengths12.8 cm and 6 cm respectively. Calculate the distance between the chordsif they are(i) on the same side of the centre,(ii) on the opposite sides of the centre.

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18.

\begin{array} { l } { \text { The lengths of the two parallel sides of a } } \\ { \text { trapezium are } 10 \mathrm { cm } \text { and } 14 \mathrm { cm } \text { . If the distance } } \\ { \text { between them is } 8 \mathrm { cm } , \text { find the area. } } \end{array}

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area= sum of parallel sides * distance between them/2. Here= 1/2*(10+14)*8= 96cm ^2

19.

10. किसी संख्या में 0-6 का गुणा करने पर गुणनफल758:42 आता है. यदि उस संख्या में 0-06 का गुणाकरें तो गुणनफल होगा-

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20.

$ १७८9 21 PO~ 20 g+ 06

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Like if you find it useful

5+/(64)_/(36) ka maan Kaya h

21.

3x +5y 72If3x-5y-3-find χ ),find x : y

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3 ( 3x + 5y) = 7 ( 3x - 5y)

9x + 15y = 21x - 35y

35y + 15y = 21x - 9x

50y = 12x

x/y = 12/50

x/y = 6/25

22.

(0 .06) WS (09— .06) 5%6505 0 पड

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sin cos------------------- + -----------------cos(90 -) sin ( 90 - )

sin cos-------- + ----------sin cos

1 + 1

2

23.

Illustration 2:External liabilities and assets of a business are70,000 and4,50,000 respectively. How much isthe capital?2018/12/27 13:06

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24.

5x + 2y = -3; x + 5y = 4

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25.

)) y 32x) 2y°—5y“—19y+42

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2y*y*y - 5y*y - 19y + 42 = 2y*y*y - 4y*y - y*y + 2y - 21y + 42 = 2y*y(y-2)-y(y-2)-21(y-2) = (2y*y-y-42)(y-2) = (2y*y+6y-7y-42)(y-2) = (2y-7)(y+6)(y-2)

26.

5x + 2y = -3; x + 5y = 4

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5x+2y=-3; x+5y=4; 5(5x+2y=3)=25x+10y=15; 2(x+5y=4)=2x+10y=8; /23x=7;x=7/23; 5x+2y=-3; 5(7/23)+2y=-3; 35/115+2y=-3; 2y=-3+35/115=-345+35/115=-310/115=-62/23; y=-62/23(2)=-31/23

27.

06 0T4 。

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546 - 400 = 146 is the answer

28.

4.05 : 20 : : 06 : ?

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5:20::06:x1:4::06::XX =6*4=24

29.

of a circle. Find out lllTuuchords of lebetween the c4. In a circle of radius 5 cm, AB and CD are two parallel8 cm and 6 cm respectively. Calculate the distance beif they are(i) on the same side of the centre,(ii) on the opposite sides of the centre.

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30.

B ol oSle 13 : Simplify: 8 2 2 5

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8 1/4 - 2 5/6 = 33/4 - 17/6 = (33×6-17×4)/(6×4) = (198-68)/24 = 130/24 = 65/12 = 5 5/12

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31.

le Findthevalueoftar.a n^{-1}\left(\tan \frac{9 \pi}{8}\right)

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= tan^-1 ( tan 9π/8)

= 9π/8

= 9*360/8

= 9*45

= 405

32.

B =0 06 =0

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33.

112. 33 + 4406

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thank you

34.

s. The height of a parallelogram is one-third of its base. I the area of the parallelogram is108 cm2, find its base and height.

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Let the base be x

Then height = x/3

Area of llgm= 108cm²

b×h= 108cm²

x*x/3= 108

x²= 108*3

x= √324

x= 18cm

Base of parallogram= 18cm

Height of the parallogram = x/3

= 18/3

= 6cm

35.

Area of a trapezium is 108 cm2. Length of one of the parallel sides is 10 cm and its height is12 cm, find the length of other side.

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Area = (1/2)×h×(a+b) 108 = (1/2)×12×(10+b) 10 + b = 108/6 b = 18 - 10 = 8cm

36.

43. Find the value of/9/9/9 ......(2) 81 (B)3 49 (d) co

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9

9

c is the right answer

37.

Simplify (x-2/5y)^3-(x+2/5y)^3

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38.

Simplify (x-2/5y)^3-(x+2/5)y^3

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39.

SIMPLIFY the following exprssse8x-(5y-3)(3 x+5 y)-(2 y-z)

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a) 8x - (5y - 3) = 8x - 5y + 3

b) (3x + 5y) - (2y - z) = 3x + 5y - 2y + z= 3x + 3y + z

40.

solve leaear equation by the substituation method 3x/2-5y-/3 =-2 x/3 + y/2=13/6

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41.

7. The adjacent sides of a parallelogram arecm and 8 cm. Find the altitudes corre-sponding to these bases if its area is 108 cm212

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thanks Nishesh

42.

7. The adjacetlonger sides is 4 cm, find the distance betweei le JllU8. The height of a parallelogram is one-third of its base. If the area of the parallelogram is108 cm2, find its base and height2llrlorram is twice its height. If the area of the parallelogram is 512 cm

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43.

25dooring tile has the shape of a parallelogram whose base is 24 cm and thea 108the corners)sponding height is 10cm. How many such tiles are required to cover floor of80 m2? (If required you can split the tiles in whatever way you want to ill uparea

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Given : base= 24 cm

Height= 20 cm

Area of floor= 1080 m²= 1080×100×100cm²

[1m²= 10000cm²]

Area of the Parallelogram = base xheight

= 24 x 10 = 240 cm²

Area of1 tile = 240 cm²

[Tiles are in the shape of a parallelogram]

Number of tiles required= Area of theFloor/Area of the Tiles

= 1080 X100 X100/240 =45000

Number of tiles required to cover a Floor=45000

thank you

44.

(12) The height of a parallelogram isone-third its base. If the area of theparallelogram is 108 cm2, find its baseand height.

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45.

The height of a parallelogram is one-third of its base. If the area of the parallelogram is108 cm2, find its base and height.

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46.

68. यदि 9 * 3 = 36; f1 * 7 = 81, तो 5 *13 का मान क्या होगा?(a) 81(b) 51(c) 65(d) 72

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9*3 = 9*3 + (9-3) = 36

11*7 = 11*7 + (11-7) = 81

Then, 5*13 = 5*13 + (5 - 13) = 65 - 8= 57

72.

Dont know the answer of this

57 is the answer......

51 is a answer of these question

47.

If 3x-5y=2x+y,find the value of 2x²+5y²/2x²-5y²

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Here 3x -5y = 2x + y => 3x-2x = y+5y = 6yso , x = 6y , putting this value below we get, { 2(6y)²+5y² }/2(6y²) -5y² = (72+5)y²/(72-5)y². y² will cancel each other. so the answer is 77/67.

48.

49 x ^ { 2 } - \frac { 81 } { 4 } \text { by } 7 x + \frac { 9 } { 2 }

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49.

12. Which term of the A.P.5, 9, 13, 17,... is 81?

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20th term is 81 of this AP

50.

3x+5y=7.5

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Bhai kya karna hai smajh nhi aa rha hai