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.

9. Shobha had 127 metres of ribbon. She usedof it to wrap a present and of the remainingribbon on a craft project. How many metresof ribbon is left with her?2019/4/12 16

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When she used 2/3hence49/4 metre is total2/3*49/4=49*2/12=49/6m usedhence remaining is49/4-49/6=49(1/4-1/6)=49/12mshe used half of thisthat is49/12*1/2=49/24m left

2.

p ics пo, Trust аtе уоur ѕwе му и сланире.14. Express each of the following numbers as the sum of two odd primes:(1) 36(ii) 42(11) 84tlv) 98

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3.

9. Shobha had 127 metres of ribbon. She usedof it to wrap a present and of the remainingribbon on a craft project. How many metresof ribbon is left with her?2019/4/12 16:2

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ribbon used to wrap up a present = 2/3* 12 1/4 = 49/6ribbon used on a craft project = 1/2*(12 1/4 - 49/6) = 49/24 = 49/12left ribbon = 12 1/4 - (49/6 + 49/12)=

4.

1.Find the volume of a cube of edge 7cm

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5.

12. What are twin primes? Write all the pairs of twin prunes DE WUCHU13. What are co-primes? Give examples of five pairs of co-primes. Are co-primes alwaysprimes? If no, Illustrate your answer by an example,14. Express each of the following numbers as the sum of two odd primes:

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6.

19. A cylindrical vessel open at the top has a basediameter 56 cm. If the total cost of painting the outercurved surface of the vessel is 352 at the rate of0.2 per 100 cm2, then the height of the vessel is(1) 15 m(3) 6 m(2) 10 m(4) 12 m

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7.

4.Express each of the following numbers as sum of two odd primes.

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8.

In Fig. if PT is a tangent of the circle with centre O and TPOmeasure of x is:25(A) 120(B) 125°(C) 110

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From which book have you taken this question? Please tell us so that we can provide you faster answer.

9.

I, If each side of a cube is 5 em. Find its volume.2. If the dimens

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Volume of cube is (side)³Volume of cube is (5 cm)³ = 125 cm³

10.

リ7cm in m

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1m=100cmhence 7cm=7/100m=0.07m

Thanks

11.

10.2H. 1, 2, 3(A) 7cm(C) 15 cm(B) 12cm(D) 24.5 cm

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(a) 7cm correct answer me

12.

Find the area of the shaded region in figure, ifABCD is a square of side 7cm and APD andBPC are semi circles.7cm

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Waste

13.

tta7cm

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14.

A bucket made-up of a metal sheet is in the form of a frustum of a cone of height 16 cmwith diameter of its lower and upper ends as 16 cm and 40 cm respectively. Find thevolume of the bucket. Also, find the cost of the bucket if the cost of metal sheet used is 20per 100 cm2.28.

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15.

8. A bucket made up of a metal sheet is in the form of a frustum of a coneof height 16 cm with radii of its lower and upper ends as 8 cm and 20 cmrespectively. Find the cost of the bucket if the cost of metal sheet used is[CBSE 2006, '08, '08C]15 per 100 cm2.[Use T 3.14.]

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16.

Find the surface arca oi a wouelT DOFind the cost ofpainting the all round outer surface ofa box with lid60 cm long, 40 cm w/ide and 30㎝high at 50 p per 20 cm2wtcar11 0m45rm Find the area of the tin sheet required for making

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17.

A bucket made-up of a metal sheet is in the form of a frustum of a cone of height 16 cmwith diameter of its lower and upper ends as 16 cm and 40 cm respectively. Find thevolume of the bucket. Also, find the cost of the bucket if the cost of metal sheet used is 20per 100 cm228.

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18.

S, A container, open at the top and made up of metal shee, is in the formof a frustum of a cone of height 16 cm with diameters of its lower andupper ends as 16 cm and 40 cm respectively. Find the cost of metal sheetused to make the container, if it costs 10 per 100 cm2. [CBSE 2013]

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19.

a frusium of aThe diameters of the lower and upper ends of a bucket in the form of24 cm, find:cone are 10 cm and 30 cm respectively. If its height is(i) The area of the metal sheet used to make the bucket.(ii) Why we should avoid the bucket made by ordinary plastic 2 [Use T3.1410

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GIVEN :Diameter of upper end of bucket =30cmRadius of the upper end of the frustum of cone( r1) = 15cmDiameter of lower end of bucket = 10 cmradius of the lower end of the frustum of cone( r2) = 5 cmH of the frustum of cone = 24 cmSlant height of bucket ( L)= √(h² + (r1- r2)²L =√24² + (15 - 5)² = √576 + 10²L =√(576+(100)= √676 = 26cmL = 26 cmArea of metal sheet require to make it = π(r1 + r2)L + πr1²

= 3.14(15 + 5) × 26 + π(5)² = 3.14 × 20 × 26 + 25 × 3.14= 3.14 (520+ 25)= 545 × 3.14= 1711.3 cm²

Hence, the Area of metal sheet used to make the bucket is 1711.3 cm².

20.

7. How many bricks of length, breadth, and height 25 cm, 15 cm, 5 cmrespectively will be required to build a wall 50 m long 6 m high and m wide:(1) 60,000(3) 1,20,000(2) 48,000(4) 80,000

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Option 4 is correct

21.

4.What will his mantle cover?

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please mention the story

Plz tell the story of the following is

22.

Vinay wants to cover a floor of 7 m wide and 4 m long by squared tiles. If each squaretile is of side 0.25 m, then how many number of tiles are required to cover the floor ofthe room?

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Area of floor = length× breadth=7m×4m =28m^2area of tile= side × side= 0.25m×0.25m =0.0625m^2no of tile= area of floor/area of tiles =28/0.0625 = 448 tiles

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23.

Date:FindtheLomof25 f 110

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The LCM of 25,110 is 550.

24.

30 per kg to? 36 per kg. Find the percentage rise in theThe price of rice rises fromprice of rice.

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25.

Date:Find the TogoA, C.S.A and volumeof the cylinder a) r=7cm h=1.4cm

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26.

Find. zes asDate

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27.

Date:Page NoFind the

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Please like the solution

28.

5. Vinay wants to cover a floor of 7 m wide and 4 m long by squared tiles. If each squaretile is of side 0.25 m, then how many number of tiles are required to cover the floor ofthe room?

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Area of floor= length×breadth= 7m×4m = 28 m^2area of tile =side×side=0.25m×0.25m =0.0625 m^2no of tile =area of floor/ area of tiles = 28/0.0625 =448 tiles

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29.

7. 1 o & 15 6६८ Lhe ,/"‘(!"vv)‘ 2 0} Ahe दिPlx) =2 4 5x + K r)ovH"{ 8‘("8welakon 2 i P 4 At L (35 5valus न ८, I

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30.

The value of a, for which the points A,B, Cwith position vectors 2i -j+k,i -3j -5kand ai -3j +k respectively are the verticesof a right angled triangle with C are

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31.

ind k, if m-iis a root of equation 5m2+2m + k

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32.

(3)r(ep)4. Determine whether the values given against each of the quadratic equation are theroots of the equation.(1)x2 + 4x _ 5 = 0,x=1,-1(2) 2m2-5m2-0, m=2,52

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33.

\frac { x ^ { 2 } + 5 x } { x ^ { 2 } - 9 } - \frac { 7 x + 3 } { x ^ { 2 } - 9 }

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34.

( i v ) \frac { x ^ { 2 } - 4 x - 9 } { ( x - 2 ) ( x ^ { 2 } + 9 ) } = \frac { A } { x - 2 } + \frac { B x + C } { ( x ^ { 2 } + 1 ) }

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35.

\frac { x ^ { 2 } - 4 x - 9 } { ( x - 2 ) \left( x ^ { 2 } + 9 \right) } = \frac { A } { x - 2 } + \frac { B x + C } { \left( x ^ { 2 } + 1 \right) }

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36.

Evaluate the following limits\lim _{x \rightarrow 3} \frac{x^{2}-9}{x^{2}\left(x^{2}-6 x+9\right)}

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37.

Mother drives a car at a uniform speed of 60 km/hr The time taken to cover 280km is

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thxx

38.

Instead we use 12 noon f 1EXERCIBE 56Fill in the blanks with a.m. or pan. in Ayushree's daily routine.Avushree's school bus arrives at the bus stop at 70sShe has a short break in the school at 11:15She comes back home at 2.30-4 She watches her favourite cartoon show at 4:305 After that she goes out to play tennis at 6:006She has her dinner at 8:00Once she was woken up in the night by the noise of some people at 2She gets up at 5.30everyday to get ready for school

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1. am2. am 3. pm4. pm5. pm6. pm7.pm8.am

39.

find the valus a

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According to the problem ,

a ) cos ( x - 20 ) = sin ( 3x - 10 )

⇒ cos ( x - 20 ) = cos [ 90 - ( 3x - 10 ) ]

⇒ x - 20 = [ 90 - ( 3x - 10 ) ]

⇒ x - 20 = 90 - 3x + 10

⇒ x + 3x = 90 + 10 + 20

⇒ 4x = 120

⇒ x = 120 / 4

∴ x = 30°

40.

Datefind the valus o

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sin2x=sin(90-30)cos30-cos(90-30)sin30sin2x=cos30×cos30-sin30×sin30sin2x=cos²30-sin²30sin2x={√3/2}²-{1/2}²sin2x=3/4-1/4sin2x=2/4sin2x=1/2

41.

ea of the13. A line formas a triangle in the first quadrant with coordinates axes. If the area oftriangle is 54y3 sq, units and the perpendicular drawn from the origin to the linemakes an angle 60° with x-axis, find the equation of the line.

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A line AB forms a triangle with coordinate axes of area 54√3 square unit and also perpendicular drawn from the origin to the line makes an angle of 60° with x -axis as shown in figure.

area of triangle = 1/2 × height × base see attachment, height = a and base = b then, area of triangle = 1/2ab = 54√3 ⇒ab = 108√3 ------(1)

Now, Let P is the length of perpendicular drawn from origin to line Then, cos60° = P/b ⇒b = P/cos60° = 2P cos30° = P/a ⇒a = P/cos30° = 2P/√3

Now, put a and b value in equation (1), 4P²/√3 = 108√3 ⇒P² = 108 × 3 P = ±18 , but we have to take only P = 18 because triangle is in 1st quadrant .a = 2P/√3 = 36/√3 = 12√3b = 2P = 36 so, equation of line : cos60°x + sin60° y= P x + √3y - 36 = 0

Hence, equation of line : x + √3y -36 = 0

42.

Exercisedinciple of mathemfatical induction prove that for all positive integral values of n:1 + 2 +3+

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43.

। (3+ 2x)99 के विस्तारण में गुणांकों का योग क्या है ?(B) 399(D) 699(A) 599(C) 299

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option A) is the correct answer

a) is the right answer of the following

44.

The number 299 is divided into two parts in the ratio 5:8. The product of the numbers is ____________. Fill in.

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45.

θ=60° ,不\frac{1}{2} \sqrt{1+\sin \theta}+\frac{1}{2} \sqrt{1-\sin \theta}0

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46.

37c43X 2 9 then what tho valus

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47.

In each of the following find the value of 'k', for which the points are collinear(7,-2),(5,1) and (3,k)(8,1) (k,-4) and ,(2,-5)

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48.

7(2) Find k, if m =-5-is a root of equation 5m2 + 2ml + k = 0.

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Put m= -7/5 and equate to 0, in given equation.

5m² + 2m + k = 05×(-7/5)² + 2×(-7/5) + k = 049/5 -14/5 + k = 035/5 + k = 0k = -7

49.

20 mm100 mm20 mmi080 mm-

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50.

Fromhefaical induction prove that for all positive integral values ofn2

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