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Let the piece of cloth is x metre and it costs Rs.y/metre. Then by the given conditions -∴, xy=200 -------------(1) and(x+5)(y-2)=200or, xy+5y-2x-10=200or, 200+5y-2x-10=200or, 5y-2x-10=0or, 5y=2x+10or, y=(2x+10)/5Putting in (1),x(2x+10)/5=200or, x(2x+10)=200×5or, 2x²+10x=1000or, x²+5x-500=0or, x²+25x-20x-500=0or, x(x+25)-20(x+25)=0or, (x+25)(x-20)=0either, x+25=0or, x=-25or, x-20=0or, x=20Since the length of the cloth can not be negative,∴, x=20 metrePutting in (1),20y=200or, y=10∴, the cloth is 20 m long and the original rate is Rs. 10/metre. Ans.

I gave20% to rakesh and suresh . suresh get more 7.5 rupees so how much money I have

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Ans. is "REGNIFYDAL" all letter reverse from right side to left side...

(sin5x-2sin3x+sinx)/(cos5x-cosx)=[{2sin(5x+x)/2cos(5x-x)/2}-2sin3x]/{2sin(5x+x)/2sin(x-5x)/2}=(2sin3xcos2x-2sin3x)/{2sin3xsin(-2x)}={2sin3x(cos2x-1)}/{-2sin3xsin2x}=-(cos2x-1)/sin2x=(1-cos2x)/sin2x=2sin²x/2sinxcosx=sinx/cosx=tanx (Proved)

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2x^4-3x^3-3x^2+6x-2 by V2; 2(V2)^4-3(V2)^3-3(V2)^2+6(V2)-2=8-6V2+6+6v2-2= 14-2=12

Bhai ek ek karke puchh lo

3x^2-5x+2=0(√3x)^2-2(√3 x)(5/2√3)+(5 2√3)^2+2-25/12=0(√3x-5/2√3)^2 + 24-25/12=0(√3x-5/2√3)^2=1/12(√3x-5/2√3)=√1/12=+- 1/2√3√3x= +1/2√3 + 5/2√3 = 1+5/2√3x= 6/2×√3√3=6/2×3=6/6=1 or √3x-5/2√3= - 1/2√3√3x= 5/2√3 -1/2√3 = 5-1/2√3x= 4/2√3√3=4/2×3=2/3

x=2/3 .. is right answerr

Radius of sphere = 3 cm

Now volume of sphere =4/3 π(3)³

Now sphere is melted and draw into wire

Volume of sphere = Volume of cylinder formed

4/3 π(3)³ = πr² × 3600

36π=r²x3600π

l=r²x100

r² = 1/100

r=1/10cm = 0.1cm

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(a) n=9(odd) Mean =sum of observation /total no. Of observation 6+5+3+5+1+6+2+7+10/9=45/9=5So mean =5

Railing length for fencing semicircular Park = 288 m

Let radius of semicircular Park = r

Then,Pi*r + r = 288r(22/7 + 2) = 288r = 288*7/36r = 8*7 = 56 m

Area of semicircular park= 1/2*pi*r*r= 1/2*22/7*56*56= 11*8*56= 4928 m^2

Let radius of the circular park = r = 105 m

Width of the road surrounded by the

Park = w = 21 m

Radius of the outer circle = R = r + w

R = 105 + 21 = 126 m

Area of the road = area of the outer

circle - area of the inner circle

A = pi × R ^2 - pi × r ^2

A = pi ( R ^2 - r ^2 )

A = pi × ( R + r ) ( R - r )

A = pi × ( R + r ) × w

Substitute the values

A = 22/ 7 × ( 126 + 105 ) × 21

A = 22 × 231 × 3

A = 15246 square meters.

Therefore,

Area of the road = 15246sq meters

y/x= 4/15x कॉमन लेने पर(1-y/x)/(1+y/x)(1-4/15)/(1+4/15)11/15/19/1511/19

A line segment has fixed value hence it will have only one midpoint.

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2(x-5) +3(x+6) =122x-10+3x+18=12x+8=12x=12-8x=4

Let z = (8+i)/(1 + 8i)

= (8+i)*(1 -8i)/ (1 + 8i)(1 - 8i)

= (8 - 64i + i + 8)/ (1 + 64)

= (16 - 63i)/65

Modulus of z= sqrt[(16/65)^2 + (-63/65)^2]= sqrt[(256 + 3969)/4225]= sqrt(4225/4225)= sqrt(1) = 1

y = 130° ( vertically opposite angle)x = 130° ( corresponding angle)

x + 50° = 180° (linear pair)x = 180° - 50° = 130°y = 130° ( vertically opposite angle)as corresponding angle are equal so AB ||CD

11/13,11/19,11/25,11/27

11/27,11/25,11/19,11/13

B because answer is on right side

answer is ( A ) 9 : 7.

=(20+12.5):(10+17.5)=32.5 :27.5 =13:11

adding 11,8,3 so we will get 20 so the final answer will be 20/11

let first part x and 2nd part 34 - x

a/c to question ,

4/7 (x ) = (34 - x )(2/5)

4x/7 + 2x/5 = 34 x 2/5 = 68/5

34x/35 = 68/5

x = 14