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let the three consecutive integers be x, x+1, x+2

(x)+(x+1)+(x+2)=51

3x+3=51

3x=51-3=48

x=48/3

x=16

the three consecutive integers are

x=16, x+1=17, x+2=18

largest number will be 7321and lowest will be 1237hence difference is 7321-1237=6084

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41x(-5)+.....=-3-205+x=-3x=-3+205x=202

2x+3y=11........ (1)2x-4y=-24........(2)

(1)-(2)

7y=35y=5

put in eq. 12x+3(5)=112x=11-152x=-4x=-2

2x + 3y = 11....(1)x = (11 - 3y)/2

2x - 4y = - 24.....(2)

Put value of x in eq(2)2*(11-3y)/2 - 4y = - 2411 - 3y - 4y = - 2411 - 7y = - 247y = 35y = 35/7y = 5

Put value of y=5 in eq(1)2x + 3*5 = 112x = 11 - 152x = - 4x = - 4/2x = - 2

Let the numbers be x, x + 1, x + 2, x + 3, x + 4, x + 5 and x + 6,

Then (x + (x + 1) + (x + 2) + (x + 3) + (x + 4) + (x + 5) + (x + 6)) / 7 = 20.

or 7x + 21 = 140 or 7x = 119 or x =17.

Latest number = x + 6 = 23.

Awesome Solution. Take the numbers as x-3, x-2, x-1, x, x+1, x+2, x+3 gives a faster answer

the right answer is 23

Let the side of an equilateral triangle is a

perimeter of equilateral triangle= area of equilateral triangle

3a= (√3/4)a^2a= (4×3)/√3a= (4×3)×√3/ √3×√3. ( by rationalising)a= 4√3side = 4√3perimeter of equilateral triangle= 3aperimeter of equilateral triangle= 3× 4√3perimeter of equilateral triangle= 12√3perimeter of equilateral triangle= 12× 1.732perimeter of equilateral triangle= 20.78 units

[ value of √3 = 1.732]

tan60= perpendicular / baseso√3= height of the tower/ distance from its footso,√3=H/15so, H=15√3m

Tanθ/2=√(1-e)/(1+e) tanФ/2or, tanФ/2=√(1+e)/(1-e) tanθ/2squaring both sides,tan²Ф/2=(1+e)/(1-e) tan²θ/2or, (1-tan²Ф/2)/(1+tan²Ф/2)={(1-e)-(1+e)tan²θ/2}/{(1-e)+(1+e)tan²θ/2}[by dividendo-componendo method]or, cosФ=(1-e-tan²θ/2-etan²θ/2)/(1-e+tan²θ/2+etan²θ/2)or, cosФ={(1-tan²θ/2)-e(1+tan²θ/2)}/{(1+tan²θ/2)-e(1-tan²θ/2)}or, cosФ=[{(1-tan²θ/2)-e(1+tan²θ/2)}/(1+tan²θ/2)]/ [{(1+tan²θ/2)-e(1-tan²θ/2)}/(1+tan²θ/2)]or, cosФ=(cosθ-e)/(1-ecosθ) [∵, (1-tan²θ/2)/(1+tan²θ/2)=cosθ] (Proved)

28x⁴ ÷ 56x

Cancel a x and 28 and 56 by 28

x³ ÷ 2

x³/2

28x⁴------56x

x³----- 2

28x^2*x^2/ 56x= x^3/2please like the solution

Given,Sin theta = 12/13

Cos theta = sqrt(1 - sin^2 theta)= sqrt(1 - (12/13)^2) = sqrt(1 - 144/169)= sqrt(169 - 144)/13= sqrt(25)/13= 5/13

tan theta = sin theta/cos theta=12/13 / 5/13= 12/5

cos theta is 5/13

tan theta is 12/5

Let theta = a

LHS:(sina+cosa)/(sina-cosa)+(sina-cosa)/(sina+cosa)=((Sina+cosa)^2)+((sina-cosa)^2)/((sina-cosa)(sina+cosa))=(Sin^2a+cos^2a+2sinacosa+sin^2a+cos^2a-2sinacosa)/(sin^2a-cos^2a)=2(sin^2a+cos^2a)/(sin^2a -cos^2a )=2/(sin^2a -cos^2a )Multiply and Divide by cos^2a=2sec^2a/tan^2a - 1= RHSHence proved

28x^4÷56x28/56(x)^[4-1]= 1/2(x)^3 ans

21x²y - 28x²y³

7x²y ( 3 - 4y)

0.6x-0.28x= 1.16-0.80.32x= 0.8x= (0.8/.32)= x= 2.5please like the solution 👍 ✔️

Sin 18°/cos 72° + √3 [tan10 tan 30 tan 40 tan 50 tan 80]

=Sin 18°/cos(90-72) + √3 [tan10 tan 30 tan 40 tan(90-40) tan(90-10)]

=Sin 18°/Sin 18 + √3 [tan10 tan 30 tan 40 cot40 cot10]

=1 + √3 [tan10 tan 30 tan 40 1/tan40 1/tan10]

=1 + √3 [tan 30]

=1 + √3 [1/√3]

=1 + 1 = 2