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Let the four consecutive numbers in AP be (a - 3d), (a - d), (a + d) and (a + 3d)Formula : an-a+(n-1) da-3d + a - d + a + d + a + 3d = 32a = 8 ......(1)Now, (a - 3d)(a + 3d)/(a - d)(a + d) = 7/1515(a^2 - 9d^2) = 7(a^2 - d^2)15a^2 - 135d^2= 7a^2 - 7d^28a^2 = 128d^2Putting the value of a = 8 in above we get.8(8)^2= 128d^2d^2= 512/128d = 2So, the four consecutive numbers are 8 - (3*2)8 - 6 = 28 - 2 = 68 + 2 = 108 + (3*2)8 + 6 = 14Four consecutive numbers are 2, 6, 10 and 14

A={2,3,4,5,6}B={7,8,9,10}A U B={2,3,4,5,6,7,8,9,10}A INTERSECTION B={}

We applied Euclid Division algorithm on n and 3.a = bq +r on putting a = n and b = 3n = 3q +r , 0<r<3i.e n = 3q -------- (1),n = 3q +1 --------- (2), n = 3q +2 -----------(3)n = 3q is divisible by 3or n +2 = 3q +1+2 = 3q +3 also divisible by 3or n +4 = 3q + 2 +4 = 3q + 6 is also divisible by 3Hence n, n+2 , n+4 are divisible by 3.

From given measures only 6cm, 5cm, 8cm,form a triangle because as per triangle property sum of length of any two sides of triangle should always be greater than the length of third side which is true for only this measure.

Reciprocal of 3/10 is 10/33. option is correct

an=n^2-n+1a1=1-1+1=1a2=4-2+1=3a3=9-3+1=7a4=16-4+1=13a5=25-5+1=21

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PD = radiusCD = small segment

Perimeter = 300m2(l+b)= 300=> l+b= 150 mheight = 6mlateral surface area = 2(l+b)*h= 2* 150*6 m²= 300*6 m²= 1800m²

Let the number of student in group second be x.. then (870/x)=(6/11);x=145×11=1595Then the total number of student=1595+870=2465

i) f(x) = [x], for all x in R ==> By the definition of greatest integer function: If x lies between two successive integers, then f(x) = least integer of them.

ii) So, at x = 2, f(x) = [2] = 2 -------- (1)

Left side limit (x ---> 2-h): f(x) = [2 - h] = 1 ----- (2) {Since (2 - h) lies between 1 & 2; and the least being 1}

Right side limit (x --> 2+h): f(x) = [2 + h] = 2 -------- (3) {Since (2+h) lies between 2 & 3; and the least being 2}

iii) Thus from the above 3 equations, left side limit is not equal to right side limit. So limit of the function does not exist. Hence it is discontinuous at x = 2 So this is not derivable at x = 2 Hence Proved.

We sum of Opposite angles if cyclic Quadrilateral is 180°

A + C = 180°

C = 180° - 105°

C = 75°

write the following in increasing order

unknown.............

opposite of known is unknown

A Computer is used for performing many Repetitive types of tasks The Process of Repeatedly performing tasks is known as looping .The Statements in the block may be Executed any number of times from Zero to Up to the Condition is True. The Loop is that in which a task is repeated until the condition is true or we can say in the loop will Executes all the statements are until the given condition is not to be false

iteration

21b-32+7b-20b

= 8b-32

= 8(b-4)

painting painted on 1 day=3/7painting painted on 2 day=2/7therefore fraction painted in 2 days =3/7+2/7=5/7( because there denominators r same)

Cost of painting for 1 cm² is ₹3So, Total cost is area of painting × per meter cost

Area = 60× 60 cm² = 3600 cm²So total cost is 3600 × 3 = ₹ 10800

Lateral Surface Area of Cuboid = 2H (L X B)Lateral surface includes four faces of the cuboid i.e. left side face, right side face, front face and back face (as shown in following four pictures):

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Let total painting work = wOn first day painting work done = (3/7)wOn second day painting work done = (2/7)w

Total painting work done in 2 days = (3/7 + 2/7)w= (5/7)w

Therefore 5/7 fraction of painting completed in 2 days

let -co of skat be X cost of bicycle = 3xcost of guitar = 10x 3x= 1437x= 1437/3=479so cost of skat = 479cost of guitar = 10x= 479×10= 4790

4790 is the answer of the following

let of skate =x, cost of bicycle =3x, cost of guitar =10x, 3x=1437, .. x=1437/3 =479, cost of skate=479, cost of guitar=10x=10×479=4790

the best answer is 4790

cost of bicycle =₹ 1437cost of skateboard = 1437/3 =₹479So cost of guitar = 479 ×10 =₹4,790

Area of door =3×2=6area of wall=10×10=100area of remaining part for painting =area of wall -area of door. =100-6=94Labour charge for painting 94sq metres =94×2.5=235so rs 235 is required to paint the wall.

There are total36 rectangle

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x=3=2x⁴ -x²+8x=2(3)⁴-3²+8(3)=2(81)-9+24=161-15=146

we know that i = √-1

i) √-27 = √27√-1 = √27.i = 3√3i

ii) -1-√-5 = -1-√5.√-1 = -1-√5i

in pentagon there 5 side.Sum of all interior angle in polygon with n side is (n-2)× 180 .In Pentagon is 3 × 180 = 540In regular pentagon all angle are equal so let each angle be x

so 5x = 540 x = 540/5 = 108

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