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Given:first term(a)= 121common difference (d)= 117- 121 = -4∵ n th term of an AP

an= a + (n – 1)d⇒121+(n-1)×(-4)⇒121-4n+4⇒12+4-4n⇒125 -4nan= 125 -4n

For first negative term , an <0⇒ 125-4n<0⇒125<4n⇒4n>125⇒n>125/4⇒n> 31 1/4

least integral value of n= 32

Hence, 32nd term of the given AP is the first negative term.

but the answer is coming 7543 cm square

this is the answer.

110 is the correct answer I think

A composite number is a positive integer that has a factor other than 1 and itself. Now considering numbers,

1) 7×11×13+13 may be written as, 13x{(7x11) + 1} i.e. 13x(78). So other than 1 and the number itself, 13 and 78 are also the factors of the number. Further, 78=39x2. So, 39 and 2 are also it's factors. So this number is definitely not prime. Hence its composite number.

2) Similarly, 7×6×5×4×3×2×1+5 can be written as 5x{(7x6x4x3x2x1)+1}, i.e. 5x(1009). So, other than the number and 1, it have 5 and 1009 as it's factors too. So it is also a composite number.

Consider a pack of rice weighing 1 Kg.Then its actual weight = 80% of 1Kg = 0.8 KgLet CP of 1 kg be Rs. xThen CP of 0.8 kg = Rs. 0.8 xSP = 105% of CP of 1 Kg= 105/100 × x = 1.05xGain = 1.05x – 0.8 x = 0.25 xGain % = [(0.25 x) / (0.8 x)] × 100 = 31.25 %.

x²-16=6-6x²-16=0x²=16x=4

the value of x 4. .

The value of x. 4

x is a natural number than x =0

x²-16=6-6x²-16=0x²=16x=+/-4but x€N x=4

X=4since, X^2=16and X=-4 is not possible according to given condition

nice but not like ache answer puso

answer for this question is 4

value x=4 it is the answer your question

x square -16 =6-6x square-16=0x square=16x=4

(X.X)-(4×4)=(3×2)-(2×3)(X^2)-16=6-6(X^2)-16=0(X^2)=16X=4

x2-16=6-6x2-16=0x2=16x=4

x=4 it's a true answer

Radius of cylindrical tank = 154cm = 1.54mheight of water =3mif height is 4.5 m then let the volume be vvolume of cylinder v = 22/7 *r *r*(H-h) =22/7*1.54*1.54*(4.5-3) =22/7*2.3716*1.5 =22*0.3388*1.5=11.1804 m^3

Given : A rectangular paper of width 14 cm is rolled along its width to form a cylinder.

Height of cylinder = h = 7 cm = width of the rectangular paper

And

Radius of cylinder ( Given ) = r = 20 cm

And

we know Volume of cylinder = π r^2 h , So

Volume of our given cylinder = 22/7 x 20 x 20 x 7 = 8800 cm^3 .Ans

Length of first part=5m 36cmLength of second part=4m 79cm

Total rope=5m 36cm+4m 79cm=9m 115 cm=10m 15cm

Given :-a rope is cut into two pieces of length 5 m 36 cm and 4 m 79 cm respectively.∴Original length of the rope=5m 36 cm+4 m 79 cm=10 m 15 cm

I don't no answer of the following questions

answer is 5% according to your questions

Length of the arc=2πr * theta/360=(2π*20*7π)/(360*4)=1.919cm

area = π*r*r or π*d*d/4 A) 22*21*21/(7*4) = 11*3*21/2 = 346.5 m*m B) 22*7*7/7 = 154 cm*cm C) 22*7*7/(7*4) = 11*7/2 = 38.5 m*m D) 22*20*20/7 = 1256 cm*cm

Radius=20cmDistance between chords =24cmLet the length of chord is 2xHalf the length of chord is x(base)Distance of each chord from center is 12cm(perpendicular)Radius is 20cm(hypotenuse)From pythagoras,20^2=x^2 + 12^2400=x^2+144400-144=x^2256=x^2x=16So, the length of chord is 2x=32cm

nice yes this answers

Let A be the event that a husband will be alive after 20 yrs.

Let B be the event that a wife will be alive after 20 yrs.

Now, PA = 0.8; PB = 0.9

Now, PA = 1-PA = 1 - 0.8 = 0.2

Now, PB 1 - PB = 1 - 0.9 = 0.1

a.Probability that both of them are alive = PA . PB

= 0.8×0.9 = 0.72

b.Probability that none of them is alive = PA . PB

= 0.2×0.1 = 0.02

c. Probability that at least one of them is alive

= 1 - 0.02 = 0.08

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thax for that..!!😍

A_loop = (π/4)36² – (π/4)20²

We can factor out the (π/4)

A_loop = (π/4)(36² – 20²)

We can factor (36² – 20²) getting (36 + 20)(36 - 20)

A_loop = (π/4)(36 + 20)(36 - 20)

A_loop = (16 • 56)(π/4)

A_loop = (896/4)π =

A_loop = 703.7 cm²

Given ratio of sum of n terms of two AP’s = (7n+1):(4n+27)Let’s consider the ratio these two AP’s mth terms as am : a’m →(2)Recall the nth term of AP formula, an = a + (n – 1)dHence equation (2) becomes,am : a’m = a + (m – 1)d : a’ + (m – 1)d’On multiplying by 2, we getam : a’m = [2a + 2(m – 1)d] : [2a’ + 2(m – 1)d’]= [2a + {(2m – 1) – 1}d] : [2a’ + {(2m – 1) – 1}d’]= S2m – 1 : S’2m – 1= [7(2m – 1) + 1] : [4(2m – 1) +27] [from (1)]= [14m – 7 +1] : [8m – 4 + 27]= [14m – 6] : [8m + 23]Thus the ratio of 11th terms of two AP’s is [14(11) – 6] : [8(11) + 23]=148:111

Sn = n/2[(2a + (n-1)d]

S12 = 12/2[2*5 + (12-1)*4]

S12 = 12/2[10 + 44]

S12 = 6*54

S12 = 324

x is the right answer

only x is the right answer

x is the correct answer of the given question

the correct answer is "x"

x is the correct answers

a+4d=41a+10d=20

a+10d-a-4d=41-206d=21d=3.5a+14=41so a=27S12=12/2(2a+11d)=6(54+38.5)=555

1 kilogram = 1000 grams

a=18a+6d=30a+6d-a=30-186d=12 so d=2so 13th term=a+12d=18+12(2)=18+24=42

x+y+z=1x^2+y^2+z^2=2x^3+y^3+z^3=3then,x^5+y^5+z^5=5

(64)^2 /3 - (125)^1/3 - (2)^5 + (27)^-2/3*(25/9)^-1/2= (4)^2 - (5) - (2)^5 + (3)^-2*(5/3)^-1= 16 - 5 - 32 + (1/9)*(3/5)= 16 - 37 + 1/15= (1/15) - 21= (1 - 21*15)/15= - 314/15 = 20.94 ans

cot x = adjacent / opposite

Hyp² = opp² + adj²

= √6²+8²

= √100

= 10

cosx = adjacent/hypotunese

= 8/10

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thanks sooo much

let the number be xAccording to questionx + 12 = 160 × (1/x)

x² + 12x - 160 = 0x² - 8x + 20x -160 = 0x(x-8) + 20(x -8) = 0(x+20)(x-8) = 0So, x = -20, 8ignoring negitive we get the number is 8

Given:Radius of the hemispherical bowl, r = 3.5 cm

Formula:Volume of a hemisphere = 2/3 π r3 = 2/3 * 22/7 * (3.5)3 = 89.83 cm3Volume of water in bowl = 89.83 cm .cube

I. did not understand

Total volume =150 cm^3

23% of 150 is.....

( 150 / 100) * 23. = 34.5

Remaining volume of ice is 150 - 34.5 = 115.5 cubic centimeter

Given,Radius of hemispherical bowl r = 3.5 cm

Volume of hemispherical bowl= 2/3*pi*r^3= 2/3 * 22/7 * (3.5*3.5*3.5)= 22/3 * 3.5 * 3.5= 89.8 cm^3

Therefore,Volume of water it would contain = 89.8 cm^3