Explore topic-wise InterviewSolutions in Current Affairs.

thanks

circle draw kase hoga

x = 133°

y = 133°

z = 133°

option A because whenever you increment with one, for even no. it becomes positive integer and when it's odd it becomes negative integer.this is satisfied only by optionA.

what you min

Let the required length be x meters

More men, More length built (Direct proportion)

Less days, Less length built (Direct Proportion)

Men20:35 }Days6 : 3 } ⋮⋮ 56 :x

=> (20 x 6 x X)=(35 x 3 x 56)

=> x = 49

Hence, the required length is 49 m.

(20×6× X)=(35×3×56)=49how will come

Total students =50Boys=24Girls=16Ratio=24:16=3:2

total no of students=50and the boys are 24then girls are 50-24=16ration of bous to grl is 24 /16=3:2.

total students =50Boys = 24Girls = 16ratio= 24:16=3:2ratio of the boys and girls = 3:2

Total students=50Boys=24Girls=16{Ratio=24:16=3:2

total=50boys=24girls=50-24=26ratio of boys to girls=26:24=13:12. ans.

4² + 5² - 2³ = 3310² + 12² - 5³ = 119 6² + 8² - 3³ = 73

thanks

Let terms are n-4, n-2, n, n+2 , n+4

Sum=n-4+n-2+n+n+2+n+4250=5nn=250/5=50

Sum of largest number and square of smallest number=n+4+(n-4)²=50+4+(50-4)²=54+46²=54+2116=2170

If 4+ 5 - 2 = 33 and 10 +12 - 5 = 119

Then6 + 8 - 3 = (10 +12 - 5) - (4 + 5 - 2) = 119 - 33 = 86 ans

Money added to money bank on first day = rs 5 on second day = rs 10 0n third day = rs 15

and so on n = 190

thus

AP = 5, 10. 15, ...........195

thus a =5 d =5 n= 190

Thus S = n/2[2a+(n-1)d] = 95[ 10 + 189*5] = 90725

thus total money is 90725 Rs

wrong solution

you are looser and.86856"*+5-.=6'"-/9

(4x)^3(-3x)=-12x^4

thanks

Let speed of the other train = S m /sec.

Speed of the first train = 30 kmph = (30 *5)/18 = 8.33 m/sec.

Relative speed = (8.33 + S) m/sec.

With this relative speed train needs to be covered (150 + 100) m in 10 seconds.

So,ST = D(8.33 + S) * 10 = 250(8.33 + S) = 25S = 16.67 m/sec

Speed = 16.67 m/sec = (16.67 * 18)/5 = 60 kmph.

Before the front of the train enters the platform, distance travelled equals to 0.

Then the front of the train enters the platform and afterwards leaves the platform, at which moment distance travelled by the train is equal to length of platform = 285 meters.at this moment, the end of the train has not yet crossed the platform and the front has just crossed the platform.

For the end of the train to cross the platform, it still needs to travel distance equal to the length of the train which is equal to 150 metres.

So, total distance travelled= 285+150= 435 metres.

10 men can reap in 10 days so 20 men can reap in 10×10/20=100/20=5 days.

Let the first term of the AP be 'a'And the common difference be 'd'Since 69 split into 3 parts such that they form an AP.Let the three parts be (a - d), (a) and (a + d).Therefore,(a - d) + (a) + (a + d) = 693a = 69a = 23The product if two smaller parts = 483So,(a)× (a - d) = 48323× (23 - d) = 483⇒ 529 - 23d = 483⇒ - 23d = 483 - 529⇒ - 23 d = - 46⇒ d = 46/23⇒ d = 2Therefore,The 3 parts are 23 - 2 = 21 ;23and 23 + 2 = 25Hence the parts of the given AP are21, 23, 25

a(11) = a + 10d = 25 ....(1)a(31) = a + 30d = 65 .....(2)eq (2) - eq(1)40 = 20dd = 2a + 20 = 25a = 5a(n) = a + (n-1)d = 5 + (n-1)2 = 5 + 2n -2 = 2n+3

Solution:-Let the four consecutive numbers in AP be (a - 3d), (a - d), (a + d) and (a + 3d)So, according to the question.a-3d + a - d + a + d + a + 3d = 324a = 32a = 32/4a = 8 ......(1)Now, (a - 3d)(a + 3d)/(a - d)(a + d) = 7/1515(a² - 9d²) = 7(a² - d²)15a² - 135d² = 7a² - 7d²15a² - 7a² = 135d² - 7d²8a² = 128d²Putting the value of a = 8 in above we get.8(8)² = 128d²128d² = 512d² = 512/128d² = 4d = 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

5^x+1 + 5^2-x = 5^3 + 1Now, on comparingx + 1 = 3. and. 2 - x = 1x = 2. x = 1

let the no.s be x and y

ATQ=> x = 5y ......(1)

when 21 added to both then

=> (x+21) = 2(y+21)....(2)

now putting the value of x in the 2nd equation=> 5y +21 = 2y+42=> 5y-2y = 42 -21=> 3y = 21=> y = 21/3 = 7

and x = 7*5 = 35.

47 = 1×4777 = 11×789 = 1×89

so, H.C.F of these will be 1

now add 5 to this number =1+5 = 6 is the number

Suppose x 2 rupee coins and y 5 rupee coins x + y = 80 2x + 5y = 220 2x + 5y - 2x - 2y = 220 - 160 3y = 60 y = 20 x = 60

thanks bro

Let length of park = 8xbreadth of park = 5x

Area of park Ap = 8x*5x = 40x^2 Area of region including path Ao = (8x + 3)*(5x + 3)

Area of path = 594 m^2 Ao - Ap = 594(8x + 3)(5x + 3) - 40x^2 = 59440x^2 + 39x + 9 - 40x^2 = 59439x = 594 - 939x = 585x = 585/39 = 15

Dimension of park:Length = 8x = 8*15 = 120 mBreadth = 5x = 5*15 = 75 m

Let the number of 2 rupee coin be x and the number of 5 rupee coin be y.

x + y = 80 . . . . . (1)2x + 5y = 220 . . . . . (2)

(1) × 5

5x + 5y = 400 . . . . . (3)

(3) - (2)

3x = 180=> x = 60

Substituting x in (1),

60 + y = 80=> y = 20

Therefore,

Number of 2 rupee coins = 60Number of 5 rupee coins = 20

Recheck:

20 + 60 = 802(60) + 5(20) = 120 + 100 = 220

Total money= Rs220Total no of coins= 80 x + y = 80 2x + 5y = 220Use diophantine equation x=60 y=202×60=rs1205×20=rs100rs100+rs120=rs22020+60=80coinsAns) Abhraya has 60 2rupee coins and 20 5 rupee coins in her purse.

denominator is √(2)+√(7)-√(5)

If 30 men can reap a field in 14 daysOne man can reap a field in 30*14 = 420 days

Then, 20 men can reap a field in420/20 = 21 days