56.25 % is the right answer.
area of square = side ^ 2
at first, it was 100% , means 1
then, a side was enlarged to 25%, means now, 125% means new side is 1.25
taking square will give 1.5625
1.5625 – 1 = 0.5625
so, 56.25% increment in area.
Let’s assume the length of each train is ‘L’ and the speeds of the two trains are ‘V₁’ and ‘V₂’ respectively.
When the trains are moving in the opposite direction, their relative speed is the sum of their individual speeds. The total distance they need to cover is the sum of their lengths. Since they cross each other completely in 5 seconds, we can set up the following equation:
(V₁ + V₂) × 5 = 2L
When the trains are moving in the same direction, their relative speed is the difference between their individual speeds. The total distance they need to cover is the difference between their lengths. Since they cross each other completely in 15 seconds, we can set up the following equation:
(V₁ – V₂) × 15 = 2L
Now, let’s solve these equations to find the ratio of their speeds.
From the first equation, we have:
(V₁ + V₂) × 5 = 2L
V₁ + V₂ = (2L) / 5
From the second equation, we have:
(V₁ – V₂) × 15 = 2L
V₁ – V₂ = (2L) / 15
Let’s add these two equations together:
V₁ + V₂ + V₁ – V₂ = (2L) / 5 + (2L) / 15
2V₁ = (6L + 2L) / 15
2V₁ = (8L) / 15
V₁ = (4L) / 15
So, the speed of the first train is (4L) / 15.
Now, let’s substitute this value back into the first equation to find V₂:
(4L) / 15 + V₂ = (2L) / 5
V₂ = (2L) / 5 – (4L) / 15
V₂ = (6L – 4L) / 15
V₂ = (2L) / 15
Therefore, the speed of the second train is (2L) / 15.
The ratio of their speeds is given by:
(V₁ / V₂) = ((4L) / 15) / ((2L) / 15)
(V₁ / V₂) = 4L / 2L
(V₁ / V₂) = 2
So, the ratio of their speeds is 2:1.
1/3
x:y=18:11
$20
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441=21^2
x/y =5/7…………………………..(1)
(x-25)/(y-25)=35/39
59x=35y+600
dividing b.t.s by y and substituting (x/y) value from equation (1)
y=84
Now, substitue y value in equation (1)
x = 60
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Product of two numbers = 1320
HCF = 6
LCM = x
Formula:
Product of two numbers =(HCF *LCM)
1320=(6*x)
x=1320/6
x=220
LCM Of the numbers is220
500
2a; 2a+2; 2a+4; 2a+6; 2a+8; 2a+10
t_2 + t_6 = 24 => 2a+2 + 2a + 10 = 24 => a = 3
therefore t_4 = 2(3) + 6 = 12
The numbers that lie between 100 and 1000 which are divisible by 14 are 112, 126,140 …,994
a = 112; l = 994, d = 14
n= (l−a)/d+1
= (994-112)/14+1
= 64
Sn=n/2(l+a)
= 64/2(994+112)
= 32*1106
= 35392
Nice