perimeter of rectangle = 2(l+b)
given l abd b as 18 cm and 26 cm
therefore perimeter of rectangle = 2(44)
perimeter of circle = 2*pi*r
given 2*pi*r=2(l+B)
i.e, 2*(22/7)*r= 2(44)
(22/7)*r=44
r=14 cms
area of the circle = pi*r*r
=(22/7)*14*14
=616 sqcm is the answer
saying 20%alcohol in 15 litre solution.
It means (20/100)*15=3 litres
Now 5 litre of water is added not alcohol.
So it means only the volume of solution is increases by 5 not alcohol.
It means 15+5=20
Now % of alcohol is
(3/20)*100=15%.
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26 rupees and 53 paise
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.
0
Speed = (45×518)m/sec=252m/sec
Total distance covered = (360+140) m = 500 m.
∴Required time= 500×225sec = 40 sec.
10m/sec
30
24.
exp: 0 is formed by multiplying 5 with 2. so first we find
how many 5’s and 2’s are there in 100!.
No of 5’s : 100/5=20/5=4/5.
20 +4 =24.
No of 2’s : 100/2=50/2=25/2=12/2=6/2=3/2=1/2.
50 +25 +12 +6 +3 +1.
It has 24 5’s and 2’s.
so the no. of zeros=24.