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6″
let the speed limit be x.
speed by first rider S1 and 2nd rider be S2.
S1=x+10;
ans S2=x+2*10; and given S2=35.
solving we get x=15mph.
Rs.265.80
1, 2, 4, 13, 31, 112,?
This series is written in base 5 numbers for decimal nos of 1,2,4,8,16,32,64
so next number is 224 .
224 in base 5 = 64 in base 10.
so 224 is next number.
15a=45
20/- per day
So, if he work one day
So,1*20=20
1/3*20=20/3
2/3*20=40/3
1/8*20=5/2
3/4*20=15
so,
20+20/3+40/3+5/2+15=57.5
So ans is 57.5 is the ans
22
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.
7500
1/24S