Assume Q can do the work alone in Q days:
work of a day by both P and Q =>
(1/15) + (1/Q ) = 1/6
solving this Q = 10.
264
Ratio: 3/4
length is 30 breadth is 12 and height is 12
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1 .
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1 10 .
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1
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1 30 .
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1
consider right angled triangle
so the distance= square root of (30*30+10*10)
Employee Performance Standards. Employee performance measurements can determine an employee’s compensation, employment status or opportunities for advancement.
Ans : – 215
Explanation :-
1^3 = 1-1=0
2^3 = 8-1 = 7
3^3 = 27-1=26
4^3 = 64 – 1 = 63
5^3 = 125-1 = 124
Hence, Ans is
6^3 = 216-1=215
sister
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.
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
7 min clock:|——-7——-|
4 min clock:|—–4—-|—-4—–|—–4—-|—–4—-|
you got 9 min: |—————9————–|