There are two drums D1 and D2, each of which is filled to the brim with water. Now, a leak is made at the bottom of each of D1 and D2, such that the leak in D1 takes 6 hours to empty it, while the leak in D2 takes 9 hours to empty it. If the capacity of D1 is more than the capacity of D2 by 60%, then find the time after which the volume of water in D2 will be 25% more than the volume of water in D1. hours hours3 hours hours
Question
There are two drums D1 and D2, each of which is filled to the brim with water. Now, a leak is made at the bottom of each of D1 and D2, such that the leak in D1 takes 6 hours to empty it, while the leak in D2 takes 9 hours to empty it. If the capacity of D1 is more than the capacity of D2 by 60%, then find the time after which the volume of water in D2 will be 25% more than the volume of water in D1. hours hours3 hours hours
Solution
To solve this problem, we need to understand the rate at which each drum is leaking.
- Drum D1 empties in 6 hours, so the rate of leakage is 1/6 of the drum per hour.
- Drum D2 empties in 9 hours, so the rate of leakage is 1/9 of the drum per hour.
We also know that the capacity of D1 is 60% more than D2. Let's assume the capacity of D2 is x litres. Then the capacity of D1 would be 1.6x litres.
We need to find the time when the volume of water in D2 is 25% more than the volume of water in D1. Let's denote this time as t.
So, we can set up the following equation:
1.6x - (1/6)1.6xt = x - (1/9)xt + 0.25*(1.6x - (1/6)1.6xt)
Solving this equation will give us the value of t.
Let's simplify the equation:
1.6 - (1/6)1.6t = 1 - (1/9)t + 0.25(1.6 - (1/6)1.6t)
Multiply through by 6*9 = 54 to get rid of the fractions:
86.4 - 14.4t = 54 - 6t + 9.6 - 1.44t
Combine like terms:
-6.84t = -32.8
Finally, solve for t:
t = 32.8 / 6.84 = 4.8 hours
So, the volume of water in D2 will be 25% more than the volume of water in D1 after approximately 4.8 hours.
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