It can take 12 minutes to fill a water tank using two pipes. If the pipe of smaller diameter is used for 9 minutes
and the pipe of larger diameter for 4 minutes, only, half the tank can be filled.
The time taken by the pipe of smaller diameter to fill the tank is
The time taken by the pipe of larger diameter to fill the tank is
If both the
pipes are working together for 10 minutes, then the tank will filled by
The ratio
of time taken by the pipes can
be
Let the rate to fill the smaller pipe be 1/a units/min
let the rate to fill the larger pipe be 1/b units/min
combined rate = 1/a + 1/b = (a+b)/(ab)
so 1/( (a+b)/(ab) ) = 12
ab/(a+b) = 12
ab = 12a + 12b
9(1/a) + 4(1/b) = 1/2 <------ (1 unit was the full tank)
9/a + 4/b = 1/2
(4a + 9b)/(ab) = 1/2
ab = 8a + 9b
then
12a + 12b = 8a + 18b
4a = 6b
a = 3b/2
sub into ab = 12a + 12b
(3b/2)(b) = 12(3b/2) + 12b
multiply by 2
3b^2 = 36b +24b
b^2 = 20b
b = 20 and a = 2b/2 = 30
It takes 30 min for smaller pipe to fill the tank, and 20 minutes for the larger.
Carry on to answer your other questions.
Let's break down the information given step-by-step:
1. We know that it takes 12 minutes for both pipes together to fill the tank.
2. If the pipe of smaller diameter is used for 9 minutes, it fills half the tank.
3. If the pipe of larger diameter is used for 4 minutes, it also fills half the tank.
Now let's find the individual time taken by each pipe to fill the tank:
1. Let's assume that the time taken by the pipe of smaller diameter to fill the tank is "x" minutes.
2. Since the pipe of smaller diameter is used for 9 minutes and fills half the tank, we can set up a proportion:
(9 minutes) / (x minutes) = 1/2
Cross multiplying, we get:
9x = 2 * x
Simplifying, we get:
9x = 2x
Subtracting 2x from both sides, we get:
7x = 0
Since x cannot be zero, there is no solution for the time taken by the pipe of smaller diameter to fill the tank. This means that the information given is inconsistent or incorrect.
3. We know that the pipe of larger diameter fills half the tank in 4 minutes. So, the larger diameter pipe takes 4 minutes to fill the tank.
4. If both pipes are working together for 10 minutes, we can calculate the fraction of the tank filled by each pipe:
Let's assume that the tank is filled by "a" fraction by the pipe of smaller diameter, and "b" fraction by the pipe of larger diameter.
According to the given information, the smaller diameter pipe fills half the tank in x minutes. So, in 10 minutes, it will fill (10/x) fraction of the tank.
Similarly, the larger diameter pipe fills half the tank in 4 minutes. So, in 10 minutes, it will fill (10/4) = 2.5 fractions of the tank.
Therefore, the tank will be filled by:
(10/x) + 2.5 fractions
5. The ratio of time taken by the pipes can be calculated by dividing the time taken by the pipe of smaller diameter by the time taken by the pipe of larger diameter. However, since we don't have a specific value for the time taken by the pipe of smaller diameter, we cannot calculate the ratio at this time.
In summary:
- The time taken by the pipe of smaller diameter to fill the tank is unknown or inconsistent.
- The time taken by the pipe of larger diameter to fill the tank is 4 minutes.
- If both pipes are working together for 10 minutes, the tank will be filled by (10/x) + 2.5 fractions, where x is the unknown or inconsistent time taken by the pipe of smaller diameter.
- The ratio of time taken by the pipes cannot be calculated without a specific value for the time taken by the pipe of smaller diameter.
To find the time taken by the pipes to fill the tank, we can use the concept of rates. Let's assume that the rate of the smaller pipe is R1 (in tanks per minute) and the rate of the larger pipe is R2 (in tanks per minute).
According to the given information, it takes 12 minutes for both pipes to fill the tank together. So, the combined rate of filling the tank is 1/12 tanks per minute.
Now, let's calculate the rates of each pipe individually.
If the smaller pipe is used for 9 minutes, it fills 9*R1 part of the tank. Similarly, if the larger pipe is used for 4 minutes, it fills 4*R2 part of the tank. Together, these two parts should equal half of the tank.
So, we have the equation:
9*R1 + 4*R2 = 1/2
If both pipes are working together for 10 minutes, they fill 10*(R1 + R2) part of the tank. We want to find the ratio of time taken by the pipes, so let's assume the time taken by the smaller pipe is t1 and the time taken by the larger pipe is t2.
So, we have the equation:
t1/t2 = (10 - t2)/(10 - t1)
We can solve these equations simultaneously to find the values of R1, R2, t1, and t2.
Let's solve the first equation:
9*R1 + 4*R2 = 1/2
To solve the second equation, we can substitute R2 = (1/2 - 9*R1)/4:
t1/t2 = (10 - t2)/(10 - t1)
t1/(10 - t1) = (10 - t2)/t2
t1*t2 = (10 - t1)*(10 - t2)
Substitute R2 = (1/2 - 9*R1)/4:
t1*(1/2 - 9*R1)/4 = (10 - t1)*(10 - (1/2 - 9*R1)/4)
t1*(1/2 - 9*R1)/4 = (10 - t1)*(10 - 1/2 + 9*R1)/4
(t1 - 1/2 + 9*R1)*(10 - t1) = 4*(10 - t1)
(t1 - 1/2 + 9*R1)*(10 - t1) - 4*(10 - t1) = 0
Now, this is a quadratic equation in terms of t1. Solve it to find the values of t1 and t2.
Once we have the values of t1 and t2, we can find the time taken by each pipe individually by substituting them in t1/t2 = (10 - t2)/(10 - t1) equation.
The ratio of time taken by the pipes can be calculated as t1/t2.