Two pumps are filling large vats of liquid. One vat is empty and the pump is filling it at a rate of 6 gallons per minute. The other vat already contains 28 gallons and is continuing to fill at a rate of 4 gallons per minute. When will the two vats have the same amount of liquid
Y'all, it is 14.
I'm not sure if you're getting question 1 and question 4 mixed up.
because for question 1 it's:
6x = 10 + 4x
and for question 4 it's:
14
Let's represent the time in minutes as 't'.
The amount of liquid in the first vat at time 't' can be represented as 6t (since it is filling at a rate of 6 gallons per minute).
The amount of liquid in the second vat at time 't' can be represented as 28 + 4t (since it already contains 28 gallons and is filling at a rate of 4 gallons per minute).
To find when the two vats have the same amount of liquid, we set them equal to each other:
6t = 28 + 4t
Simplifying, we get:
2t = 28
Dividing both sides by 2, we get:
t = 14
Therefore, the two vats will have the same amount of liquid after 14 minutes.
To find out when the two vats will have the same amount of liquid, we can set up an equation based on the rates of filling.
Let's assume that the time required for the two vats to have the same amount of liquid is "t" minutes.
For the first vat, the rate of filling is 6 gallons per minute. So, after "t" minutes, the first vat will have a total of 6t gallons.
For the second vat, the rate of filling is 4 gallons per minute. Since the vat already contains 28 gallons, after "t" minutes, the second vat will have 28 + 4t gallons.
We can now set up an equation to find when the two vats will have the same amount of liquid:
6t = 28 + 4t
Simplifying the equation, we get:
6t - 4t = 28
2t = 28
Dividing both sides of the equation by 2, we find:
t = 14
Therefore, the two vats will have the same amount of liquid after 14 minutes.
To find out when the two vats will have the same amount of liquid, we need to set up an equation based on the information given.
Let's represent the time in minutes as 't', and the amount of liquid in the first vat as 'x' (in gallons). We know that the first vat is empty and filling at a rate of 6 gallons per minute. Therefore, the amount of liquid in the first vat after time 't' can be represented as 6t.
Now, let's represent the amount of liquid in the second vat as 'y' (in gallons). Since the second vat already contains 28 gallons and is filling at a rate of 4 gallons per minute, the amount of liquid in the second vat after time 't' can be represented as 28 + 4t.
To find the time when the two vats have the same amount of liquid, we can set up the equation:
6t = 28 + 4t
Now, let's solve the equation for 't':
6t - 4t = 28
2t = 28
t = 28/2
t = 14
Therefore, the two vats will have the same amount of liquid after 14 minutes.