A leaky value on the water meter overcharges the residents for one gallon of water every 1 1/3 months. The overcharged amount w varies directly with time t find the equation that models this direct variation how many months it will take for the residents to be overcharged for 9 gallons of water?
4/3 months/ gallon or 1 gallon/ (4/3) months is 3/4 gallons/month
w = ( 3/4 ) * t
t is in months , w in gallons
9 =(3/4) t
3 = t/4
t = 12 months which is one year.
To model the direct variation between the overcharged amount (w) and time (t), we can use the formula:
w = kt,
where k is the constant of variation.
We know that the residents are overcharged for one gallon of water every 1 1/3 months. To find the value of k, we need to determine how many gallons of water are overcharged in that time period.
Let's convert the time of 1 1/3 months into a mixed number fraction:
1 1/3 months = 4/3 months
Now, we can set up a proportion to find the value of k. Since the residents are overcharged by 1 gallon of water in 4/3 months:
1 gallon / (4/3 months) = k gallons / t months
To simplify the proportion, we divide 1 by 4/3:
1 ÷ (4/3) = 3/4
So, the proportion becomes:
3/4 = k/t
Now, we need to find the value of k in terms of t. We can rearrange the equation to do that:
k = (3/4)t
Therefore, the equation that models the direct variation between the overcharged amount (w) and time (t) is:
w = (3/4)t
To find the number of months it will take for the residents to be overcharged for 9 gallons of water, we can substitute w = 9 into the equation:
9 = (3/4)t
Now, we can solve for t:
(3/4)t = 9
Multiply both sides of the equation by 4/3 to isolate t:
t = (9 × 4/3)
Simplifying, we have:
t = 12
Therefore, it will take the residents 12 months to be overcharged for 9 gallons of water.