Answer:
7 gallons
Step-by-step explanation:
<u>The first method </u>
You will start with 38 gallons of water on 1st day then as you count the days, subtract 0.75 gallons of water in every odd number day, and add 0.5 gallons of water for everyday that is a multiple of 3 up to the 31st day.Then find the difference of the value you get with the initial amount of water that the tank could hold.
<u>The second method</u>
Given 0.75 gallons of water are lost in every odd numbered day and 0.5 gallons of water are added in every day that is a multiple of 3 then;
⇒ identify all days that are odd in this month.
They are 16 days
⇒Find the amount of water that will be lost in 16 odd number days of that month
16×0.75=12 gallons
⇒Find the number of days that are multiple of 3
They are 10 days
⇒Find amount of water that can be added in 10 days
10×0.5=5 gallons
⇒The water that will be in tank on 31st will be
38-12+5
(38+5)-12
43-12=31 gallons
⇒Amount of water to add
38-31=7gallons
