Evan bought two plants. he decided to water his first plant every 3 days and his second plant every 4 days. if he watered both plants on june 1, how many days passed before he watered both plays again

A.    If he watered both plants on june first the next time he watered them on the same day 12 days would have passed. To find out how many days would have passed i found the lcm (lowest common multiple) by multiplying 3 and 4.

B. if he watered both plants on the same day 8 days would have passed before they where watered on the same day again. i found this answer by finding the lcm(lowest common multiple)by multiplying 2 and 4 .

Step-by-step explanation:

step-by-step explanation:

12: 00 p.m. to 7: 00 p.m. is a total of 7 hours.

4: 00 p.m. to 7: 00 p.m. is a total of 3 hours.

this means that from 4: 00 p.m to 7: 00 p.m. this represents 3/7 of what was lost from 12: 00 p.m. to 7: 00 p.m.

after knowing this, after doing 11,600 - 6,800 = 4,800

4,800 represents 3/7 of what was lost for 7 hours.

we then divide 4,800 by 3 which gets 1,600.

because we know that 1,600 represents 1/7 of how much water was lost, we just have to multiply by 7, which gets 11,200.   after adding 11,200 to how much water was left at 7: 00 p.m. which was 6,800. 11,200 + 6,800 = 18,000

this means that the pool originally had 18,000 gallons in the pool.

you can also check your work by doing 18,000 - 4(1600), which becomes           1800 - 6400 which gets you 11,600.