I have two questions about math problems. SO pls help and explain it step by step

Question

) A rectangle shape dog park opens every morning at 7:00 a.m. A new dog enters the dog park every 3 minutes and a dog leaves every 5 minutes. If the park has a length of 30 feet and a width of 20 feet and the maximum capacity for the park is 1 dog for every 10 square feet, then what time will it be when the first dog who normally would enter would need to wait.

2) 2 years ago, Sara was 53 the age of Jayne.  In 6 years, Sarah will be 45 the age of Jayne.In 6 years, Sarah will be 54 the age of Jayne. How old is Jayne today?How old is Jayne today?

Answer 1

The park is a rectangle with L = 30 and W = 20.

Area = L x W = 30 x 20 = 600 square feet.

We can fit a dog for every 10 square feet, so

600/10 = 60 dogs

We can fit a maximum of 60 dogs in the park. When there are 60 dogs in the park, the next dog would have to wait. Let’s find out when this occurs.

To understand how the dogs come and go, let’s make a little chart:

Entering (every 3 minutes)            

7:03 am  1 dog  

7:06 am  2 dogs

7:09 am  3 dogs

7:12 am  4 dogs

7:15 am  5 dogs

Leaving (every 5 minutes)

7:05 am – 1 dog

7:10 am – 2 dogs

7:15 am – 3 dogs

By 7:15 am, 5 dogs have entered and 3 dogs have left, leaving 2 dogs in the park.

Every 15 minutes, there are 2 dogs in the park.

Every hour, there are 8 dogs in the park. (since each hour has 4 periods of 15 minutes).

60 dogs / 8 dogs per hour = 7.5 hours



It takes 7.5 hours for the park to reach maximum capacity of 60 dogs. 

7 AM + 7.5 hours = 2:30 PM Answer

Answer 2

Each hour, the population of dogs increases by (60 ÷ 3) - (60 ÷ 5) = 20 - 12 = 8

The capacity of the park is (30)(20) ÷ 10 = 60 dogs

60 dogs ÷ (8 dogs/hour) = 7.5 hours after 7 AM = 2:30 PM

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Answer 3

7:00

dog/3m

-dog/5min

600sq/ft

1dog/10sq/ft/max

60dogs/max

t/3=1

t/5=-1

t/3-t/5=60

5t/15-3t/5=60

2t/5=60

2t=300

t=150min=2.5hrs=9:30