Jimmy walks to his school every morning. He leaves home everyday at the same time. And the school starts at the same time everyday at 8 am. If he walks at 40 meters per minute, he will reach his school 5 minutes late. If he walks at 60 meters per minute, he will reach his school 3 minutes before the bell rings.

The bell rins at 8:19 AM, and the distance between Jimmy's house and the school is 960 meters.

Step-by-step explanation:

Let's define the variables:

T = Number of minutes after 8 AM such that the school's bell sounds.

D = Distance between jimmy's house and the school, in meters

And remember the relation:

Distance = Speed*Time.

We know that:

"If he walks at 40 meters per minute, he will reach his school 5 minutes late."

D = (T + 5)*40

"If he walks at 60 meters per minute, he will reach his school 3 minutes before the bell rings."

D = (T - 3)*60

Then we have a system of equations:

D = (T + 5)*40

D = (T - 3)*60

We can take the quotient of these two equations and get:

D/D = ((T + 5)*40 )/((T - 3)*60)

1 = (4/6)*(T + 5)/(T - 3)

Now let's solve this for T.

6/4 = (T + 5)/(T - 3)

(6/4)*T - (6/4)*3 = T + 5

(6/4)*T - T = 5 + (6/4)*3

T*(2/4) = 5 + 18/4 = 5 + 9/2

T = (4/2)*(5 + 9/2) = 19

Then the bell sounds 19 minutes after 8AM, at 8:19 AM

And with this, we can find the distance D:

D = (T + 5)*40 = (19 + 5)*40 = 960

The distance between Jimmy's house and the school is 960 meters.

i used to know how to do this in 7th but i forget easily

step-by-step explanation: