Andrew is about to leave for school. if he walks at the speed of 50 meters per minute he will arrive 3 minutes after the bell rings. if he runs at the speed of 80 meters per minute he will arrive 3 minutes before the bell rings. in how many minutes will the bell ring?

If Andrew walks at a speed of 50 meters per minute, then he'll be 3 minutes late. This means he took x+3 minutes to get to class. The x represents the amount of time it takes if he arrived on time, then the +3 is those 3 minutes he is late. We then write

d = 50(x+3)

d = 50x+150

If he runs to school, then he's 3 minutes early. So he took (x-3) minutes to get to class. His rate is 80 meters per minute

d = 80(x-3)

d = 80x-240

We're referring to the same distance (d) value, which allows us to perform substitution to solve the two equations we formed

80x-240 = 50x+150

80x-50x = 150+240

30x = 390

x = 390/30

x = 13

Andrew has 13 minutes to get to class.

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Checking the answer:

If he walks at a speed of 50 m per minute, then he'll be 3 minutes late. So he really took 13+3 = 16 minutes to arrive. His distance traveled is d = r*t = 50*16 = 800 meters.

If he ran 80 m per minute, and is 3 minutes early, then he took 13-3 = 10 minutes to get to class. His distance is the same as before d = r*t = 80*10 = 800 meters. The fact we get 800 both times helps confirm we have the right answer.

sheenuvt12 sheenuvt12 · Answer 2 · 2022-07-29T11:46:04+02:00

Andrew has 13 minutes to get to class.

What is Algebra?

Algebra is the part of mathematics that helps represent problems or situations in the form of mathematical expressions.

In algebra, we use numbers like 2, −7, 0.068 etc., which have a definite or fixed value. In algebra we use variables like x, y, and z along with numbers.

Now, let x be the time it takes for the bell to ring.

let d be the distance from his house to the school, in meters

distance = speed*time

If Andrew walks at a speed of 50 meters per minute, then he'll be 3 minutes late.

So, Andrew took x+3 minutes to get to class.

Here in the expression +3 shows that she is late.

So,

d = 50(x+3)

d = 50x+150

Now, If he runs to school, then he reached 3 minutes early.

So, toe get in class he will take (x-3) minutes

Here -3 show the child reached early

d = 80(x-3)

d = 80x-240

Since the distance is same.

So, the two equations are equal.

80x-240 = 50x+150

80x-50x = 150+240

30x = 390

x = 390/30

x = 13

Hence, Andrew has 13 minutes to get to class.

Learn more about algebra here:

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