A bear loves visiting his friend, the piglet. The distance between their houses is 300m, and the bear always walks from his house to the piglet's house in exactly the same amount of time. But today, the bear stops at a beehive after walking 1/4 of the distance. He eats honey for 2.5 minutes and then continues to the piglet's house at a speed 15 m/min faster than usual. The total journey to the piglet's house takes the exact same amount of time as it usually does, when the bear does not stop for honey. How long does the journey take?

Answer:

10 min.

Step-by-step explanation:

Remember d = rt

The distance from bear's house to beehive = 75m and 225m = distance from beehive to piglet's house

Let r = normal rate of speed

The normal time is 300/r without stopping

75/r = time from bear's house to beehive

225/(r + 15) = time from beehive to piglet's house

So, time from beehive to piglet's house + time at beehive + time from bear's house to beehive = normal time without stopping

225/(r + 15) + 2.5 + 75/r = 300/r

225r + 2.5r(r + 15) + 75(r + 15) = 300(r + 15)

225r + 2.5r^2 + 37.5r + 75r + 1125 = 300r + 4500 Let's remove the deceimal by multiplying thru by 10

2250r + 25r^2 + 375r + 750r + 11250 = 3000r + 45000

25r^2 + 3375r + 11250 =  3000r + 45000

25r^2 + 375r - 33750 = 0

25(r^2 + 15r - 1350) = 0

25(r - 30 )(r + 45) = 0

r = 30 or -45     But, the rate cannot be negative, therefore r = 30

The journey takes 300/30 = 10 min.

By solving a system of equations, we will see that the journey takes 10 minutes.

Let's define the variables:

We have that usually, the bear travels 300 meters in the time T, then we have:

S*T = 300m.

And in this situation, the bear also travels in the same time T, then we have:

First, the bear walks 1/4 of the distance, this is:

S*T/4

Then the bear stops for 2.5 minutes

Then the bear walks at 15m/min faster than usual for a time t..

S*T/4 + (S + 15 m/min)*t = 300m

Such that:

t + T/4 + 2.5 min = T.

Then we can write:

t = T - T/4 - 2.5 min = (3/4)*T - 2.5 min

Replacing that we get:

S*T/4 + (S + 15 m/min)*( (3/4)*T - 2.5 min) = 300m

Then we have a system of equations:

S*T = 300m.

S*T/4 + (S + 15 m/min)*( (3/4)*T - 2.5 min) = 300m

To solve this, we isolate one of the variables in one of the equations, I will isolate S in the first one:

S = 300m/T

Now we can replace this in the other equation:

(300m/T)*T/4 + (300m/T + 15 m/min)*( (3/4)*T - 2.5 min) = 300m

300m/4 +  (300m/T + 15 m/min)*( (3/4)*T - 2.5 min) = 300m

Now we can solve this for T.

75m + 225m + (11.25 m/min)*T - (750m/min)/T - 37.5m = 300m

(11.25 m/min)*T - (750m/min)/T - 37.5m = 0m

If we multiply both sides by T, we get:

(11.25 m/min)*T^2 - (750m/min) - 37.5m*T = 0m

This is a quadratic equation, the solutions are given by:

[tex]T = \frac{37.5m \pm \sqrt{(-37.5m)^2 - 4*(11.25 m/min) (-750m/min)} }{2*(11.25 m/min)} \\\\T = (37.5m \pm 187.5m)/( 22.5 m/min)[/tex]

We only care for the positive solution, which is:

T = (37.5 m + 187.5m)/(22.5 m/min) = 10 min

So the journey takes 10 minutes.

If you want to learn more about systems of equations, you can read:

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