Word Problems

[Dillinger4]

One of my friends need help with a problem so i decided to give them a hand. Im not too sure about the answer given by his text book. Here is the problem.

Gail and Bill drove to a beach at an average speed of 50mi/h. They returned home over the same road at an average speed of 55mi/h. The trip home took 30 min less time. How far is the beach from their home.

Now i came up with the equation. Going 50(t)
Return 55(t - 0.5)

Now what i don't understand is, how are we able to find the distance if the amount of driving time has not been given? The book says 275mi.

[wy125]

This is how I set it up.

You know that the average speed during the first trip was 50 MPH. So, that means


t1 = x/V1

where,

t1 is the time the first trip took

x is the distance to the beach

and V1 is the average speed


The next bit of information is that the return trip, which is the same distance, was at a speed of 55 mph and took 30 minutes less than the first trip.

So we have,

t2 = x/V2

and

t2 = t1 - 1/2 (1/2 hr = 30 minutes)

sustituing for t2 in the second equation leaves us the two following equations


Eq. 1 t1 = x / V1

Eq. 2 t1 - 1/2 = x / V2

Divding Eq. 1 by Eq. 2 gives,


t1/(t1 - 1/2) = V2 / V1


Solving for t1 we get

t1 = V2/ ( 2 V1 (V2/V1 - 1))

Now we can solve for x, the distance to the beach...

x = V1 t1

x = V2 / (2 (V2/V1 - 1))

When you plug in all the values you get

x = 275 miles



=========

Your equations were correct. You just needed to set them equal to each other and solve for t. Then multiply the value you get for t by 50.

[Dillinger4]

Thanks to your help i have a better understanding of how to solve the problem. Here is what i came up with.

50(t) = 55(t - 0.5)
50t = 55t -2.75
50t + (-55t) = 55t + (-55t) - 2.75
-5t / -5 = t
-27.5/ 5 = 5.5
t = 5.5

d = V1(t)
275 = 50mph(5.5)

d = V2(t - 0.5)
275 = 55(5)

So they seem to match. But i wanted to try and figure out the distance each way so i tried this. But the problem is that when the two trips are added togther the total mileage comes out to be 272.25 not 275. Am i doing somthing wrong again?

137.5 = 50mph(2.75)
134.75 = 55mph(2.45)

[wy125]

What do you mean by the distance each way? 275 miles is the distance each way. The time you calculated, 5.5 hours is the time it took to get to the beach while 5 hours is the time it took on the return trip. The distance each way is 275 miles.

[Dillinger4]

Right, Im sorry. I see now. Thanks.

[Guv]

A very easy equation is as follows.

Distance / 50 = Distance / 55 + 1 / 2

Multiply both sides by 550, resulting in

11 * Distance = 10 * Distance + 275

Subtract (10 * Distance) from both sides.

Distance = 275

[Dillinger4]

Looking at the way in which your example is laid out, i guess am over complicating things. May i gain your insight on another problem? It is presented as such......

Enrique cycled to his grandmothers house at 15km/h and returned home over the same route at 18/kmh . If the whole trip took 7hr and 20 min. How far does enrique live from his grandmother's house?


Now knowing what the answer is (60km) i can see how they get it by 60 km = 15km/h(4hr) and 60km = 18km/h(3.66) but i have no clue how to set up an equation which would produce the answer.

It's obvious that the time it takes for him to get home will be less since he is travling at a higher rate of speed. So i can't just simply divide 7hr 20 min in half. So how would i figure out the time each way? Thanks.

[Guv]

Distance / Rate = Time

Distance / 15 + Distance / 18 = 22 / 3

Multiply by 90, resulting in the following.

6 * Distance + 5 * Distance = 22 * 30

11 * Distance = 22 * 30

Distance = 60

[nishantp]

When i first saw these, i thought they were way over my head. But i sat down and well...solved it wooo

7h20min = 22/3 hrs
t = total time
d = distance

t = (d/15)+(d/18)
22/3 = (18d+15d)/270
22/3 = (33d)/270
1980=33d
d = 60

Im so proud

[nishantp]

And naturally Guv solved it in half the time I had to write it all out.


Got any more of those?

[Dillinger4]

Very good nishantp. Your math teacher would be proud. I noticed a couple of things. In your first line you have 7h20min = 22/3 hrs what is the break down? I think i am missing somthing.

[nishantp]

Yeah me and Guv both did that. It just makes things easier.

7hrs20mins = 7 and 1/3hrs = 22/3hrs

Remember fractions? all that suffering in grade 4-5? lol


you have the 7 and 1/3, so instead of saying 7.333333333... we leave it as an improper fraction to be more accurate. 3 * 7 + 1 = 22, and the denominator, 3, remains.

[Dillinger4]

Ahhh ok. It's just an improper fraction. Now where is the 270 comming from? You end up going from t = (d/15)+(d/18) to
22/3 = (18d+15d)/270.

[nishantp]

270 is a common denominator for 15 and 18. I was too lazy to find the real LCD like Guv, so i multiplied 15 and 18 and got 270. So you cross multiply, and voila.


Think you can dig up any more of those problems?

[Dillinger4]

Dam i forgot about this thread. Here's an easy one.

On their 1200km trip to Texas, the Wallaces first took a train and later a plane. The train travling at 48 km/h, took 2 h longer than the plane, travling at 240 km/h. How long did the whole trip take.

[sql_lall]

d = 1200
s₁ = 48
s₂ = 240
t₁ = t₂ + 2

d = s * t

=>
1200 = s₁*(t₂ + 2) + s₂ * t₂
1200 = 48*(t₂+2) + 240 * t₂
1200 = 96 + (240+48)*t₂
1104 = 288* t₂
=> t₂ = 1104/288 = 23/6hr = 230min = 3hr50min
=> t₁ = t₂ + 2 = 5hr50min

=> total time = 9hr40min

Also, thankyou to whoever said how to do subscripts!

[Dillinger4]

Very Good. Here's my attempt.

48km/h(t + 2) + 240km/h = 1200km
48t + 96 + 240t = 1200
288t + 96 - 96 = 1200 - 96

288t / 288 = t
1104/288 = 3.83


Distance 1: 48(3.83 + 2) = 279.84
Distance 2: 240(3.83) = 919.2
279.84 + 919.2 = 1199.04 km


3.83 + 3.83 + 2 = 9.66 or 9hr 40min