Time and Distance Problem 13 Solution

This is the solution for a algebra time and distance word problem asked by a anonymous user: "Two trains left a station at the same time. One traveled north at a certain speed and the other traveled south at twice that speed. After 3 hours , the trains were 400 miles apart. How fast was each train traveling?"


The speed of the 1st train is x and the 2nd train is twice that or 2x.  Ok the trick here is lets say someone is going 50 mph north and the other person is going twice that 100 the opposite direction.   This distance separating them would be the same thing as if you added their speeds and it was going one direction -- in other words if someone was just going 150mph they would cover the same ground.    Also rate multiplied by time equals distance.  We know the time is 3 hours and the distance is 400.  So

(x + 2x) (3) = 400

3x(3) = 400

9x = 400

x = 400/9  = 44 and 4/9ths  mph is the 1st train and 88 and 8/9ths (800/9) mph is the 2nd train.

Time and Distance Problem 11 Solution

This is the answer to the algebra time and distance word problem #11 which was asked by another anonymous user on the "Algebra Word Problem Questions" post. The question read:

"Alexis left Miami and drove at a speed of 20 kph. Thomas left 3 hours later, from the same point, and drove at a speed of 30 kph. How long will it take Thomas to catch up to Alexis?"


Ok non-algebraically the easiest way to figure this one out is to figure how far kilometers Alexis has gone before Thomas even leaves. Since Thomas left 3 hours later and Alexis drove 20 kph then Alexis has gone 20(3) or 60 k or kilometers. Thomas is going faster but how much faster than Alexis? (30 - 20)kmh = 10 So, the question is if he's creeping up at 10 kmh how long would it take him to go the distance that Alexis is ahead which we figured out to be 60 kilometers. Well at 10 kph it would take 6 hours to to 60 k... 60 k / 10 kph = 6 h

For the algebra way you just set up the equation where you have :

20(3 + x) = 30x --> the time is x + 3 because she had a 3 hour head start

then you have 60 + 20x = 30x

60 = `10x

x = 60/10 = 6 --> 6 hours

Tags: Time and Distance Problem Solutions, algebra word problem solutions

Time and Distance Problem 10 Solution

This is the solution to the algebra time and distance word problem # 10 which was submitted under the ask a question post anonymously which read..

"Junior's boat will go 15 miles per hour in still water. If he can go 12 miles downstream in the same amount of time as it takes to go 9 miles upstream then what is the speed of the current?"


distance = rate X time ... d = r*t we know the speed of the boat in still water 15 mph

we know he can go 12 miles downstream in the same time he can go 9 miles upstream-- so times are equal --- so since time is equal we can set the distances and rates equal to one another...

d1/r1 = d2/r2

The rate downstream will be 15 + x and upstream will be 15 - x so we have:

12/(15+x) = 9 /(15-x)

cross multiply and you get 12(15-x) = 9 (15+x)

180 - 12x = 135 + 9x

45 = 21x

x = 45/21 = 15/7 This is the answer the current is 15/7 or 2 and 1/7ths mph..

Tags: Time and Distance Problem Solutions, algebra word problem solutions

Time and Distance Problem 9 Solution

This is the solution for the time and distance algebra word problem #9 that read:


"if I walked 9 laps around a track. One lap is 1312 ft. I walked at a rate of 4 miles an hour. how many minutes did it take me to walk 9 laps?"

The easiest way to think about this problem is to think about an easier problem and then copy the steps. What if the problem read:

if I walked 9 laps around a track -- 1 lap is 1 mile and i walked at a rate of 4.5 miles and hour. how many minutes did take me to walk 9 laps?

well the total distance would be the length of 1 lap: 1 mile * the # of laps which is 9 --> 9 * 1 = 9 miles

the rate was 4.5 miles an hour -- distance = rate * time

so 9 miles = 4.5miles/hour* time

9 miles/4.5 miles/hour = time or 2 hours would be how long it would take--->

So this is not a hard math problem the arithmetic is what's hard in this problem, but if you just follow the same steps you will get the answer easily enough. So, again we have:


"if I walked 9 laps around a track. One lap is 1312 ft. I walked at a rate of 4 miles an hour. how many minutes did it take me to walk 9 laps"

ok so take 9 * 1312 1st to get the total distance in feet which is : 11808

4 miles an hour would be (5280 ft in a mile) 5280 * 4 = 21120 ft/hour

But we dont want hours we want minutes-- there are 60 minutes in an hour

so (21120 ft*1hour)/(hour*60minutes) --> you're multiplying by 1hour/60 minutes. so divide by 60..
362ft/minute --- the hours(1 hour = hour) cancel out..
*** Correction -- user Donaldson pointed out I made a division error -- I said 362 feet should be 352 feet/min...*** Thanks!

ok so now we have a total distance in feet of 11808 feet -- divided by the rate of 352 ft/minute

11808 ft/(352 ft/minute) = 11808 ft *minute/ 352 ft

the ft cancel out you divide 11808 by 352 and get 33.54545 = 33.55 minutes --> which is a good enough answer but if you want it more exact:



.545454 of 60 seconds would be 60 *.5454 = 32.72 seconds == approx 33 seconds --- so 32 minutes 33 seconds would be a more recognizable (as far as common usage of time) answer.

You can kind of double check the answer by making sure its reasonable when you see that the total distance of 11808 feet that you walked is roughly 2 miles and your dividing by 4 miles an hour=== 1/2 an hour or 30 minutes approximately

Tags: Time and Distance Problem Solutions, algebra word problem solutions

Time and Distance Problem 8 Solution

This is the solution to the algebra time and distance word problem # 8:

Distances in this problem are equal, and rate multiplied by time is equal to distance. You have what the times are--- 4 and 5 hours and and the rates are ( x + 15 ) for with the wind of 15 miles per an hour and ( x - 15) for against the wind at 15 miles per an hour.

So we have...

4( x + 15 ) = 5( x - 15 )

4x + 60 = 5x - 75

x = 135 --> This is how fast the vehicle is moving. But with the wind its 135 + 15 or 150 miles an hour going and 135 - 15 or 120 miles an hour coming back.

To check this 4(150) = 5(120) 600 = 600 ---> You also now know the distance was 600 miles, which was not asked in the problem.

Technorati Tags: Time and Distance Problem Solutions, Algebra Word Problem Solutions

Time and Distance Problem 7 Solution

Here's is the answer to the algebra word time and distance problem # 7:

If Bill is averaging 60 miles per an hour and Craig doesn't leave until 2 hours at 90 mi/hour, then that means that Bill traveled 60 mi/hr * 2 hr = 120 miles before Craig even left. Now how much faster is 90 miles an hour than 60 ? 90 - 60 = 30(this is the net gain of Bill which is what were worried about since were trying to figure out when he will catch up). So the question then becomes how long would it take you going 30 miles per an hour to go 120 miles --- 120/30 =4 So it takes 4 hours hours for Craig to catch him and since he was going 90 miles an hour he went a total of 90 * 4 or 360 miles.

Algebraically>..

90x = 60x + 120

30x = 120

x = 4

90(4) = 360

Technorati Tags: Time and Distance Problem Solutions, Algebra Word Problem Solutions

Time Distance Problem 6 Solution

In this Algebra word problem solution, an non-algebraic approach would be looking at far Bob goes in 1 hour, since he has an hour head start. He's going 60 mi/hr so he went 60 miles in 1 hour. Steve is going 90 mi/hr or 30 mi/hr faster. So how long would it take someone going 30 mi/hr to go 60 miles? Well the 1st hour he went 30 miles, the 2nd hour he goes the other 30 or 2 hours total until he catches up to Bob. If your going 90 miles an hour for 2 hours-- you have gone 90 * 2 or 180 miles.

Algebraic approach:

x = time

rate * time = distance

60 = # of miles per hour Bob is going

90 = # of miles per hour Steve is going

x + 1 = the hour head start Bob had

Bob = 60(x+1)

Steve = 90x

Since Bob & Steve are going the same distance we can set their respective distance * time equal to one another.

60(x+1) = 90x
60x + 60 = 90x
60 = 30x
30x = 60
x =60/30
x = 2 or 2 hours

check
60(2+1) = 90(2)
180 = 180 --- rate times time = distance which is the answer to the 2nd part of the problem they went 180 miles total.....

Time and Distance Problem 5 Solution

This is similar to the previous problem. The only difference is that they left at a certain time and you have to figure out what time they end up being 2100 miles apart.

Once again, they're both covering distance individually that adds to the total distance were looking for 2100, therefore we can just add their respective speeds to figure out how long it will take to cover that distance.

So with x being used as "time", 300 and 600 the respective speeds or rates to be added, and 2100 the distance --- rate X time = distance we have....

600x + 300x = 2100

900x = 2100

x = 2100/900

x = 2 and 1/3 hours or 7/3 of an hour

So to get the time in hours from when they left we can either figure out a what a third (1/3) of an hour is an add it to 2 (2 and 1/3 hours) or we can figure out how many minutes (7/3) X 60 would be..

1/3 of an hour is (1/3) X 60 or 60/3 which is 20-- so 20 minutes added to 2 is 2 hours and 20 minutes which is the answer but the alternate approach would be taking
(7/3) X 60 and you end up with 420/3 which is 140 or 140 minutes --- with 60 minutes in an hour you can see that 2 hours is 120 minutes with 20 minutes left over and once again 2 hours and 20 minutes

This is still not the answer the problem is looking for. It's asking what time were they that far apart. If they left at 12 (noon) , 2 hours and 20 minutes later would be 2:20 pm--- So 2:20 pm is the answer.

Time & Distance Problem 4 Solution

Bob & Steve are both covering their respective distances (in this case the total distance to find is how far apart they are 1750 miles) individually with their given speeds. It's the same distance covered if they were going their added speeds, which is 500 mi/hr + 200 mi/hr = 700 mi/hr -- So, how long would it take to go 1750 miles at 700 mi/hour.. Distance = Rate X Time .... Time = Distance / Rate

So time = 1750 / 700 = 2.5 hours

Algebraically you would have

500x + 200x = 1750

700x = 1750
x = 1750/700
x = 2.5

Time & Distance Problem 3 Solution

In this problem its important to know that if you're moving a certain speed and someone else is moving another speed, the distance being covered between the both of you is the same as if you were going the added speeds yourself. In other words, if you're moving 20 miles per an hour and someone else moving towards you is going 60 miles an hour then the amount of distance being crosses is at the rate of 80 miles an hour.

Another way of looking at is if there's 800 miles between you and someone else, the 1st hour you went 20 miles (20 mi/hr) and they went 60 miles (60 mi/hr) or 80 miles total of the distance. Which you can see 2 things from. 1) It would take 10 of those hours 80 X 10 = 800 to go 800 miles and 2) You can just add the speeds 60 mi/hr + 20 mi/hr to get 80 mi/hr and figure out how long it would take you to go 800 miles.

So algebraically: 20x + 60x = 800
80x + 800
x = 800/80 = 10

So, in this problem it would take 10 hours for Bob and Steve to meet.

Time and Distance Problem 2 Solution

 Here's the answer to the time and distance problem # 2 that read, "Bob leaves in his car driving at a constant speed of 40 miles/hour. 4 hours later Steve leaves going the same direction at 60 miles/hour. How long will it take until Steve catches up (is side by side) with Bob?"

A non algebra solution would be to realize that Steve is moving 20 miles/hour faster than Bob and has to cover 160 miles (40 * 4) to catch up to Bob (40 miles/hour X 4 hours = 160 miles) So how long until does it take to cover 160 miles going 20 miles/hour-- 8 hours

Algebraically you got 40(x+4) = 60x where x is time and x + 2 means 2 hours ahead

So, 40x + 160 = 60x which makes

160 = 20x ... x = 160/20 = 8 ----

So in 8 hours they meet.

Time and Distance Problem 1 Solution

I set this problem up so you can almost stumble upon the solution accidentally by playing with the numbers I gave. If you see that in 1 hour's time they both cover the same amount of miles as their speed in miles per in an hour. In this case Bill traveled 3.5 miles in 1 hour and Susie 1.5 miles in 1 hour. You might have also stumbled upon the fact that this number when added equals the distance of the track which was 5 miles. This means that if they together covered a distance of 5 miles then that's when they "meet"-- so, coincidentally this is the answer for the 1st & 2nd parts of the problem. Bill goes 3.5 miles (distance) and Susie 1.5 miles (distance) when they meet each other in the 1 hour (time). Whether or not they kept going after their argument is another story.

A pure algebraic solution would use this scenario and just add the respective speeds together so 3.5x + 1.5x = 5 , where x is the speed in miles/hour and 5 is the distance in miles. You can do this when you realize that if something is moving towards you at a given speed say 30 miles and your moving 60 miles/hour its as though you're moving towards an object standing still at 90 miles/hour. In other words, you can figure the time you cover a respective distance in relation to another person moving towards you just by adding the 2 speeds and figuring out the time it would take you to cover that distance with that new "added up" speed. So, 5x = 5 which means x = 1 hour*. Substituting back in the problem 3.5(1) + 1.5(1) = 5 =.... 5 = 5, which is correct. So, Bill went 3.5 miles and Susie 1.5 miles when they met in 1 hours time.



*When you divide miles by miles/hour the miles cancel and your left with the hours.....