Half-Life Problems #1 - 10
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Problem #1: The half-life of Zn-71 is 2.4 minutes. If one had 100.0 g at the beginning, how many grams would be left after 7.2 minutes has elapsed?
Solution:
7.2 / 2.4 = 3 half-lives(1/2)3 = 0.125 (the amount remaining after 3 half-lives)
100.0 g x 0.125 = 12.5 g remaining
Problem #2: Pd-100 has a half-life of 3.6 days. If one had 6.02 x 1023 atoms at the start, how many atoms would be present after 20.0 days?
Solution:
20.0 / 3.6 = 5.56 half-lives(1/2)5.56 = 0.0213 (the decimal fraction remaining after 5.56 half-lives)
(6.02 x 1023) (0.0213) = 1.28 x 1022 atoms remain
Problem #3: Os-182 has a half-life of 21.5 hours. How many grams of a 10.0 gram sample would have decayed after exactly three half-lives?
Solution:
(1/2)3 = 0.125 (the amount remaining after 3 half-lives)10.0 g x 0.125 = 1.25 g remain
10.0 g − 1.25 g = 8.75 g have decayed
Note that the length of the half-life played no role in this calculation. In addition, note that the question asked for the amount that decayed, not the amount that remaning.
Problem #4: After 24.0 days, 2.00 milligrams of an original 128.0 milligram sample remain. What is the half-life of the sample?
Solution:
The decimal fraction remaining:
2.00 mg / 128.0 mg = 0.015625
2) How many half-lives must have elaspsed to get to 0.015625 remaining?
(1/2)n = 0.015625n log 0.5 = log 0.015625
n = log 0.5 / log 0.015625
n = 6
3) Determine the half-life:
24 days / 6 half-lives = 4.00 days
Problem #5: A radioactive isotope decayed to 17/32 of its original mass after 60 minutes. Find the half-life of this radioisotope.
Solution:
17/32 = 0.53125 (this is the decimal amount that remains)(1/2)n = 0.53125
n log 0.5 = log 0.53125
n = 0.91254 (this is how many half-lives have elapsed)
60 min / 0.91254 = 65.75 min Microsoft word 2016 free download mac.
n = 66 min (to two sig figs)
Problem #6: How long will it take for a 40.0 gram sample of I-131 (half-life = 8.040 days) to decay to 1/100 its original mass?
Solution: Bootstrap lightbox gallery.
(1/2)n = 0.01n log 0.5 = log 0.01
n = 6.64
6.64 x 8.040 days = 53.4 days
Problem #7: Fermium-253 has a half-life of 0.334 seconds. A radioactive sample is considered to be completely decayed after 10 half-lives. How much time will elapse for this sample to be considered gone?
Solution:
0.334 x 10 = 3.34 seconds
Problem #8: At time zero, there are 10.0 grams of W-187. If the half-life is 23.9 hours, how much will be present at the end of one day? Two days? Seven days?
Solution:
24.0 hr / 23.9 hr/half-life = 1.0042 half-livesOne day = one half-life; (1/2)1.0042 = 0.4985465 remaining = 4.98 g
Two days = two half-lives; (1/2)2.0084 = 0.2485486 remaining = 2.48 g
Seven days = 7 half-lives; (1/2)7.0294 = 0.0076549 remaining = 0.0765 g
Problem #9: 100.0 grams of an isotope with a half-life of 36.0 hours is present at time zero. How much time will have elapsed when 5.00 grams remains?
Solution:
5.00 / 100.0 = 0.05 (decimal fraction remaining)(1/2)n = 0.05
n log 0.5 = log 0.05
n = 4.32 half-lives
36.0 hours x 4.32 = 155.6 hours
Problem #10: How much time will be required for a sample of H-3 to lose 75% of its radioactivity? The half-life of tritium is 12.26 years.
Solution:
If you lose 75%, then 25% remains. Use 0.25 rather than 25%.(1/2)n = 0.25
n = 2 (remember (1/2)2 = 1/4 and 1/4 = 0.25)
12.26 x 2 = 24.52 years
Comment: the more general explanation follows:
(1/2)n
n log 0.5 = log 0.25
n = log 0.25 / log 0.5
n = 2
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by Rick GarlikovRate x Time = Distance Problems
One of the students on a math-help forum wrote:
> Plane A leaves los Angeles for York @ 500 mph, at the same time
> plane b leaves from York for Los Angeles @650mph. Assume the distance from
> los Angeles to York is 3000 miles, find how long it will take them to meet.
>
> (2.6 hours)
>
> I know the answer, but I don't understand the reasoning or how to get it.
This was his second attempt at sending us this problem. The first time, he left out the numbers and just wanted us to help with the concept, which was okay; but, more importantly, he stated the problem mistakenly that first time. What he wrote then was:
'Two planes leave at the same time going in opposite directions at different speeds. How long will it take them to meet? They both travel the same distance.'
We wrote back that having the numbers might be helpful, and that it wasn't clear whether they were flying toward each other from different locations or whether they were initially flying away from each other from the same place and then going around the world to cross paths on the way, or what. But more importantly, he had stated a contradiction - since there was no way the planes could fly the same distance if they flew for the same amount of time at different rates. For example, if I drive 90 mph and you drive 45 mph, then no matter how long we drive, as long as we drive the same amount of time as each other, I will cover twice the distance you do. The faster vehicle will always go further over the same amount of time.
So he sent back the problem, as above, in its entirety. Someone then pointed out a way for him to look at the specifics of the problem, but I wanted to address the more general issue of his lack of understanding of rate-time-distance problems, and wrote much of the following. For my purpose here, I am adding to what I wrote, by including a section on other kind of rate-time and rate-quantity problems -- the ones that involve doing work at different rates and ones that involve things like dilution rates or unit prices:
When you first sent us your problem, you included in it an assumption (same distances covered by both planes) that showed, I think, that you have a fundamental misunderstanding of these sorts of problems, so I want to address that. Someone else addressed some specifics about this problem, but I want to talk about the rt=d (rate multiplied by time = distance) formula itself first, because I think you don't really understand it. I think you possibly just see it as a bunch of symbols that don't really have much meaning for you -- a formula to plug numbers into in order to crank out answers.
I assume you drive or at least have taken trips in a car. The rt=d formula is just a way of saying that the distance you travel depends on the speed you go for any length of time. If you go 30 miles per hour for 1 hour, you will go 30 miles. 2 hours at that speed would take you 60 miles; 3 hours would be 90 miles, etc. If you were to cut the speed in half, you would go only 15 miles the first hour and you would have gone 30 miles after two hours, etc. Do you see this part of it?
Notice also that understanding this helps you figure out any one of the components from knowing the other two. E.g., if you have 60 miles to drive and the road conditions are such that you will be driving 30 mph the whole way, then you know it will take you 2 hours to get there. And if your cousin comes to visit you from 100 miles away and he tells you the trip took two and a half hours, you know he AVERAGED 40 miles per hour. Right?
Now there are a couple of things you need to see about all of this, and some of that is going to be counter-intuitive. First, the easy part:
the calculations above all depend on AVERAGE speed. So in the last example of your cousin's driving 100 miles in two and a half hours, you might act surprised and say 'But you have a lead foot; what took you so long?' And he might say, 'Well I drove 80 mph while we were actually able to move, but there was some construction where we were at a standstill.' So you understand that will mess up his AVERAGE speed for the whole trip (just as making pit stops does), since part of the time he was not moving at all.
Now if you think about it (forget the formula for a minute), you can actually tell from what he says how long he was driving and how long he was standing still (assuming, what is false but what they never worry about in these problems -- acceleration and deceleration getting from 0 to 80 and back; just assume he was either going 80 or 0 the whole time).
For example, to look at a slightly simpler modification of this first, suppose he said he drove 100 mph while he was moving. If it took him two and a half hours to drive 100 miles, but the whole time he was actually moving he drove 100 mph, then you can tell he was stuck at the construction for one and a half hours, since otherwise the trip should have only taken him one hour. Well, similarly, since he says he drove 80 while he was moving, you know that it should have taken him 1 hour and a fourth (i.e., one hour and 15 minutes) to cover the 100 miles since he did 80 miles the first hour and then only had to go another 1/4 of that distance at the same speed. Since the trip took him 2 hours and 30 minutes all together, that means half of it was driving and half of it was waiting.
To represent this idea by means of the formula, you have to think about how to get the formula to represent the two different PARTS or COMPONENTS of the trip and that is where these problems start to get a bit tricky. But the idea is to think about it conceptually, in words or ideas, first, not to try to go right to the formula first and automatically plug in numbers. That is, you have to think about it conceptually WHEN it is a problem you don't already know how to do mechanically. Once you figure out how to do a particular type of problem, and figure out the mechanics, then you can do all other exactly similar problems mechanically. The rest of this message is about thinking your way through problems you don't immediately know how to do mechanically. Most problems on your test you will hopefully do (mostly) mechanically.
In the example of your cousin, we know the whole distance is 100 miles. And we know the whole time is 2.5 hours. BUT his rates were different AND they were different at different times. However, can you see that no matter how many different rates he drove for various different time periods, his TOTAL distance depended simply on the SUM of each of the different distances he drove during each time period? E.g., if you drive a half hour at 60 mph, you will cover 30 miles. Then if you speed up to 80 mph for another half hour, you will cover 40 miles, and then if you slow down to 30 mph, you will only cover 15 miles in the next half hour. But if you drove like this, you would have covered a total of 85 miles (30 + 40 + 15). It is fairly easy to see this looking at it this way, but it is more difficult to see it if we scramble it up and leave out one of the amounts and you have to figure it out going 'backwards'. That is what word problems do.
Further, what makes them difficult is that the components they give you, or ask you to find can involve variable distances, variable times, variable speeds, or any two or three of these. How you 'reassemble' all this in order to use the d=rt formula takes some reflection (and sometimes some trial and error as you reflect) that is 'outside' of the formula itself. You have to think about how to use the formula.
So the trick is to be able to understand EXACTLY what they are giving you and EXACTLY what it is that is missing, but you do that from thinking, not from the formula, because the formula only works for the COMPONENTS of any trip where you are going an average speed for a certain amount of time. ONCE the conditions deal with different speeds or different times, you have to look at each of those components and how they go together. And that can be very difficult if you are not methodical in how you think about the components and how they go together. The formula doesn't tell you which components you need to look at and how they go together. For that, you need to think, and the thinking is not always as easy or straight forward as it seems like it ought to be.
In the case of your cousin above, if we call the time he spent driving 80 mph, T1; then the time he spent standing still is (2.5 - T1) hours, since the whole trip took 2.5 hours. So we have
100 miles = (80 mph x T1) + (0 mph x [2.5 - T1])
which is equivalent then to: 100 miles = 80 mph x T1
So T1 will equal 1 and a quarter hours. And, since the time he spent going zero is (2.5 - 1.25), it also turns out to be 1.25 hours.
Now sometimes the right answers will seem counter-intuitive, so it is really important to think about your components methodically and systematically. There is a famous trick problem: To qualify for a race, you need to average 60 mph driving two laps around a 1 mile long track. You have some sort of engine difficulty the first lap so that you only average 30 mph during that lap; how fast do you have to drive the second lap to average 60 for both of them? I am going to go through THIS problem with you because, since it is SO difficult, it will illustrate a way of looking at almost all the kinds of things you have to think about when working any of these kinds of problems FOR THE FIRST TIME (i.e.,before you can do them mechanically because you recognize the TYPE of problem it is). Intuitively it would seem you need to drive 90, but this turns out to be wrong for reasons I will give in a minute.
The answer is that NO MATTER HOW FAST you do the second lap, you can't make it. And this SEEMS really odd and that it can't possibly be right, but it is. The reason is that in order to average at least 60 mph over two one-mile laps, since 60 mph is one mile per minute, you will need to do the whole two miles in two minutes or less. But if you drove the first mile at only 30, you used up the whole two minutes just doing IT. How to download apps in pc windows 10. So you have run out of time to qualify.
To see this with the rt=d formula, you need to look at the overall trip and break it into components, and that is the hardest part of doing this (these) problem(s), because (often) the components are difficult to figure out, and because it is hard to see which ones you need to put together in which way. The race car one is a perfect example, because you can't just add the average speed of the two laps together and divide them by two, BECAUSE the time you need for each lap will be different. You made a similar error in your first statement of the airplane problem-- you assumed the distances were the same for two planes flying the same time (even though they were going different rates). You have to be really be careful about what you assume, and that is often difficult, and takes a lot of thinking. In the race car problem, the trick is in seeing that the average rate for the whole qualifying run will be equal to the (sum of the distances of each different part of the qualifying run) divided by (the total time it took to go all those distances) --which, I know, looks like a mouthful when you first read it, but we will look at the problem piecemeal:
You set it up giving all the parts you know in terms of COMPONENTS:
You know all about the first lap, but only the distance of the second lap; but you also know what the total distance and the average speed for the total distance have to be. Now you have to mess around trying to put all that together somehow, and sometimes that might take some trial-and-error.
Lets call the distance of the first lap D1, the rate R1, and the time T1;
the second lap we will call all those things D2, R2, and T2;
and for the whole run, we will call the time T, the rate R, and the distance D. Ntfs write mac high sierra. We know what some of these things have to be:
D1 and D2 are each one mile, so we know D = 2 miles
We know R must equal 60 mph (or more, but we will skip the 'more' part).
We know R1 = 30
We need to find R2
So I tried the following first, and it got me nowhere, and I include it here because I want you to see that it is possible, and easy, to do this stuff mechanically, and even thoughtfully, and get nowhere even though it looks like a promising way to go. This stuff is not as easy as the books and some teachers make it look; so you don't need to think you are stupid or ignorant just because you take a wrong path on one of these, or any algebra problem. All of us can work types of problems we recognize how to do without thinking, but that is quite different from being able to do types of problems we meet for the first time or have forgotten how to do automatically:
We know that the total average rate will be the total distance divided by the total time: R will be D/T, by solving the D=RT equation for R (or by just realizing that rate simply is distance divided by the time; e.g., 80 miles per hour, 186,000 feet per second, 800 kilometers/day, 2,000,999 miles per lunar month, or whatever).
and since D = (D1 + D2), if we can figure out D1 and D2 in terms of their own rates and times, we might have something going here.
[The stuff (in blue and in green) down to the next blank line will be a futile, though reasonable, attempt.]
Well, D1 = R1 x T1
(since the distance for the first lap will be the speed of that lap times the time),
and D2 = R2 x T2,
so just simply plugging all this stuff in, we will get
D = (R1 x T1) + (R2 x T2)
That means
2 miles = (30 mph x T1) + (R2 x T2)
But now we are stuck again, unless we can figure out how to represent some of this in terms of things we know.
Well, since R2 is what we are looking for, let's mess with T1 and T2.
T1 will equal D1/R1, and since we know D1=1 mile and R1=30 mph, we know that T1 must be 1/30 of an hour.
So we can plug that back in, and get:
2 = (30 x 1/30) + (R2 x T2)
And that means 2 = 1 + (R2 x T2)
But now I am stuck because first I don't know what either R2 or T2 is, and second, we already know
R2 x T2 = D2, which is 1 mile, so we end up with the trivial statement 2 = 1 + 1.
This is not good.
So now I go back to thinking again, instead of just plugging in.
Let's see if we can't just set up something algebraically closer to what the problem states. The average rate has to equal 60, and we have part of the components, so:
R = D/T which is what I had before, before I went off track.
Hence, 60 mph = 2 miles/T
which means, if you solve for T, that T = 2/60 or 1/30 of an hour.
But we also know that T = T1 + T2 since the total time will equal the sum of the times for each lap. So 1/30 = T1 + T2
Now, T1 = D1/R1, which is 1 mile/30 mph, which is 1/30 of an hour (two minutes), so that means: 1/30= 1/30 + T2
That means T2 must be zero seconds. Which means he has to do the second lap in NO time at all. And that is impossible.
So let's go back and look at your airplane problem now, which should be easier than this:
We'll call them plane 1 and plane 2, and their distances, times, and rates also 1 and 2 respectively.
So we know R1 = 500 mph and R2 = 650 mph
We know D (the total distance they cover together) is 3000 miles
And we can know that T1=T2, since they will fly for the same amount of time.
Further those times each are equal to the total time, so we can call all the times simply T, and T is also what the problem asks us to solve.
So the question is how we can express all these components in some way that will get us a solution.
Well, we know that the total distance = the sum of the distances both planes travel:
D = D1 + D2
and we know that the distance each plane travels is its rate times its time.
We know the total distance and we know the rates of each plane, so we can say that
3000 = D1 + D2 = (R1T1 + R2T2) = (R1T + R2T)
since the times are all the same. Which means after you plug in all the numbers you know that
3000 = (500T + 650T) = 1150T,
so T will = 3000/1150, which is 2.60869.
(-- which is a number psychologically likely to make you feel you must have made a mistake since it doesn't come out a whole or simple number).
Let me give three quick examples to show that none of this stuff is necessarily easy to figure out, even in cases when it actually is easy to do. There is a problem that is really difficult for people who know calculus and lots of high level math, but is easy for people who don't know calculus or higher level math:
1) Two trains start toward each other at the same time just like your two planes do. One is going 90 and the other 60. The distance is the same as above, 3000 miles. A really fast and tireless bee starts when they do, at the front of one of them, and flies 300 mph toward the other. When it gets to the second train, it immediately turns and flies back toward the first, and when it gets to it, it turns and flies back again, always flying back and forth between the two trains as they speed toward each other. How far will the bee travel altogether before it is squished between the two trains, which are, unfortunately, on the same track? Notice the bee makes a shorter trip each time, because the trains are getting closer and closer all the time. In calculus, this is really hard, because you have to 'sum an infinite series', but that is the most obvious way to see this problem if you know calculus. If you don't, it is a simple problem, not unlike your plane problem -- with an added feature:
Notice the two trains are covering 150 mph between them. There are3000 miles to cover altogether, so it will take them 20 hours. The bee is flying 300 mph then for twenty hours, so the bee flies 6000 miles. Ta da!
When I was taking algebra, I was given the following problem:
2) a train leaves NY for LA, going 65 mph; and 8 hours and 25 minutes later a train leaves LA for NY going 88 mph. When they pass, which one will be closer to NY?
I worked on it forever and finally just to try to see how to do it, I changed the numbers to make them easier to work with. I used my new problem to figure out how far from NY each train would be, and then I found out that they would be the SAME distance from NY, and I thought'what a coincidence!' UNTIL I then re-read the problem in light of that answer. OF COURSE, they would be the same distance from NY -- they are passing each other AT THE SAME PLACE, which NO MATTER WHERE THAT IS, will be the same distance from NY, even if they are in Nevada. The problem did not ask for which train travels the farthest nor for whether they will meet closer to NY or to LA or any of the standard things. It was a trick problem that didn't need ANY algebra. Algebra only made it hard, if you were going to plug into formulas without carefully paying attention to what was being asked.
Finally, Richard Feynman, the physicist that one author has referred to as 'no ordinary genius' in a book by that title, one time got one right in just a few seconds during a high school math competition. Everyone else was still picking up their pencils to start figuring:
3) A river is flowing downstream at 15 mph relative to the shore. A rowing team is practicing rowing and at first they row upstream (against the current). They can only go 1.5 mph relative to the shore at this rate. The guy at the back end of the boat is wearing a hat when they begin, but after a while his hat falls into the water (and floats) and it is 15 minutes before they notice it. They then instantaneously reverse direction and row back to catch up with the hat, rowing with the same strength or power they were rowing with before. How long will it take them to catch up with the hat as it is pushed downstream by the current?
The answer is 15 minutes, because all those various rates cancel each other out and are irrelevant -- just as if the water was still water. But the easiest way to see this is that suppose you travel west in your car for an hour and twenty minutes, and realize you left something at home that you need. How long will it take you to get home, traveling at the same power? One hour and twenty minutes, of course, (assuming no lights or traffic, etc., of course). Yet the earth has been turning eastward at 1000 miles per hour (that is how fast the earth turns at the middle, since it is 24,000 miles around it takes 24 hours to make one complete rotation), and you were going against that as you drove west, and now you are going with that as you drive back east. Yet we never need to take that into account. Nor do we need to take into account the distance or rate the earth has been traveling in its orbit around the sun.
Don't feel bad when you can't do any of these KINDS of problems immediately and automatically the first time you see one. Just think about them and think about the formula d=rt and try to figure out the essential components that will give you a solution. That is not easy the first time you see a given type of problem. But once you see how a given type works, you can do other problems of that type fairly quickly and mechanically.
One thing to BE VERY CAREFUL ABOUT: keep your 'units' straight, and preferably with their numbers in your work as you go. Sometimes you will have a problem where one rate is given in miles per hour and another rate is feet per second, or they will give you a rate in miles per hour and ask you about something in minutes. Or they will ask about distances in miles after giving rates in feet per second. Almost anything can vary 'unit'-wise, so be careful, and keep your units with your numbers. Otherwise you could end up adding, subtracting, multiplying, or dividing 'unlike' things you don't realize are inappropriate to put together. E.g., you can't subtract 25 feet/second from 50 miles/hour and get '25' feet/second or miles/hour. (In physics, and elsewhere, you can multiply or divide various unlike units of certain sorts, to get foot-seconds or gram-centimeters or whatever, but again you have to keep all that straight or you will mix some inappropriately, and get wrong numbers with wrong units.)
Keeping your units with their numbers also helps you keep your multiplication and division straight if you realize you can multiply and divide (and 'cancel out') units the same way you can numbers. I.e., if you multiply miles/hour times hours, the 'hours' cancel out and you get 'miles', which you should. If you divide miles by seconds, you get miles/second. But sometimes lining up your units is not so obvious, and it is difficult to know whether you need to multiply or divide quantities. So if you have some numbers with rates and some numbers with distances, and aren't sure which way you need to divide in order to get times, just line up your quantities with their units so that you divide the distances by the rates in order to end up with times. I.e., if you have 30 miles/hour and 400 miles, you divided the number of miles by the number of miles/hour, which in order to get 'hours' -- 400 miles/30 (miles/hour) = 13 and a third hours.
(Remember dividing by a fraction is the same as multiplying by its reciprocal, so when you divide 'miles' by 'miles/hour', it is the same as multiplying 'miles' x 'hours/mile', and the 'miles' 'cancel out' leaving just 'hours'. So if you ever end up with something like 40/hour and there is no unit in the numerator, you have likely done something wrong, since there should be 40 something/hour. In the above example, however, notice that when you divided 400 miles by 30 miles/hour, it is only the units that 'reciprocate'; not the numbers; the 400 is divided by the 30, and the miles are divided by the miles/hour, to yield hours -- 13 and a third hours in this case.)
Rate x Time = Quantity of Work Problems
These are the sorts of problems that go something like: 1 man does a job in 3 hours, another does it in 4 hours and a third does it 6 hours; how long will it take them to do the job working together? Or, 1 pipe lets in water that will fill a pool in two days, another will fill it by itself in a day and a half, and a third pipe will empty the pool in two days. If all three pipes are open, how long will it take to fill the pool?These are just like the rate x time = distance problems except that instead of distances, you are getting a quantity of work, and you have to be sure to keep your units straight in these. So, looking at the first problem, there are some things to notice: 1) if they all work together the time they work will all be the same, since they are working from the start of the job until it is finished. 2) What they will be achieving is, not distance, but work, and that work is usually, but not necessarily, in the form of
'1 complete job'.
So if we use the first case as an example, what you have is A works at the rate of 1 job/3 hours, B works at the rate of 1 job/4 hours, and C does 1 job/6 hours. These are our components, and we 'just' have to assemble them in a rate x time = job form as we did the distance problems. So, again, here is where the 'thinking' part comes in. First, we know that their rates will combine somehow so that the quantities of their individual labor will produce the whole product, and second we know that they will each work the same amount of time, call it t.
So if we call the whole job or product Q, and each man's contribution to it Q1, Q2, and Q3, we know that: Q = Q1 + Q2 + Q3.
Further, we know that each man's contribution will be the product of his rate of work times the amount of time he works, so Q1 = tR1, Q2 = t R2, and Q3 = tR3
so Q = tR1 + tR2 + tR3
Now we know the rates: 1 job/3 hours, 1 job/4 hours, and 1 job/6 hours.
So Q = t(1 job/3 hours) + t(1 job/4 hours) + t(1 job/6 hours) = t(1/3 + 1/4 + 1/6 job/hour)
That appears to leave us with two variables, Q and t, but actually we know Q, because
Q is 1 COMPLETE JOB.
Remember the question was 'how long, (t), does it take for all the people working together to do the (i.e., one whole) job'.
So then we have:
1 job = t(1 job/3 hours) + t(1 job/4 hours) + t(1 job/6 hours) = t(1/3 + 1/4 + 1/6 job/hour)
1 job = t(4/12 + 3/12 + 2/12 job/hour)
1 job = t(9/12 job/hour)
1 job/(9/12 job/hour) = t
t = 4/3 hour, or 1 hour and 20 minutes
To check the answer, you just add up how much work each person would do in that time, and see whether it comes out to one complete job. So:
Person 1 does 4/3 hours x 1/3 job/hour = 4/9 job = 16/36 job
Person 2 does 4/3 hours x 1/4 job/hour = 4/12 job = 12/36 job
Person 3 does 4/3 hours x 1/6 job/hour = 4/18 job = 8/36 job
which adds up to 36/36 of the job, or 1 complete job.
Other Rate x Something = Something Else Problems
The same ideas apply to any sorts of problems that have to do with anything like rates, whether of speeds, monitary fees (costs/unit), dilution rates of chemicals, proportional rates of quantities, percentages, batting averages (batting average x official 'at bats' = hits), etc. E.g, apples are $12/bushel and oranges are $20/bushel, so how much of each will make up 6 bushels of fruit at an average of $18/bushel?If b = the number of bushels of apples in the mix, then (6-b) will equal the number of bushels of oranges.
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And we know that the total cost of the final mix will be 6 bushels x $18/bushel, which will be $108.
This total will be what the apples and the oranges sell for together:
(b bushels of apples x $12/bushel) + [(6-b) bushels of oranges x $20/bushel] = $108
$12b + $120 - $20b = $108
$12 = $8b
b = 1.5 bushels of apples, and therefore 6-b = 4.5 bushels of oranges.
Checking the answer, we get:
1.5 bushels of apples x $12/bushel = $18 worth of apples
4.5 bushels of oranges x $20/bushel = $90 worth of oranges
Which gives 6 bushels of fruit worth a total of $108, or $18/bushel.
And Now a Really Difficult and Strange One As a Brain-Teaser
Someone sent me the following rate, time, distance problem that took me forever to solve, and at first I was certain there was not sufficient information to solve it, but it turned out there was. I give it to you here as a kind of supreme test, and also to torment you as it tormented me in a brain-teaser kind of way. By the way, I sent it to someone I had met on the Internet and he solved it much too quickly. I hope it takes all of you at least as long as it took me:There are two boats that start out on opposite sides of a river at the same time. Each one is heading across the river to the other side. They each go a constant speed throughout the entire problem (so ignore having to slow down to turn around, and ignore current, etc.), but they are not necessarily the same speed as each other. When each boat reaches its opposite bank, it immediately turns around and heads back to where it started. The boats thus pass each other twice. The first time they pass, they are 700 yards from one of the banks of the river. The second time they pass, they have each turned around after reaching their respective opposite shores and have started back toward where they each began. When they pass the second time, they are 300 yards from the other bank of the river. How wide is the river? The answer is below, if and when you are ready to look at it.
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A Solution To the Two Boats On the River Problem
I was getting nowhere trying to solve it by setting up equations because I could get all kinds of equations, but nothing that helped me get anywhere. And I finally decided I didn't really understand what was involved in this problem, since I couldn't even see what difference different distances would make. It seemed intuitively to me that the rates could be adjusted to make almost any distance work out okay. There just did not seem to be sufficient information to limit the answer to one distance. So in order to see what factors were at work in this mess, to see what I needed, and to see what I seemed to be missing, I first picked an arbitrary distance, 1500 yards, just to see what would happen. When I did that, I saw a problem with the result, and what caused that problem -- what the limiting factors were-- so I then was able to see what was happening in the situation and then to figure out the equations I needed to solve it. Here's the deal:We know that the boats do not change their speeds, and we know that the time each boat travels to the first meeting place is the same. We also know that the time each boat travels from the first meeting place to the second meeting place is the same. Since their speeds do not change, the ratio of their speeds does not change. We can also represent all that. So here it goes:
Call the width of the river in yards X.
I am calling the rate of the boat that travels 700 yards from its shoreline to the first meeting point R1, and the rate of the other boat R2. I am calling the amount of times for each boat T1 and T2 respectively then. The boats then are also referred to as boat1 and boat2.
The distance boat1, D1, goes to the first meeting is 700 yards (given in the problem).
The distance boat2 goes to the first meeting then is (X - 700) yards.
The first boat's speed during the first part of the trip is R1 = (D1/T1)
And the second boat's speed during the first part of the trip R2 = (D2/T2)
So the ratio of their speeds during the first part of the trip (i.e., to the first meeting place) is (D1/T1)/(D2/T2)
That will also be the ratio of their speeds to each other during the second part of the trip (i.e., from the first meeting place to the
second meeting place) because their speeds do not change.
Notice, however, that in each part of the trip, the boats travel for the same amount of time as each other. (They don't necessarily travel the same amount of time in each trip as they did in the other trip, but they travel the same amount of time as each other in the same trips -- i.e., they each travel for a same amount of time to meet each other the first time, and then they each travel a [possibly different] same amount of time for the part from there till they meet again.)
That means the T's in the denominators cancel out and we can disregard them.
Therefore the ratio of their speeds to each other will be the same ratio as their distances they each travel. That is R1/R2 = (D1)/(D2)
Some of you may have seen this directly, but I didn't. It makes sense when you come to think of it, of course, because they are each going a constant speed for the same time and that means the faster one will go further by the ratio of its speed. For example, if it is going twice as fast, it will go twice as far in the same amount of time. If it is going one and a half times faster, it will go one and a half times farther if they both travel the same amount of time.
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Since their speeds do not ever change during the entire trip, we know that the ratios of their speeds during the first part of the trip will be the same as the ratio of their speeds during the second part of the trip. This means that the ratio of their distances for each part of the trip will be the same, which is to say (D1)/(D2) for the first part of the trip is equal to (D1)/(D2) for the second part of the trip. And we have the numbers for those distances, expressed in X, the width of the river.
So we can plug in the numbers we have:
In the trip to the first meeting, boat1 goes 700 yards.
Boat2 goes (X - 700) yards.
So the ratio of their distances (which is the same as the ratio of their rates, is):
700/(X-700) for the first part of their trip (i.e., from starting out till the time they first meet).
Boat1 then continues on to the bank, which is (X - 700) yards away, and then it turns around and goes back 300 yards. Therefore after the first meeting, boat1 travels (X - 700) + 300, which is (X - 400) yards.
After the first meeting Boat2 travels 700 yards to the bank and then turns around and travels back (X - 300) yards, which is (X + 400) yards, where it then meets Boat1 again.
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But the ratio of the speeds of both boats to each other is the same as it was before, and the ratio of the distances to each other is thus the same as it was before, so now we finally get the equation that will yield the value of X.
The ratio of the distances of boat1/boat2 the first trip = the ratio of the distances of boat1/boat2 the second trip. Or, symbolically, that is (D1)/(D2) for the first trip = (D1)/(D2) for the second trip, and we can substitute all our known D's to get:
Ratio of distance in the first leg of the trip of boat1 to boat2 is 700/(X-700)
Ratio of distance in the second leg of the trip of boat1 to boat2 is [(X-400)/(X+400)]
and since these ratios are equal to each other: Apple wireless keyboard with numeric keypad.
Cross-multiplying gives us:
700X + 280,000 = X2 - 1100X + 280,000
Which gives us 1800X = X2
And dividing both sides by X gives us 1800 = X
So the width of the river is 1800 yards.
To check that, we see that the first boat goes 700 yards in the same time it takes the second boat to go 1100 yards for the first meeting. So the ratio of their speeds (and the ratio of their distances) is 700/1100 or 7/11 in the first part of the trip. In the next part of the trip, the first boat goes the rest of the way to the other side (1100 yards) and then back 300 yard, a total of 1400 yards in the second leg, and the second boat goes 700 to the shore and then has to go back 1500 yards to get within 300 yards of its home shore, which is a total distance of 2200 yards for its second leg. Hence, the ratio of their speeds (and the ratio of their distances) is 1400/2200, or also 7/11 in the second part of their trip, so it all checks out.
