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## Practice Questions

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## Table of Contents

## Transcription

## Related Books

### Applications of Linear Equations, Motion & Mixtures

- Distance is equal to rate times time. (d = r t) The amount of pure substance in a mixture is equal to the percent times the total amount. (a = p t)
- In the table method each row represents a different situation. The columns can be used for distance, rate, and time.
- To create the equation that connects all the information in the table method, multiply across, and add down.
- In the beaker method each beaker stands for a substance. Record the percent and the amount for each.
- To create the equation that connects all the information in the beaker method, multiply the percent by the amount, and add beakers.
- The table method often works well for motion problems, and the beaker method works well for mixture problems.

### Applications of Linear Equations, Motion & Mixtures

- percentage change = [(final value − original value)/original value] ×100
- original temperature = 30

final temperature = 60 - percentage ∆ = [(60 − 30)/60] ×100
- percentage ∆ = [30/60] ×100 percentage ∆ = 0.5 ×100 = 50

- percentage change = [(final value − original value)/original value] ×100
- original temperature = 62 final temperature = 75
- percentage ∆ = [(75 − 62)/75] ×100
- percentage ∆ = [13/75] ×100
- percentage ∆ = [1300/75] = 17.―3

- Price of coat + sales tax = total price
- sales tax = (.07)(35) = 2.45

- Price of computer + sales tax = total price
- sales tax = (.08)(472) = 37.76

- Original price - amount of discount = final price

discount = (.15)(112) - 112 − (.15)(112) =
- 112 − 16.8 =

- Original price - amount of discount = final price

discount = (.10)(69) - 69 − (.10)(69) =
- 69 − 6.9 =

What is the final price of the suit?

To the nearest percent, what is the percent of change between the original price and the final price paid, including sales tax?

- Original price - discount = discounted price

160 - (.30)(160) = discounted price - 160 − 48 = $ 112
- Discounted price + amount of sales tax = final price

112 + (.075)(112) = final price - 112 + 8.4 = $ 120.4
- percentage change = [(final value − original value)/original value] ×100
- [(120.4 − 160)/160] ×100
- [( − 39.6)/160] ×100
- − 0.2475 ×100 =

25% decrease

What is the final price of the camera?

To the nearest percent, what is the percent of change between the original price and the final price paid, including sales tax?

- Original price - discount = discounted price

225 − (.15)(225) = discounted price - 225 − 33.75 = 191.25
- Discounted price + amount of sales tax = final price

191.25 + (.075)(191.25) = final price - 191.25 + 14.34 = $ 205.59
- percentage change = [(final value − original value)/original value] ×100
- [(205.59 − 225)/225] ×100
- [( − 19.41)/225] ×100

9% decrease

- price of chair + sales tax = total price
- sales tax = (.0825)(12) = 0.99

- original price - amount of discount = final price
- discount = (.20)(540) = $108

- percentage change = [(final value − original value)/original value] ×100
- original temperature = 30

final temperature = 60 - percentage ∆ = frac60 − 3060 ×100
- percentage ∆ = frac3060 ×100
- percentage ∆ = 0.5 ×100 = 50

- percentage change = [(final value − original value)/original value] ×100
- original temperature = 62 final temperature = 75
- percentage ∆ = [(75 − 62)/75] ×100
- percentage ∆ = [13/75] ×100
- percentage ∆ = [1300/75] = 17.―3

- Price of coat + sales tax = total price
- sales tax = (.07)(35) = 2.45

- Price of computer + sales tax = total price
- sales tax = (.08)(472) = 37.76

- Original price - amount of discount = final price

discount = (.15)(112) - 112 − (.15)(112) =
- 112 − 16.8 =

- Original price - amount of discount = final price

discount = (.10)(69) - 69 − (.10)(69) =
- 69 − 6.9 =

What is the final price of the suit?

To the nearest percent, what is the percent of change between the original price and the final price paid, including sales tax?

- Original price - discount = discounted price

160 - (.30)(160) = discounted price - 160 − 48 = $ 112
- Discounted price + amount of sales tax = final price

112 + (.075)(112) = final price - 112 + 8.4 = $ 120.4
- percentage change = [(final value − original value)/original value] ×100
- [(120.4 − 160)/160] ×100
- [( − 39.6)/160] ×100
- − 0.2475 ×100 =

25% decrease

What is the final price of the camera?

To the nearest percent, what is the percent of change between the original price and the final price paid, including sales tax?

- Original price - discount = discounted price

225 - (.15)(225) = discounted price - 225 - 33.75 = 191.25
- Discounted price + amount of sales tax = final price

191.25 + (.075)(191.25) = final price - 191.25 + 14.34 = $ 205.59
- percentage change = [(final value − original value)/original value] ×100
- [(205.59 − 225)/225] ×100
- [( − 19.41)/225] ×100

**8**

**.**

**627**

**9**

**%**decrease

- price of chair + sales tax = total price
- sales tax = (.0825)(12) = 0.99

- original price - amount of discount = final price
- discount = (.20)(540) = $108

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

Answer

### Applications of Linear Equations, Motion & Mixtures

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

- Intro 0:00
- Objectives 0:21
- Motion and Mixtures 0:46
- Motion Problems: Distance, Rate, and Time
- Mixture Problems: Amount, Percent, and Total
- The Table Method
- The Beaker Method
- Example 1 5:05
- Example 2 9:44
- Example 3 14:20
- Example 4 19:13

### Algebra 1 Online Course

### Transcription: Applications of Linear Equations, Motion & Mixtures

*Welcome back to www.educator.com.*0000

*In this lesson we are going to continue on with our examples of applications of linear equations.*0003

*In this we are going to look at some trickier applications, the ones involving lot of motion and mixtures.*0008

*It usually gives students a lot of headaches so pay close attention on how we approach this with some nice methods that will organizer our information.*0013

*Our two big goals, motion and mixture problems is what we want work at.*0026

*And the two methods we will use to attack these will be using the table method and the beaker method.*0031

*You will see that the table method usually works pretty good for a lot of motion problems*0037

*and the beaker method works pretty good when you are dealing with those mixtures.*0040

*As I said earlier, motion and mixture problems tend to be a weak area for many people.*0048

*The problem comes from being able to take all the information and organize it in some sort of way so they can actually build your equation.*0053

*In terms of equation, there is only two equations that work in the background.*0061

*For a lot of motion problems, it is the distance, the rate at the time, that all need to be related together.*0066

*The equation for that is distance = rate × time.*0073

*Let us go ahead and put it in there.*0077

*We have our distance, we have our rate, and we have our time.*0078

*With our mixture problems, there is a very similar formula that we are after and that is the one that deals with the pure substance in the mixture.*0086

*For that one we are looking at the amount of pure substance, multiply it by some sort of percent and the amount in the total mixture.*0094

*This guy over here is just the pure amount.*0112

*To help us organize these we will begin by using what is known as the table method.*0120

*In the table method, you build a small table usually has about three rows and about four columns in it*0125

*and we end up putting into this table is information about your distance, rate, and time.*0132

*The information about different things that is moving in the problem.*0137

*Every row in this table here usually represents a different object.*0142

*If I’m talking about to bicyclists, I might talk about bike number 1 and bike number 2.*0147

*That way I can keep track of each of their motions separately.*0157

*In each of the columns, we will put information about the rate, distance, and the time.*0160

*The way we usually do that is we have information about the rate here, the time here, and the distance over here.*0166

*What we are looking at is that equation of distance = rate × time, in fact it is hidden in the background.*0179

*rate × time = distance.*0186

*For that reason it will be important on how we pull the information out of this table.*0191

*We want to remember that we want to multiply across.*0196

*In this direction we will multiply the rate and time in order to get our distance and what we want to do from there is add down to get our total.*0200

*It will help us build that equation and even if it does involve a couple of objects moving.*0211

*The other method which is good for mixtures is known as the beaker method.*0219

*The way this method works is you end up creating a beaker for each of the mixtures you are putting together.*0224

*Maybe this one over here is my 30% solution, this guy over here is my 90% solution and I'm looking to create a new solution over here on this side.*0230

*You will notice I have already put in the addition and equals signs so I can see that I'm mixing two beakers together to get a third one over here.*0241

*There are two things that we want to record in each of these beakers.*0250

*First we will record the amount, how much is in this beaker.*0254

*Is it 50 L? Is it 40 L?*0260

*We want to know what the exact amount in each of these will be.*0263

*We will also want to know what percentage is each of the solution. 30%, 20%, we will definitely keep track of that for each of these beakers.*0269

*In order to create the equation from these we will multiply the amount and the percentage together from each of the beakers.*0279

*Watch for me to use this method when we get to those mixture problems.*0300

*Let us go ahead and see this method in action as we are looking at example 1.*0307

*This says that building an equation to represent the following problem.*0313

*We have two airplanes and they are leaving a city at the same exact time and they are flying in opposite directions.*0317

*If one flies at 410 mph and the other one 105 mph faster, how long will it take them to be a total of 3290 miles apart.*0323

*You can see what I'm saying, there is a lot of information in this problem that is very confusing to try and keep it all straight.*0333

*It will help us out, let us keep track of both of these planes.*0339

*We have plane 1 and plane 2, what do we want to know about these planes?*0343

*We want to know how fast they are going so we will keep track of their rate.*0357

*We want to know how long they have been traveling, their time.*0361

*Of course we want to know how far they have traveled so we can get their distance.*0364

*Let us see what we can now figure out about these planes and put it into the table.*0372

*They are leaving a city at the same time and fly in opposite direction and one flies at 410 mph.*0378

*Let us say that is plane 1, they are traveling at 410 mph and the other one is 120 mph faster, if I was to add 120 to 410 that will give me 530 mph.*0384

*I know how fast each of these planes are going, that is definitely good.*0401

*How long have they been traveling now?*0404

*I’m not sure I got a lot of information about that because that is what I'm looking for in the problem.*0408

*It says how long will it take them to be 3290 miles apart?*0413

*My time here is completely unknown.*0418

*I do not how you know long each of them have been travelling.*0424

*To think about how the distance is working to all of this, remember the distance is equal to the rate multiplied by time.*0428

*I have multiplied across to get my distance is 410x for plane 1.*0436

*My distance for the other one is 530x.*0445

*I get a sense of how far each of them had gone and I want to know when they will reach a total of 3290 miles.*0452

*I’m looking at their total distance.*0460

*We will add this last column down.*0462

*The distance from plane 1+ the distance from plane 2 must equal a total distance of 3290 miles.*0466

*You can see that the table helps out so we can build this equation.*0479

*It organizes that information and we can crunch it down.*0485

*Now that we have this equation, we can not necessarily stop.*0490

*We have to move forward and figure out what the solution is.*0492

*Let us see if we can solve this.*0495

*3290 miles and we also have the 410x + 530x = 3290.*0503

*The first thing one I want to do is get my x’s together.*0518

*Let us see what that will be, 940x = 3290.*0521

*We will divide both sides by 940 and this will give me x = 3 ½.*0533

*In the context of the problem, what was that mean?*0548

*x was our time so this tells us that the planes traveled for 3 hours and ½ of an hour or 30 minutes.*0551

*When we use this table again to set up another problem and you will see how it helps us organize that information.*0577

*In this example, we are dealing with two trains and they are leaving the downtown station at the same time and they are traveling in opposite directions.*0587

*One had an average speed of 10 mph more than the other one and at exactly 1/5 of an hour they were 12 miles apart.*0595

*What are their two speeds?*0603

*We are going to start this one often much the same way we did the last one.*0605

*We have two trains, we will label our rows.*0607

*I have train 1 and train 2.*0611

*I want to keep track of how fast they are going, their rates.*0618

*How long they have been traveling, their time?*0623

*How far they have gone, their distance?*0626

*Let us go to the problem over here to even pull out the information and put in the right spot in the table.*0631

*One have an average speed of 10 mph more than the other one.*0636

*That does not give me a definite speed for any of these trains.*0642

*I just know that one is traveling faster than the other one.*0644

*Let us say we have no idea how fast train 1 is going we will use our unknown for that.*0650

*Since the other one was traveling 10 mph faster, I can take its rate and add 10.*0655

*Onto time, it says at exactly 1/5 of an hour and that is pretty good.*0666

*At least I know how long each of them have been traveling so 1/5 of an hour for each of them in terms of the time.*0671

*Let us put these together and see if we can get our distance.*0683

*Multiply rate and time, I will get 1/5x, multiplying rate and time here 1/5x + 10.*0686

*After 1/5 an hour they were a total of 12 miles apart.*0702

*I'm looking at both of these distances added together so 1/5x + 1/5 x + 10 so when will their distances be a total of 12 miles apart?*0706

*There is my equation and I'm going to use and try and solve from here 1/5x + 1/5x + 10 = 12.*0727

*We will use our techniques for solving linear equations and see if we can pick this apart.*0748

*The one is coming to mine right now since I'm staring at those fractions is we will multiply by a common denominator to clear at those fractions.*0753

*Let us go ahead and multiply everything through by 5.*0760

*I will do that on the left side and I will do it on the right side x + x + 10 and 5 × 12 = 60.*0764

*We can start combining our like terms 2x + 10 = 60.*0781

*I will go ahead and subtract 10 from both sides and finally we will divide both sides by 2.*0791

*I will get that x = 25 but again what exactly does this represent?*0810

*If you remember when we hunted down our variables and we said what x was, this is actually the speed of our first train.*0815

*Let us write that down, train 1 is moving at 25 mph and since the other train is going 10 mph faster, we will say that train 2 is moving at 35 mph.*0822

*Now we have the speed of both trains.*0856

*Now that we have seen a couple of those the table methods, let us get into using the beaker method with some good mixture problems.*0861

*In this problem we want to know how many mL of a 40% acid solution must be mixed with 80 mL of a 70% acid solution*0868

* in order to get a new acid solution that is only 50%.*0875

*A lot of information and it is tough to visualize this one unlike the motion problem where you can actually see two trains moving.*0881

*This is why I use the beaker method so I have something a little bit more visual that I can grab onto each of our beakers.*0887

*Let us go ahead and start recording the amount and the percentage for each of these things.*0894

*Let us start with the amount so it says how many mL of a 40% acid solution must be mixed with 80 mL of a 70% acid solution?*0902

*It looks like it does not tell me how much of the 40% I have, so we will leave that as an unknown amount.*0912

*We will be mixing it with 80 mL of this other one.*0922

*The percentage of this beaker is our 40% and the other beaker is our 70%.*0928

*This new beaker on the, we are trying to mix the two together, how much will it have in terms of their amount?*0942

*I’m mixing some unknown amount + some 80 mL so this new beaker at the end will have a total of x + 80 mL.*0949

*It is important that the last beaker should have a total of both the amounts.*0960

*What is its percentage? This will have a 50% solution.*0965

*We have recorded the amounts, we have recorded the percentages, and we will multiply those two quantities together and actually get our equation.*0971

*(40% is the same as .4 multiplied by x) + (70% is the same as .7 multiplied by 80) all of this must be equal to .5 multiplied by x + 80.*0981

*By multiplying those two together and putting in the appropriate addition and equals I have now set up my equation and I can solve from here.*1000

*This one is like our equation that has fractions only this one has in decimals.*1008

*I’m going to multiply through by 10 to get rid of a lot of those decimals.*1013

*.4 multiplied by 10 = 4x, .7 multiplied by 10 will be 7 and .5 multiplied by 10 would be 5.*1024

*This is the equation that I need to solve in order to figure out the amount of that 40% solution.*1041

*Let me just copy this onto the next page and we will go from there.*1049

*I’m going to start combining together each side of this equation using my distributive property and then I can work on getting those x’s isolated.*1072

*4x + 560 = 5 × 8 another 0.*1082

*We will subtract 4x from both sides 560 = 400 + x subtract the 400 from both sides, so 160 = x.*1100

*What this is telling me is since x represents the amount of 40% solution, I know that we need 160 mL of the 40% solution.*1127

*It is always a good idea to keep track of what your unknown represents.*1148

*I have one last example and this one is using the beaker method again but this one involves a lot of coins.*1154

*This is to demonstrate that these two methods are flexible and you do not always have to use them*1162

*with just a pure mixture or even just with distance, motion and travel.*1167

*They still can apply to a lot of other types of problems.*1171

*This one says a man has $2.55 in quarters and nickels.*1176

*He has 9 more nickels than he does quarters.*1180

*How many nickels and quarters does he have?*1183

*We have to figure out what the two are mixed together.*1185

*It is like a mixture problem that is why we are going to use the beaker method to attack it.*1190

*The two things that I will try to keep track of is how much of quarters and nickels do I have and what would the amounts of each of these be.*1194

*I better label my beakers to make it a little bit easier.*1205

*We will say that this first one contains just nickels and the second one contains quarters.*1208

*Let us see what we know about the nickels.*1219

*He has 9 more nickels than quarters.*1221

*We do not know a lot about those quarters, do we?*1223

*The amount of quarters we will leave that as x and the amount of nickels there is 9 more than however many quarters he had, x + 9.*1226

*I can write down what each of them represent.*1239

*A quarter is the .25 and nickel.05.*1245

*What we are looking to do is to combine these together, the amount and how much each of them are to get a total amount in that last beaker.*1253

*He has a total of $2.55 and the reason why I’m not writing down a specific amount or specific amount per coin in there*1267

* is because this one represents the two already put together just the $2.05.*1277

*Let us go ahead and build our equation.*1283

*.05 multiplied by the amounts x + 9.*1286

*.25x = $2.55 this one is similar to the previous one.*1292

*It has lots of decimals running around and I want to get rid of a lot of those decimals.*1302

*I’m going to multiply this one by hundreds and then we can do it just some nice numbers and no decimals.*1307

*This would make it 5 × x + 9 and 25x = 2.55 that is the equation that we will solve and figure out how many quarters and nickels we actually have.*1316

*5x + 9 + 25x = 2.55 working to simplify this equation I will distribute through on the left side of the equation giving me 5x + 45.*1342

*I will go ahead and combine my like terms 5x + 25x that will give 30x.*1376

*Then subtract 45 from both sides and you can get that x a little bit isolated.*1390

*Finally let us go head and divide it by 30 to see if we can figure out what that x is, so x = 7.*1404

*In the context of this problem, x represents the number of quarters we have.*1416

*We have 7/4 if you remember we have exactly 9 more nickels than quarters so we can add 9 to this number and figure out the total nickels.*1423

*We have 16 nickels.*1440

*We have figured out the total amount of both coins.*1447

*In each of these types of methods, they are great ways that you can get your information organized and into a good spot.*1453

*It will be handy with motion and with mixtures.*1461

*Thanks for watching www.educator.com.*1465

1 answer

Last reply by: Professor Eric Smith

Wed Oct 26, 2016 6:21 PM

Post by Pauline Nunn on August 29, 2016

omg these techniques make this all make sooo much more since!

Wow- thanks for sharing this. Ive been convoluted and confusing myself for so long where i didn't need to- this will help sooo much for testing and keeping calm.

0 answers

Post by Paul Cassidy on February 23, 2014

would someone like to confirm that the practice questions for the percentage of change are not correct. The equation statement is stated

as thus: percentage change = [(final value âˆ’ original value)/original value] Ã—100. However the solution demonstrates the following:

percentage change = [(final value âˆ’ original value)/final value] Ã—100.

I am pretty sure that answers that are being delivered are incorrect. could some please verify. Thank you

1 answer

Last reply by: Paul Cassidy

Sun Feb 23, 2014 8:06 AM

Post by Michael Heath on February 21, 2014

Hi Erick

If I set an equation like this: .3(40) + .2(60) = x(40 + 60), do I multiply the expression on the right by 10, like I will do to the expression on the left hand side?