Question

$1000 was invested altogether. The first investment yielded a 15% profit, but the second investment yielded a 10% loss. Altogether this was equivalent to a 6% loss. How Much was in each of these investments?

Answer 1

x + y = 1000
(1.15)x + (0.90)y = 940
Two equations in two unknowns. Just solve for x and y now. First equation is self-explanatory. In the second equation, (1.15)x is the expression for a 15% profit on the first investment. Similar for second term. 940 shows a 6% loss on 1000. Easy to solve at this point now that the equations are set up.

Answer 2

Thanks. Can I ask you another question too? I need to know if I set it up correctly.

Answer 3

Would it be, 1.15x for the first invest or .15x ?

Answer 4

ok, but did you see what I did for the question above? No problems understanding?

Answer 5

Kind of...

Answer 6

where do you get the .9 from though

Answer 7

I get everything except where you get the .9y, and why it is 1.15x as opposed to .15x

Answer 8

You'll have to get that methodology down cold before you are able to do these on your own. 0.90y is the expression for a 10% loss because 90% of the investment remains.

Answer 9

Whew! Back now, (for now!)

Answer 10

Ok I understand now actually. It's common sense just not seeing it.

Answer 11

1.15 because you retain the first investment (1) and show a profit (0.15), so principal + profit gives the 1.15

Answer 12

These questions can be hard at first, but they will get easier with practice.

Answer 13

Ahh I see. Are you going to be online?

Answer 14

I'm going to see help with 3 more.

Answer 15

Just post new question in a new thread. I'll look for you.

Answer 16

I just need to know if I set up the equation right, should I make a thread for that?

Answer 17

I can look here if you like.

Answer 18

Ok. I'll make it fast. Thanks.

Answer 19

CLyde has 900 invest at 8%. How much money must Clyde invest at 3% so that he will assume a 6% return altogether on the two investments.

Answer 20

I put this equation...

Answer 21

.06 = 900(.08) + x (.03)

Answer 22

I came up with: 972 + (1.03)x = (900 + x)(1.06). That will work. I still have to check your method out. But first, I'm going to solve mine for x.

Answer 23

Did you get 972 by multiplying 900 by .08

Answer 24

Not exactly. I muliplied it by 1.08. Principal + profit

Answer 25

By the way, my equation worked because I took my answer and went through the problem statement. You should get 600 for the principal to be invested at .03

Answer 26

I believe you. I just need help setting it up.

Answer 27

why do you add 1 to the left of the decimals. i'm confused on what you're referring to as the principal.

Answer 28

We can either stick with my method and show why it works, and more importantly, get you into being able to set it up like that, or we can look at your equation and see what went wrong. I sort of suggest just looking at what I did and we can go through it more closely.

Answer 29

Principal is what I'm calling the amount to be invested.

Answer 30

lets do that.

Answer 31

I add 1 to the decimal to show the "afterwards" on the investment, when there is a profit. When there is a loss, I subtract the loss percentage.

Answer 32

ultimately saving time... good idea.

Answer 33

So, starting with this last of the two problems, he invests 900 at 8%, so we know that from that 900, he will end up with 972 because he retains the 900 and gets a profit 900 x (1.08) = 972. 72 is the profit from 900 x 0.08. Okay so far?

Answer 34

so why is it (at the end of the equation) multiplied by 1.06 as opposed to added.

Answer 35

oh nevermind. because its the 6% of the return

Answer 36

im with u

Answer 37

i'll let u continue until the answer, then ask questions

Answer 38

Well, we'll get to that, but that's not the next step. Gotta do this in order or it won't make sense. So, you're ok on the 972?

Answer 39

yea. i under stand it.

Answer 40

the 1 on the left represents the original investment.

Answer 41

(left of decimal) yea I get that part. thank u alot.

Answer 42

same rule applies for the rest because they're all an increase of return? right?

Answer 43

Now, about the other investment, we don't know what it is yet, so we'll just call it x, but we do know that we will get a 3% profit from it. We will also get our investment back, so we will end up from the x with (1 + 0.03)x at the end of the investment period. Same as saying (1.03)x. Yes, same rule, just saw your last question.

Answer 44

so the reason you put the 900 on the right side of the = is because it represents the original investment?

Answer 45

So, he ends up with 900(1.08) + (1.03)x when all is said and done and the investment period is over. That is simplified to 972 + (1.03)x. Answering your last question, 972 is on the same side as the (1.03)x and represents the total amount he will have.

Answer 46

That amount is the "afterward". What he starts with is 900 + x, two separate investments that he will look at together as far as Principal or amount to be invested as a "package" (conceptually). He wants that 900 + x as a hypothetical (for comparison purposes) amount invested at 6% to equal the "afterward" . So, he sets the hypothetical (900 + x)(1.06) as equal to the 972 + (1.03)x. Now we have one equation in one variable which is easier to solve than the first problem.

Answer 47

So, that's basically it. You can feel free to ask questions about it, but that's basically all that I did.

Answer 48

Okay, I see how you did that. Now, with any investment problems, you can use this same method. You just have to do the math for the difference in the investment, and represent properly if its a loss or gain... right?

Answer 49

Bear in mind that though these two questions were investment problems, the way we eventually solved each was very different. There were some similarities in thinking about "before and after" and that profit gave multiplying principal by 1 + % returned (and loss by 1 - % lost). But notice how the first problem required 2 equations in 2 unknowns, and the second problem required 1 equation in 1 unknown.

Answer 50

Right, that is just with what their asking. But essentially for investments, and when they say gain or loss you can use the 1 + .05 or whatever the percent is. or 1 - .05 if it's a loss.

Answer 51

i have another question. where i need to see if I set it up correctly. can I ask you here? i dont want someone to answer with a different method and confuse me. if not its ok.

Answer 52

Yes, that will be a great help conceptually and mathematically. And a little bit of a mental shortcut.

Answer 53

On last question, sure, go ahead. But before we get into it, I just want to say that, once we are all done, you might want to go back and review this whole thread at some time, because I think it will help.

Answer 54

Great. Got that embedded now, excellent method.

Answer 55

Yea. I'm new to this site, do the chat logs stay saved?

Answer 56

You click on your name near the top right and that will open a second window. Then you go to where you have "questions asked". It will be there but I don't know how long it is saved.

Answer 57

There is a total of $7000 in two investments. The annual income from the 11% investment is $50 more than the annual income from the 9% investment. How much is invested at each rate? I did two equations.

x + y = 7000
.11x = .9y + 50

correct?

Answer 58

I 've been here only under 2 weeks, so maybe they go away after a while.

Answer 59

Oh okay. Thank you for your help. It's really appreciated.

Answer 60

On that second equation, be careful. It would be .09, not .9 .9 is 90%

Answer 61

Ahhhhh... I need to observe! How was the actual set up, minus that?

Answer 62

On a first look, it looks good, but I always work it through before giving a final answer. So, I'm going through it now.

Answer 63

Ok. great. I primarily need to get the steps of setting up the equation more, as opposed to doing the actual arithmetic behind it. The setup is very difficult to me.

Answer 64

There's something wrong here. Is there something left out of the problem statement? There is reference to annual income. Is there a investment length period overall?

Answer 65

As far as the question, that is all that is provided.

Answer 66

OK, then it works out just fine. After doing the math, you should get 3600 for the .09 investment and 3400 for the .11 investment.

Answer 67

So, good job! You came up with those equations on your own!

Answer 68

Yup. It takes a lot of re-practicing for these. Not as easy as it was in high school lol. May I ask you another? And is there any more feedback I can give you to boost your profile on this site? Nothing but A+++ feedback lol. (If you're familiar with eBay)

Answer 69

I don't go in so much for the recognition. I joined because I had one stumper of a question which I eventually figured out myself. Then I hung around because I like helping out and I like math. So, you don't have to give any kudos, really. There is some provision somewhere for what they call fan testimonial, but I don't know too much about it. You can ask another question, but then I have to go.

Answer 70

Alright I'll look into that. Thank you again, I'll make this last one fast...

Answer 71

$5000 is in three investments. Twice as much is invested at 8% as at 6% and the remainder is invest at 9%. If the annual return is $390, how much is invested at each rate?

Answer 72

You can take your time, don't rush, these things can get tricky if rushed.

Answer 73

I understand, but I don't want to hinder your schedule. Now, I'm really confused on the "remainder" into 9% I don't know how to represent that. So far i've came up with x + y + z = 5000 and (.08)x and y(.06)(2)

Answer 74

So you can get an idea of what I understand, I'll let you explain and I'll ask questions after you're done.

Answer 75

Ok, I've got it.

Answer 76

Mathematically, it's 3 equations in 3 unknowns which on the surface makes this problem at first appear harder than it really is. It really simplifies down to 2 equations in 2 unknowns which are solved first, and then the last equation in 3 unknowns is actually very easily solved. Ok, so much for the math analysis, we'll get to it now.

Answer 77

You're off to a good start with x + y + z = 5000, so let's go with those variables and that as the first equation. Also, let's call x the amount we invest at .06, y the amount at .08, and z the amount at .09.

Answer 78

Im with you...

Answer 79

"Twice as much is invested at 8% as at 6%" is a big key to simplifying. It tells us that y = 2x. So, equation 2 becomes x + 2x + z = 5000 and simplifies to 3x + z = 5000. That's going to be a big help.

Answer 80

It helps simplify equation 3 tremendously. Equation 3 starts out as (0.06)x + (0.08)y + (0.09)z = 390 and ends up as (0.06)x + (0.08)(2x) + (0.09)z = 390 which ends up as (0.22)x + (0.09)z = 390 . So, this eventual equation 3 ends up as being in the same two variables as equation 2. So, you solve these two equations first for x and z, and then simply plug those into equation 1.

Answer 81

So, all that's left is to do the math, but that's easy now. If you want, I can try to get the actual numbers.

Answer 82

now to plugin the 3 equation to the 1st. I would have to solve for x or z. in the second equation i solved for Z, so I would have to solve for X. i this case i will subtract the .09z from both sides, then divide the .22 from both sides to isolate the X ?

Answer 83

You should get x = 1200, y = 2400, and z = 1400.

Answer 84

The method is to solve equations 2 and 3 together for x and z. You will get actual numbers for x and z. Take the x and z (not equations 2 or 3) and put the numerical x and z into equation 1 and you will get y. Or just remember that y is twice x.

Answer 85

so .22x + y + .09z = 5000 ?

Answer 86

Remember, equation 2 is 3x + z = 5000. Equation 3 is (0.22)x + (0.09)z = 390. Easy way to solve that is to take equation 2 and that z = 5000 - 3x. Take THAT z and put it into the z in equation 3. No "y" in your equation above.

Answer 87

so. (0.22)x + (0.09)(5000 - 3x) = 390 ?

Answer 88

Equation 3 will become (0.22)x + (0.09)(5000 - 3x) = 390. NOW, you've got one equation in one variable, so you can solve for x and you should get 1200 for x. Just saw your post. Looks like you figured out that part that I just mentioned. You're getting it.

Answer 89

Yup. Thanks to you. I'll need to review some of these at my lab tomorrow.

Answer 90

They're actually fun once you get the hang of it. After you review, you'll be surprised at how easy it gets.

Answer 91

Yea I can see how it becomes easier. It's all the set up, the arithmetic is relatively easy

Answer 92

Yes. Well, take care now. It's been great working with you. And good luck in all your studies. Bye for now.

Answer 93

Thank you for you're time. Have a great evening...