Question

You invested $8000 in stocks and bonds, paying 6% and 9% annual interest, respectively. If the total interest earned for the year was $540, how much was invested in stocks and how much in bonds?

Answer 1

okay, we need to write two equations here.

one shows how much we've got invested in each class, and the other how much money we made.

let S be the amount invested in stocks. let B be the amount invested in bonds.

can you write an equation showing the relationship between B, S and 8000?

Answer 2

actually...yes :O)
.06x + .09(8000-x) =540

Answer 3

now that i look at it, it's EXACTLY the same formula we worked before

Answer 4

so I came up with 6000 at .06 and 2000 at .09

Answer 5

okay, let's check the result. 6000 * 0.06 is the stock interest = 360
2000 * 0.09 is the bond interest = 180

total interest is 540 = 360+180
looks like a valid solution, good job!

Answer 6

mind if I hit you with a couple more problems? This class I am taking is required for my degree and unfortunately, is online only. The teacher is on a different campus than the one I attend so I don't get to attend lecture/interact personally with the instructor

Answer 7

if we'd done the two equation route, it might look like

S + B = 8000
0.06S + 0.09B = 540
same solution, of course.

Answer 8

sure, fire away

Answer 9

Two gasoline distributors, A and B, are 228 miles apart on Interstate 20. Distributors A charges $3.02/gal and B charges $2.93/gal. Each charges .2 cents/gal per mile for delivery. Where on Interstate 20 is the cost to the customer the same?

Once again, I need help setting it up

Answer 10

This question is instructor generated, so I am assuming that it isn't possible to charge .2 cents per gal per mile

Answer 11

well, let's imagine it looks like this:

A ------------C----------B
we can call the distance between A and C (C = customer) x, and then the other chunk is 228-x

the customer buys the same amount of gas regardless of who she buys from, right? let's assume that to be true.

so the price for customer to buy one gallon from A is distance from A * delivery charge rate for A + cost of one gallon from A

price for customer to buy one gallon from B is distance from B * delivery charge rate for B + cost of one gallon from B

sound good so far?

Answer 12

if you agree, then just write out those two expressions and set them equal to each other; the price is the same at the spot where the customer wants to be

Answer 13

*eyes glaze over*

Answer 14

sorry :-)

okay, the customer is in the middle, right? let's say it is x miles from A to the customer. from B to the customer it must be 228-x miles, yes?

Answer 15

it's me being ADHD. I tend to overthink the simple stuff and "get" complex things

Answer 16

okay, well, we can make this simple or complicated, your choice :-)

Answer 17

right now, i'm trying to take the knowns and plug them into the variables in the correct positions.

Answer 18

so yes, the customer is in the middle, x miles from A and 228-x miles from B.

cost of a gallon of gas is the cost for the gas plus the delivery charge. the delivery charge is 0.002 dollars * the distance from the distributor.

a gallon bought from A costs $3.02 + x*$0.002
a gallon bought from B costs $2.93 + (228-x)*$0.002

Answer 19

geez. i'm going to have to start paying you by the hour, hahaha.

Answer 20

we just set 3.02 + 0.002x = 2.93 + 0.002(228-x)

Answer 21

okay give me a moment to pencil this out

Answer 22

sure!

Answer 23

I'd suggest multiplying both sides by 1000 as a first step to get rid of the pesky decimals

Answer 24

3020 + 2x = 2930 + 2(228-x)

Answer 25

if i am working this through correctly, i should currently see 90 = 465 - 2x?
x = 187.5?

Answer 26

but if we bumped each side by 1k, do we in the end need to reduce it by 1k?

Answer 27

3020 + 2x = 2930 + 2(228-x)
3020 + 2x = 3386 - 2x
4x = 3386 - 3020
4x = 366
x = 91.5

Answer 28

no, you can add, subtract, multiply, divide, take square roots, logs, you name it, so long as you do the same thing to both sides of the equation, it won't change the solution.

Answer 29

if you have 2 dollar bills and I give you 3 more dollar bills you have 5 dollar bills, right? if we then discover that we didn't have our glasses on and they were actually 10 dollar bills, you still have 5 of them, right?

Answer 30

lol right

Answer 31

one thing gets me: why don't we apply the changes to what is enclosed in the ( ) ?

Answer 32

oh, that's easy — we applied it to the outside. a*b*1000 = a*1000 * b, right?

Answer 33

right

Answer 34

here, we can expand it first, then apply the *1000:

3.02 + 0.002x = 2.93 + 0.002(228-x)
3.02 + 0.002x = 2.93 + 0.002*228 - 0.002*x
3020 + 2x = 2930 + 2*228 - 2*x
3020 + 2x = 3386 - 2x
same as we had...

Answer 35

but a bit more work this way

Answer 36

I'd rather the shorter path. less opportunity to miscalculate.

Answer 37

so let's check our answer.

91.5 miles from A, a gallon of gas costs $3.02 + $0.002*91.5 = $3.203
228-91.5 = 136.5 miles from B, a gallon of gas costs $2.93 + $0.002*136.5 = $3.203

Answer 38

this makes sense — if it costs 2/10 of a cent to carry the gas a mile, then every 10 miles the price changes by 2 cents. we need to have a difference of $3.02-$2.93 = 9 cents, so that's 4.5*10 = 45 miles. if we take 228-45 and then divide the result in half, that's the distance to one of the endpoints.

Answer 39

This should be simple enough to do in my sleep. Why does this particular setup seem to stonewall me? Where is the difficult part? What am I overlooking?

Answer 40

228-45 = 183, 183/2 = 91.5 :-)

Answer 41

actually, I had to think about it for a minute or two before I wrote down the right thing. don't beat yourself up too much :-)

Answer 42

where'd you get the 4.5 from the 9 cents?

Answer 43

other than 4.5 being 9/2

Answer 44

difference of 9 cents per gallon in the underlying price of gas, right? every 10 miles the purchase location changes = 10 * 0.002 cents per mile = 2 cents change in the charge to deliver the gas. if we want a 9 cent change, and we know the distance to change the price by 2 cents, then 9/2 = 4.5 is the number of those distances end to end to change it by 9 cents.

Answer 45

right on

Answer 46

5 miles changes it by 1 cent, I should have done that to cause less confusion, but I like multiplying by 10 better than I do multiplying by 5, especially with small numbers like this.

so 5 miles change on the distance to deliver the gas changes the delivery charge by 1 cent. to overcome the 9 cent per gallon difference in price, we need to move the delivery point by 9 cents/ (1 cent/5 miles) = 45 miles * cents / cents = 45 miles.

Answer 47

sorry - while working this problem, i was pencilling another. gimme sec to submit answer and see if i got it right

Answer 48

you do math like i do. you round to the nearest manageable 5 or ten then +-/* as is needed

Answer 49

what I did was 9 cents / (2 cents/10 miles) = 90 miles * cents / 2 cents = 45 miles.

Answer 50

next problem? :-)

Answer 51

and a tpath to boot.

Answer 52

jussasec.

Answer 53

In a new development, 57 one and two bedroom condominiums are sold. Each 1-bedroom condo sold for $130,000 and each 2-bedroom condo sold for $150,000. If sales totaled $8,250,000, how many of each type was sold?

we're back with our old friend equation

Answer 54

yep. set it up and knock it down :-)

Answer 55

130,000x + 150,000(57-x) = 8250,000

Answer 56

gah i want to knock off a few zeroes.

Answer 57

here's how you could do it in your head...

assume all 57 sales are 1 bedroom. 57*130000 = 7410000. that's shy of what we need by 8250000-7410000 = 840000. each 2 bedroom condo instead of a 1 bedroom condo nets us an extra 150000-130000 = 20000. we need 840000/20000=42 condos to be 2 bedroom rather than 1.

Answer 58

if we did this in thousands of dollars instead of dollars, it would be a bit more tractable in the old skull cavity, to be sure

Answer 59

15 1br/42 2br?

Answer 60

yep. 15+42 = 57, check
15*130 + 42*150 = 1500 + 450 + 6000 + 300 = 7500 + 750 = 8250
tack on 3 zeros to get the original scale back

Answer 61

You're a lifesaver. The only way I know to repay your kindness is to help others on here.

Answer 62

that would be just fine with me :-)

Answer 63

hah yeah :) I laid hands on a router years ago and fell head over heels in love

Answer 64

just need to get used to working with ipv6 and subnetting in v6

Answer 65

dear God, I've hit the jackpot :)

Answer 66

now i just need to get more comfortable with VLSM. i can do the addressing while in class, send me home with homework and i falter

Answer 67

v6 would require a lecture since it's in hex rather than binary

Answer 68

meaning that i would have to pull out lecture notes to give you the skinny on it

Answer 69

for now, though, most of the world is still v4, so we're not facing any immediate issues with it, other than cisco just pushed it down into core cisco curriculum last semester

Answer 70

if you started there in 88, you had to be 20-ish (give or take) so that makes you about my age

Answer 71

lol i'm 40 going on peter pan

Answer 72

makes me wonder if you ever met some of the guys from ARPA

Answer 73

Thankee-sai! I will most definitely look you up again. I really enjoy being shown how to think it through. One last question

Answer 74

sure. make it a doozie :-)

Answer 75

What is it about this formula I have been using throughout this assignment is so important that my prof is drilling us on it?

Answer 76

oh, it's just the process of recognizing how to write the equation(s) describing the relationships between the various items.

Answer 77

but they are all ax + b(c-x) = d

Answer 78

and then of course to solve them, but right now it sounds like you're doing it with just one variable, so that's not such a big deal.

say I have 85 cents in my pocket, nickels and dimes only, 10 coins. how many of each do I have?

the way you're being taught, you would choose either nickels or dimes as x, and the other would be 10-x. then you have 5x + 10(10-x) = 85, 3 nickels and 7 dimes

Answer 79

holy pellet. you're right. i DID do that one in my head

Answer 80

but that doesn't work so well if you have more than 2 quantities. here's another way you can write it where you can expand to more than 2 quantities without trouble:

let n be the number of nickels and d the number of dimes.

n + d = 10
5n + 10d = 85

now we have a system of two equations in two unknowns (n and d). we can solve this by substitution or elimination or with matrices.

you can see that if we had a problem where I had pennies in the mix as well, it wouldn't be possible to write it with just one variable, right?

Answer 81

pellet hahahahaha! it changed my expletive into a funny

Answer 82

yeah, keeping us safe from ourselves :-)

Answer 83

yeeeahhhhh....i'm 40, not 4. I ha.....meh. not worth the diatribe

Answer 84

so I'll solve that system first by substitution:

n + d = 10
5n + 10d = 85

well, we pick one of the equations and solve it for one variable. here it's easy: we'll use the first one, solve it for n:
n+d = 10
n = 10-d
now we substitute that in the other eq, giving us an eq in just one variable, which we solve. then we get the value of the other variable from that and either equation (or most likely, the substitution eq).

n = 10-d
5n + 10d = 85
5(10-d) + 10d = 85
50 - 5d + 10d = 85
50 + 5d = 85
5d = 35
d = 7
n = 10-d = 10 -7 = 3

that should look a lot like what you do with one variable, right?

Answer 85

more or less, yeah. so how'd you get started coming here to tutor?

Answer 86

now elimination:

n + d = 10
5n + 10d = 85

here we try to find some combination of multipliers for one or both equations so that we can get the coefficients of one of the variables to be equal but opposite in sign. then we can add the two equations together, giving us an equation in one variable.

here I'll multiply the first one by -5:
-5n -5d = -50
5n + 10d = 85
--------------
0n + 5d = 35
5d = 35
d = 7
now plug that back into any of the equations and solve for the other variable
n+d = 10
n +7 = 10
n = 3

this approach can be extended to do 3 equations in 3 unknowns, 4 in 4, etc.

Answer 87

i've got a memory leak somewhere. too much lag in my typing

Answer 88

I don't remember how I stumbled across OpenStudy. it was a few years ago. I'm sort on and off. worked my way up to 99 this summer, then went off on a trip, and after my return didn't spend any time here until just a few weeks ago, I think.

Answer 89

it's fun getting people who think they can't do this stuff to realize that they can, and it's actually fun, not complete tedium. it's even more fun helping people who are really enthusiastic about learning, and I learn new things here just about every day.

Answer 90

I learned to love math WAAAAAAYYYYYY back when I was in 3rd grade and they taught us "new" math.

Answer 91

unfortunately, there are also a lot of kids who just want answers, don't want to work at all...

Answer 92

do you know the wonderful song by that name by Tom Lehrer?

Answer 93

pfft it ain't worth it if you don't work for it. don't learn a damn thing

Answer 94

I DO! i laughed my bum off watching it the first time

Answer 95

my 10 year old son was infected by the Tom Lehrer bug years ago. I keep expected to get a call from the school chiding me about the inappropriate songs he's singing, because he's memorized quite a bit :-)

Answer 96

fortunately for me, my own two cubs are grown. they can handle their own inappropriate behavior :P

Answer 97

what really piqued my interest was seeing Donald in MathMagic Land http://www.youtube.com/watch?v=AJgkaU08VvY

Answer 98

it was the highlight of each school year :-)