Questions on Algebra in Finance answered by real tutors!

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Okay, we have Amortization formula: ------------> EQN 1
Given---->
Subst.

, cross multiply

---->
, Answer
.
thank you,
Jojo

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--->where x is the length of loan in days
240x=95(365)
days
150 is closest to that figure

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Let = the amount he started out with
After the Tropicana, he had
dollars
After Ceasar's Palace, he had
dollars
After Aladdins, he had
dollars
Then he returned to his room with $48

Now solve for




He had $29 before he went to the Casinos
You can go back over it step by step again
He started with $29
After the Tropicana, he had
dollars
After Ceasar's Palace, he had
dollars
After Aladdins, he had
dollars
Then he returned to his room with $48
OK

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Let ----> amount of money he has before going to the casinos
At Tropicana:
Double the money but lost $30: , amount he got
.
At Ceasar's Palace:
Tripled the money and lost $54:
------> , amount he got
.
At Aladdins: Quadrupled the money & spent $72:
-----> , amount he got
.
Back to Room:, what's remaining.
Continuing by solving :
--->
, Answer ---> Original Money he has before gambling
Let's check by going back,
At Tropicana: =$28.00
At Ceasar's Palace: =$30.00
At Aladdins: =$48.00 Amount he got going back to his room
Thank you,
Jojo

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that's not an answer i can choose from the answers that i can choose from are: $18,021.23; $19,280.50; $19,642.08 or $42,142.08 can you please rework it for me or give me the formula so i can try. thank's.
-------------------------------------------------------------------------
P(1+R/T)^T*Y WHERE: P=PRINCIPLE (900), R=INTEREST RATE (.08), T=TIMES THE INTEREST IS CALCULATED EACH YEAR (2) & Y=NO. OF YEARS (8).
900(1+.08/2)^2*8
900(1.04)^16
900*1.873=$1,685.70 ans.
------------------------------------------
SORRY I CANNOT GET ANY NUMBERS CLOSE TO YOUR CHOICES EVEN STARTING WITH $9,000.
9,000(1+.08/2)^2*8
9,000(1.04)^16
9,000*1.873=$16,856.83

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15,000(1+.08*110/365)^110/365
15,000(1.0241)^.30137
15,000*1.0072=$15,108.04 ANS.

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invert the bottom term and multiply
=

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Use Google to find these answers on the internet.
Cheers,
Stan H.

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(14+H)/5=2
14+H=2*5
14+H=10
H=10-14
H=-4 ANSWER.
PROOF:
(14-4)/5=2
10/5=2
2=2

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multiply both sides by c and then divide both sides by .018814.

...multiplication property of equality...

...division property of equality...

.
If you would like a personal tutor to help you progress through your math education then click on my name or email me at justin.sheppard.tech@hotmail.com

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Start with the given equation


Multiply both sides by "c"


Divide both sides by 0.018814 to isolate "c"


Divide


Rearrange the equation


So the answer is (rounded to the nearest hundredth)

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(1)+(2)+3=6
(-6)+(0)+6=6
(-54)+40+20=6

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first account = .075x
second account = .04(6500-x)
438.5 = .075x + 260 - .04x
438.5 = .035x + 260
178.5 = .035x
5100 = x
==================
first account = $5100.00
second account = 6500 - 5100 = $1400.00
:)

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Let n be the number of nickels and d dimes.
A) .05n+.10d=4.75
B) .10n+.05d=3.50
.
A) .05n+.10d=4.75
B)-.20n-.10d=-7.00 multiply each side by -2
---------------------add
==>-.15n=-2.25
n=15
d=40
.
Ed

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Let the amounts in three different accounts be x,y,z respectively.
Then x + y + z = 25000 ------------(1)
and annual interest of x is 8%
y is 6%
z is 4%
The amount deposited in the 8% account is twice the amount in the 6% account.
That is x=2y --------------------(2)
the three accounts earns total annual interest of $1520
that gives 8%x + 6%y + 4%z = 1520
8x + 6y + 4z = 152000
4x + 3y + 2z = 76000 --------------(3)
Solving the 3 eqns for x,y,z
plug x=2y in eqns 1 and 3
3y + z = 25000 ----------(4)
11y + 2z = 76000 ----------(5)
subtract eqns (5) and 2* eqn (4)
5y = 26000
y = 5200
x = 10400 from eqn (2)
and z = 9400 from eqn (1)
The money deposited in the 3 accounts is 10400, 5200 and 9400

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Let = amount invested @ 8%
Let = amount invested @ 6%
Let = amount invested @ 4%
It is given that
(1)
(2) annually
(3)
This is 3 equations and 3 unknowns, so it should
be solvable
Multiply both sides of (2) by
(2)
Multiply both sides of (1) by and
subtract from (2)
(1)
(4)
Since
(4)
(4)



and
(1)
(1)


$10,400 was invested @ 8%
$5,200 was invested @ 6%
$9,400 was invested @ 4%
check:
(1)
(1)

(2)
(2)
(2)

(3)
{3)
(3)
OK

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Let the age of the water color be w and the oil be p.
A)
p-w=35
p=35+w
B)
p+5=2(w-5)
p+5=2w-10
35+w+5=2w-10 substitute for p.
w+40=2w-10
50=w
p=85
.
Ed

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We have 2 unknowns we will call O-oil painting and W-water color. we need 2 equations to solve these unknows. the difference between O and W is 35 written as O-W=35. the age of O five years from now written as O+5 is equal to twice the age of W five years ago written as 2(W-5). so O+5=2(W-5). that gives us our 2 equations. solve for O in the first and you get O=35+W now substitute this into the second equation and you get (35+W)+5=2(W-5) multiply out the right side and you get 35+W+5=2W-10 solve for W which equals 50. Now we know that O=35+W(50) so O=85
answer Oil painting 0=85
Water color W=50

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8/3:4/12
8/3:1/3
8:1

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Purchasing: a contractor buys 16yds of nylon carpet and 20 yds of wool carpet for $1840. A second purchase, at the same prices, includes 18 yds of nylon carpet and 25 ydss of wool carpet for $2200. Find the cost per yard of the wool carpet?
:
Let w = cost/yd for wool
Let n = cost/yd for nylon
:
Write an equation for each statement:
;
"contractor buys 16yds of nylon carpet and 20 yds of wool carpet for $1840."
16n + 20w = 1840
:
" 18 yds of nylon carpet and 25 yds of wool carpet for $2200"
18n + 25w = 2200
:
Find the cost per yard of the wool carpet?
:
The want the price of the wool, so let's eliminate n
Multiply the 2nd equation by 16, multiply the 1st equation by 18; results;
:
288n + 400w = 35200
288n + 360w = 33120
---------------------subtraction eliminates n, find w
40w = 2080
w =
w = $52 a yd for wool
:
:
In order to check our solution we need to find the price of nylon
Using the 1st equation, substitute 52 for w
16n + 20(52) = 1840
16n + 1040 = 1840
16n = 1840 - 1040
16n = 800
n =
n = $50 a yd for nylon
:
Check our solution in the 2nd equation:
18(50) + 25(52) = 2200
900 + 1300 = 2200; confirms our solution
:
:
Did these steps make sense to you? Any questions?

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you need to solve 3 equations in 3 unknowns.
4*i + 6*v + 10*x = 114000 (original equation 1)
8*i + 3*v + 5*x = 72000 (original equation 2)
2*i + 9*v + 5*x = 81000 (original equation 3)
-----
there are several ways to solve.
one way is as follows:
multiply first equation by -1
this will allow you to add up all equations and eliminate the x from the equation.
-----
-4*i - 6*v - 10*x = -114000 (equation 1)
8*i + 3*v + 5*x = 72000 (equation 2)
2*i + 9*v + 5*x = 81000 (equation 3)
adding them up gets
6*i + 6*v = 39000
dividing both sides by 6 gets
i + v = 6500
solve for i gets
i = 6500 - v
-----
take original first and second equation and substitute 6500 - v for i.
they are reproduced here for easy reference
-----
4*i + 6*v + 10*x = 114000 (original equation 1)
8*i + 3*v + 5*x = 72000 (original equation 2)
-----
substituting 6500 - v for i gets
26000 - 4*v + 6*v + 10*x = 114000 (equation 1)
52000 - 8*v + 3*v + 5*x = 72000 (equation 2)
combining like terms gets
26000 + 2*v + 10*x = 114000 (equation 1)
52000 - 5*v + 5*x = 72000 (equation 2)
multiplying equation 2 by 2 gets
26000 + 2*v + 10*x = 114000 (equation 1)
114000 - 10*v + 10*x = 144000 (equation 2)
subtracting equation 2 from equation 1 gets
-88000 + 12*v = -30000
adding 88000 to both sides gets
12*v = 48000
v = 4000
-----
original equations 1 and 2 are reproduced here for easy reference.
4*i + 6*v + 10*x = 114000 (original equation 1)
8*i + 3*v + 5*x = 72000 (original equation 2)
substituting 4000 for v in original equations.
4*i + 24000 + 10*x = 114000 (equation 1)
8*i + 12000 + 5*x = 72000 (equation 2)
multiply equation 2 by 2 gets
4*i + 24000 + 10*x = 114000 (equation 1)
16*i + 24000 + 10*x = 144000 (equation 2)
subtract equation 1 from equation 2 gets
-12*i = -30000
-i = -2500
i = 2500
you could have used original equations 2 and 3 to solve for i once you knew v.
you would have gotten the same answer (i = 2500).
-----
original equation 3 reproduced here for easy reference
2*i + 9*v + 5*x = 81000 (original equation 3)
substituting 2500 for i and 4000 for v in original equation 3 gets
5000 + 36000 + 5*x = 81000
combining like terms.
41000 + 5*x = 81000
subtracting 41000 from both sides.
5*x = 40000
dividing both sides by 5.
x = 8000
-----
answers are:
i = 2500
v = 4000
x = 8000
-----
equation 1 reproduced here for easy reference.
4*i + 6*v + 10*x = 114000 (original equation 1)
substituting in original equation 1 gets
4*2500 + 6*4000 + 10*8000 = 114000
10000 + 24000 + 80000 = 114000
114000 = 114000
answer is good.
works for other equations as well.
-----
this is one way to solve it.
when you get into matrix algebra you'll see other ways to solve it as well.
i assumed you weren't there yet which is why i chose this solution.

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What'd you get???
.
.
.
Let S be the number of small pizzas, L the number of large pizzas.
The profit equation is

The bounding lines for the feasible region are,



.
.
.
The x-axis is the number of small pizzas.
The y-axis is the number of large pizzas.
The black lines are the min and max values for each pizza type.
The red line is the max number of pizzas able to be made total.
.
.
.

The vertices of the feasible region are shown as circles.
The min and max of the profit equation will occur at the vertices.
The coordinates for the vertices are
1.(70,100)
2.(70,140)
3.(90,100)
4.(90,120)
.
.
.




.
.
.
Max profit is $406 with 70 smalls and 140 larges.
Min profit is $320 with 70 smalls and 100 larges.

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Ok, the first thing we need to do is set up our formulas.
Let's use:
r = regular admission tickets = $10
m = members admission tickets = $7
s = student admission tickets = $5
So we know that 750 tickets were sold so:

and
We know the price of the tickets and the total dollar amount sold was $5400 so:
Price of each ticket times the number of that ticket sold equals $5400 or:

We also know that 20 more student tickets than regular tickets were sold so:

Now for the fun part :/)
Use the elimination method to solve:
10r+7m+5s=5400
r+ m+ s=750
We want to eliminate m because it is the only variable that is not in the s=r+20 equation.
so:
Multiply everything in the bottom equation by 7.
10r+ 7m+ 5s=5400
(7)r+(7)m+(7)s=(7)750
OR
10r+7m+5s=5400
7r+7m+7s=5250
Now we can subtract and get

combine like terms:

Now it is time to use the equation:
Substitute r+20 for s so:

Distribute

combine like terms

add 40 to both sides and get
now substitute 190 for r in the equation and solve for s:
and you get
go back to the first equation and plug in 190 for r and 210 for s:


Solve for m:
combine like terms:

subtract 400 from both sides:

Elimination and substitution are fun. It gets better as you get used to them.

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let x = number of mountain bikes to be manufactured.
let y = number of trail bikes to be manufactured.
-----
total cost of material is $2500.
cost of material for mountain bike is 70.
cost of material for trail bike is 50.
equation for total material is:
70*x + 50*y = 2500
-----
total cost of labor is $2600.
cost of labor for mountain bike is 80.
cost of labor for trail bike is 40.
equation for total labor is:
80*x + 40*y = 2600
-----
solve these equations simultaneously.
70*x + 50*y = 2500 (first equation)
80*x + 40*y = 2600 (second equation)
multiply both sides of first equation by 4 and multiply both sides of second equation by 5. this will allow you to eliminate the y from the equation so you can solve for x.
-----
280*x + 200*y = 10000
400*x + 200*y = 13000
subtract first equation from second equation.
120*x = 3000
divide both sides by 120.
x = 25
-----
substitute 25 for x in first equation.
70*(25) + 50*y = 2500
1750 + 50*y = 2500
subtract 1750 from both sides
50*y = 750
y = 15
-----
substitute 25 for x and 15 for y in second equation.
80*25 + 40*15 = 2600
2000 + 600 = 2600
2600 = 2600
x = 25 and y = 15 are good in second equation.
-----
answer is:
x = 25
y = 15
number of mountain bikes to manufacture = 25
number of trail bikes to manufacture = 15
-----

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Rate x Time = Distance
Rate with wind = Speed of Plane in calm air + Speed of Wind=
Rate against wind = Speed of Plane in calm air - Speed of Wind=
Distance is the same in both cases.
Time with :
Time against :
Distance with :
Distance against :
Generate your two rate equations ( and ) in terms of P and W and the known quantitites.
Solve for P and W.

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let x = amount of grams in first alloy.
let y = amount of grams in second alloy.
-----
question is how many grams of each alloy is needed to make 60 grams of gold and 88 grams of lead.
-----
number of grams of gold in first alloy is .10 * x.
number of grams of gold in second alloy is .30 * y.
-----
number of grams of lead in first alloy is .15 * x.
number of grams of lead in second alloy is .4 * y.
-----
total number of grams of gold in third alloy is 60.
total number of grams of lead in third alloy is 88.
-----
total grams of gold from first and second alloy must = 60.
equation looks like.
.10*x + .30*y = 60
-----
total grams of lead from first and second alloy must = 88.
equation looks like.
.15*x + .4*y = 88
-----
solve these equations simultaneously.
.10*x + .30*y = 60 (first equation)
.15*x + .4*y = 88 (second equation)
multiply both sides of first equation by 1.5
.15*x + .45*y = 90
subtract first equation from second eqution.
.05*y = 2
divide both sides by .05
y = 40
-----
substitute 40 for y in first equation.
.10*x + .30*(40) = 60
.10*x + 12 = 60
.10*x = 48
x = 480
-----
substitute 40 for y and 480 for x in second equation.
.15*480 + .45*40 = 90
72 + 18 = 90
90 = 90
x = 480 and y = 40 looks good in second equation.
-----
answer is you need 480 grams of first alloy and 40 grams of second alloy.
-----

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The Acme Class Ring Company designs and sells two types of rings: the VIP and the SST. They can produce up to 24 rings each day using up to 60 total man-hours of labor. It takes 3 man-hours to make one VIP ring, versus 2 man-hours to make one SST ring.
How many of each type of ring should be made daily to maximize the company's profit, if the profit on a VIP ring is $60 and on an SST ring is $20?
------------------------------------
Quantity Inequality: P + T <= 24
Man-hrs Inequality: 3P + 2T <= 60
------------------------
Solve each inequality for P and graph in the 1st quadrant:
P = -T + 24
P = (-2/3)T + 20
--------------------

--------------------
Check the T/P values of the P intercept, the T intercept, and the
intersection of the two equation lines: (0.20), (24,0), and (12,12)
--------------------
Evaluate the profit from each production combination:
(0,20): 20*0 + 60*20 = 1200
(24,0): 20*24 + 60*0 = 480
(12,12): 20*12 + 60*12 = 12*80 = 960
--------------
Ans: Best profit with 0 T rings and 20 P rings
================
Cheers,
Stan H.
========

a. 24 VIP and 4 SST
b. 20 VIP and 0 SST
c. 20 VIP and 4 SST
d. 24 VIP and 0 SST

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Percent=(New-Old)/Old x 100




The closest whole percent is d.17%.

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Compounding interest follows this formula where e is the mathematical constant approximately equal to 2.718..., t is the number of periods (years) and r is the nominal interest rate . Why this is requires some knowledge of calculus. However, calculating interest rates using the formula is simple!


For your problem, we simply plug in the values: . Therefore, your effective interest rate is about 22.14%.

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THE ONLY EASY WAY IS TO USE A CALCULATOR.
P(1+R/M)^MT
5000(1+.05/12)^12*10
5000(1+.004167)^120
5000(1.004167)^120
5000*1.647=8235.05

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9,000-.28*9,0000=9,000-2,520=$6,480 is the depreciated value of the car affter 1 year.

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A school had 900 students at the start of a school year. If there is an enrollment increase of 7% by the beginning of the next year, what is the new enrollment?
---------------
New enrollment = 900 + 0.07*900 = 1.07*900 = 963 students
==================================================================
Cheers,
Stan H.

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2100 is a Real Number.
Cheers,
Stan H.

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63) x = amount invested
They invested a total of $16000 but we do not know how much in each account. Most likely they invested the larger amount in the 8% interest. So we will make x times 8%;
.08x; the amount invested at 8%
The remainder was invested at 4.5%, so we need to subtract that from $16000;
.045(16000-x)
They earned a total of $1070, so we add the two amounts together to equal 1070;
.08x + .045(16000-x)=1070
Now we solve for x;
.08x + 720 - .045x = 1070
collect like terms;
.035x+720=1070
Subtract 720 from both sides;
.035x=350
divide .035 to both sides;
x=10000
Now this is the amount at 8% in the corporate bond, so we subtract that from the 16000 - 10000 = $6000
61)We do not know how much interest was earned, but we do know that they are equal. So we will make these two equal to each other;
.045x = .035(42000-x)
.045x = 1470-.035x
.045x+.035x=1470
.08x=1470
x=$18375 - this would be the amount at 4.5%
42000-18375 = $23625 - amount at 3.5%
Hope you understand, let me know if you do not.
:)

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Please help me with this problem. Find the percent of change in price if the old selling price was $20 and the new selling price is $15.
.
15/20 * 100 = 0.75 * 100 = 75%
.
So, you could think of it as:
75% of the original price

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Let x represent the amount invested at 9%.
Then the amount invested at 12% is 5000 - x.
*******************************************************
The interest from the 9% investment is x*9%
The interest from the 12% investment is (5000 - x)*12%
*******************************************************
Now use the following relationship to set up an equation:
"The interest from the investment at 9% was $198 more than the interest from the investment at 12%."
x*9% - (5000 - x)*12% = 198
*******************************************************
Solve the equation for x.
9x - 12(5000 - x) = 19800 (after multiply both sides by 100)
9x - 60000 + 12x = 19800
21x = 79800
x = 3800
*******************************************************
So she invested $3800 at 9%.

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.05x=2[.08(33,600-x)]
.05x=2[2,688-.08x]
.05x=5,376-.16x
.05x+.16x=5,376
.21x=5,376
x=5,376/.21
x=25,600 is the amount invested @ 5%.
33,600-25,600=8,000 invested @ 8%.
Proof:
.05*25,600=2*.08*8,000
1,280=.16*8,000
1,280=1,280

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First subtract 3 on both sides
32=5x
Next divide 5 on both sides
6.4=x

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35 = 3 + 5x
Subtract 3 from 35. you should get 32 = 5x. Then divide 5 from x and 32. You should get x = 6.4

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They would share in the same ratio as they put in.
20:25:30
In lowest terms, that would be,
4:5:6
which happens to sum to 15.
.
.
.
The person who put in $20, would share 4/15ths of the jackpot, or $4,000,000.
The person who put in $25, would share 5/15ths of the jackpot, or $5,000,000.
The person who put in $30, would share 6/15ths of the jackpot, or $6,000,000.

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20/(20+25+30)=20/75=.2667 or 26.67% of the amount bet.
.2667*15,000,000=$4,000,000 should be his share of the winnings.

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It clearly shpws the depreciation as:

So in 9 years: AMOUNT IT DEPRECIATED
Therefore,
, VALUE after 9 years.
Thank you,
Jojo

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Use the quadratic equation to get the roots

Solved by pluggable solver: SOLVE quadratic equation with variable

Quadratic equation (in our case ) has the following solutons: For these solutions to exist, the discriminant should not be a negative number. First, we need to compute the discriminant : . Discriminant d=9 is greater than zero. That means that there are two solutions: . Quadratic expression can be factored: Again, the answer is: 4, 1. Here's your graph:

In order to find the vertex point you need to complete the square. You take the middle term half it and then square it.
y=x^2 - 5x + 4
Subtract 4 from each side
y-4=x^2 - 5x take -5 and divide by 2 -5/2 and then square this term 25/4
y-4+25/4=(x-5/2)^2
y-16/4+25/4=(x-5/2)^2
y+9/4=(x-5/2)^2
Subtract 9/4 from both sides
y=(x-5/2)^2 -9/4
The vertex o=is (5/2,-9/4)
We know that since a is positive the parabola is facing upward
the directrix is below the vertex at distance -c on the x axis
the focus is a distance +c above the vertex
a=coefficient on x^2 term
It is 1 The equation for determining the c value is a=1=1/4c


4c=1
c=1/4
Y value for vertex+1/4= Focus, x value for vertex
the directrix is a line y=8/4=2

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formula is given as:

since x = 257, equation becomes

which becomes

which becomes

which becomes

rounded to the nearest hundredth of a milligram equals
1.41
which is close to 1.42 but not quite right on.
since that's the only choice real close, i would assume 1.42 is the answer they are looking for.
answer is b. 1.42

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Suppose the amount of a radioactive element remaining in a sample of 100
milligrams after x years can be described by A(x) = 100e^-0.01657x.
How much is remaining after 257 years?
Round to the nearest hundredth of a milligram.
:
A(x) =
:
Substitute 257 for x;
A(x) =
:
A(x) =
:
on a good calc find:
A(x) = 100 * .01414
:
A(x) = 1.41 milligrams