A man invested three equal amounts of money at 5%,16% and 25% annual interest rates,respectively.if his total annual income is 552,000pesos how much did he invest at each rate.

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A positive number is 5 times another number.If 21 is added to both the numbers,then one of the new numbers becomes twice the other number.What are the numbers

Posted on Jun 30, 2008

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A man invested three equal amounts of money at 5%,16% and 25% annual
interest rates,respectively.if his total annual income is 552,000pesos
how much did he invest at each rate.

The answer here is 1,200,000pesos or 1.2 million pesos.

Solution:

[x + (x) (0.05)] + [x + (x) (0.16)]+ [x + (x) (0.25)] = 3x + 552,000

(x + 0.05x) + (x + 0.16x) + (x + 0.25x) = 3x + 552,000

1.05x + 1.16x + 1.25 x = 3x + 552,000

3.46x = 3x + 552,000

3.46x - 3x = 552,000

0.46x = 552,000

x = 552,000/0.46

x = 1,200,000

Answer: 1,200,000 pesos

Posted on Aug 01, 2010

Find the two consecutive positive integers such that 1/4 the smaller is 3 more than 1/5 the larger?

Posted on Oct 12, 2009

159,537.57 pesos

Posted on Dec 04, 2007

Four less than three times a number is 5.Find the number.

Posted on Dec 05, 2010

You guys..i know math is fun n all..but personally i think i **** but my parents ask me not to think that way..n well..its working! i mean i got this really easy chapter in math n i am able to do it so fast..n getting the right answers...i love algebra..i h8 linear equations!!!(don't think that way)!!!????!?!!!?!?!!

Posted on Sep 19, 2008

This problem deals with two numbers (ages of boys, or girls, or unicorns - this is quite inessential.) One of the numbers is known, another is not. The key word in the problem is "times". One of the numbers at hand exceeds the second number by a given factor.

The problem at hand belongs to a class of problems described by the equation

(4)
ax = b,
where x is a variable that denotes the unknown, while a and b are constant (but arbitrary) coefficients.

Equation (4) says that two entities are (or expected to be) equal. One is the number on the right - b. The other (incidentally) a product of two numbers - ax. A quick solution to the equation is obtained following a rule similar to Euclid's

*if equals be divided into equals, the results are equal.*
Thus we are prompted to conclude that (ax)/a = b/a, or

(5)
x = b/a,
that apparently asserts that the unknown is actually equal to b/a. The problem is solved. Or is it?

Do not forget that in (4) *coefficients* a and b are arbitrary. As far as the equation (4) is concerned, they may be anything. Solving a *general* equation like (4) is different from solving specific equations like 4x = 12, whose solution is readily obtained as x = 12/4, x = 3. In (4) we are obligated to account for all possible *specific* cases. Most of the cases are, indeed, handled in the same manner, as in (5). The exception is when a = 0. This is one of characteristic properties of zero that multiplied by another number, any number, it does not change. We may not know x, but if a = 0, then ax is bound to be 0! So unless b is also zero, the equality in (4) is not possible. We arrive at the following cases:

a
b
Not zero
Any
x = b/a
Zero
Not zero
(4) has no solutions
Zero
Zero
any x solves (4)
The original problem imposes additional (*semantic*) constraints. First of all, no one's age can be 0 or negative. Secondly, you would probably be very much surprised to hear a reply "13.5" to a question of yours, "How old are you?" Somewhere in the grade school, where kids of about the same age learn, work, and play together, the difference of a few month loses its significance. From that time on, we count years of our life with integers, discarding the fractional part. This means that in the original word problem it is very natural to assume that all quantities involved are positive integers such that a divides b evenly. (However, this particular fact is not carved in stone. Faced with a similar problem, you may want to check with your teacher.) The same goes for the comon usage of the word *times*. We never say "1 time as young", let alone "1 times as young."

However, assume the problem reads

*A 40 years old father is 7 times older than his son. What is the son's age? *
The formal answer is 40/7 years. Which does not look quite right. For, one would never hear such year count in the context of age determination. Is there a better answer? There might. For example, rounding to the nearest integer, we may suggest that the son's age is 6. This will not mean that 6 = 40/7, but rather that in our opinion the boy is big enough to count his age by years without the fractional part or months. In situations like this, it's the context rather than mathematics that determines the expected answer.

To summarize, above we have looked into three classes of problems. One - that of solving an abstract equation ax = b - has a unique solution for a different from 0. When a = 0, the equation either has infinitely many solutions (b = 0), or no solutions at all (b different from 0). Problems of the second kind deal with solving the same equation ax = b but subject to some constraints. For example, we may be only interested in positive integer solutions. In which case, fractional solutions of the equation ax = b will not solve the restricted problem and thus must be discarded. The third kind of problems - certain word problems - when formalized, lead to problems ofof the second kind with constraints determined from the word problem context.

The names "a" and "b" for the constants in the equation (4) are as arbitrary as they were in the equation a + x = b and are as arbitrary as the name "x" for the unknown. The latter equation is verbalized as "a constant plus the unknown equals another constant", while (4) is expressed as "a constant times the unknown equals another constant". Bearing in mind the arbitrariness of the names given to constants, we combine two equations a + x = b and ax = b into one:

On the one hand, both a + x = b and ax = b are specializations of the equation ax + b = c. (The first is obtained from ax + b = c when a = 1, the second when b = 0.) On the other hand, both a + x = b and ax = b emerge as intermediate steps when solving a more general equation ax + b = c.

Think of the term ax as another unknown. ax + b = c is then a shorthand for "the (new) unknown plus a constant equals another constant" which is solved by subtracting the first constant from both sides. The result of this step is an equation like (4): "the unknown times a constant equals a constant". The latter is solved, as in (5), by dividing both sides by the same constant.

Posted on Aug 13, 2008

I was at the Inet and noticed there an one quite good tool, which helped me to resolve my old problems with word files, what is more it was the first tool, which might help in this situation as well as relieved me - recover word.

Posted on Dec 17, 2010

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Posted on Jan 02, 2017

$15,000 at 3% and $6,000 at 7%.

If this is homework, be sure to show your work.

If this is homework, be sure to show your work.

Sep 06, 2014 | Office Equipment & Supplies

With simple interest, there is no compounding. To earn 2200.50 interest in six years, you need to earn 2200.50/6 = 366.75 each year. In order to earn 366.75 at 10.5%, you need a principal of 366.75/10.5% = 3492.86.

If the interest is compounded annually then you only need to invest 2682.13.

If the interest is compounded annually then you only need to invest 2682.13.

May 03, 2014 | Casio FX-115ES Scientific Calculator

Please see images below. The first one is 500,000 over 4 years with a 4% annual return. The second one is 500,000 over 4 years with a 6% monthly return. The second one is effective right now. More can be found at http://www.acorn2oak-fx.com/managedforexaccounts.html

Apologies but for some reason I cannot upload image 2.

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Apologies but for some reason I cannot upload image 2.

Compare Managed Forex Accounts Providers

Feb 17, 2014 | Sharp EL-738 Scientific Calculator

Invest R10000 in a bank investing at 14% compounded twice a year.

A = P(1+i)^n, where A is the amount, P is the principal or initial investment, i is the interest rate per period, and n is the number of periods.

If the annual rate is 14%, the semi-annual rate is 7%. One year is now composed of 2 6-month periods.

So after one year, we have A = 10 000 (1.07)^2 or 11,449.

Good luck,

Paul

A = P(1+i)^n, where A is the amount, P is the principal or initial investment, i is the interest rate per period, and n is the number of periods.

If the annual rate is 14%, the semi-annual rate is 7%. One year is now composed of 2 6-month periods.

So after one year, we have A = 10 000 (1.07)^2 or 11,449.

Good luck,

Paul

Nov 19, 2013 | Sharp EL-738 Scientific Calculator

What you want to do is solve the linear equation 2x=10 to find the value of the unknown x that makes the equality true. Solving an equation usually involves rearranging therms, factors and so on. However your calculator was not designed to handle the solution of equations, no matter how simple they are.

Some Casio scientific calculators can solve some types of equations (polynomials of degree 2 or 3, simultaneous linear equations in up to 3 unknowns). These are the FX-115 ES (Plus) and FX-991 ES (Plus C). Other equivalent models are sold in the world under different names.

Some Casio scientific calculators can solve some types of equations (polynomials of degree 2 or 3, simultaneous linear equations in up to 3 unknowns). These are the FX-115 ES (Plus) and FX-991 ES (Plus C). Other equivalent models are sold in the world under different names.

Oct 16, 2013 | Casio FX350MS Scientific Calculator

Hi,

Jane starts with 1200$ at the beginning of the first year, and at the end of the fourth year she has 1200$+300$=1500$

Use x for her annual interest rate, that means at the end of the first year she will have 1200$*[(100+x)/100]. At the end of the second year her first-year money earns at the same rate, so she will have 1200$*[(100+x)/100]*[(100+x)/100]=1200$*[(100+x)/100]^2 at the end of the second year.

At the end of the third year she will have 1200$*[(100+x)/100]^2 *[(100+x)/100]=1200$*[(100+x)/100]^3

At the end of the fourth year she will have

1200$*[(100+x)/100]^3 *[(100+x)/100]=1200$*[(100+x)/100]^4 which is equals to 1500$

1200$*[(100+x)/100]^4=1500$ divide both sides by 1200$

[(100+x)/100]^4=1,25 take the fourth root of both sides

(100+x)/100=1,05737 both sides*100

100+x = 105,737 both sides -100

x=5,737

So Jane's annaual interest rate was 5,737%.

Hope it helps you.

Jane starts with 1200$ at the beginning of the first year, and at the end of the fourth year she has 1200$+300$=1500$

Use x for her annual interest rate, that means at the end of the first year she will have 1200$*[(100+x)/100]. At the end of the second year her first-year money earns at the same rate, so she will have 1200$*[(100+x)/100]*[(100+x)/100]=1200$*[(100+x)/100]^2 at the end of the second year.

At the end of the third year she will have 1200$*[(100+x)/100]^2 *[(100+x)/100]=1200$*[(100+x)/100]^3

At the end of the fourth year she will have

1200$*[(100+x)/100]^3 *[(100+x)/100]=1200$*[(100+x)/100]^4 which is equals to 1500$

1200$*[(100+x)/100]^4=1500$ divide both sides by 1200$

[(100+x)/100]^4=1,25 take the fourth root of both sides

(100+x)/100=1,05737 both sides*100

100+x = 105,737 both sides -100

x=5,737

So Jane's annaual interest rate was 5,737%.

Hope it helps you.

Mar 10, 2011 | Sharp EL-738 Scientific Calculator

GOOD QUESTION, Patweetyp...

I've got you covered.

Seeing how it's been three weeks since you posted this there's a chance you've already gotten your answer, but let me go ahead solve this for those out there who may have had the same problem.

Background: There are FIVE TVM Keys; and as you would assume, you MUST input FOUR of them in order for the BA II Plus to solve for the fifth. Now let's get down to business.

1. Hit CLR TVM. [This is just a cleanup maneuver]

1a. Hit ENTER [yes, BA II Plus always needs to be told to store the value, ALWAYS]

2. Type in -2400

3. Hit PV [Since you are investing money at time 0, your present value is negative 2400]

3a. Hit ENTER

4. Type 6

5. Hit I/Y [Your annually compounded interest rate is 6 percent]

5a. Hit ENTER

6. Type 0

7. Hit PMT [you do not have any recurring deposits*]

7a. Hit ENTER

8. Type 1

9. Hit N [there is one year until expiration/liquidation/termination]

9a. Hit ENTER

10. Hit CPT

11. Hit FV [this is your ANSWER = 2544]

----- From here, all you would need to do is change N in order to get your other answers (5 years, 10 years, etc) ----

*If your calculation does not require a recurring payment then you really just have a basic equation of value which would be solved faster by hand. (IE, 2400*(1.06)^1=FV=2544).

You'll notice that I underlined "annually compounded" as well as "one year". The reason for this is because you ALWAYS need your interest term to match your time interval. For instance, if you had monthly payments of which you wanted the year-end total you would need a monthly effective interest rate, and N would be 12.

Okay, I hope that helped. The BAII Plus is the best calculator for time-value-money calculations I've come across. When things get more advanced, you will start using the amortization table which cannot be found in any other TI Calculator (from what I know). TVM is perfect for annuities, mortgages, loans, bonds, and more.

411@themathcheetah.com for more questions.

TEXAS INSTRUMENTS = 1-800-TI-CARES...they are friendly.

I've got you covered.

Seeing how it's been three weeks since you posted this there's a chance you've already gotten your answer, but let me go ahead solve this for those out there who may have had the same problem.

Background: There are FIVE TVM Keys; and as you would assume, you MUST input FOUR of them in order for the BA II Plus to solve for the fifth. Now let's get down to business.

1. Hit CLR TVM. [This is just a cleanup maneuver]

1a. Hit ENTER [yes, BA II Plus always needs to be told to store the value, ALWAYS]

2. Type in -2400

3. Hit PV [Since you are investing money at time 0, your present value is negative 2400]

3a. Hit ENTER

4. Type 6

5. Hit I/Y [Your annually compounded interest rate is 6 percent]

5a. Hit ENTER

6. Type 0

7. Hit PMT [you do not have any recurring deposits*]

7a. Hit ENTER

8. Type 1

9. Hit N [there is one year until expiration/liquidation/termination]

9a. Hit ENTER

10. Hit CPT

11. Hit FV [this is your ANSWER = 2544]

----- From here, all you would need to do is change N in order to get your other answers (5 years, 10 years, etc) ----

*If your calculation does not require a recurring payment then you really just have a basic equation of value which would be solved faster by hand. (IE, 2400*(1.06)^1=FV=2544).

You'll notice that I underlined "annually compounded" as well as "one year". The reason for this is because you ALWAYS need your interest term to match your time interval. For instance, if you had monthly payments of which you wanted the year-end total you would need a monthly effective interest rate, and N would be 12.

Okay, I hope that helped. The BAII Plus is the best calculator for time-value-money calculations I've come across. When things get more advanced, you will start using the amortization table which cannot be found in any other TI Calculator (from what I know). TVM is perfect for annuities, mortgages, loans, bonds, and more.

411@themathcheetah.com for more questions.

TEXAS INSTRUMENTS = 1-800-TI-CARES...they are friendly.

Feb 19, 2011 | Texas Instruments BA-II Plus Calculator

4 5 0 0 0 +/- PV (investment amount, negative because you're paying it out)

2 5 0 0 0 0 FV (desired amount, positive because you're receiving it)

2 0 SHIFT xP/YR (20 years)

I/YR (calculate annual interest rate)

2 5 0 0 0 0 FV (desired amount, positive because you're receiving it)

2 0 SHIFT xP/YR (20 years)

I/YR (calculate annual interest rate)

Jan 23, 2011 | HP 10bII Calculator

x=7

y=-1

y=-1

Jan 30, 2009 | Bagatrix Algebra Solved! 2005 (105101) for...

i dont know the solution...

Jul 04, 2008 | Computers & Internet

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A certain amount of money was invested at 6% per year, and another amount at 8.5%, with a total return of $1250. If the amounts invested at each rate were switched, the yearly income would have been $1375. To the nearest whole dollar, how much was invested at each rate?

fgh!

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