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.

Re: solving word problems involving linear equation

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

Re: solving word problems involving linear equation

Mrs.tan invested P10,000 more than her husband did.

if they both invested the money at 5% per year and their combined incoin from investments was P4,000 HOW MUCH DID THEY INVEST in each?

Posted on Nov 06, 2008

Re: solving word problems involving linear equation

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

Re: solving word problems involving linear equation

The total reciepts for the show were $70,500.some tickets were sold for $85 each and the rest at $95 each. if 800 tickets were sold how many $95 tickets were sold

Posted on Aug 17, 2008

Re: solving word problems involving linear equation

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

Re: solving word problems involving linear equation

159,537.57 pesos

Posted on Dec 04, 2007

Re: solving word problems involving linear equation

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

Posted on Dec 05, 2010

Re: solving word problems involving linear equation

Tang ina ina

Posted on Jan 19, 2012

Re: solving word problems involving linear equation

888,888,999,999 dollars for you

Posted on Dec 01, 2012

Re: solving word problems involving linear equation

Give three consecutive integer is 36 ?find the largest integer

Posted on Nov 29, 2012

Re: solving word problems involving linear equation

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

Re: solving word problems involving linear equation

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

Re: solving word problems involving linear equation

This must be a trick question. The amount of his annual income is listed. Not the amount of his annual INVESTMENT.

Problem can't be solved without the amount of investment.

Posted on Jun 14, 2008

Re: solving word problems involving linear equation

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

upon boot-up the system automatically detects the amount of memory installed, if for some reason the amount of memory installed it's not the same with the amount detected there might be a problem with the placement of the modules or the system simply does not support either the module, quantity or type of memory. (if you're talking about system memory eg. SDRAM/DDRAM).

May 25, 2010 | Dell LATTITUE C- SERIES FDD MODULE 1.44MB...

I devised these simple steps that tend to remedy quite a few issues with most portable / external hard drives and devices (though not always)

A few things to check but assumes USB and Windows for other interfaces / operating systems similar steps may be adapted to suit.

1. Ensure it is connected directly to the computer to a USB 2.0 port not a USB 1.0 port as this can have effects on performance and reliability

2. Use only the cables that came with it NOT one that fits that may have been lying around or is longer. Not all USB cables are equal even though they should be)

3. Do not connect through an external USB HUB unless that hub is USB 2.0 AND has its own power supply.

4. Use ONLY the power supply that came with it if it has an external power supply Don’t use any other unless you know it has both the same voltage and current rating e.g. 12V 500mA anything rated below that would not work properly.

5. Always use the same port for connecting your devices. Some devices do not like being switched about. If switched they may want to install software / drivers again. This can be especialy true if you move a HUB to another port

If you checked and fixed anything there and still have issues then check your hardware from CONTROL PANEL / SYSTEM / HARDWARE.

Any exclamation marks by hardware need fixing before you investigate any further

A few things to check but assumes USB and Windows for other interfaces / operating systems similar steps may be adapted to suit.

1. Ensure it is connected directly to the computer to a USB 2.0 port not a USB 1.0 port as this can have effects on performance and reliability

2. Use only the cables that came with it NOT one that fits that may have been lying around or is longer. Not all USB cables are equal even though they should be)

3. Do not connect through an external USB HUB unless that hub is USB 2.0 AND has its own power supply.

4. Use ONLY the power supply that came with it if it has an external power supply Don’t use any other unless you know it has both the same voltage and current rating e.g. 12V 500mA anything rated below that would not work properly.

5. Always use the same port for connecting your devices. Some devices do not like being switched about. If switched they may want to install software / drivers again. This can be especialy true if you move a HUB to another port

If you checked and fixed anything there and still have issues then check your hardware from CONTROL PANEL / SYSTEM / HARDWARE.

Any exclamation marks by hardware need fixing before you investigate any further

Aug 22, 2009 | Targus Slimline USB External Floppy Drive...

I devised these simple steps that tend to remedy quite a few issues with most USB portable / external hard drives and devices (though not always)

A few things to check but assumes USB and Windows for other interfaces / operating systems similar steps may be adapted to suit.

1. Ensure it is connected directly to the computer to a USB 2.0 port not a USB 1.0 port as this can have effects on performance and reliability

2. Use only the cables that came with it NOT one that fits that may have been lying around or is longer. Not all USB cables are equal even though they should be)

3. Do not connect through an external USB HUB unless that hub is USB 2.0 AND has its own power supply.

4. Use ONLY the power supply that came with it if it has an external power supply Don’t use any other unless you know it has both the same voltage and current rating e.g. 12V 500mA anything rated below that would not work properly.

5. Always use the same port for connecting your devices. Some devices do not like being switched about. If switched they may want to install software / drivers again. This can be especialy true if you move a HUB to another port

If you checked and fixed anything there and still have issues then check your hardware from CONTROL PANEL / SYSTEM / HARDWARE.

Any exclamation marks by hardware need fixing before you investigate any further

For MAC users I am sorry but apart from the above steps that is all I can offer.

A few things to check but assumes USB and Windows for other interfaces / operating systems similar steps may be adapted to suit.

1. Ensure it is connected directly to the computer to a USB 2.0 port not a USB 1.0 port as this can have effects on performance and reliability

2. Use only the cables that came with it NOT one that fits that may have been lying around or is longer. Not all USB cables are equal even though they should be)

3. Do not connect through an external USB HUB unless that hub is USB 2.0 AND has its own power supply.

4. Use ONLY the power supply that came with it if it has an external power supply Don’t use any other unless you know it has both the same voltage and current rating e.g. 12V 500mA anything rated below that would not work properly.

5. Always use the same port for connecting your devices. Some devices do not like being switched about. If switched they may want to install software / drivers again. This can be especialy true if you move a HUB to another port

If you checked and fixed anything there and still have issues then check your hardware from CONTROL PANEL / SYSTEM / HARDWARE.

Any exclamation marks by hardware need fixing before you investigate any further

For MAC users I am sorry but apart from the above steps that is all I can offer.

Jul 21, 2009 | Memorex (32023239) External Floppy Drive

Mr Tan invested 35,000 and Mrs Tan invested 45,000

The equation to solve this is given by:

mr Tan's money x

Mrs Tan's money x + 10000

x times .05 + (x + 10000) times .05 = 4000

solve for x

Loringh

The equation to solve this is given by:

mr Tan's money x

Mrs Tan's money x + 10000

x times .05 + (x + 10000) times .05 = 4000

solve for x

Loringh

Nov 06, 2008 | Floppy Drives

This question is strangely worded and does not pertain to floppy drives. However, taking all information here as given and sufficient enough to form the equations, you arrive at the following:

X=Y+60

60(3.25) = 195 One tenant must pay $195 more for his extra 60square ft.

(975-195)/2=390 The remaining rent (assuming the given info is all that is needed to solve)

390/3.25=120 is divided between the two and represents the rent of the smaller bedroom.

y=120 sq ft The rent of the smaller room holder is divided by $3.25 to reach the sq ft.

so,

x=180 sq ft

X=Y+60

60(3.25) = 195 One tenant must pay $195 more for his extra 60square ft.

(975-195)/2=390 The remaining rent (assuming the given info is all that is needed to solve)

390/3.25=120 is divided between the two and represents the rent of the smaller bedroom.

y=120 sq ft The rent of the smaller room holder is divided by $3.25 to reach the sq ft.

so,

x=180 sq ft

Sep 17, 2008 | Floppy Drives

i dont know the solution...

Jul 04, 2008 | Floppy Drives

x=7 Therefore, first is 7, second is 35

I like numbers but as I get older, they get harder!

I like numbers but as I get older, they get harder!

Jun 30, 2008 | Floppy Drives

I would need the capacity and average attendance for concerts to be able to make this decision. There are other issues that would have a bottom line impact that are not discussed.

Apr 13, 2008 | Floppy Drives

the figure that is given for songs you can load to your MP3 player are calculated using 128Mb/s for 3 min songs. But the amount of memeory occupied by each song depends on the bit rate (KB/s) and the duration of the song (time in sec). So if are going to load songs with a longer duaration and want to load more songs then load them with reduced bit rate.

However note reducing the bit rate of your songs wilol also compromise the quality of your songs, it will be good too use variable bit rate setting a low bit rate like 64-96Kb/s.

To convert your songs you can try using Easy CD-DA extractor convertor it offers a variety of output formats at different bitrates and qualities.

However note reducing the bit rate of your songs wilol also compromise the quality of your songs, it will be good too use variable bit rate setting a low bit rate like 64-96Kb/s.

To convert your songs you can try using Easy CD-DA extractor convertor it offers a variety of output formats at different bitrates and qualities.

Jan 12, 2008 | Floppy Drives

Oct 26, 2016 | Panasonic Technics Keyboard Model...

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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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