Questions on Word Problems: Linear Equations And Systems Word Problems answered by real tutors!

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Question 1210642: 4.
You have $39 to spend at the music store. Each cassette tape costs $5 and each CD costs $11. Write a linear inequality that represents this situation. Let x represent the number of tapes and y the number of CDs.

Answer by ikleyn(53937) About Me  (Show Source):
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.
You have $39 to spend at the music store. Each cassette tape costs $5 and each CD costs $11.
Write a linear inequality that represents this situation. Let x represent the number of tapes and y the number of CDs.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~


Write inequalities as you read the problem


    5x + 11y <= 39,

    x >= 0,  y >= 0.



Actually, there are 3 (three) inequalities, describing the situation, not one.


Plus an additional condition that x and y are integer numbers.

Happy learning (!)




Question 962822: A plane takes about 4.5 hours to fly 1800 miles from chicago to sacramento. At the same time another plane flies from chicago to sacramento. This plane travels with the same average airspeed, but the flight to chicago takes 3.75 hours. Find the average airspeed of the planes. Find the average wind speed.
Answer by ikleyn(53937) About Me  (Show Source):
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.
A plane takes about 4.5 hours to fly 1800 miles from chicago to sacramento.
At the same time another plane flies from chicago to sacramento.
This plane travels with the same average airspeed, but the flight to chicago takes 3.75 hours.
Find the average airspeed of the planes. Find the average wind speed.
~~~~~~~~~~~~~~~~~~~~~~~~~~


As the problem is worded in the post, it is total gibberish
unfit for consumption.




Question 1157924: The transport department is collecting information about traffic levels on the SouthEast Freeway. Yesterday they placed a traffic data collection instrument at a particular location along the motorway. The instrument reported that the number of cars passing each minute varied linearly between 6:30am and 7:30am, and then quadratically (of the form at2 + bt) from 7:30am until 8am. The datashowedthat90carspassedthedatacollectorat7am, whileat7:30am150carspassedin1minute; at 8am, there were 360 cars passing in 1 minute. (The problem text probably means that 90 cars passed in the 1 minute between 7:00am and 7:01am)
(a) Assign a variable to be the number of cars passing the data collection instrument in any minute, and a variable to represent the number of minutes after 6:30am.
(b) Find the function that relates these two variables, and graph the function.
(c) How many vehicles were passing the data collection instrument at 6:30am? At 7:50am?
(d) For the linear part of the graph, explain the meaning of the slope in the context of how many cars were passing.

Answer by KMST(5396) About Me  (Show Source):
You can put this solution on YOUR website!
EDIT #1: Quick edit of unintentionally posted unfinished, uncorrected answer.
EDIT #2: Completed missing parts of the answer.

The data showed that 90 cars passed the data collector at 7am, while at 7:30am 150 cars passed in 1 minute.
At 8am, there were 360 cars passing in 1 minute.
I interpret that as the number if cars passing in 1 minute was considered as passing at 7:00am (or any other time listed) if they passed during a 1-minute period that had the exact time stated at the beginning, the end, or somewhere in the middle.
The text says that from 7:30am until 8am the number of cars passing per minute will vary as at%5E2%2Bb , using t as the variable to represent time. (The letters a, and b obviously represent constants)

(a) Assign a variable to be the number of cars passing the data collection instrument in any minute, and a variable to represent the number of minutes after 6:30am.
t%22=%22 number of minutes after 6:30am. (The student is invited to assign a variable, but part (b) suggest that the teacher would prefer to use t instead).
y%22=%22 number of cars passing per minute

(b) Find the function that relates these two variables, and graph the function.
It is obviously what they call a piecewise function:
linear between 6:30am and 7:30am, and then
quadratic (of the form at2 + bt) from 7:30am until 8am.
The information we have can be tabulated as

We will need to find the constant a and b for the quadratic part of the function.
Linear functions are often written a of the for y=mx+b , but to avoid confusion, I would not use the letter "b" and say the linear part will follow the form y=mt%2Bp .

For the linear part we have two data points:
y=90 for t=30 , and y=150 for t=60 .
From those two points we can calculate the slope as
m=%28150-90%29%2F%2860-30%29=60%2F30-->highlight%28m=2%29
Then, we can substitute into y=mt%2Bp the value found and the (t,y) values for one of the points used, to find the constant p.
Using point (30,90), with t=30, y=90, we get
90=3%2A20%2Bp-->90=60%2Bp-->90-60=p-->highlight%28p=30%29
NOTE: When you type that calculation, or when you key it into a calculator,
it must be written as "(150-90)/(60-30)" because
when we see a horizontal line separating "150-90" from "60-30",
we know we are supposed to calculate first the top and bottom parts,
but in math "150-90/60-30" means 150-90%2F60-30=118.5 ,even if your calculator has another symbol for "divided by".

For the quadratic part we have two data points:
y=150 for t=60 , and y=360 for t=90 .
If we substitute each pair of values into the quadratic function y=at%5E2%2Bb , we can find a and b.
We get:
150=a%2A60%5E2%2Bb%2A60-->150=3600a%2B60b , which obviously simplifies to
15=360a%2B6b and further to 5=120a%2B2b or highlight%28120a%2B2b=5%29
and
360=a%2A90%5E2%2Bb%2A90-->360=8100a%2B90b-->4=90a%2Bb or highlight%2890a%2Bb=4%29
system%2890a%2Bb=4%2C%22+%22%2C120a%2B2b=5%29 is a system of linear equations in a and b.
Solving it, we find highlight%28system%28a=0.05%2C+b=-0.5%29%29
Putting both pieces of the function together, we have
y=system%282t%2B30%2C%22++%22%2C0.05t%5E2-0.5t%29 .


(c) How many vehicles were passing the data collection instrument at 6:30am? At 7:50am?
At 6:30am, t=0 and we use y=2t%2B30 to find y=2%2A0%2B30=30
At 7:50am, t=80 and we use y=0.05t%5E2-0.5t to find
y=0.5%2A80%5E2-0.5%2A80=0.05%2A6400-40=320-40=280

(d) For the linear part of the graph, explain the meaning of the slope in the context of how many cars were passing.
The slope of the linear part of the graph m=2 means that each minute the number of cars passing per minute would increase by 2.


Question 1210625: A group of farmers bought 28 tractors and trucks. There were 1.8 times more tractors than trucks. How many tractors and how many trucks were bought by the farmers?

Answer by ikleyn(53937) About Me  (Show Source):
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.
A group of farmers bought 28 tractors and trucks. There were 1.8 times more tractors than trucks.
How many tractors and how many trucks were bought by the farmers?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~


This problem's description in your post is bad English presentation.
A normal English presentation/translation would be

    A group of farmers bought 28 tractors and trucks. There were 1.8 times as many tractors as trucks. 
    How many tractors and how many trucks were bought by the farmers?

Below is my solution for this modified version.

Let x be the number of trucks.
Then the number of tractors is 1.8x.


The total equation is

    x + 1.8x = 28,

      2.8x   = 28,

         x   = 28/2.8 = 10.



ANSWER.  10 trucks and 10*1.8 = 18 tractors.

Solved.

Enjoy the solution.


From which language did you translate your original version ?


//////////////////////////////////////////


Looking at your original post, I see that you are a beginner when it comes to translating math problems into English.

Keep in mind that English differs from many other languages — even in terms of proper word order within sentences.

Therefore, word-to-word translation often does not work properly.
You should know standard patterns.

Making mistakes typical for beginners is not a catastrophe.

Simply learn using guidance from tutors, and make sure to read from high-quality textbooks and problem sets.

Also, be aware that the internet is full of sheer rubbish that is completely unsuitable for learning purposes.

Finding a reliable learning resource is truly no easy task.


Happy learning (!)




Question 1210624: A group of farmers bought 28 tractors and trucks. There were 1.8 times more tractors than trucks. How many tractors and how many trucks were bought by the farmers?

Answer by josgarithmetic(39838) About Me  (Show Source):
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u trucks
a tractors
"1.8 times more tractors than trucks. "
u%2B1.8u, how many tractors

%28u%2B1.8u%29%2Bu=28
3.8u=28
u=28%2F3.8
u=7.3684....

The exercise is nonsense.


Were you expecting 10 trucks and 18 tractors?


Question 997721: Ms.Vangie Dimaano wishes to realize a return of 12% on her total investments. If she has P40,000 invested at 10%, then how much additional money should she invest at 14%?
Answer by ikleyn(53937) About Me  (Show Source):
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.
Ms.Vangie Dimaano wishes to realize a return of 12% on her total investments.
If she has P40,000 invested at 10%, then how much additional money should she invest at 14%?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~


        Calculations in the post by @lwsshak3 are fatally incorrect and lead to wrong answer.
        For correct solution see what follows in my post.


let x=additional money invested at 14%
0.1*40000 + 0.14x = 0.12*(40000+x)
4000 + 0.14x = 4800 + 0.12x
0.02x = 800
x = 800/0.02 = 80000/2 = 40000.

how much additional money should she invest at 14% ? P40,000.         ANSWER

Solved correctly to teach you in a right way.


////////////////////////


Since 12% is midway between 10% and 14%, this answer is intuitively obvious
without any calculations, so the problem can be easily solved/answered MENTALLY.




Question 53108: Help solve:
act exam rose from 20.6 in 1990 to 21.0 in 1998. What percent increase was this.

Answer by ikleyn(53937) About Me  (Show Source):
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.
Help solve:
act exam rose from 20.6 in 1990 to 21.0 in 1998. What percent increase was this.
~~~~~~~~~~~~~~~~~~~~~~~~~


        Calculations and the answer in the post by @consc198 are incorrect.
        For a correct solution see what follows.


21.0 - 20.6 = 0.4.
0.4 divided by 20.6 is approximately 0.01942.
The percent increase is 1.942%.




Question 53109: Help solve:
average tuition in 1990 was $2035.00. By 2000 it had risen approx 85%. What was the approx. cost in 2000

Answer by ikleyn(53937) About Me  (Show Source):
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.
Help solve:
average tuition in 1990 was $2035.00. By 2000 it had risen approx 85%. What was the approx. cost in 2000
~~~~~~~~~~~~~~~~~~~~~~~~~


        The solution by @consc198 in his post is, OBVIOUSLY, wrong.
        For a correct solution see what follows.


The cost in 2000, according to the problem is


    2035 + 0.85*2035 = 1.85*2035 = 3764.75 dollars.   


Since they ask "approx.", you may round it to 3765 dollars.

Solved correctly.

Do not trust to this person, @consc198, since he regularly produces wrong solutions without any hesitation.


Simply saying, a person, whose level in Math is about 3 of 5, decided that he can and he should teach others.




Question 264864: an airplane travels 900 miles from houston to miami in 6 hors against the wind.on its return trip, with the wind, it takes only 5 hours. find the rate of the airplane with no wind. find the rate of the wind
Answer by ikleyn(53937) About Me  (Show Source):
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.
an airplane travels 900 miles from Houston to Miami in 6 hors against the wind.
on its return trip, with the wind, it takes only 5 hours.
find the rate of the airplane with no wind. find the rate of the wind
~~~~~~~~~~~~~~~~~~~~~~~~


        The solution in the post by @mananth is correct: it leads to correct answer.
        But the @Mananth' solution is badly organized.

        One of the goals of such problems is to teach students to present their solution in perfect
        form with straightforward logic.

        Therefore I place my solution here.


Let the ground speed be x
And the wind   speed be y


AGAINST WIND

Distance = 900 miles.
Rate = Distance/time
x-y = 900/6 = 150 miles per hour.


WITH WIND

Distance = 900 miles.
Rate = Distance/time
x + y = 900/5 = 180  miles per hour.


Thus we have a system of two equations for unknowns x and y

    x - y = 150    (1)
    x + y = 180    (2)


To find 'x', add the equations.  You will get

    2x = 150 + 180 = 330  --->  x = 330/2 = 165.


To find 'y', substitute x = 165 in equation (2)

    165 + y = 180  --->  y = 180 - 165 = 15.


ANSWER.  The rate of the plane with no wind is 165 miles per hour.

         The rate of the wind is 15 miles per hour.

Solved.




Question 1008376: At an altitude of 12,250 feet,a hot air balloon springs a leak and begins to be descending at a rate of 450 ft. Per minute. The last opportunity for the crew to safely jump out with a parachute is 1900 feet above the ground. How long from the time they Sprung the leak will they have before they are required to jump, use an inequality and solve.

Thank You!

Found 2 solutions by josgarithmetic, ikleyn:
Answer by josgarithmetic(39838) About Me  (Show Source):
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Linear relationship of points, ( time, altitude )
Time units in minutes and altitude unit of feet

Described is (0, 12250) and slope -450 feet per minute.
An unknown limiting point is some (x, 1900).

Slope-intercept form equation
y=-450x%2B12250

The described requirement is y%3E=1900
so -450x%2B12250%3E=1900

-450x%3E=1900-12250
x%3C=%281900-12250%29%2F%28-450%29
highlight%28x%3C=23%29

Answer by ikleyn(53937) About Me  (Show Source):
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.
At an altitude of 12,250 feet, a hot air balloon springs a leak and begins to be descending at a rate
of 450 ft per minute. The last opportunity for the crew to safely jump out with a parachute is 1900 feet
above the ground. How long from the time they Sprung the leak will they have before they are required to jump,
use an inequality and solve.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~


        What is offered as an solution in the post by @mananth, is impenetrable nonsense.
        Even do not try to understand what is written there - simply ignore it.
        My correct solution is presented below.


As you read the problem, write the inequality for the safety jump

    12250 - 450t >= 1900,


which implies

    t <= %2812250-1900%29%2F450,

or

    t <= 23 minutes.    <<<---===  ANSWER

Solved correctly.




Question 1163456: In an online shopping rebate program, customers recieve a rebate of a given percent of their purchase amount until they reach a certain minimum threshold in a month. They recieve a larger percentage rebate on any additional purchases beyond this threshold. If a customer spends more than his minimun threshold in a month, his total rebate is given by the function R=4x + (x+5)(0.01z-4). Which of the following statements is true?
a) the initial threshold is 1,000.
b) x represents the total amount the customer spends.
c) x represents the percentage discount for the purchases over the threshold.
d) the initial threshold is 400.

Answer by KMST(5396) About Me  (Show Source):
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R=4x%2B%28x%2B5%29%2A%280.01z-4%29
R=4x%2B0.01zx-4x%2B0.05z-20%29
R=4x%2B0.01zx-4x%2B0.05z-20%29
R=0.01zx%2B0.05%28z-20%2F0.05%29
R=0.01xz%2B0.05%28z-400%29
The term 0.01xz represents %22x%25%22 of the amount z
The term 0.05%28z-400%29%7D%7D%5D+represents+an+extra+%7B%7B%7B%225%25%22 of the z in excess over 400
The initial threshold is highlight%28400%29
z represents the total amount the customer spends in a month
x represent the percentage of rebate for purchases up to 400%2Fmonth
Any excess over that 400 threshold of monthly expenses results in an additional %225%25%22 rebate.


Question 1154870: Beach Hotel in Cancun is offering two weekend specials. One includes a 2-night stay with 3 meals and cost $195. The other includes a 3-night stay with 5 meals and cost $300.
Found 2 solutions by josgarithmetic, ikleyn:
Answer by josgarithmetic(39838) About Me  (Show Source):
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Would the question be, "What is the price for a night and what is the cost of a meal"?

Answer by ikleyn(53937) About Me  (Show Source):
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.
Beach Hotel in Cancun is offering two weekend specials. One includes a 2-night stay
with 3 meals and cost $195. The other includes a 3-night stay with 5 meals and cost $300.
~~~~~~~~~~~~~~~~~~~~~~~~~~

To the problem's creator:


        This doesn't look like as a Math problem.

        This looks like promotional information about the hotel.




Question 17643: If possible, I would love some help with this problem. I need to be able to solve it with substitution. I am able to do substitution but the only problem I have is finding the equations within the problem. Thanks so much in advance! Here is the question:
Kelly invested her savings of $4800. She invested part in mutal funds, at 9% per year an the rest in GIC's, at 10% per year. At the end of the yera ,the interest from the mutual funds investment was $43 less than the interest from the GIC investment. How much was invested in each type of investment?

Answer by MathTherapy(10858) About Me  (Show Source):
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If possible, I would love some help with this problem. I need to be able to solve it with substitution. I am able to do 
substitution but the only problem I have is finding the equations within the problem. Thanks so much in advance! Here is the question: 
Kelly invested her savings of $4800. She invested part in mutal funds, at 9% per year an the rest in GIC's, at 10% per year. At the end of the yera ,the interest from the mutual funds investment was $43 less than the interest from the GIC investment. How much was invested in each type of investment?
=============================================================
Post from the other person:
1.09y - part of the money invested at 9% per year
1.1x - part of money invested at 10% per year
1.09y-1.1x=$43 - the difference is $43

What the other person posted (see above), makes absolutely NO SENSE!! 

Let amount invested at 9%, be N, and amount invested at 10%,T
Then we get the following TOTAL-INVESTMENT equation: N + T = 4.800 ---- eq (i)
In addition, interest EARNED from the 9% investment = .09N, while 10% earned .1T
We then get the following equation for the DIFFERENCE in INTEREST EARNED from each investment: .09N = .1T - 43 ---- eq (ii)

We now have the following system of equations:
     N + T = 4.800 ------ eq (i)
         T = 4.800 - N -- eq (i)
      .09N = .1T - 43 --- eq (ii) 
      .09N = .1(4,800 - N) - 43 ---- Substituting 4,800 - N for T, in eq (ii)
      .09N = 480 - .1N - 43
      .09N = 437 - .1N
.09N + .1N = 437
      .19N = 437
Amount invested at 9%, or N+=+437%2F.19+=+%2243%2C700%22%2F19+=+highlight%28%22%242%2C300%22%29
Amount invested at 10%: $4,800 - $2,300 = $2,500


Question 452157: bayside insurance offers two health plans. under plan a, giselle would have to pay the first $120 of her medical bills, plus 30% of the rest. under the plan b, giselle would pay the first $160 , but only 20% of the rest. for what amount of medical bills will plan b save giselle money? assume she has over $160 in bills giselle would save with plan b if she had more than $ in bills.

Answer by ikleyn(53937) About Me  (Show Source):
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.
bayside insurance offers two health plans.
under plan a, giselle would have to pay the first $120 of her medical bills, plus 30% of the rest.
under the plan b, giselle would pay the first $160 , but only 20% of the rest.
for what amount of medical bills will plan b save giselle money?
assume she has over $160 in bills giselle would save with plan b if she had more than $ in bills.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~


        In the post by @mananth,  his governing inequalities do not correspond to the problem,
        so they are  INCORRECT,  making the whole his solution incorrect.

        I came to solve the problem in a right way,  as it should be.


Plan A:  120 + 0.3(x - 120),  where x is the whole medical bill value.

Plan B:  160 + 0.2(x - 160),  where x is the whole medical bill value.


The question is to find at what value of 'x' will be  plan B < plan A.


So, we should solve this inequality

    160 + 0.2(x-160) < 120 + 0.3*(x-120).


Simplify step by step

    160 + 0.2x - 32 < 120 + 0.3x - 36

    160 - 32 - 120 + 36 < 0.3x - 0.2x

             44         <     0.1x

              x         >      44/0.1 = 440.


So, plan B is more cheap than plan B at x > 440.    ANSWER

Solved correctly.




Question 445895: 7 audio cassettes and 3 video cassettes cost rs 1110,while 5 audio cassettes and 4 video cassettes cost rs 1350 find the cost of an audio cassette and a video casette
Found 2 solutions by josgarithmetic, ikleyn:
Answer by josgarithmetic(39838) About Me  (Show Source):
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price of audio cassette, a
price of video cassette, v

system%287a%2B3v=1110%2C5a%2B4v=1350%29

Multiplying each to get coefficient 12 for v,
system%2828a%2B12v=4440%2C15a%2B12v=4050%29
Subtract second from first
13a=390
highlight%28a=30%29

Substitute a found into first equation
7%2A30%2B3v=1110
3v=1110-210
v=370-70
highlight%28v=300%29

Answer by ikleyn(53937) About Me  (Show Source):
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.
7 audio cassettes and 3 video cassettes cost rs 1110,while 5 audio cassettes and 4 video cassettes cost rs 1350
find the cost of an audio cassette and a video cassette
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~


        In the post by @mananth the solution is incorrect, since equations are written incorrectly.
        I came to do the job accurately.


Audio --- x Cost
Video --- y Cost

Equations

    7x + 3y = 1110    (1)

    5x + 4y = 1350    (2)


I will solve using the determinant method (= Cramer's method),
sinse it is easier to write (no need to write excessive words).


D (the determinant) is         7*4 - 5*3 = 28 - 15 = 13.

Dx (the determinant for x) is  1110*4 - 1350*3 = 390.

Dy (the determinant for y) is  7*1350 - 5*1110 = 3900.


Hence,  x = Dx/D = 390/13 = 30;  y = Dy/D = 3900/13 = 300.


ANSWER.  Audio-cassette costs rs 30;  video-cassette costs rs 300.


CHECK. First  equation, left side is  7*30 + 3*300 = 1110.   ! correct !

       Second equation, left side is  5*30 + 4*300 = 1350.   ! correct !

Solved correctly.




Question 439419: A swimming pool holds 540,000 liters of water. THe pool has two drainage pipes. When the pool is completely full, the first pipe alone can empty it in 180 min, and the second pipe alone can empty it in 120 min. When both pipes are draining together, how long does it take them to empty the pool?
Found 2 solutions by josgarithmetic, ikleyn:
Answer by josgarithmetic(39838) About Me  (Show Source):
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Both drainage pipes open at the same time, drain rate is
1%2F180%2B1%2F120 pool per minutes.

Choose denominator 360.
%282%2F2%29%281%2F180%29%2B%283%2F3%29%281%2F120%29
2%2F360%2B3%2F360
5%2F360
1%2F72

Reciprocal of that will be minutes per pool
72 minutes to drain the pool.

Answer by ikleyn(53937) About Me  (Show Source):
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.
A swimming pool holds 540,000 liters of water. The pool has two drainage pipes.
When the pool is completely full, the first pipe alone can empty it in 180 min,
and the second pipe alone can empty it in 120 min.
When both pipes are draining together, how long does it take them to empty the pool?
~~~~~~~~~~~~~~~~~~~~


        The technology, which @mananth uses in his solution, is based on using fractions,
        so it is the  Pre-Algebra  level of about 6th or 7th grade.

        Meanwhile,  the problem itself is pure Arithmetical and can be solved by a  4-th grade student.
        Moreover,  this problem is a standard problem of this type and is specially designed for the  4th grade level
        and is expected to be solved via Arithmetic.

        So, I will give an Arithmetic solution, as it is designed/assumed by the problem.


First  pipe drains  540000%2F180 = 3000 liters of water per minute.

Second pipe drains  540000%2F120 = 4500 liters ow water per minute.


Working together, the two pipes drain 3000 + 4500 = 7500 liters of water per minute.


So, the time for 2 pipes to drain the full amount of water from the pool is

    540000%2F7500 = 72 minutes.


ANSWER.  72 minutes for two pipes, or 1 hour and 12 minutes.

Solved at the level for which this problem is really designed.




Question 419129: A glue company needs to make some glue tat it can sell at $120 per barrel.It wants to use 150 barrels of glue worth $100 per barrel, along with glue worth $150 per barrel and and glue worth $190 per barrel.It must use the same number of barrels of $150 and $190 glue.How much of the $150 and $190 glue will be needed?How many Barrels of $120 glue will be produced?
Found 3 solutions by josgarithmetic, greenestamps, ikleyn:
Answer by josgarithmetic(39838) About Me  (Show Source):
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PRICE        VOLUME-barrels         COST
 100           150                   15000
 150            v                      150v
 190            v                      190v
-----------------------------------------------
 120         2v+150              340v+15000

Result mix must be 120 dollars per barrel.
%28340v%2B15000%29%2F%282v%2B150%29=120
.
.
highlight%28v=30%29

Answer by greenestamps(13367) About Me  (Show Source):
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Another tutor has provided a standard formal algebraic solution to the problem. But the numbers in this problem make it a good one for solving informally using logical reasoning instead of formal algebra.

Since equal amounts of the glues worth $150 per barrel and $190 per barrel are to be used, the average cost of those two ingredients is halfway between those two prices, which is $170 per barrel.

Then the problem becomes mixing glue worth $100 per barrel with glue worth $170 per barrel to get glue worth $120 per barrel. The ratio in which those two ingredients must be mixed is exactly determined by where the $120 price lies between the $100 and $170 prices.

The difference between $100 and $120 is $20; the difference between $120 and $170 is $50. Those differences mean the two ingredients must be mixed in the ratio 20:50 = 2:5.

Because $120 is closer to $100 than it is to $170, the larger portion must be the ingredient worth $100 per barrel. So the mixture needs to be 5 parts of the glue worth $100 per barrel and 2 parts of the glue worth $170 per barrel.

Given that there are 150 barrels of glue worth $100 per barrel, use a proportion to find that we need 60 barrels of the glue worth $170 per barrel:

150%3A5=x%3A2
5x=300
x=60

Then, since the glue worth $170 per barrel is equal amounts of the glues worth $150 and $190 per barrel, there must be 30 barrels each of the $150 per barrel glue and the $190 per barrel glue.

ANSWERS:
(1) 30 barrels each of the $150 per barrel and $190 per barrel glue will be used
(2) The total number of barrels of glue will be 150+30+30 = 210


Answer by ikleyn(53937) About Me  (Show Source):
You can put this solution on YOUR website!
.
A glue company needs to make some glue tat it can sell at $120 per barrel.
It wants to use 150 barrels of glue worth $100 per barrel, along with glue worth $150 per barrel
and glue worth $190 per barrel.
It must use the same number of barrels of $150 and $190 glue.
How much of the $150 and $190 glue will be needed? How many Barrels of $120 glue will be produced?
~~~~~~~~~~~~~~~~~~~~~~~~


        The solution and the answer in the post by @mananth both are incorrect.
        I came to bring a correct solution.


$100 -----150barrels
$150------- x
$190-------x

Mix $120 = 150+2x
100*150 + 150x + 190x = 120*(150 + 2x)
15000 + 340x = 240x + 18000
100x = 3000
x = 30 barrels at $150 and $190.     <<<---===   ANSWER

Solved correctly.




Question 424727: Children's tickets to a movie cost $4. Adult tickets cost $7. If 275 tickets were sold for a total cost of $1174, how many of each type were sold?
Found 3 solutions by greenestamps, josgarithmetic, ikleyn:
Answer by greenestamps(13367) About Me  (Show Source):
You can put this solution on YOUR website!


This kind of problem can be solved quickly using logical reasoning and simple arithmetic.

If all 275 tickets were children's tickets, the total cost would be 275($4) = $1100; the actual total is $1174, which is $74 more.

The difference in the cost of an adult ticket and a children's ticket is $3, so the number of adult tickets should be $74/$3.

But that is not a whole number....

That means the numbers given in the problem cannot be correct.


Answer by josgarithmetic(39838) About Me  (Show Source):
You can put this solution on YOUR website!
x children
y adults

system%28x%2By=275%2C4x%2B7y=1174%29

Easiest seems try 4 times the people count.
system%284x%2B4y=1100%2C4x%2B7y=1174%29
Subtract.
3y=74
y=24%262%2F3
This can not be. Whole number values are necessary.

Recheck the problem description from your source.

Answer by ikleyn(53937) About Me  (Show Source):
You can put this solution on YOUR website!
.
Children's tickets to a movie cost $4. Adult tickets cost $7.
If 275 tickets were sold for a total cost of $1174, how many of each type were sold?
~~~~~~~~~~~~~~~~~~~~~~~


        The solution by @mananth in his post is  INCORRECT.
        The problem is posed  INCORRECTLY  and has no solution.
        I came to explain  WHY  it is so.


Let x be the number of adult tickets.

Then the number of children tickets is (275-x).


Write the total cost equation (same as the total revenue)

    7x + 4*(275-x) = 1174  dollars.


Simplify and find x

    7x + 4*275 - 4x = 1174,

    7x - 4x = 1174 - 4*275

       3x   =     74

        x   =     74/3 = 242%2F3  <<<---=== ? ? ? ? ? ? ?  

We got a  CONTRADICTION:  the number of tickets must be integer,  but we obtained non-integer answer.

It means that the problem is posed  INCORRECTLY  and describes a situation
which  NEVER  may happen.

The solution by @mananth in his post is  INCORRECT.
Simply  IGNORE  his solution.




Question 427227: a radiator contains 25 quartz of water and antifreeze solution, which of 60 percent (by volume) is antifreeze. how much of this solution should be drained and replaced with water for the new solution to be 40 percent antifreeze.?
Found 3 solutions by timofer, greenestamps, ikleyn:
Answer by timofer(159) About Me  (Show Source):
You can put this solution on YOUR website!
drain and replace some v quarts
volume to be unchanged

%280.6%2A25-0.6v%2B0%2Av%29%2F25=0.4
0.6%2A25-0.6v=0.4%2A25
-0.6v=0.4%2A25-0.6%2A25
v=%280.6-0.4%2925%2F0.6
v=0.2%2A25%2F0.6
v=%281%2F3%2925
highlight%28v=8%261%2F3%29

Answer by greenestamps(13367) About Me  (Show Source):
You can put this solution on YOUR website!


Here are two informal solutions using logical reasoning instead of formal algebra.

(1)

At the start, the 25-quart radiator is full of 60% antifreeze, so it contains .60(25) = 15 quarts of antifreeze.

In the end, we want it to be full of 40% antifreeze, so it will contain .40(25) = 10 quarts of antifreeze.

The amount of antifreeze we want to finish with (10 quarts) is 2/3 of the 15 quarts we started with; since we are adding water which contains no antifreeze, we want the radiator to finish with 2/3 of the original antifreeze mixture, which means we want to drain 1/3 of the original mixture and replace it with water. 1/3 of 25 quarts is 8 1/3 quarts.

ANSWER: 8 1/3 quarts

(2)

We are mixing 60% antifreeze with 0% antifreeze to obtain a mixture that is 40% antifreeze. 40% is twice as close to 60% as it is to 0%, so the amount of the original antifreeze mixture must be twice as much as the added water -- i.e., 2/3 of the final mixture must be the original 60% antifreeze and 1/3 must be the added water. Again, 1/3 of 25 quarts is 8 1/3 quarts.

ANSWER (again, of course): 8 1/3 quarts


Answer by ikleyn(53937) About Me  (Show Source):
You can put this solution on YOUR website!
.
a radiator contains 25 quartz of water and antifreeze solution, which of 60 percent (by volume) is antifreeze.
how much of this solution should be drained and replaced with water for the new solution to be 40 percent antifreeze?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~


        The solution in the post by @mananth is incorrect.
        I came to bring a correct solution.


Let x be the volume of the original 60% antifreeze solution to partly drain
and to replace by pure water to get the 40% antifreeze solution.


After draining, we then have (25-x) quartz of the 60% antifreeze solution.
It contains 0.6*(25-x) quartz of the pure antifreeze.


Adding water does not change the amount of the antifreeze in the solution.


At the end, the volume of the pure antifreeze in the radiator after adding x quartz of water is 0.4*25 quartz.


So, we equate these two expressions for the pure antifreeze amount

    0.6*(25-x) = 0.4*25  quartz.    (1)


Simplify and find x

    15 - 0.6x = 10,

    15 - 10 = 0.6x,

       5    = 0.6x

       x    = 5%2F0.6 = 50%2F6 = 81%2F3.


ANSWER.  81%2F3  quartz of the original 60% solution should be drained and replaced by pure water to get the 40% antifreeze solution.

Solved correctly.




Question 428486: mr. sharma ivested a total of 30,000 in two ventures for a year. the annual return from one of them is 8% and the other paid 10.5% for the year. he recieved a total income of 2250 from both. how much did he invest at each rate???
i need help solving this problem.

Found 2 solutions by josgarithmetic, ikleyn:
Answer by josgarithmetic(39838) About Me  (Show Source):
You can put this solution on YOUR website!
Two quantities invested.
x at 10.5%
30000-x at 8%
One year of total increase on balance, 2250 units

0.105x%2B0.08%2830000-x%29=2250

0.105x%2B0.08%2A30000-0.08x=2250
%280.105-0.08%29x=2250-0.08%2A30000

x=%282250-0.08%2A30000%29%2F%280.105-0.08%29
-150%2F0.025--------------------------This is bad. Negative number here
can not be allowed.

Problem description is wrong.

Answer by ikleyn(53937) About Me  (Show Source):
You can put this solution on YOUR website!
.
mr. sharma ivested a total of 30,000 in two ventures for a year. the annual return from one of them is 8%
and the other paid 10.5% for the year. he recieved a total income of 2250 from both.
how much did he invest at each rate?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~


The problem is  SELF-CONTRADICTORY  and is posed  INCORRECTLY.
It describes a situation which  NEVER  may happen.

Indeed,  if the entire amount of  30,000  is invested at  the smaller interest rate of  8%,
it will generate the interest of   0.08*30000 = 2400,   which is  GREATER  (!)  than  2250.


*************************************************************

        So,  this problem is a typical  highlight%28highlight%28FAKE%29%29   (= FATALLY  WRONG).

*************************************************************


The solution in the post by @mananth is incorrect.

It doesn't stand up to scrutiny.




Question 729988: At a fund raising service in a church,there were twice as more women as there were men.The men paid $7.5 each and the women paid $5 each.The total cash collected was $3850.How many people were in the church that day?
Found 2 solutions by timofer, ikleyn:
Answer by timofer(159) About Me  (Show Source):
You can put this solution on YOUR website!
Something seems wrong with the wording. If you mean twice as many women as men, then could use
men, x
women, 2x
total participants, 3x


7.5x%2B5%282x%29=3850
17.5x=3850
x=220

You want three times that, so 660 people gave in the fund raiser.

Answer by ikleyn(53937) About Me  (Show Source):
You can put this solution on YOUR website!
.
At a fund raising service in a church, there were twice as more women as there were men.
The men paid $7.5 each and the women paid $5 each. The total cash collected was $3850.
How many people were in the church that day?
~~~~~~~~~~~~~~~~~~~~~


        This problem ideally suits to be solved mentally.


In accordance with the problem, we can group men and women in the church in sets,
each containing two women and one men each.


Every such group contributes 2*5 + 7.5 = 17.5 dollars into the total 3850 dollars.


Hence, the number of such groups is  3850/17.5 = 220.


It gives 220 men and 2*220 = 440 women, or 220 + 440 = 660 persons, in total.    ANSWER

Solved.

You can do algebra in the same way - it will lead you to the same answer.




Question 1165246: Owen is signing up for a gym membership with a one-time fee to join and then a monthly fee to remain a member. The one-time fee to join the gym is $100 and the total cost of membership, including the joining fee, would be $200 for 2 months. Write an equation for C, in terms of t,
representing the total cost of the gym membership over t months.

Answer by mananth(16949) About Me  (Show Source):
You can put this solution on YOUR website!
Owen is signing up for a gym membership with a one-time fee to join and then a monthly fee to remain a member. The one-time fee to join the gym is $100 and the total cost of membership, including the joining fee, would be $200 for 2 months. Write an equation for C, in terms of t,
representing the total cost of the gym membership over t months.
one-time fee -$100
Monthly fee - $ x
C = total cost =$ 200 for 2 months
The equation
C= 100+2x
200= 100+2x
200-100=2x
100= 2x
x= 50
General Equation
c= 100 + 50t
t is number of months




Question 731825: 9) The cost of 8 muffins and 2 quarts of milk is $18. The cost of 3 muffins and 1 quart of milk is $7.50. What is the cost of each item?
10) The length of a rectangle is 3 meters more than the width. The perimeter is 26 meters. What are the dimensions of the rectangle?
11) A veterinarian examined twice as many cats as dogs. She examined a total of 30 cats and dogs. How many of each animal did she examine?
12) In one season, Naomi made four times as many goals as Kennedy. Together, they made 15 goals. How many goals did each girl make?

Found 2 solutions by timofer, ikleyn:
Answer by timofer(159) About Me  (Show Source):
You can put this solution on YOUR website!
I will start the first two but you do the rest of what are needed yourself.

The muffins and milks
x muffins y quarts milk
system%288x%2B2y=18%2C3x%2By=7.5%29
Solve the system of simultaneous equations. Choose either substitution or elimination.

The rectangle
w for width
w+3 for the length
Given, perimeter is 26 meters.
Find w and w+3.
2%28w%2B3%29%2B2w=26, and obvious what to do.

Answer by ikleyn(53937) About Me  (Show Source):
You can put this solution on YOUR website!
.

Please, ONE and ONLY ONE problem per post.



Question 1166455: Eric earns a weekly salary and a commission on each item that he sells. The equation y = 10x + 50 represents the amount of money that Eric earns weekly. Bailey earns a greater weekly salary than Eric but the same commission rate. Which graph could represent the amount of money that Bailey earns weekly, y, based on the number of items sold, x?
Answer by CPhill(2264) About Me  (Show Source):
You can put this solution on YOUR website!
This is a comparison problem based on the properties of linear equations, specifically the slope and the y-intercept.
The correct graph representing Bailey's weekly earnings will be a line that is **parallel to Eric's graph** but **shifted vertically higher** on the coordinate plane .
---
## 📈 Analysis of the Earnings Equations
Both Eric's and Bailey's earnings can be modeled by a linear equation in the slope-intercept form: $y = mx + b$.
* $y$: Total weekly earnings.
* $x$: Number of items sold.
* $m$: **Commission rate** (slope).
* $b$: **Weekly salary** (y-intercept).
### 1. Eric's Earnings
Eric's equation is: $y = 10x + 50$
* **Slope ($m_{Eric}$):** $10$. This is his commission rate per item sold.
* **Y-intercept ($b_{Eric}$):** $50$. This is his weekly salary.
### 2. Bailey's Earnings
The prompt gives two conditions for Bailey's equation:
* **Same commission rate:** This means Bailey's slope ($m_{Bailey}$) must be the same as Eric's slope ($m_{Eric}$).
$$m_{Bailey} = 10$$
* In graphical terms, the two lines must be **parallel**.
* **Greater weekly salary:** This means Bailey's y-intercept ($b_{Bailey}$) must be greater than Eric's y-intercept ($b_{Eric} = 50$).
$$b_{Bailey} > 50$$
* In graphical terms, Bailey's line must cross the y-axis at a **higher point** than Eric's line.
## 🖼️ Conclusion for Bailey's Graph
The graph that represents Bailey's earnings must satisfy these two conditions:
1. **It must be a line with the same steepness (slope of 10) as Eric's line.**
2. **It must be a line that starts at a higher y-intercept (a point above 50 on the y-axis).**
Therefore, the graph that represents Bailey's earnings is a **parallel line that lies entirely above Eric's line**.


Question 1166495: Experts fear that without conservation efforts, tigers could disappear from the wild by 2022. Just one hundred years ago, there were at least 100,000 wild tigers. By 2010, the estimated world tiger population was 3,200. The table below shows the estimated world tiger population for selected years from 1970 through 2010.
Year Wild Tiger Population
1970 37,500
1980 28,000
1990 12,500
2000 6,300
2010 3,200
The polynomial function f(x)=0.76x^3-30x^2-882x+37,807 models the population of wild tigers, where x represents the number of years since 1970.
Using the website https://www.desmos.com/calculator, graph the polynomial using the settings 0 to 40 (step 5) for the x-axis and 0 to 40000 (step 5000) for the y-axis. Do not use commas in any of the numbers. Share the graph by exporting and downloading the image. Include the graph as part of your solution.
Find and interpret the function’s value at 40. Does this overestimate or underestimate the actual data shown in the table? By how much?
Find and interpret the function’s value at 10. Does this overestimate or underestimate the actual data shown in the table? By how much?
Based on the graph alone, what is the lowest population of wild tigers and in what year does this take place?

Answer by CPhill(2264) About Me  (Show Source):
You can put this solution on YOUR website!
I cannot directly access the external website (Desmos) or include an exported image from it. However, I have generated a representation of the function and the data points based on your specifications, and performed all the required calculations.
## Graphing the Model
The polynomial function is $f(x) = 0.76x^3 - 30x^2 - 882x + 37,807$.
The graph of this function, plotted for the specified ranges ($x$-axis: 0 to 40, $y$-axis: 0 to 40,000), shows a curve that decreases sharply until a minimum point near the end of the domain and then begins to slightly increase.
[Image of wild tiger population model]

---
## Analysis of the Function's Value at $x=40$ (Year 2010)
The value $x=40$ represents the year $1970 + 40 = \mathbf{2010}$.
1. **Find the function's value $f(40)$:**
$$f(40) = 0.76(40)^3 - 30(40)^2 - 882(40) + 37,807$$
$$f(40) = 48,640 - 48,000 - 35,280 + 37,807$$
$$f(40) = \mathbf{3,167}$$
2. **Interpretation and Comparison:**
* The model estimates the wild tiger population in 2010 to be **3,167** tigers.
* The actual data for 2010 is **3,200** tigers.
* **Comparison:** $3,167 - 3,200 = -33$.
The function **underestimates** the actual population by **33** tigers.
---
## Analysis of the Function’s Value at $x=10$ (Year 1980)
The value $x=10$ represents the year $1970 + 10 = \mathbf{1980}$.
1. **Find the function's value $f(10)$:**
$$f(10) = 0.76(10)^3 - 30(10)^2 - 882(10) + 37,807$$
$$f(10) = 760 - 3,000 - 8,820 + 37,807$$
$$f(10) = \mathbf{26,747}$$
2. **Interpretation and Comparison:**
* The model estimates the wild tiger population in 1980 to be **26,747** tigers.
* The actual data for 1980 is **28,000** tigers.
* **Comparison:** $26,747 - 28,000 = -1,253$.
The function **underestimates** the actual population by **1,253** tigers.
---
## Lowest Population Based on the Graph (Model)
The lowest population on the graph for the range $x=[0, 40]$ occurs at the local minimum of the function.
1. **Year of Lowest Population:** The function reaches its minimum at $x \approx 36.79$ years since 1970.
$$1970 + 36.79 \approx \mathbf{2006.79}$$
This occurs during the year **2006** (or 2007).
2. **Lowest Population:** The population at this point is found by calculating $f(36.79)$:
$$f(36.79) \approx \mathbf{2,769}$$
The lowest modeled population of wild tigers is approximately **2,769**, which takes place around the end of the year **2006**.


Question 736206: In a a heated game of kickball, the third graders won by 16 points over the fourth graders.
The total points scored were 38 how many points did each team score?

Answer by ikleyn(53937) About Me  (Show Source):
You can put this solution on YOUR website!
.
In a heated game of kickball, the third graders won by 16 points over the fourth graders.
The total points scored were 38 how many points did each team score?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~


        In this post,  the problem is stated with a certain enthusiasm.
        This suggests that the response should be presented with equal enthusiasm.


Let  x  be the points of third graders, and
let  y  be the points of fourth graders.


Write equations as you read the problem

    x + y = 38,     (1)

    x - y = 16.     (2)


Add equations (1) and (2).  The terms with 'y' will cancel each other, and you get

    2x = 38 + 16 = 54  --->  x = 54/2 = 27.


Subtract equation (2) from equation (1).  The terms with 'x' will cancel each other, and you will get

    2y = 38 - 16 = 22  --->  y = 22/2 = 11.


ANSWER.  Three-graders scored 27 points;  four-graders scored 11 points.

Solved.




Question 740606: A theater will sell 500 tickets to a play. Adult tickets cost $10 per ticket and children tickets cost $7 per ticket. If they sold $3,860 in tickets, what system of equation will that represent.
Answer by ikleyn(53937) About Me  (Show Source):
You can put this solution on YOUR website!
.
A theater will sell 500 tickets to a play. Adult tickets cost $10 per ticket and children tickets cost $7 per ticket.
If they sold $3,860 in tickets, what system of equation will that represent.
~~~~~~~~~~~~~~~~~~~~~~~~~~

    x +  y =  500     (total tickets)

  10x + 7y = 3860     (total money),


where x is the number of adult tickets, y is the number of children tickets.

Answered.




Question 612808: a two digit number is 5 times the sum of its digits and is also equal to 5 more than the product of its digit find the number

Found 2 solutions by greenestamps, ikleyn:
Answer by greenestamps(13367) About Me  (Show Source):
You can put this solution on YOUR website!


Here is a response using different logical reasoning to show that the problem has no solution.

(1) The 2-digit number is 5 times the sum of its digits, so the units digit is either 5 or 0.

(2) The 2-digit number is also 5 more than the product of its digits. If the units digit is 0, then the product of the digits is 0, and 5 more than 0 is 5, which is not a 2-digit number.

(3) From (1) and (2), the units digit of the 2-digit number must be 5.

(4) From (1) and (3), the sum of the digits must be odd. Since the units digit is 5, the tens digit must be even.

That reasoning leaves only these possibilities for the 2-digit number: 25, 45, 65, and 85. Of those, only 45 satisfies the condition that the number is 5 times the sum of its digits; and 45 does not satisfy the condition that it is 5 more than the product of its digits.

ANSWER: No solution


Answer by ikleyn(53937) About Me  (Show Source):
You can put this solution on YOUR website!
.
A two digit number is 5 times the sum of its digits and is also equal to 5 more than the product of its digits.
Find the number.
~~~~~~~~~~~~~~~~~~~~~~~~~~~


            In this my post,  I will  PROVE  that the analysis and the answer by @Theo is  INCORRECT.

            I also will prove that the problem is posed incorrectly and that such a number as described in the post does not exist.


Let N be a two-digit integer number equal to 5 times the sum of its digits.

Let's write N as a two-digit number (ab), where 'a' is a tens digit and 'b' is the ones digit.



Then from the problem, 'b' is either 0 or 5.

     We will consider these cases below, separately.



Also from the problem we can write  N = 10a + b = 5(a+b).


From this equation, we have

    10a + b = 5(a+b),

    10a + b = 5a + 5b,

    5a = 4b.    (*)


Now,  from this last equality  (*),   if b = 0,  then a = 0, and this case doesn't work, 
producing the number 00, which we are not going to interpret as a two-digit number.



If, on contrary,  b = 5,  then  from (*)  4b = 20  --->  a = 20/5 = 4,

    hence the number  N  is  45.


But then the product of its digits is  4*5 = 20,  and  N = 20+5 = 25, according to the second condition
"N is 5 more than the product of its digits".


But 25 = 45 is the contradiction, which ruins the problem itself into the dust and the analysis by @Theo, also into the dust.

Thus the problem is solved completely,  i.e.  DISPROVED.

I don't know how and from which source such idiotic problems come to the forum.




Question 1166937: Your teacher's savings account has a $10 a month fee if her balance falls below $1,000. Her account has a balance of $1,283.78 and she withdrawals $290. What is her balance 6 months later?
Found 3 solutions by timofer, ikleyn, josgarithmetic:
Answer by timofer(159) About Me  (Show Source):
You can put this solution on YOUR website!
That looks like could be highlight%28923.78%29 orhighlight%28933.78%29, maybe.

Answer by ikleyn(53937) About Me  (Show Source):
You can put this solution on YOUR website!
.
Your teacher's savings account has a $10 a month fee if her balance falls below $1,000.
Her account has a balance of $1,283.78 and she withdrawals $290. What is her balance 6 months later?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~


        Unfortunately, the problem missed to disclose a criterion, according to which it is determined
        that the balance is below $1,000. It can be the balance on the last day of a month, or, alternatively,
        the average balance per month.

        But since the problem does not disclose a precise criterion, I will assume that the balance on the last day
        of the month is the criterion. It is very natural and very consistent with the problem to think so.

        OK. Armed with this assumption, let' start the solution.


Let's assume that the teacher withdrawals $290 on February, 25.


So, the day 6 months later is August, 25,  and we should find out the teacher's balance on August, 25.


Let's make a timebook with balances, as shown below.


Feb.25  $1283.78 - $290 = $993.78.  At the end of February, $10 subtracted. The rest is  $983.78.


Mar.25.  No withdrawals.            At the end of Match,    $10 subtracted. The rest is  $973.78.


Apr.25.  No withdrawals.            At the end of April,    $10 subtracted. The rest is  $963.78.


May 25.  No withdrawals.            At the end of May,      $10 subtracted. The rest is  $953.78.


June 25. No withdrawals.            At the end of June,     $10 subtracted. The rest is  $943.78.


July 25. No withdrawals.            At the end of July,     $10 subtracted. The rest is  $933.78.


Aug. 25. No withdrawals.            On August, 25, the teacher has  $933.78 at her account.


ANSWER.  The teacher's balance 6 months later is $933.78.

Solved.

----------------------------------------------

I was taught as a child that it is bad manners to count money in someone else's pocket.
Especially when the other person is so reputable and respectful person as a teacher  (my teacher !)
Therefore,  I didn't even attempt to solve this problem.
But when I saw another tutor's incorrect solution,  giving  $923.78  as the answer,
I  couldn't contain myself anymore.   May my teachers forgive me . . .



Answer by josgarithmetic(39838) About Me  (Show Source):
You can put this solution on YOUR website!
1283.78-290%3C1000


-

Balance SIX MONTHS LATER only if no other transactions:
1283.73-290-10-6%2A10
The extra 10 taken out as fee is because the current discussed month has the first fee. Counting the next six months will count six more of these ten dollar fees.


Older solutions: 1..45, 46..90, 91..135, 136..180, 181..225, 226..270, 271..315, 316..360, 361..405, 406..450, 451..495, 496..540, 541..585, 586..630, 631..675, 676..720, 721..765, 766..810, 811..855, 856..900, 901..945, 946..990, 991..1035, 1036..1080, 1081..1125, 1126..1170, 1171..1215, 1216..1260, 1261..1305, 1306..1350, 1351..1395, 1396..1440, 1441..1485, 1486..1530, 1531..1575, 1576..1620, 1621..1665, 1666..1710, 1711..1755, 1756..1800, 1801..1845, 1846..1890, 1891..1935, 1936..1980, 1981..2025, 2026..2070, 2071..2115, 2116..2160, 2161..2205, 2206..2250, 2251..2295, 2296..2340, 2341..2385, 2386..2430, 2431..2475, 2476..2520, 2521..2565, 2566..2610, 2611..2655, 2656..2700, 2701..2745, 2746..2790, 2791..2835, 2836..2880, 2881..2925, 2926..2970, 2971..3015, 3016..3060, 3061..3105, 3106..3150, 3151..3195, 3196..3240, 3241..3285, 3286..3330, 3331..3375, 3376..3420, 3421..3465, 3466..3510, 3511..3555, 3556..3600, 3601..3645, 3646..3690, 3691..3735, 3736..3780, 3781..3825, 3826..3870, 3871..3915, 3916..3960, 3961..4005, 4006..4050, 4051..4095, 4096..4140, 4141..4185, 4186..4230, 4231..4275, 4276..4320, 4321..4365, 4366..4410, 4411..4455, 4456..4500, 4501..4545, 4546..4590, 4591..4635, 4636..4680, 4681..4725, 4726..4770, 4771..4815, 4816..4860, 4861..4905, 4906..4950, 4951..4995, 4996..5040, 5041..5085, 5086..5130, 5131..5175, 5176..5220, 5221..5265, 5266..5310, 5311..5355, 5356..5400, 5401..5445, 5446..5490, 5491..5535, 5536..5580, 5581..5625, 5626..5670, 5671..5715, 5716..5760, 5761..5805, 5806..5850, 5851..5895, 5896..5940, 5941..5985, 5986..6030, 6031..6075, 6076..6120, 6121..6165, 6166..6210, 6211..6255, 6256..6300, 6301..6345, 6346..6390, 6391..6435, 6436..6480, 6481..6525, 6526..6570, 6571..6615, 6616..6660, 6661..6705, 6706..6750, 6751..6795, 6796..6840, 6841..6885, 6886..6930, 6931..6975, 6976..7020, 7021..7065, 7066..7110, 7111..7155, 7156..7200, 7201..7245, 7246..7290, 7291..7335, 7336..7380, 7381..7425, 7426..7470, 7471..7515, 7516..7560, 7561..7605, 7606..7650, 7651..7695, 7696..7740, 7741..7785, 7786..7830, 7831..7875, 7876..7920, 7921..7965, 7966..8010, 8011..8055, 8056..8100, 8101..8145, 8146..8190, 8191..8235, 8236..8280, 8281..8325, 8326..8370, 8371..8415, 8416..8460, 8461..8505, 8506..8550, 8551..8595, 8596..8640, 8641..8685, 8686..8730, 8731..8775, 8776..8820, 8821..8865, 8866..8910, 8911..8955, 8956..9000, 9001..9045, 9046..9090, 9091..9135, 9136..9180, 9181..9225, 9226..9270, 9271..9315, 9316..9360, 9361..9405, 9406..9450, 9451..9495, 9496..9540, 9541..9585, 9586..9630, 9631..9675, 9676..9720, 9721..9765, 9766..9810, 9811..9855, 9856..9900, 9901..9945, 9946..9990, 9991..10035, 10036..10080, 10081..10125, 10126..10170, 10171..10215, 10216..10260, 10261..10305, 10306..10350, 10351..10395, 10396..10440, 10441..10485, 10486..10530, 10531..10575, 10576..10620, 10621..10665, 10666..10710, 10711..10755, 10756..10800, 10801..10845, 10846..10890, 10891..10935, 10936..10980, 10981..11025, 11026..11070, 11071..11115, 11116..11160, 11161..11205, 11206..11250, 11251..11295, 11296..11340, 11341..11385, 11386..11430, 11431..11475, 11476..11520, 11521..11565, 11566..11610, 11611..11655, 11656..11700, 11701..11745, 11746..11790, 11791..11835, 11836..11880, 11881..11925, 11926..11970, 11971..12015, 12016..12060, 12061..12105, 12106..12150, 12151..12195, 12196..12240, 12241..12285, 12286..12330, 12331..12375, 12376..12420, 12421..12465, 12466..12510, 12511..12555, 12556..12600, 12601..12645, 12646..12690, 12691..12735, 12736..12780, 12781..12825, 12826..12870, 12871..12915, 12916..12960, 12961..13005, 13006..13050, 13051..13095, 13096..13140, 13141..13185, 13186..13230, 13231..13275, 13276..13320, 13321..13365, 13366..13410, 13411..13455, 13456..13500, 13501..13545, 13546..13590, 13591..13635, 13636..13680, 13681..13725, 13726..13770, 13771..13815, 13816..13860, 13861..13905, 13906..13950, 13951..13995, 13996..14040, 14041..14085, 14086..14130, 14131..14175, 14176..14220, 14221..14265, 14266..14310, 14311..14355, 14356..14400, 14401..14445, 14446..14490, 14491..14535, 14536..14580, 14581..14625, 14626..14670, 14671..14715, 14716..14760, 14761..14805, 14806..14850, 14851..14895, 14896..14940, 14941..14985, 14986..15030, 15031..15075, 15076..15120, 15121..15165, 15166..15210, 15211..15255, 15256..15300, 15301..15345, 15346..15390, 15391..15435, 15436..15480, 15481..15525, 15526..15570, 15571..15615, 15616..15660, 15661..15705, 15706..15750, 15751..15795, 15796..15840, 15841..15885, 15886..15930, 15931..15975, 15976..16020, 16021..16065, 16066..16110, 16111..16155, 16156..16200, 16201..16245, 16246..16290, 16291..16335, 16336..16380, 16381..16425, 16426..16470, 16471..16515, 16516..16560, 16561..16605, 16606..16650, 16651..16695, 16696..16740, 16741..16785, 16786..16830, 16831..16875, 16876..16920, 16921..16965, 16966..17010, 17011..17055, 17056..17100, 17101..17145, 17146..17190, 17191..17235, 17236..17280, 17281..17325, 17326..17370, 17371..17415, 17416..17460, 17461..17505, 17506..17550, 17551..17595, 17596..17640, 17641..17685, 17686..17730, 17731..17775, 17776..17820, 17821..17865, 17866..17910, 17911..17955, 17956..18000, 18001..18045, 18046..18090, 18091..18135, 18136..18180, 18181..18225, 18226..18270, 18271..18315, 18316..18360, 18361..18405