Cost Formula Word Problem
The cost formula for a manufacturer’s product is C = 5000 + 2 x , where C is the cost (in dollars) and x is the number of units manufactured. (a) If no units are produced, what is the cost?
(b) If the manufacturer produces 3000 units, what is the cost? (c) If the manufacturer has spent $16,000 on production, how many units were manufactured? Answer these questions by substituting the numbers into the formula. (a) If no units are produced, then x = 0, and C = 5000 + 2 x becomes C = 5000 + 2(0) = 5000. The cost is $5,000. (b) If the manufacturer produces 3000 units, then x = 3000, and C = 5000 + 2 x becomes C = 5000 + 2(3000) = 5000 + 6000 = 11,000. The manufacturer’s cost would be $11,000. (c) The manufacturer’s cost is $16,000, so C = 16,000. Substitute C = 16,000 into C = 5000 + 2 xto get 16,000 = 5000 + 2 x .

There were 5500 units produced.
Profit Formula Word Problem
The profit formula for a manufacturer’s product is P = 2 x – 4000 where x is the number of units sold and P is the profit (in dollars). (a) What is the profit when 12,000 units were sold?
(b) What is the loss when 1500 units were sold?
(c) How many units must be sold for the manufacturer to have a profit of $3000? (d) How many units must be sold for the manufacturer to break even? (This question could have been phrased, “How many units must be sold in order for the manufacturer to cover its costs?”) (a) If 12,000 units are sold, then x = 12,000. The profit equation then becomes P = 2(12,000) – 4000 = 24,000 – 4000 = 20,000. The profit is $20,000. (b) Think of a loss as a negative profit. When 1500 units are sold, P = 2 x – 4000 becomes P = 2(1500) – 4000 = 3000 – 4000 = – 1000. The manufacturer loses $1000 when 1500 units are sold. (c) If the profit is $3000, then P = 3000; P = 2 x – 4000 becomes 3000 = 2 x – 4000.

A total of 3500 units were sold.
(d) The break-even point occurs when the profit is zero, that is when P = 0. Then P = 2 x – 4000 becomes 0 =...

...*What nominal amount rate compounded semi-annually is equivalent to 5% effective?
-Continuous Compounding
*Accumulate P6,000 for 3 ½ years at 12.6% compounded continuously.
-Amount Ordinary Annuity
*Jacob borrowed P25,000 for enrollment of his son in college in a cooperative. He agreed to pay the principal and the interest by paying monthly for 6 months at 12% interest compounded monthly. How much is the monthly installment?
-Present Value of Ordinary Annuity
*If money is worth 5% compounded quarterly, find the present value of P5,500 for 10years and 9 months.
-Deferred Ordinary Annuity
*Find the present value of a deferred annuity of P1,050 a year for 8 years that is deferred 2 years, If money is worth 7% effective.
-Problem on Amortization
*Arlette owes P10,000 with interest at 12% compounded semi-annually for three years. Find her periodic payment payable semi-annually and construct the amortization table.
-Sinking Fund
*Paula borrowed P5,000 from the Baranggay Tanod’s cooperative. In order to pay her dept, she decided to place eight equal semi-annual deposits in fund which pays 3% compounded semiannually. Find the periodic deposit.
-Bonds
*A P1,000, 6% bonds pays coupon semi-annually and will be redeemed on September 1, 2005. If the bond is redeemable at part.. Find the purchase price if the bond is bought on September 1, 2002 to yield 4% compounded semiannually.
-Depreciation
*A machine costing P45,000 is estimated to have...

...The problems of definition of a word
Nobody doubts that word is a basic unit of lexicology, alongside with morphemes and phraseological units. Each word is a small unit within a vast, efficient and well-balanced system. It is immediately understood by every native speaker. A word is a dialectical unity of form and content.
The definition of the word is one of the most difficult in linguistics because the simplest word is a multi-aspect unit possessing a sound form (a certain arrangement of phonemes). Besides, it has morphological structure (a certain arrangement of morphemes).
When it is used in actual speech, it may occur in different word forms and signal different meanings. The word is a sort of focus for the problems of phonology, lexicology, syntax, morphology and also for some other sciences dealing with the language and speech. Each of these sciences suggests its own definition of the word but in fact none of these definitions can be considered totally satisfactory in all aspects. If we try to divorce two facets of the word (form and content), the word will lose its identity. The word is further complicated by the existence of variants and word forms. ‘Write - writing - has been writing - wrote - written’ - these are word forms. They are...

...AAS – Angle/Angle/Side
-Given: One Side & Two Angles
-Side which is an opposite of one of the angle.
Problem: Points A and B are 42.75 yards apart on one side of a river and point C across the river with angle A = 75°13’ and angle B = 42°34’. Find the length of point C and its angle.
Solution:
Given: a = 42.75yards b = c =
α = 75°13’ β = 42°34’ У =
Case II. SSA – Side/Side/Angle
-Ambiguous Case -Given: Two sides and an angle -Can represent: 1Δ, 2Δs or no triangle.
-For 1Δ: a. Angle given is <90°. b. Opposite side of the given angle is greater than the other given side.
-For 2Δ: 1. a<b 2. Β>α 3. α<β 4. У<β
Problem: A high-powered sniper rifle was fired, releasing the bullet at 12.43m diagonally to the left forming an angle of 94°. As the bullet landed on 12.43m it deflected 19.4m to the right. Find the angle made by the deflection and the nearest distance of the shooter from where the bullet landed after it deflected.
Solution:
Given: a = 19.4m b = 12.43m c =
α = 94° β = У =
Case III. SAS – Side/Angle/Side
-Given: 2 side lengths and the measured angle between them.
Problem: A triangular field is surveyed. The length of one side of the...

..."Age" WordProblems (page 1 of 2)
In January of the year 2000, I was one more than eleven times as old as my son William. In January of 2009, I was seven more than three times as old as him. How old was my son in January of 2000?
Obviously, in "real life" you'd have walked up to my kid and and asked him how old he was, and he'd have proudly held up three grubby fingers, but that won't help you on your homework. Here's how you'd figure out his age for class:
First, name things and translate the English into math: Let "E " stand for my age in 2000, and let "W " stand for William's age. Then E = 11W + 1 in the year 2000 (from "eleven times as much, plus another one"). In the year 2009 (nine years after the year 2000), William and I will each be nine years older, so our ages will be E + 9 and W + 9. Also, I was seven more than three times as old as William was, so E + 9 = 3(W + 9) + 7 = 3W + 27 + 7 = 3W + 34. This gives you two equations, each having two variables:
E = 11W + 1
E + 9 = 3W + 34
If you know how to solve systems of equations, you can proceed with those techniques. Otherwise, you can use the first equation to simplify the second: since E = 11W + 1, plug "11W + 1 " in for "E " in the second equation:
E + 9 = 3W + 34
(11W + 1) + 9 = 3W + 34
11W – 3W = 34 – 9 – 1
8W = 24
W = 3
Remember that the problem did not ask for the value of the variable W; it asked for the age of a person. So the answer is: William...

...Examples of Age wordproblems :
1. Six years ago Jun was 4 times as old as Joan.In four years, he would be twice as old as Joan. How old are they now?
6 years ago
joan x
jun 4x
..
after 4 years
joan x+10
Jun 4x+10
4x+10=2(x+10)
4x+10=2x+20
4x-2x=20-10
2x=10
x=5 joan 6 years ago. Now age = 11
Jun 4x, = 20, 6 years ago. Now age =26 years now
2. At present, Pedro's age is 25percent of his father's age.Twenty years from now Pedro's age would be 50percent of his fathers age. How old are they?
Let pedro's age be x ( pedro's age is 1/4 of father'e age OR father is 4 times pedros age)
father's age = 4x
..
20 years later
pedro = x+20
Father = 4x+20
2(x+20)=4x+20
2x+40 =4x+20
2x=20
x=10 pedro's age now
father= 10*4 = 40 years
3. Kevin is 4 years older than Margaret.
Next year Kevin will be 2 times as old as Margaret.
How old is Kevin?
Kevin x
Margaret x-4
(X+1)(X-4+1=x-3),x+1=2*(x-3),x+1=2x-6
1+6=2x-x
7=x
4. Ann is 2 years older than Betty.
Last year Ann was 2 times as old as Betty.
How old is Ann?
Ann x
Betty x-2
x-1=2*(x-3)
x-1=2x-6
6-1=2x-x
5=x
5. Mark is 3 years older than Philip. Two years ago Susan was twice as old as Tom.
Find their present ages.
Philip x
Mark 2x
x+2=2x+2-3
x+2=2x-1
2+1=2x-x
3=x
Mixture wordproblems:
1.A store owner wants to mix cashews and almonds. Cashews cost 2 dollars per pound and almonds cost 5 dollars per pound. He plans to sell 150...

...+ 10000*8=55000+80000=$135000
Vp=V*P=10 000*21=$210 000
Z=VP-TC=210 000 – 135 000=$75 000
b. Determine the annual break-even volume for the Retread Tire Company operation.
V=Cf/(P-Cv)=55000/(21-8)=4230.77
2. Evergreen Fertilizer Company produces fertilizer. The company’s fixed monthly cost is $30,000, and its variable cost per pound of fertilizer is $0.16. Evergreen sells the fertilizer for $0.40 per pound. Determine the monthly break-even volume for the company.
Cf $30 000
Cv $0.16
P $0.40
V=30 000/(0.40-0.16)=125 000
3. If Evergreen Fertilizer Company in Problem 2 changes the price of its fertilizer from $0.40 per pound to $0.60 per pound, what effect will the change have on the break-even volume?
V=30 000/(0.60-0.16)=68181.82 change with 56818.18
4. If Evergreen Fertilizer Company increases its advertising expenditures by $14,000 per year, what effect will the increase have on the break-even volume computed in Problem 2?
48 000/0.24=200 000, which is 75 000 increase
5. Annie McCoy, a student at Tech, plans to open a hot dog stand inside Tech’s football stadium during home games. There are seven home games scheduled for the upcoming season. She must pay the Tech athletic department a vendor’s fee of $2,500 for the season. Her stand and other equipment will cost her $3,100 for the season. She estimates that each hot dog she sells will cost her $0.35. She has talked to friends at other universities who sell hot dogs at games. Based on their...

...Concussion conundrum
Read and write a review of the following article.
http://www.afl.com.au/tabid/208/default.aspx?newsid=140857
ADELAIDE admits it doesn't know what it would do in a Grand Final if it was down to 17 fit players and a player with suspected concussion wanted to go back onto the ground.
Crows football operations manager Phil Harper told SEN radio on Monday morning Adelaide's coaching staff had briefly considered that very dilemma during Saturday night's Showdown when forward Kurt Tippett wanted to return to the field against the club doctor's advice.
Tippett suffered a heavy head knock in a collision with Port defender Tom Logan early in Saturday night's game and was taken from the ground. Shaun McKernan had already been substituted with a broken jaw.
Harper said Crows coach Brenton Sanderson had been adamant the club had to follow its doctor's advice that Tippett was not fit to return despite the key forward's subsequent desire to resume.
"By half-time Kurt was feeling a bit better and wanted to go back out and play and the doctors were saying no," Harper said.
"We were saying, 'He wants to come back on, what do we do?' And the coach was quite adamant 'No, no, the doctor says he doesn't come on'."
But Harper admitted a similar call might not be so clear-cut in a Grand Final.
"We had a good philosophical discussion in the box as to well, what if it was a Grand Final and we're down to 17 guys and he wants to go on, what do we do?"...

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