NEXT CENTURY MATHEMATICS BY FERNANDO ORINES

NEXT CENTURY MATHEMATICS BY FERNANDO ORINES2014

AGE PROBLEMS

Age problems are algebra word problems that deal with the ages of people currently, in the past or in the future.If the problem involves a single person, then it is similar to an Integer Problem. Read the problem carefully to determine the relationship between the numbers. This is shown in the following example involving a single person.If the age problem involves the ages of two or more people then using a table would be a good idea. A table will help you to organize the information and to write the equations.Age Problems Involving A Single Person

Example 1:Five years ago, Johns age was half of the age he will be in 8 years. How old is he now?Solution:Step 1:Letxbe Johns age now. Look at the question and put the relevant expressions above it.

Step 2:Write out the equation.

Isolatevariablex

Answer:John is now 18 years old.Example 2:Ten years from now, Orlando will be three times older than he is today. What is his current age?Solution: let x= current agex + 10 = 3xx- 3x = -10-2x = -10x = 5Example 3:In 20 years, Kayleen will be four times older than she is today. What is her current age?Solution: let x= current agex + 20 = 4xx - 4x = -20-3x = -20x = 6 2/3

Age Problems Involving More Than One PersonExample 4:John is twice as old as his friend Peter. Peter is 5 years older than Alice. In 5 years, John will be three times as old as Alice. How old is Peter now?Solution:Step 1: Set up a table.age nowage in 5 yrs

John

Peter

Alice

Step 2:Fill in the table with information given in the question.John istwice as oldas his friend Peter. Peter is5 years olderthan Alice. In 5 years, John will be three times as old as Alice. How old is Peter now?Letxbe Peters age now. Add 5 to get the ages in 5 yrs.age nowage in 5 yrs

John2x2x+ 5

Peterxx+ 5

Alicex5x 5 + 5

Write the new relationship in an equation using the ages in 5 yrs.In 5 years, John will bethree timesas old as Alice.2x+ 5 =3(x 5 + 5)2x+ 5 = 3xIsolatevariablexx= 5Answer:Peter is now 5 years old.Example 5:Johns father is 5 times older than John and John is twice as old as his sister Alice. In two years time, the sum of their ages will be 58. How old is John now?age nowage in 2 yrs

Johns father

John

Alice

Solution:Step 1:Set up a table.

Step 2:Fill in the table with information given in the question.Johns father is5 times olderthan John and John is twice as old as his sister Alice. In two years time, the sum of their ages will be 58. How old is John now?Letxbe Johns age now. Add 2 to get the ages in 2 yrs.age nowage in 2 yrs

Johns father5x5x+ 2

Johnxx+ 2

Alice

Write the new relationship in an equation using the ages in 2 yrs.In two years time, the sum of their ages will be 58.

Answer:John is now 8 years old.

aVERAGE WORD PROBLEMS

There are two main types of average problems commonly encountered in school algebra: Average (Arithmetic MeanandAverage Speed.Average (Arithmetic Mean)

The average (arithmetic mean) uses the formula:

The formula can also be written as

Example1 :The average (arithmetic mean) of a list of 6 numbers is 20. If we remove one of the numbers, the average of the remaining numbers is 15. What is the number that was removed?Solution:Step 1:The removed number could be obtained by difference between the sum of original 6 numbers and the sum of remaining 5 numbers i.e.sum of original 6 numbers sum of remaining 5 numbersStep 2:Using the formula

Sum of original 6 numbers = 20 6 = 120sum of remaining 5 numbers= 15 5 = 75Step 3:Using the formula from step 1Number removed = sum of original 6 numbers sum of remaining 5 numbers120 75 = 45Answer:The number removed is 45.

Average Speed ProblemsExamples of Average Speed ProblemsComputation of average speed is a trickier type of average problems. Average speed uses the formula:

Example2:John drove for 3 hours at a rate of 50miles per hourand for 2 hours at 60 miles per hour. What was his average speed for the whole journey?Solution:Step1: Theformulafor distance isDistance = RateTimeTotal distance = 50 3 + 60 2 = 270Step 2:Total time = 3 + 2 = 5Step 3:Using the formula

Answer:The average speed is 54 miles per hour.Be careful!You will get the wrong answer if you add the two speeds and divide the answer by two.

DIGIT WORD PROBLEMS

Digitword problemsare problems that involve individualdigitsin integers and howdigitsare related according to the question.Some problems would involvetreatingthe digits as individual numbers to be related. This would make it similar to aninteger problem, except that the integers are between 0 and 9,inclusive.Example1:The tens digit of a number is three times the ones digit. The sum of the digits in the number is 8. What is the number?Solution:Step1:Sentence: The tens digit of a number is three times the ones digit.Assign variables:Letx=ones digit

3x=tens digit

Sentence: The sum of the digits in the number is 8.x+ 3x =8Step 2:Solve the equationx +3x= 8Isolatevariablex4x= 8x= 2The ones digit is 2. The tens digit is 3 2 = 6Answer:The number is 62.

DISTANCE WORD PROBLEMS

Distance Problems: Traveling In Different DirectionsExample1:A bus and a car leave the same place and traveled in opposite directions. If the bus is traveling at 50 mph and the car is traveling at 55 mph, in how many hours will they be 210 miles apart?Solution:Step 1:Set up artdtable.rtd

bus

car

Step 2:Fill in the table with information given in the question.If the bus is traveling at50mph and the car is traveling at55mph, in how many hours will they be 210 miles apart?Lett= time when they are 210 miles apart.rtd

bus50t

car55t

Step 3:Fill in the values fordusing the formulad = rtrtd

bus50t50t

car55t55t

Step 4:Since the total distance is 210, we get the equation:50t+ 55t= 210105t= 210Isolatevariablet

Answer:They will be 210 miles apart in 2 hours.Distance Problems: Given Total TimeExample2:John took a drive to town at an average rate of 40 mph. In the evening, he drove back at 30 mph. If he spent a total of 7 hours traveling, what is the distance traveled by John?Solution:Step 1:Set up artdtable.rtd

Case 1

Case 2

Step 2:Fill in the table with information given in the question.John took a drive to town at an average rate of40mph. In the evening, he drove back at30mph. If he spent a total of7hours traveling, what is the distance traveled by John?Lett= time to travel to town.7t =time to return from town.rtd

Case 140t

Case 2307t

Step 3:Fill in the values fordusing the formulad = rtrtd

Case 140t40t

Case 2307 t30(7 t)

Step 4:Since the distances traveled in both cases are the same, we get the equation:40t= 30(7 t)Usedistributive property40t= 210 30tIsolatevariablet40t+ 30t= 21070t= 210

Step 5:The distance traveled by John to town is40t= 120The distance traveled by John to go back is also 120So, the total distance traveled by John is 240Answer:The distance traveled by John is 240 miles.Wind and Current Word ProblemsThere is another group of distance-time problems that involves the speed of the water current or the speed of wind affecting the speed of the vehicle. Example3:Flying against a headwind, Abby Lee took 5.5 hours to fly 2750 km to Cebu City. With no wind change in the wind, the return flight took 5 hours. What is the wind speed and the speed of the plane?Solution:Let x= planes speedy= wind speedStep 1:Set up artdtablePLANErtd

Withx+y55 (x+y)

Against x-y5.5 5.5 (x-y)

Step 2:5 (x+y) = 2750x + y = 55055.5 (x-y) = 2750x y = 5005.5Step 3: add the two equations to cancel out one variable.(x + y = 550) + (x y = 500)2x = 1050x= 525So, the Planes speed is 525km/hAnd the wind is 25km/hExample4:Traveling downstream, Elmo can go 6 km in 45 minutes. On the return trip, it takes him 1,5 hours. What is the boat's speed in still water and what is the rate of the current?Solution:Let x= Boats speedy= current

Step 1:Set up artdtableBOATrtd

Withx+y3 / 43/4 (x+y)

Againstx-y1 . 53/2 (x-y)

Step 2:FIND THE EQUATION[ ( x+y) = 6] 4/3 x + y = 8[3/2 ( x+y) = 6] 2/3 x y = 4Step 3: add the two equations to cancel out one variable.(x+y=8) + (x-y=4)2x = 12x = 6Step 4: SUBSTITUTE THE VALUE OF X IN THE EQUATION x + y = 86 + y = 8y = 2So, the boats speed is 6km/hr and the water is 2km/hr.

Distance Word Problems - Traveling at Different RatesDistance problems are word problems that involve thedistancean object will travel at a certain averageratefor a given period oftime.The formula for distance problems is:distance=ratetimeord = r t.Things to watch out for:Make sure that you change the units when necessary. For example, if the rate is given in miles per hour and the time is given in minutes then change the units appropriately.It would be helpful to use a table to organize the information for distance problems. A table helps you to think about one number at a time instead being confused by the question.

Example5:A bus traveling at an average rate of 50 kilometers per hour made the trip to town in 6 hours. If it had traveled at 45 kilometers per hour, how many more minutes would it have taken to make the trip?Solution:Step 1:Set up artdtable.rtd

Case 1

Case 2

Step 2:Fill in the table with information given in the question.A bus traveling at an average rate of50kilometers per hour made the trip to town in6hours. If it had traveled at45kilometers per hour, how many more minutes would it have taken to make the trip?Lett= time to make the trip in Case 2.rtd

Case 1506

Case 245t

Step 3:Fill in the values fordusing the formulad = rtrtd

Case 150650 6 = 300

Case 245t45t

Step 4:Since the distances traveled in both cases are the same, we get the equation:45t= 300Isolatevariablet

Step 5:Beware- the question asked for how manymoreminutes would it have taken to make the trip, so we need to deduct the original 6 hours taken.

Answer:The time taken would have been 40 minutes longer.Distance Word Problems - Given the Total TimeDistance problems areword problemsthat involve thedistancean object willtravelat a certain averageratefor a given period oftime.The formula for distance problems is:distance=ratetimeord = r t.Things to watch out for:Make sure that you change the units when necessary. For example, if the rate is given inmiles per hourand the time is given in minutes then change the units appropriately.It would be helpful to use atableto organize the information for distance problems. A table helps you to think about one number at a time instead being confusedby the question.

Example6 :John took a drive to town at anaverage rateof 40 mph. In the evening, he drove back at 30 mph. If he spent a total of 7 hours traveling, what is the distance traveled by John?Solution:Step1:Set up artdtable.rtd

Case 1

Case 2

Step 2:Fill in the table with information given in the question.John took a drive to town at an average rate of40mph. In the evening, he drove back at30mph. If he spent a total of7hours traveling, what is the distance traveled by John?Lett= time to travel to town.7t =time to return from town.rtd

Case 140t

Case 2307t

Step 3:Fill in the values fordusing the formulad = rtrtd

Case 140t40t

Case 2307 t30(7 t)

Step 4:Since the distances traveled in both cases are the same, we get the equation:40t= 30(7 t)Usedistributive property40t= 210 30tIsolatevariablet40t+ 30t= 21070t= 210

Step 5:The distance traveled by John to town is40t= 120The distance traveled by John to go back is also 120So, the total distance traveled by John is 240Answer:The distance traveled by John is 240 miles.

MIXTURES WORD PROBLEMS

Mixture problems are word problems where items or quantities of different values are mixed together.Adding to the SolutionExample1:John has 20ouncesof a 20% ofsalt solution, How much salt should he add to make it a 25% solution?Solution:Step1:Set up a table forsalt.originaladdedresult

concentration

amount

Step 2:Fill in the table with information given in the question.John has20ounces of a20%of salt solution. How much salt should he add to make it a 25%solution?The salt added is100%salt, which is 1 in decimal.Change all thepercentto decimalsLetx= amount of salt added. The result would be 20 +x.originaladdedresult

concentration0.210.25

amount20x20 +x

Step 3:Multiply down each column.originaladdedresult

concentration0.210.25

Amount20x20 +x

Multiply0.2 201 x0.25(20 +x)

Step 4:original + added = result0.2 20 + 1 x= 0.25(20 +x)4 +x= 5 + 0.25xIsolatevariablexx 0.25x= 5 40.75x= 1

Answer:He should addounces of salt

Removing From the SolutionExample2:John has 20 ounces of a 20% of salt solution. How much water should he evaporate to make it a 30% solution?Solution:Step 1:Set up a table forwater. The water is removed from the original.originalremovedResult

concentration

Amount

Step 2:Fill in the table with information given in the question.John has20ounces of a20%of salt solution. How much water should he evaporate to make it a30%solution?The original concentration of water is 100% 20%=80%The resulted concentration of water is 100% 30%=70%The water evaporated is 100% water, which is1in decimal.Change all thepercentto decimals.Letx= amount of water evaporated. The result would be 20 x.originalremovedresult

concentration0.810.7

amount20x20 x

Step 3:Multiply down each column.originalremovedresult

concentration0.810.7

amount20x20 x

multiply0.8 201 x0.70(20 x)

Step 4:Since the water is removed, we need to subtractoriginal removed = result0.8 20 1 x= 0.70(20 x)16 x= 14 0.7xIsolatevariablexx 0.7x= 16 140.3x= 2

Answer: He should evaporate 6.67 ounces of water

Replacing the SolutionExample3:A tank has a capacity of 10 gallons. When it is full, it contains 15% alcohol. How many gallons must be replaced by an 80% alcohol solution to give 10 gallons of 70% solution?Solution:Step 1:Set up a table foralcohol. The alcohol is replaced i.e. removed and added.originalremovedaddedresult

concentration

amount

Step 2:Fill in the table with information given in the question.A tank has a capacity of10gallons. When it is full, it contains15%alcohol. How many gallons must be replaced by an80%alcohol solution to give10gallons of70%solution?Change all thepercentto decimals.Letx= amount of alcohol solution replaced.originalremovedaddedresult

concentration0.150.150.80.7

amount10xx10

Step 3:Multiply down each column.originalremovedaddedresult

concentration0.150.150.80.7

amount10xx10

multiply0.15 100.15 x0.8 x0.7 10

Step 4:Since the alcohol solution is replaced, we need to subtract and add.original removed + added = result0.15 10 0.15 x+ 0.8 x= 0.7 101.5 0.15x+ 0.8x= 7Isolatevariablex0.8x 0.15x= 7 1.50.65x= 5.5

Answer:8.46 gallons of alcohol solution needs to be replaced.

Mixing Quantities Of Different CostsExample 4:How many pounds of chocolate worth $1.20 a pound must be mixed with 10 pounds of chocolate worth 90 cents a pound to produce a mixture worth $1.00 a pound?Solution:Step 1:Set up a table for differenttypes of chocolate.originaladdedresult

cost

amount

Step 2:Fill in the table with information given in the question.How many pounds of chocolate worth$1.20a pound must be mixed with10pounds of chocolate worth90cents a pound to produce a mixture worth$1.00a pound?Letx= amount of chocolate added.originaladdedresult

cost0.91.21

amount10xx+ 10

Step 3:Multiply down each column.originaladdedresult

cost0.91.21

amount10xx+ 10

multiply0.9 101.2 x1 (x+ 10)

Step 4:original + added = result0.9 10 + 1.2 x= 1 (x+ 10)9 + 1.2x=x+ 10Isolatevariablex1.2xx= 10 - 90.2x= 1

Answer:5 pounds of the $1.20 chocolate needs to be added.

MONEY WORD PROBLEMS

First, we will look at a money word problem involving calculating Simple Interest. Simple Interest word problems are based on the formula for Simple Interest and the formula for Amount. Then, we will look at a money word problem that involves coins and dollar bills.Formula for Simple Interesti = prtirepresents the interest earnedprepresents the principal which is the number of dollars investedtrepresents the time the money is invested which is generally stated in years or fractions of a year.Formula for AmountA = p + iArepresents what your investment is worth if you consider the total amount of the original investment (p) and the interest earned (i)Example1:James needs interest income of $5,000. How much money must he invest for one year at 7%? (Give your answer to the nearest dollar)Solution:5,000 =p(0.07)1p= 71,428.57He must invest $71,429

In this chapter, we will learn how to solve algebra word problems that involve motion.ALGEBRA MOTION PROBLEMS

Motion (Distance) FormulaMotion problems are based on the formulad = rtwhered= distance,r= rate andt= time.When solving motion problems, a sketch is often helpful and a table can be used for organizing the information.Example1:John and Philip who live 14 miles apart start at noon to walk toward each other at rates of 3 mph and 4 mph respectively. In how many hours will they meet?Solution:Letx= time walked.rtd

John3x3x

Philip4x4x

3x+ 4x= 147x= 14x= 2They will meet in 2 hours.

Example2:In still water, Peters boat goes 4 times as fast as the current in the river. He takes a 15-mile trip up the river and returns in 4 hours. Find the rate of the current.Solution:Letx= rate of the current.rtd

down river4x+x15 / 5x15

up river4x-x15 / 3x15

ANSWER: The rate of the current is 2 mph.

NUMBER SEQUENCE PROBLEMS

Number Sequence Problems are word problems that involves a number sequence. Sometimes you may be asked to obtain the value of a particular term of the sequence or you may be asked to determine the pattern of a sequence.

Number Sequence Problems: Value Of A Particular TermThe question will describe how the sequence of numbers is generated. After a certain number of terms, the sequence will repeat through the same numbers again. Try to follow the description and write down the sequence of numbers until you can determine how many terms before the numbers repeat. That information can then be used to determine what a particular term would be.For example,If we have a sequence of numbers: x, y, z, x, y, z, ....that repeats after the third term.If we want to find out what is the fifth term then we get the remainder of 5 3, which is 2.The fifth term is then the same as the second term, which isy.Example1:The first term in a sequence of number is 2. Each even-numbered term is 3 more than the previous term and each odd-numbered term, excluding the first, is 1 times the previous term. What is the 45th term of the sequence?

Solution:Step 1:Write down the terms until you notice a repetition2, 5, -5, - 2, 2, 5, -5, -2, ...The sequence repeats after the fourth term.Step 2:To find the 45th term, get the remainder for 45 4, which is 1Step 3:The 45th term is the same as the 1st term, which is 2Answer:The 45th term is 2.Number Sequence Problems: Determine The Pattern Of A Sequence

Example2:6, 13, 27, 55, In the sequence above, each term after the first is determined by multiplying the preceding term bymand then addingn. What is the value ofn?Solution:Method 1:The fastest way to solve this would be if you notice that the pattern:6 2 + 1 = 1313 2 + 1 =27Answer: The value ofnis 1.Method 2:If you were not able to see the pattern then you can come with two equations and then solve forn.6m+n=13 (equation 1)13m+n= 27 (equation 2)Use substitution methodIsolateninequation 1n =13 6mSubstitute intoequation 213m+ 13 6m= 277m= 14m= 2Substitutem= 2 intoequation 16(2) +n= 13n= 1Answer:n= 1

PROPORTION AND RATIO WORD PROBLEMS

Proportion problems are word problems where the items in the question are proportional to each other. In this chapter, we will learn the two main types of proportional problems: Directly Proportional ProblemsandInversely Proportional Problems.

Directly Proportional ProblemsThe question usually will not tell you that the items aredirectly proportional. Instead, it will give you the value of two items which are related and then asks you to figure out what will be the value of one of the item if the value of the other item changes.Proportion problems are usually of the form:Ifxtheny. Ifxis changed toathen what will be the value ofy?

For example,If two pencils cost $1.50, how many pencils can you buy with $9.00?The main difficulty with this type of question is to figure outwhich values to divideand which values to multiply.The following method is helpful:Change the word problem into the form:Ifxtheny. Ifxis changed toathen what will be the value ofy?which can then be represented as:

For example,You can think of the sentence:If two pencils cost $1.50, how many pencils can you buy with $9.00?asIf$1.50thentwopencils. If$9.00then how many pencils?Write the proportional relationship:

Example 1:Jane ran 100 meters in 15 seconds. How long did she take to run 1 meter?Step 1:Think of the word problem as:If 100 then 15. If 1 then how many?Step 2:Write the proportional relationship:

Answer:She took 0.15 secondsExample 2:Ifof a tank can be filled in 2 minutes, how many minutes will it take to fill the whole tank?Step 1:Think of the word problem as:Ifthen 2. If 1 then how many? (Whole tank is)Step 2:Write the proportional relationship:

Answer:It took 3.5 minutesExample 3:A car travels 125 miles in 3 hours. How far would it travel in 5 hours?Step 1:Think of the word problem as:If 3 then 125. If 5 then how many?Step 2: Write the proportional relationship:

Answer:He traveledmiles.Inversely Proportional ProblemsInversely Proportional questions are similar to directly proportional problems, but the difference is that whenxincreaseywill decrease and vice versa - which is theinverse proportionrelationship. The most common example of inverse proportion problems would be themoremenon a job thelesstimetaken for the job to completeAgain, the technique is to change the proportion problems into the form:Ifxtheny. Ifxis changed toathen what will be the value ofy?and then write the inverse relationship (take note of the "inverse" form):

Example4:It takes 4 men 6 hours to repair a road. How long will it take 7 men to do the job if they work at the same rate?Step 1:Think of the word problem as:If 4 then 6. If 7 then how many?Step 2:Write out the inverse relationship:

Answer:They will takehours.

Ratio Word Problems

Ratio problems areword problemsthat useratiosto relate the different items in the question.Themainthings to be aware about for ratio problems area: Change thequantitiesto the sameunitif necessary. Write the items in the ratio as afraction. Make sure that you have the same items in the numerator and denominator.Ratio problems: Two-term RatiosExample 1:In a bag of red and green sweets, the ratio of red sweets to green sweets is 3:4. If the bag contains 120 green sweets, how many red sweets are there?Solution:Step1:Assign variables :Letx= red sweetsWrite the items in the ratio as a fraction.

Step 2:Solve the equationCross Multiply3 120 = 4 x360 = 4xIsolatevariablex

Answer:There are 90 red sweets.Example 2:John has 30 marbles, 18 of which are red and 12 of which are blue. Jane has 20 marbles, all of them either red or blue. If the ratio of the red marbles to the blue marbles is the same for both John and Jane, then John has how manymoreblue marbles than Jane?Solution:Step 1: Sentence: Jane has 20 marbles, all of them either red or blue.Assign variables:Letx= blue marbles for Jane20 x= red marbles for JaneWe get the ratio from JohnJohn has 30 marbles, 18 of which are red and 12 of which are blue.

Direct VariationThere are many situations in our daily lives that involvedirect variation.For example, a worker may be paid according to the number of hours he worked. The two quantitiesx(the number of hours worked) andy(the amount paid) are related in such a way that whenxchanges,ychanges proportionately such that the ratioremains a constant.We say thatyvaries directlywithx. Let us represent the constant byk, i.e. ory = kxwherek 0Ifyvaries directly asx, this relation is written asyxand read asyvaries asx. The sign is read varies as and is called thesign of variation.

Example3:Ifyvaries directly asxand giveny= 9 whenx= 5, find: the equation connectingxandy the value ofywhenx= 15 the value ofxwheny= 6Solution:a)yxi.e.y=kxwherekis a constantSubstitutex= 5 andy= 9 into the equation:

y=xb) Substitutex= 15 into the equationy== 27c) Substitutey= 6 into the equation

Example4:The cost of a taxi fare (C) varies directly as the distance (D) travelled. When the distance is 60 km, the cost is $35. Find the cost when the distance is 95 km.Solution: i.e.C=kD, wherekis a constant.SubstituteC= 35 andD= 60 into the equation35 = 60kk=Therefore,C=DSubstituteD= 95 into the equation:C=55.42Answer : The cost for 95 km is $55.42Joint and Combined Variation Word Problems

Joint variationis a variation where a quantityvaries directlyas the product of two or more other quantities. For example, the area of a rectangle varies whenever its length or its width varies. We say that, whereAis the area,lis the length andwis the width.Combined variationis a variation where a quantity depends on two (or more) other quantities, andvaries directlywith some of them andvaries inverselywith others.

Example 5: A quantity varies inversely as two or more other quantities.The figure below shows a rectangular solid with a fixed volume. Express its width,w, as a joint variation in terms of its length,l, and height,h.

Solution:

Answer: In other words, the longer the lengthlor the heighth, the narrower is the widthw.Example 6: A quantity varies directly as one quantity and inversely as another.The speed,s, of a moving object varies directly as the distance travelled,d, and varies inversely as the time taken,t. Expresssas a joint variation in terms ofdandt.

Solution:

Answer: In other words, the longer the distance or the shorter the time, the faster is the speed.Inverse Variation Word ProblemsIn general, when two variablesxandyare such thatxy=kwherekis a non-zero constant, we say thatyvaries inverselywithx.In notation, inverse variation is written as

Example7:Suppose thatyvaries inversely asxand thaty= 8 whenx= 3.a) Form an equation connectingxandy.b) Calculate the value ofywhenx= 10.Solution: i.e.xy=kwherekis a non-zero constanta) Substitutex= 3 andy= 8 into the equation to obtaink3 8 =kk= 24The equation isxy= 24b) Whenx= 10, 10 y= 24 y=

Example:Suppose thatyvaries inversely asx2 and thaty= 10 whenx=.a) Find the equation connectingxandy.b) Find the value ofywhenx= 3.Solution:i.e.yx2=ka) Substitutex=andy= 10 into the equation to obtaink

The equation isyx2=b) Whenx= 3,

WORK WORD PROBLEMS

"Work" Problems: Two PersonsThe formula for Work Problems that involvetwo personsis

This formula can be extended formore than two persons. It can also be used in problems that involve pipes filling up a tank.Example 1:Peter can mow the lawn in 40 minutes and John can mow the lawn in 60 minutes. How long will it take for them to mow the lawn together?Solution:Step 1:Assignvariables:Letx= time to mow lawn togetherStep 2:Use the formula:

Step 3:Solve the equationTheLCMof 40 and 60 is 120Multiply both sides with 120

Answer:The time taken for both of them to mow the lawn together is 24 minutes.Work Problems: More than Two Persons

Example 1:Jane, Paul and Peter can finish painting the fence in 2 hours. If Jane does the job alone she can finish it in 5 hours. If Paul does the job alone he can finish it in 6 hours. How long will it take for Peter to finish the job alone?Solution:Step 1:Assignvariables:Letx= time taken by PeterStep 2:Use the formula:

Step 3:Solve the equationMultiply both sides with 30x

Answer:The time taken for Peter to paint the fence alone ishours.Work Problems: Pipes Filling up a Tank

Example 1:A tank can be filled by pipe A in 3 hours and by pipe B in 5 hours. When the tank is full, it can be drained by pipe C in 4 hours. if the tank is initially empty and all three pipes are open, how many hours will it take to fill up the tank?Solution:Step 1:Assignvariables:Letx= time taken to fill up the tankStep 2:Use the formula:Since pipe C drains the water it is subtracted.

Step 3:Solve the equationTheLCMof 3, 4 and 5 is 60Multiply both sides with 60

Answer:The time taken to fill the tank ishours.

60