365 problems and solutions (day 70)

8/6/2019 365 Problems and Solutions (Day 70)

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365 Math Problemsand SolutionsMastering Math the Sage Way

Sage Review GroupMs. Cris Anne Joy Martin Perez


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365 Math Problems and Solutions

CAJMP Page 2

DAY 1P ROBLEM

Her age two years from now is one half of her current age more than 12. Whatis her present age?

SOLUTIONS In de aling with age problems, the construction of the past, present and futuretable never fails. Once you get a stable grip on problems like this, you caneliminate this first step (construction of time table) and create a workingequation right away, but if you are still quite confused and not very confident, I

strongly encourage you to do this unfailing step. Here are the procedures insolving this problem:

1. Construct a time table.PAST PRESENT FUTURE (two years from

now)

2. Construct variables.Let be the present age of the person.After this, we update the table as follows.

PAST PRESENT FUTURE (two years fromnow)

X X + 2

Note that we dont have anything in the problem which tells us aboutthe past age of the person, thus, we leave the row under past blank.Also, we put + 2 under the future. We add 2 because the problemsaid that the future is referring to the time two years from now.

3. Interpret the problem.In other words, the problem says that:

= 12 + 12 In mathematical terms, the problem says that:

+ 2 =12

+ 12

Similar Problems You MigWant to Work On:

1. Half of Fabcurrent age is more than one-foof her age ten yfrom now. WhaFabrizias cuage?

2. One-seventh Tiffanys presentplus one-half ofage ten years agthirty less than current age. Howis Tiffany current

3. One-sixth Legendres age years from nowtwenty less thanhis current age. Wis one-half of

current age?4. Baby Cyrels

eighteen years fnow is five timeage two years now. What is current age?

5. One- third of Cheage plus one-haher age seven yfrom now is current age. How

is she now?


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CAJMP Page 3

Note that 12

+ 12 is not the same as 12

( + 12) . If the problem wants us to write 12

( + 12) , it

will say one -half of her age twelve years from now, but it didnt.

4. Solve the problem.

Given:+ 2 =

12

+ 12

A problem-solving strategy is we transpose all the terms with s in one side and the otherterms in the other:

12 = 12 2 12

= 10

= 20.

5. Interpret the answer you got.Since we let be the present age of the person, her present age is 20. This is also what theproblem is asking for.

Therefore, her present age is 20 years old.

6. Check the result you got by recalling the problem.During entrance examinations, usually, checking your answer is not very advisable since theexam is time-pressured, but we are not just talking about college entrance examinations here.

We are talking about how you can improve your Math skills so we check the answer we got.

The problem said that Her age two years from now is one -half of her current age more than12. If we let the present age to be 20, the age two years from now is 22. Also, if we let thepresent age to be 20, one-half of her current age is 10, plus 12 is 22. Therefore, it is right thatthe present age is 20.


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CAJMP Page 4

DAY 2P ROBLEM

The difference between two numbers is 12. Twice the smaller plus four timesthe larger is 108. What are the numbers?

SOLUTIONS A problem like this one is very straightforward. Unlike other kinds of problem,we only need to assign variables then start solving right away. Here are theprocedures in solving this problem:

1. Assign variables.

For this problem, we assign two variables because that is what theproblem gave us. Let and be two numbers, being the larger of thetwo. Note that it is advisable to assign variables which will give you clueson what it represents. In this problem, is the larger number and isthe smaller.

2. Translate words into mathematical symbols.The difference between two numbers is 12 only says that:

= 12 EQUATION A Note that the difference is 12, a positive number, so it should be .If in case, the problem said that the difference is 12, we shouldinterpret it as because a smaller number minus a larger one yieldsa negative number. Next, the problem said that twice the smaller plusfour times the larger is 108. In mathematical symbols, this is:

2 + 4 = 108 EQUATION BA rule of the thumb says that we should get 2 equations if we have 2variables or 2 unknowns. After acquiring equation A and B, we are now

ready to solve.

3. Solve the problem using the data you gathered earlier.In solving systems with two variables, we have several ways. For thisproblem, I will use substitution. I will be using the other ways in otherproblems in this compilation.


1. The diffebetween numbers is Twice the smallthe numbers pluslarger is 72. Whathe numbers?

2. The diffebetween numbers is 28. Stimes the smallerthan twice the lais 56. What arenumbers?

3. The diffebetween numbers is 60. Tthe smaller plus tthe larger is What are numbers?

4. The diffebetween numbers is Three times the lanumber plus times the smnumber is 201. Ware the numbers?

5. Two numbers hadifference of 23.times the smaller twice the large130. What are two numbers?


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CAJMP Page 5

Equation A: = 12 Rewriting this: = 1 2 + EQUATION C

We then plug in equation C to Equation B:2 + 4 12 + = 108

Distributing: 2 + 4 12 + 4 = 108 Simplifying: 2 + 4 8 + 4 = 108

2 + 4 = 108 48 6 = 60

= 10 .Plugging this in Equation C: = 12 + 10

= 22 .

4. Interpret the answer you got.Since the problem is asking for the two numbers, we answer both the values of and .

Therefore, the two numbers are 10(smaller) and 22(larger).

5. Evaluate the result.We let the two numbers to be 10 as the smaller, and 22 as the larger. Their difference is22 10 = 12 . Then twice the smaller number (10) is 2 10 = 20 . Four times the larger number(22) is 4 22 = 88 , and 88 + 20 = 108 . Well, it seems like we got the correct answer.


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CAJMP Page 6

DAY 3P ROBLEM

On a bank account of $1600, how much yearly interest will be paid annually if the rate is 5%?

SOLUTIONS Investment problems complicatedness can range from very easy to quitedifficult. This one is quite easy. We only need to remember the equation forsimple interest which is = , where is the interest, is the principalmoney, is the rate and is the time that your money was left in the bank. Acommon mistake in solving this problem is students substitute right away thevalues they have in hand. We still need to convert the time and rate to a similarunit (they should be both in terms of year, months or even days) if they differ.Here are the procedures in solving this problem:

1. Identify the given information and identify the unknown.The problem is asking for the yearly interest. It means that in = must be missing. The money left in the bank to grow must be $1600. Thisis the principal money. So = $1600. The rate is often easily identified.It is usually the one with the percentage sign. So the rate must be 5%.

= 5% 0.05

1 . Note that 5% =5

100 or 0.05 . Alwaysremember that 0.05 is not equal to 0.5. This is also a common mistakethat you should be careful of. Lastly, the problem asks about the yearlyinterest. It is also a way of saying how much interest will be paid aftera year? So = 1 . The and are both in terms of year so weneed not convert. Now that we identified the values we need, we arenow ready to solve.

2. Substitute the given and solve.=

= $16000.05

1 (1 ) = $1600 0.05

= $80 3. Interpret the answer you got.

Since we are looking for the interest, and we found the value of , this must be the answer.

Therefore, the yearly interest is $80.


1. Gerald inv$3400 in a bank. much interest wigain after two yeathe rate is annually?

2. How much moneMr. Hull investedbank with rate semiannually ifinterest he got aone year is $60,00

3. How long did invested her $50in a bank which y7% insemiannually ifinterest she got athat span of tim$3,500?

4. Moira decided

invest her $15,00a bank. After a and a half she$3,375 interest. Wis the rate of bank?

5. On a bank accou$45,000, how minterest will be after four monththe rate is 4%year?


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CAJMP Page 7

4. Evaluate the result.The easiest way to check this is to perform the multiplication again and see if you got the sameresult as your answer. Take note that the year unit was canceled while we were solving. Thisleaves us the unit $, which is right because the interest must have a unit of a money. Also,remember that the interest is always less than the principal money. So if ever you got a value for

greater than , there must be something wrong in your solutions.


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CAJMP Page 8

DAY 4P ROBLEM

How many peanuts should be mixed with a nut mixture that is 40% peanuts toproduce 36 ounces of a 60% peanut mixture?

SOLUTIONS Just like in age problems, the unfailing step in solving mixture problems isconstructing a diagram that will contain the data you need in solving. Trueenough, there are some difficulties in solving mixture problems because youneed the skill to identify certain values just by analyzing the problem, yet thisskill can be acquired from continuous practice. Here are the procedures insolving this problem:

1. Construct a diagram.

We construct a diagram which iscomposed of mixture 1, mixture 2 and the product. The sign betweenmixtures 1 and 2 depends on what the problem said. In this case, theproblem said that mixture 1 will be mixed with mixture 2. So we addthem. If in case the problem said that mixture 2 will be removed frommixture 1, we use subtraction instead.

2. We fill in the diagram with the necessary information.A convenient way for this is we put both the amount and percentage inthe diagram, careful enough as not to put the data in the wrong place.Also, we should always remember that the amount can be added andsubtracted, but not th e percentage. It doesnt follow that if mixture 1 is20% and mixture 2 is 30%, the product is 50%. It is NOT correct to thinkof it that way. In the problem, it said, How many peanuts, so mixture1 must be pure peanut or 100% peanut. The problem continued withmust be mixed with a nut mixture that is 40% peanuts so mixture 2must be 40% peanuts. The problem ended with to produce 36 ounces


1. How much wmust be added 100% alcohol in to produce 10 lite60% alcohol?

2. Milas juice staso famous forSummer-flavored juice. Its recipethat its 72.5% is mup of pineapple j

How much pinea juice must be mwith a 45% pineamixture in ordeproduce 2 liters opopular fruit juice

3. One alloy con80% gold and anoalloy contains gold. The two aare combined make 40 grams o

alloy that is 70% How many gramthe 80% alloyused?

4. How much salt be added to 20%solution in ordeproduce 20 ounce25% salt solution

5. Each mixture composed of gumworms and gubears. How mgummy worms be added to a camixture compose40% gummy wororder to producegrams of 60% guworms mixture?

MIXTURE 1 MIXTURE 2 PRODUCT+ =


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CAJMP Page 9

of a 60% peanut mixture? so we are given both how much and what percentage the product is.We then update the diagram:

3. Assign variables.It is always best to assign the variable to the value we want to find. Let be the amount presentin the first mixture. Like what Ive said in the previous steps, we can add or subtract the amountpresent in the mixtures. So +? = . The second mixture must have 36 ounces of peanut. If we are going to check it, + 36 = + 36 = 36 , which agrees to theamount in the product. We then update the diagram again:

4. Solve the problem using the data above.We multiply the percentage of each mixture to their respective amounts:

100% + 36 40% = 36 60% 1.00 + 36

0.40 = 36 0.60

+ 36 0.40 0.40 = 36 0.60 + 14.4 0.40 = 21.6 0.40 = 21.6 14.4 0.60 = 7.2

= 12 5. Interpret the answer you got.

Since we let be the amount present in the first mixture which contain pure peanuts, this is theanswer.

Therefore, 12 ounces of peanuts must be added.6. Evaluate the result.

We let = 12 and we check if the equation below will be satisfied:1 + 36 0.4 = 36 0.60

12 1 + 36 12 0.4 = 36 0.60 12 + 24 0.4 = 36 0.60

12 + 9.6 = 36 0.60 21.6 = 21.6

?

100%

?

40%

36 ounces

60%+ =

X ounces

100%

(36 x) ounces

40%

36 ounces

60%+ =


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CAJMP Page 10

DAY 5P ROBLEM

Lori starts jogging from a certain point and runs 5kph. Jeffrey jogs from thesame point 15 minutes later at a rate of 8kph. How long will it take Jeffrey tocatch up to Lori?

SOLUTIONS Motion problems range to very easy to quite difficult ones. This belongs to theaverage questions. In dealing with this, we only need to remember that = ,where is the total distance traveled, is the rate of the one in motion and isthe time that it took the traveling object to complete the distance. It is alwaysbest to write the given data down and analyze the problem, translate it to alanguage which you master in order to come up with the right working equation,and always remember that you wont be given a problem with no sufficientgiven data. The data in the problem will always be enough. Here are theprocedures in solving this problem:

1. Write down the data given in the problem. = 5

= 8

15 . 2. Analyze the problem.

Although Lori left earlier, Jeffreys rate is bigger. That means, Jeffrey isfaster than Lori. But Lori has this 15-minute advantage because he leftearlier. This means that we add 15 to his time. Also, since we are lookingfor the time that Jeffrey will catch up Lori, we set their distances to beequal, since if you catch someone, it means that you have the samedistance with him/her.

3. Assign variables.Let be the time after Jeffrey left, that he will catch up Lori.

This is not the same for the both of them. + 15 will be the time afterLori left that Jeffrey will catch him, since he has that 15-minuteadvantage.

4. Create a working equation and solve.Since we agreed earlier that Lori and Jeffrey have the same distances, westart by equating them:

=


1. Bapor Tabo traveling at 7 Star Ferry startesail 5 hours Bapor Tabo leftport. Star Ferry;sis 12 kph. Howwill Star Ferry up Bapor Tabo?

2. Ella and Yori hlittle misunderst

ding. Ella stwalking away Yori at a rate of 2After 4 minutethinking and mahis mind up, realized that he rwas wrong andshould chase Elfast as possiblewalked at a rate okph. How long w

take Yori to catcto Ella?

3. A delivery trutraveling at a ratekph. It was 7 minafter it left that owner realized thwas deliveringpackage to the wrplace so he semotorcycle to cthe truck as faspossible. motorcycle travfor 75 kph. Howwill it take motorcycle to cthe truck?


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CAJMP Page 11

= 5 + 15 = 8 5 + 5 15 = 8

5 + 7 5 = 8 7 5 = 8

5

7 5 = 3

= 753


Since we let be the time after Jeffrey, such that he will catch Lori, we can consider the answerwe got to be the final answer.

Therefore, Jeffrey will catch Lori 25 minutes after Jeffrey left.Also, we can consider + 15 to be the final answer, taking necessaryprecautions that this time is in terms of Lori, unlike the other one which

is in terms of Jeffrey. Since = 25 , + 15 = 25 + 15 = 40 .Therefore, Jeffrey will catch Lori 40 minutes after Lori left.

6. Evaluate the result.Suppose, = 25. Then, after 25 minutes of traveling in 8 , Jeffreytraveled:

=

=8

6025

= 103

.

Jeffrey traveled103 .

On the other hand, after 40 minutes of traveling in 5 , Lori traveled:=

=5

6040

=103

.

Lori traveled 103

.

They have the same distance traveled. Meaning, Jeffrey really caught upLori after the 25 minutes for Jeffrey and after 40 minutes for Lori.

4. A criminal traveling at a ratekph. After 2 minthe police wenchase him. They traveling at a rat15 kph. How lonit take the policcatch the crimina

5. Pepe and Neneplaying hideseek. Nene runs rate of 3 kph. Aft

minutes of siTagutaguanMaliwanag Buwan, Pepe bsearching for herrate of 8 kph. long will Pepe Nene if Nene dichange her positi


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CAJMP Page 12

DAY 6P ROBLEM

I have both cows and chickens and their sum is 30. I found out that the totalnumber of head is 5/2 of the number of chickens legs less than the number of the legs of the cows. How many cows do I have?

SOLUTIONS This problem only needs you to remember that chickens have two legs and cowshave four. If you take this in consideration, you will get the answer. Also, this isan example of a problem which needs two variables in solving, unlike the otherswhich require only one, but there is no extra step in solving this. We assignvariables right away and we dont need to construct any diagram or table. Hereare the procedures in solving this problem:

1. Assign variables.Let be the number of cows and let be the number of chickens. Sincechickens and cows both have a single head, the number of chickensheads is and the number of cows heads is . On the other hand, cowshave 4 legs. So in order to look for the number of cows legs, we multiply

by 4. Similarly, the number of chickens legs is multiplied by 2, sincechickens have two legs.

2. Construct the working equations.Like what Ive said from the previous days, since we have two variables,there must also be t wo working equations. First, it said that I have bothcows and chickens and their sum is 30. In mathematical terms, thismust be:

+ = 30 + = 30 EQUATION A

Then it said that the total number of head ( + ) is 52

of the number of

chickens legs (2 ) less than the number of the legs of the cows (4 ) . Inmathematical terms, this must be:

= 52 + = 4 52 2 EQUATION B


1. The total numbedogs and traine30. Twice the number of headtwice the numbehuman legs

than 46

of the nu

of dogs. How dogs are there?

2. I have both hand chickens. T

are 34 legs. 5 tthe number chickens head less than 8 timesnumber of hoheads. How mhorses are there?

3. I have both bears and penguThe total numbetheir heads is Thrice the numbpenguins legs four times number of bears legs is 9 tthe total numbetheir heads mthan 53. How mpenguins do I hav

4. The total numbemy ducks and ca35. The total num

of their legs isHow many duckthere?

5. The total numbemy pigs and duc16. The total numof their legs isHow many pigs have?


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CAJMP Page 13

3. Simplify the equations and solve.We first simplify Equation B because it is the more complicated expression:

+ = 4 52 2 + = 4

10

2

+ = 4 5 5 + = 4

6 = 3 63

=

2 = EQUATION C Equation A is + = 30 , we can also write is as = 30 and we plug it in to Equation C.

2 = 2 = 30 2 + = 30

3 = 30 = 10

Since = 30 and = 10 as we found it, we substitute it to .= 30

= 30 10 = 20

4. Interpret the answer you got.Since we have two values, and , we have to choose between them. The problem is asking forthe number of cows, and in step 1, we let that to be . Thus, we choose .

Therefore, the number of cows is 20.5. Evaluate the result.

If we let 20 to be the number of cows and 10 to be the number of chicken, we try to substitute itto the problem. The sum of 10 and 20 is 30, so that agrees with the first statement of theproblem. The total number of heads is also 30. The number of cows legs is 4 = 4 20 = 80 ,

minus the 52

of the number of chickens legs will be 52

2 10 = 102

10 = 1002

= 50 . 80 50 =30 . Our acquired values agree with the problem, making the result we got correct.


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CAJMP Page 14

DAY 7P ROBLEM

A discount store prices its blank videotapes by raising the wholesale price by40% and a dding $0.20. What must be the tapes wholesale price be if the tapesells for $3.00?

SOLUTIONSThis problem is just a simple percentage one but with a twist. The only crucialpart here is the identification of the given data. We always have to be verycareful so as not to mismatch the data we have at hand. We use theformula = , where is the percentage or the discounted price, is therate or the one easily identified because of the percentage symbol and is thebase or the original money. If is less than 1 or less than 100%, is less than ,but if is greater than 1 or greater than 100%, is greater than . Here are theprocedures in solving this problem:

1. Assign variables.Since the problem is asking for the wholesale price or the original price,we let be the whole price of the tape. This assigning is quite easy forthis problem because the problem is straightforward and is asking foronly one value.

2. Translate the problem into mathematical terms.The problem said that the wholesale price is raised by 40%. If is thewholesale price of the tape, that will be 40% + . Take note that it isnot just 40% since this is just a discounted price and lesser than . Theproblem said that the price was raised, so we still have to add 40% to

. Next, it said that the raised price is then added to $0.20 . So we add40% + to $0.20, then we equate all of this to $3.00 because theproblem said that $3.00 is the selling price, meaning, this is the productof all the increments. From all of these, the working equation will be:

40% + + $0.20 = $3.00 3. Solve.

40% + + $0.20 = $3.00 0.40 + + $0.20 = $3.00 0.40 + = $3.00 $0.20

1.40 = $2.80 = $2.00


1. Mr. Go sells gaby raising wholesale price30% and aP1500. If a laptopP14500, what musits wholesale price

2. What will be the of 4 shirts ifwholesale price P200 and the seprice is gained

raising the wprice by 50% adding P50?

3. Beatrice spent Pon an online sShe bought necklace. The osaid that the seprice of her itemacquired by rathe wholesale pric30% and adding

handling fee. Wmust be wholesale price onecklace?

4. Another store pits dresses by rathe wholesale pric40% and adding What must be dress s whoprice be if the dsells for $73?

5. What will be the of an iPod ifwholesale price P18000 and selling price is by raising the wprice by 20% adding P500?


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CAJMP Page 15

4. Interpret the answer you got.Since we let be the wholesale price, this must be the final answer.

Therefore, the wholesale price is $2.00.5. Evaluate the result.

We work backwards in order to check if we got the correct answer. For instance, $2.00 is thewholesale price of a blank videotape. 40% of $2.00 is 0.8. So if we raise the price by 40%, thiswill be $2.00 + $0.8 = $2.8 . Then, if we add $0.2 like what the problem is saying, the new pricewill be $2.8 + $0.2 = $3.00 . This agrees with the value given in the problem, so our answermust be correct.


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CAJMP Page 16

DAY 8P ROBLEM

Simplify:+ [ + ].

SOLUTIONS Unlike word problems which we commonly solve, numerical problems like the one given above are verystraightforward and requires less analysis and less steps in solving than word problems. We only need tobe extra careful because those problems often confuse students. Here are theprocedures in solving this problem:

1. Simplify.We simplify this by means of writing the original problem thenperforming the operation little by little until we arrive at the answer.Take note that we also solve using MDAS (Multiplication or Division thenAddition or Subtraction). We begin solving by solving the innermost partof the expression, regardless if it is inside a brace, parentheses orbrackets:

2 8 + 7 2 [ 8 + 6 3 2 2 4 42 8 2 3 ]43

2 8 + 7 2 [ 8 + 6 3 2 2 4

1 6 8

2 3 ]

43

2 8 + 7 2 [ 8 + 6 3 2 2 4 2 2 3 ]43

2 8 + 7 2 [ 8 + 6 3216 2 2 3 ]43

2 8 + 7 2 [ 8 + 6 2 2 2 3 ]43

2 8 + 7 2 [ 8 + 6 0 2 3 ]43 2 8 + 7 2 8 + 0 2 3

43

2 8 + 7 2 8 + 0 643


Simplify the folloexpressions:

1. 28 37 +16 6 4 4 2 23 4 7

3 + 7

2. {56 7 6 + 5 4

10 8 } + 83

3. 47+13[96+123{8 3 16 2

44 128 9]

4. 19+21[87+5{64 4 2 44 128

3]

5. 7 + 2 [3 +7{44 128 63 216

5]


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CAJMP Page 17

2 8 + 7 2 243

28 + 14443

17243

= 4

2. Interpret the answer you got.Since there is no variable and the problem is asking us to simplify the expression, our resultmust already be the final answer.

Therefore, the answer is 4.3. Evaluate the result.

In checking the result you got, there is no other way except to run through your solutions andcheck one-by-one if all operations are perfectly performed. We cannot go backwards so this isthe only way.


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Now, since the second quantity in Ratio A is the same with the first quantity in Ratio B, we canput into ratio the first quantity of Ratio A (24) and the second quantity of Ratio B (80).

24:80 .4. Interpret the answer you got.

We got the ratio 24: 80 for hamsters to fishes, but always remember that ratios should alwaysbe in their lowest term. So we divide 24:80 by their GCF which is 8. So 24:80 will be 3:10.

Therefore, the ratio of hamsters to fishes is 3:10.5. Evaluate the result.

Like the previous number, the best way to check this is to go over your solution and check eachstep if all operations are perfectly executed and all logic applied are valid. If yes, then your resultmust be correct.


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CAJMP Page 20

DAY 10P ROBLEM

Jake takes 45 minutes to mow the lawn. His friend, Mark, takes 30 minutes tomow the lawn. If they work together, how long will it take for them to mow thelawn?

SOLUTIONS The problem above seems like the most common variation of work problems. If they are helping each other, work problems solutions are usually in the

form 1 + 1 = 1 , where 1 is the rate of person 1, 1 is the rate of person 2 and

is the time of completion if they work together. We equate this to 1 becausethere is only one job. In this case, there is only one lawn to mow. Here are theprocedures in solving this problem:

1. Identify , and .

If we let Jake be the person 1, his rate is 145

. If we let Mark be the

person 2, his rate is 130

. The problem is asking for the time that

they will finish mowing if they work together. This must be .2. Assign variables.

Let be the time that Jake and Mark will finish mowing together.3. Solve.

We begin by substituting the values we acquired earlier to 1 + 1 = 1 .

145

+1

30= 1

145

+1

30= 1 90

45

+

30= 1 90

90

45+

90

30= 90

2 +

3 = 90

5 = 90

= 90 5

= 18


1. Savannah can paroom in 105 minJordan can paintsame room inminutes. How will it take for Savannah and Joto paint the roothey are wotogether?

2. It takes Shelby28 minutes to cu

grass. Abigail 70 minutes to cugrass. If they together, how will it take themcut the grass?

3. Joshua can dparticular typingin 24 hours andcan do the samein 40 hours. If work together,

many hours wtake them to fthe job?

4. Peter can mowlawn in 40 miand John can the lawn in minutes. How will it take for to mow the together?

5. Shawna can po

large condriveway in minutes. Dan pour the sdriveway in minutes. Find long it would them if they wotogether.


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CAJMP Page 21

Some remarks: In the second line of the solving process, we multiplied everything by 90 because it is the

LCM of 45 and 30 which are in the denominator of the fractions. We then distribute the to the rates of Jake and Mark. We cancel the unit lawn, leaving us with the unit minutes which seems appropriate

because the problem is asking how long so our answer must contain a unit for time.4. Interpret the answer you got.

Since we are looking for the value which will answer how long will it take Jake and Mark tomow the lawn if they work together? and apparently, we let that value to be and we foundthe value for , this must be the answer that we are solving for.

Therefore, if Jake and Mark work together, it will take them 18 minutes to mow the lawn.5. Evaluate the result.

18 minutes is less than the time it takes Jake to mow the lawn and also less than the time ittakes Mark to mow the same lawn. One indication that your answer is correct is that the time

you obtained must be less than the time of the two people who worked together. They workedtogether, meaning, the work will be faster, thus, the time will be lesser. We check by workingbackwards and letting = 18 . We simplify and we should get 1 as the answer.

145

+1

3018

1 1845

+1 18

30

1845

+18

30

9 2

9 5+

6 3

6 5

25

+3

5

2 + 35

55

= 1 We know from the first part of the solution that this is equal to 1. We got the same result fromworking backwards. So our answer must be correct.


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CAJMP Page 22

DAY 11P ROBLEM

Simplify: + + + + + . SOLUTIONS

Because this is, again, a numerical problem, there is no step which will involve analysis. The problem isvery straightforward and requires only carefulness in order to arrive at the correct answer. Here are theprocedures in solving this problem:

1. Solve.We first write the given expression down and perform the operationsone-at-a-time.

12

62+88 75 + 13 62+88 75 + 16 62+88 75 Note that the expressions inside the parentheses are just the same foreach term: (62 + 88 75) . This is just equal to 75 , after evaluation, sowe write it:

12

75 +13

75 +16

(75)

Now, we surely can perform addition, but to solve easier, it is better tofactor out 75 first so that we will only be left with simpler fractions:

75 12

+ 13

+ 16

= 753 + 2 + 1

6

Note that we acquired the GCF of 2, 3 and 6 in order to simplify thefractions. Now, we can add them:

= 75 66

= 75 1 = 75 .

2. Interpret the answer you got.Since the problem only asked us to simplify the given expression andthere are no variables or any of the like, our final answer must be 75.

Therefore, 75 is the answer.3. Evaluate the result.

The best way to check this is to go over your solutions and look for mistakes. You run througheach part and each operation, making sure that all operations are carried out smoothly. Also,instead of factoring out 75, you can add right away, and check if you will arrive at the sameanswer.

Similar Problems You Mig

Want to Work On:


1. 17

47+31 5712

47+31 571

14(47+ 31 5

2. 17

78+35 5013

78+35 401

9(78+ 35

50

3. 16

26+58 4913

26+58 4912

(26+ 58 494. 1

441+63 74

16

41+63 741

1241+63 7

5. 13

76+47 4316

76+47 431

4 (76+ 47 43


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CAJMP Page 23

DAY 12P ROBLEM

In a class of 35 students, 20 like Math and 22 like Chemistry. How manystudents like Chemistry only?

SOLUTIONS This is a problem which can be answered through logic, but there is an algebraicway of solving this problem. We create a Venn diagram and fill it in with theimportant data found in the problem. Here are the procedures in solving thisproblem:

1. Construct a Venn diagram.We label those who like both Math and Chemistry. Now, we are lookingfor those who like Chemistry only, meaning, the part of that 22 whichdoesnt include those who like Math. Note that there are only 35students in all.

2. Assign variables.Let be the number of students who like both Math and Chemistry. Inthe Venn diagram, this is the portion where the two circles overlap. Let be the number of students who like Chemistry only, and let be thenumber of students who like Math only. We then update the Venndiagram:

Similar Problems You Mig

Want to Work On:

1. In a class ostudents, 25 Math and 23 Philosophy. many like both Mand Philosophy?

2. In a caf, there25 customers. 1them like their coice cold while 1

their coffee hot. many of them their coffee hot o

3. In a PE class students, 40 Table Tennis whilike Swimming.many of them Table Tennis only

4. 50 respondents wasked in a suabout their drvacation. 42 saidthey want to spenin Baguio. 25 waspend it in BatanHow many ofrespondents likespend their vacaon both places?

5. In a pet shop, twere 20 custom13 of them

Hamsters for pets. 15 of themdogs for their How many of like dogs only aspets?

2220 35

Math Chemistry

x35

Math Chemistry

z


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CAJMP Page 24

Take note that we have values for those who like Math and those who like Chemistry. Sincethere are 20 students who like Math, + = 20 . So = 20 (EQUATION A). On the otherhand, there are 22 students who like Chemistry, + = 22 . So = 22 (EQUATIONB).Since there are 35 students in all, it means that + + = 35 (EQUATION C). It shouldoccur to you that since there are three unknowns, there must also be three equations.

3. Solve.+ + = 35

20 + + = 35 20 + + = 35

20 + 22 = 35 (x+y=22 from Equation B)20+22 = 35

42 = 35 42 35 =

= 7

We plug = 7 to Equation A:= 20

= 20 7 = 13 = 13

We plug = 7 to Equation B:= 22

= 22 7 = 15 = 15

4. Interpret the answer you got.We arrived at three answers, , and . The problem is asking for the number of students wholike Chemistry only. We assigned it to be .

Therefore, there are 15 students who like Chemistry only.5. Evaluate the result.

We check our answer by means of updating the Venn diagram and check each value:

Those who like Math is 20 according to the problem. According to our Venn diagram, there are13+7 students or 20. Correct. According to the problem, there are 22 students who likeChemistry. According to our Venn diagram, there are 7+15 students or 22. Correct. Lastly, theproblem said that there are 35 students in all. According to our Venn diagram, there are13+7+15 students or 35 students. All of the values be obtained are correct.

157

35

Math Chemistry

13


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CAJMP Page 26

DAY 14P ROBLEM

( ) is equivalent to:SOLUTIONS

Again, this numerical problem requires no additional analysis. It is verystraightforward. It only asks us to simplify the expression or in other words,eliminate the negative exponents of the factors. Aside from being careful indealing with this kind of problem, we only need to remember some law of

exponents which says, = 1 . We get the reciprocal whenever there is anegative exponent. Here are the procedures in solving this problem:

1. Solve.We first reciprocate the term in order to make the outermost exponentpositive:

= Then, we distribute 2 to each of the factors and simplify those who havepositive exponents:

1

3 2 2 =

1(

3) 2 4 =

19 ( 4 )

In here, note that (3) 2 became 9 and not 9. It is because theexponent 2 is distributed even to the negative sign, making it positive.Now, note that we have a negative exponent again (4 in ) . Since it isin the denominator, we simply put it to the numerator in order to makethe exponent positive:

19 ( 4 ) =

4

9

2. Interpret the answer you got.Since we did not have any variables, our result must already be the finalanswer.

Therefore, the answer is .

3. Evaluate the result.The only way to evaluate this is, again, to go over the solutions and checkif all operations are carried out perfectly and see if none of the steps is missing. Also, check forirregularities. For instance, a number was raised to a negative exponent and yet, it wasevaluated. If all of these are considered, then the answer must be correct.



1.

2 3 3

equivalent to? 2. 3 3 4

equivalent to?

3. 7 5equivalent to?

4. 7 6equivalent to?

5. 6 4 equivalent to?

6. 6 4 equivalent to?

7. 4 (7

2

)equivalent to?

8. 5 (2 ) 1equivalent to?

9. 5 (2 ) equivalent to?

10. (5 2 )2equivalent to?

11. 6 3 2equivalent to?

12. 2)( 3 7equivalent to?

13. 2)( 3 8equivalent to?


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CAJMP Page 27

DAY 15P ROBLEM

In four exams, Angels scores are as follows: 85, 79, 91 and 87. What shouldshe get in her fifth exam to have an average of 87?

SOLUTIONS This is a simple algebraic problem which requires some Statistical information.First, the problem spoke about average. The average being referred to in thisproblem is the common average which we add all the terms (in this case, theterms are scores) then we divide the sum by the number of terms which weadded (in this case, there are 5). In the problem, the average is given and alsofour out of five of the scores. We are asked to find the fifth score. Here are theprocedures in solving this problem:

1. Assign variables.Since we are looking for the fifth score, and it seems that all the othervalues which we need is present, we let be the fifth score of Angel.

2. Construct a working equation.87 =

87 =

5

5

87 =8 5 + 7 9 + 9 1 + 8 7 +

5

3. Solve.We simply simplify the working equation we obtained and look for .

87 =8 5 + 7 9 + 9 1 + 8 7 +

5

87 =342 +

5

We then cross-multiply five to the other side:87 5 = 342 +

Simplify:435 = 342 +

We transpose:435 342 =



1. In four exRoberts scores are93, 83 and 85. Wshould be his gradhis fifth exam itarget average is 9

2. In order to qualifthe next roundteam must have

average of 85 in ten games. Tscores in the firgames are 70, 7590, 85, 80, 87, 8482. What musttheir score in tenth game in ordequalify to the round?

3. In order to pass Mone must haveaverage of 60. Vagot 57, 60, 61 anin her first four exWhat must be grade in her 5th eif she wants to her Math subject?

4. In two games, tAs scores are 6370. What should third score be if

want an average68?

5. In three exams, Nscores are 65, 7570. What should4th score be ifwants an averag80?


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Since we are looking for Angels fifth score and we let this to be equal to , this must be theanswer.

Therefore, Angel needs to get 93 in her fifth exam in order to get an average of 87.5. Evaluate the result.

To check this, we let 93 be the grade of Angel in her fifth exam. We get the average and see if it

will match the given average in the problem which is 87.

=8 5 + 7 9 + 9 1 + 8 7 + 93

5

=435

5

= 87 This is matched with the given average in the problem. Therefore, our answer must be correct.


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DAY 17P ROBLEM

Jenny used 2/3 of her savings in paying for her tuition fee and of theremaining money in buying books. What part of her savings was left?

SOLUTIONS This is a typical algebraic expression but very confusing. Most of the studentswill solve this without thinking carefully and without interpreting the problemcorrectly. Here are the procedures in solving this problem:

1. Assign variables.Let be the part of Jennys savings that was left, after all, this is whatwe are asked to solve.

2. Construct a working equation.

Jenny used 23

of her savings in paying for her tuition fee. This means that13

is left for her other expenses. Then, the problem said that one-fourth

of the remaining money was used to buy books. We then write thisinformation:

23

+14

13

+ = 1

We equate it to 1 because 23, one-fourth of 1

3and are all parts of a

whole. Meaning, if we add them, they will be equal to a whole or to one.3. Solve.

We then solve for from the working equation:23

+14

13

+ = 1

23

+1

12+ = 1

23

+1

12+ = 1 12

2 123

+ 1212

+ 12 = 12

243

+ 1 + 1 2 = 12

8 + 1 + 1 2 = 12 9 + 1 2 = 12 12 = 12 9


1. Rachel used 14

o

salary to buy

clothes. 14

of

remaining moneyput to investmWhat part of salary was left?

2. Crista spent 1/3 ofallowance to paytuition fee. She s

of the remamoney to buy sbooks. What parher allowance left?

3. Jhett won oncompetition. 2/5 ocash prize donated to a cha of the remamoney was savedhis future educaWhat part of his pwas left?

4. Matthew is a thieof his stolen mwas spent for the fof his family. oremaining moneyfor his childeducation. He then arrested by police. What par

the money was lef5. Michelle used 1

3

allowance to

some clothes. 14

remaining moneyspent for shoes. Wpart of her allowwas left?


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CAJMP Page 32

12 = 3

= 312

=14

4. Interpret the answer you got.Since we let to be the part of Jennys savings thaw was left and this is precisely what theproblem is asking us, this must already be our final answer.

Therefore, of Jennys savings was left.

Note that we chose 14

instead of 312

because although the latter is also correct, we prefer answers

in lowest term. In that case, we choose 14.

5. Evaluate the result.

We let the part which was left of Jennys savings to be equal to 14. Then we add all the parts and

check if the sum will be equal to 1.

= 23

+ 14

13

+ 14

=23

+1

12+

14

=8

12+

112

+3

12

=1212

= 1 Our obtained value agrees with what the problem said. Therefore, our answer must be correct.


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CAJMP Page 33

DAY 18P ROBLEM

What is the solution set of the equation + =0?SOLUTIONS

Asking about the solution set of an equation is just another way of asking for thevalue/s of that will satisfy the given equation. Often, students use trial andmethod in determining the solution set, but this can be very misleading. Youmight miss out some of the other values which are also part of the solution set.Here are the procedures in solving this problem:

1. Look for special functions.Special functions are those which can be radicals, absolute values orgreatest integral factor. In this case we have an absolute value function.So we isolate it in one side.

2 + 5 = 3 2. Solve.

There are two parts in solving this problem. First we set 2 + 5 equal to3, and the other is we set 2 + 5 equal to 3. We do this because if 3 = 3 and 3 = 3 . We first set it to be equal to 3:

2 + 5 = 3 2 = 3 5 2 = 2 = 1

Next, we set 2 + 5 to be equal to 3.2 + 5 = 3 2 = 3 5

2 = 8 = 4

3. Interpret the answer you got.

Since the problem is asking for the solution set of the values of , both 1 and 4 are answers.We write the solution set in such a way that we enclose them in braces.Therefore, the solution set of the equation is {,}.

4. Evaluate the result.We evaluate by means of checking the values and see if they will satisfy 2 + 5 3 = 0 . Wetry 1, 2 1 + 5 3 = 3 3 = 0 . We also try 4, 2 4 + 5 3 = 3 3 = 3 3 =0.


Determine the solution setthe following equations:

1. 1 = 4 .2. 3 = 3 .3. 6 + 9 = 04. 1 + 5 = 05. 6 + 5 1 4 =6. 5 2 = 0 .7. 9 4 = 86 .8. 3 + 7 7 3 =9. 5 + 4 = 110. +4

10= 1

11. 39

+ 7 = 8 .

12. 4 2 9 =13. 10 7 + 3 = 0 .14. 8 3 8 = 4015. 10 10 8 + 410 .16. 4 9 6 = 17. 3 9 + 2 + 1

10 .18. 6 3 8 915 .19. 2 + 5 + 3 = 020. 2 + 5 + 3 = 3


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CAJMP Page 34

DAY 19 P ROBLEM

What is the last digit of ?

SOLUTIONS At a glance, the problem seemed very difficult to solve. It seems that the onlyway to solve this one is to evaluate the expression by expanding it, butremember that this kind of problem only needs a pattern. Remember that weare just looking for the last digit of 31351 . Here are the procedures in solving thisproblem:

1. Look for a pattern.We look for a pattern by evaluating the first few terms:

30 = 1 31 = 3 32 = 9

33 = 27 (last digit is 7)

34 = 81 (last digit is 1)35 = 243 (last digit is 3)36 = 729 (last digit is 9)

37 = 2187 (last digit is 7)

It seems like the last digit just revolves from 1, 3, 9, and 7. After that, itwill go back at 1.

2. Solve.What we do now, after discovering the pattern, is we divide 1351 by 4. If it is divisible by 4, then the last digit of the expression is 7. If it is notdivisible by 4, we count the remainder.

1351 4 = 337 3. We count 3 in (1, 3, 9, 7). Making the first number 1, second number 3and the third number 9. This must be the last digit of 31351 .

3. Interpret the answer you got.The problem is asking for the last digit of 31351 and this is what we got from the previous step.

Therefore, the last digit of is 9.4. Evaluate the result.

Aside from evaluating the expression, the best way to check if you got the correct answer is togo over your solutions and check if the division and counting were performed flawlessly. Also,make sure that you included 30 in the process of looking for the pattern and you did not start at31 right away. After checking these things and still arriving at the same answer, then it must becorrect.


What is the last digit of following:

1. 32435 2. 31981 3. 31000 4. 298

5. 2192 6. 4107 7. 43281 8. 59864 9. 56651 10. 6743 11. 69854 12. 71351 13. 72342 14. 78654 15. 89865 16. 87543 17. 81122 18. 9100 19. 93245 20. 9540


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CAJMP Page 35

DAY 20 P ROBLEM

of the number plus the number is 3. What is the number?

SOLUTIONS The problem above is an example of an easy finding the number problem. Theonly crucial part here is the construction of a working equation , which isnt thatdifficult in this case. Here are the procedures in solving this problem:

1. Assign variables.Let be the number referred in the problem.

2. Translate the problem to mathematical terms.The problem said that of the number plus the number is 3. In otherwords, it says that:

12

+ = 3

Since we let be the number, we insert in the expression above:12

+ = 3

3. Solve.

We solve for from the working equation above:12

+ = 3

12

+ = 3 2

22

+ 2 = 2 3

+ 2 = 6 3 = 6


The problem is asking for the number which if added to the half of it is 3. We let that number be.

Therefore, the number is 2.5. Evaluate the result.

We let the number be 2. We then check if we will get 3 if we add 2 and the half of 2:12

2 + 2 = 1 + 2 = 3 . Therefore, the answer we obtained must be correct.


1. Six less than times the number What is the numbe

2. Two more than number is twicenumber. What isnumber?

3. One-third of number plus one-of the numberseven less than number. What isnumber?

4. One-eighth of number plus one-t

of the number one-sixth of number is 15. Whthe number?

5. One-third of number plus one-of the numberfourteen less thannumber. What isnumber?


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CAJMP Page 37

After all, this is what the problem is asking. Therefore, the time it will take the car to complete

his return trip is = = 600

. The total time will be = 4 + 600 .

3. Create a working equation.We simply plug all the values we got from the previous step to the general form of the average

speed: =

240 =1200

4 + 600

240 =1200

4 + 600

240 =1200

4 + 600

4. Solve.We cross multiply the working equation:

240 4 + 600 = 1200 960 + 144000 = 1200 144000 = 1200 960

144000 = 240 122000

240=

240

240


Since we let be the rate of the car for its return trip (town A to town B), and we let that valuehave as its unit, this must be the answer.

Therefore, the average rate of the car for its return trip is 600 kph.6. Evaluate the result.

We let = 600 and we substitute it to 12004+ 600

and check if it will yield an answer which

agrees with the value we have for the average speed of the entire trip:1200

4+ 600600=

12004+1 =

12005 = 240 . According to the problem, the average speed for the

entire trip is, indeed, 240 kph. The answer we got it correct.

5. A river connects city Acity B which measures 3000apart. A boat traveled from A to city B at an average rat600 kph. At what speed ditravel on the way back if average of its entire trip is

kph?


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CAJMP Page 38

DAY 22 P ROBLEM

How much 60% acid solution should be added to 8 liters of 25% acid solution to produce a 40% acidsolution?

SOLUTIONS Like what Ive said in Day 4, mixture problems are always in the form:

The signs and the labels are only the ones which vary. Here are the proceduresin solving this problem:

1. Identify the data in the problem.First of all, the problem said that a solution will be mixed with anothersolution. This means that the operation between mixtures 1 and 2 is,indeed, addition. We then update the diagram:

2. Assign variables.In the diagram above, we have two unknown quantities: the amount of the second mixture and the amount of the product. Since we are lookingfor the amount of the 60% acid solution, we assign a variable to it.

60% . If this is so,

the amount present in the product must be 8 + . Rememberthat the amount can be added or subtracted but not the percentage. Wethen update the diagram above:


1. How many liter70% alcohol shouadded to 20% oliters alcohol in oto produce a 50%

the solution?2. How many quart

70% salt solshould be added tquarts of a 25%solution in ordeproduce a mixthat is 60% salt?

3. How many gallon70% milk-solution should added to a 10-lite

10% solution in to produce a soluwhich is 40% solution?

4. How many gram80% red candy shbe added to a 25-g8% red candy in oto produce a swwhich is 40%candy?

5. How many gallon70% acid somust be mixed 50 gallons of a solution to obtasolution that is acid?

MIXTURE 1 MIXTURE 2 PRODUCT+ =

8 liters

25%

?

60%

?

40%+ =

8 liters

25%

x

60%

(x+8 liters)

40%+ =


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CAJMP Page 39

3. Create a working equation.We create a working equation by means of multiplying the amount by its percentage:

8 25% + 60% = + 8 (40%) 8 0.25 + 0.60 = + 8 (0.40)

4. Solve.From the working equation above, we solve for the unknown value which is :

8 0.25 + 0.60 = + 8 (0.40) 2 + 0.60 = 0.40 + 3.2 2 + 0.60 0.40 = 3.2 0.60 0.40 = 3.2 2

0.20 = 1.2 0.20 = 1.2 10

2 = 12 22

= 122


The problem is asking for the amount of the 60% acid solution. We let this to be equal to , sothis must already be the answer.Therefore, 6 liters of a 60% solution should be added to an 8-liter 25% acid solution in order to

obtain a solution which is 40% acid.6. Evaluate the result.

In order to check the result that we obtained, we plug in 6 liters to the working equation:8 0.25 + 0.60 = + 8 (0.40) and check if it satisfies the equation. If it does,then our answer must be correct.

8 0.25 + 0.60 = + 8 (0.40) 8 0.25 + 6 0.60 = 6 + 8 0.40

2 + 3.6 = 14 0.40 5.6 = 5.6

We verified that our value satisfies the working equation. Therefore, our answer must becorrect.


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CAJMP Page 40

DAY 23 P ROBLEM

The sum of three numbers is 124. The second number is 4 more than twicethe first. The third number is 2 less than half of the second number. Amongthe three numbers, which is the greatest and what is its value?

SOLUTIONS The problem is very straightforward and requires no additional step likeconstructing a diagram or a table. We only need to master the skill of representing values in order to answer problems like this. Here are theprocedures in solving this problem:

1. Assign variables.Since the first part of the problem told us that there are three numbers,we assign 3 variables. ,

and . Note that weare still not sure about which among the numbers is the greatest.

2. Translate to mathematical terms.The second number is 4 more than twice the first is:

= 2 + 4 = 2 + 4

The third number is 2 less than half of the second number:

=12 2

=12 2

=12

2 + 4 2 =

12

2 +12

4 2 = + 2 2

= We were able to represent both and in terms of .

3. Create a working equation.The problem said that the sum of the three numbers is 124. So wesimply add the three numbers:

+ + = 124 + + = 124

Similar Problems You MighWant to Work On:

1. The sum of integers is 60. second number is 8than half of the The third numbefour more than second. What is value of the senumber?

2. The sum of

numbers is 46. second number is more than half third. The first numis 11 more than the second. What isthird number?

3. The sum of numbers is 89. second number is 7than four times first. The third num

is 33 less than tthe second. What isfirst number?

4. The sum of numbers is 240. second number isless than twice first. The third numis 52 more than 1/the second. What the three numbers?

5. The sum of numbers is 64. second number is less than five timesfirst. The third imore than half second. What is value of the number?


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CAJMP Page 41

4. Solve.From the working equation above and from step 2, we solve for first:

+ + = 124 + 2 + 4 + = 124

+ 2 + = 124

4

4 = 120 44

= 1204

= 30 Now, since we got , we solve for . Take note that = 2 + 4 :

= 2 + 4 = 2 30 + 4

= 6 0 + 4 = 64

Now, note that = . Since = 30, = 30 .

5. Interpret the answer you got.The problem is asking for the largest value among the three numbers. We found out that

= 64 is larger than both and . Since is equivalent to the second number, this must be theanswer.

Therefore, among the three values, the second number is the largest. Its value is 64.6. Evaluate the result.

We let the first, second and the third number be equal to 30, 64 and 30, respectively. We addthose three and check if the sum will be the same as what the problem told us:

+ + = 30 + 64 + 30 = 124 which agrees with the value given in the problem. We also check if the second number is 4 morethan twice the first number which is 30. 2 30 + 4 = 60 + 4 = 64 . Correct. We check if thethird number is 2 less than half of the second. Half of the second number (which is 64) is642

= 32 , subtract 2 from it, we get 30, which is the same as our value. Therefore, our answer

must be correct.


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CAJMP Page 43

DAY 25 P ROBLEM

If the endpoints of the diameter of a circle is at (5, 0) and (-1, 8), then the radius of the circle measureswhat?

SOLUTIONS A typical student will plot the points and measure the diameter then divide it by 2 in order to obtain theradius of the circle. Note that this is always confusing and very harmful. We cannot draw accurately,most of the times. So we should have alternatives in solving this problem.Whenever we are given two points, remember to consider the possibility of

using the distance formula or = ( 1 2 )2

+ ( 1 2 )2

. Here are theprocedures in solving this problem:

1. Assign variables.Let be the diameter of the circle. We can acquire because this is thedistance between the two points, and so we use the distance formula.Let 5 and 0 be 1 , 1 , respectively and let 1 and 8 be 2 , 2 respectively.

2. Solve.

= 1 2 2 + 1 2 2 = 5 1 2 + 0 8 2

= 6 2 + 8 2 = 36+64

= 100 = 10

3. Interpret the answer you got.We got = 10 , which is the length of the diameter of the circle, butnote that the problem is not asking for the diameter. It is asking for theradius of the circle. We should always remember that =

2. = 10

2= 5 .

Therefore, the radius of the circle is 5 linear units.4. Evaluate the result.

The best way to evaluate this is to go over the solutions and check each step if all the operationsare perfectly performed. Also, note that the linear units is attached to 5. The unit for thediameter and radius is always in some linear units while the area is in square units.


1. If the endpoints ofdiameter of a circat (8, 4) and (-8,then what is measure of the raof the circle?

2. If the endpoints ofdiameter of a circat (27, -4) and (23what is the length oradius?

3. If the endpoints ofdiameter of a circat (5, -15) and (-10what the radius ofcircle measures wh

4. If the diameter ocircle ends in the p(-14, 10) and (10what is the lengththe radius of the cir

5. If the endpoints ofdiameter of a circat (2, -8) and (2what is the measurits radius?


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CAJMP Page 44

DAY 26 P ROBLEM

Twice the smaller of two numbers is four less than twice the larger. Three sumof three times the smaller and four times the larger is 99. What are thenumbers?

SOLUTIONS This finding the number problem is quite confusing than the ordinary findingthe number problem where there is only one unknown number. In here, thereare two numbers, in which one is greater than the other. Nevertheless, the skillneeded here is the translation to mathematical terms. Here are the proceduresin solving this problem:

1. Assign variables.Since it seems that we have two unknowns, we assign two variables.Let be the larger of the two numbers and let be the smaller of twonumbers. Note that we assigned variables in a way which will help usremember what value they are representing.

2. Translate to mathematical terms.Because we have two unknowns, we should also develop two workingequations. To say that twice the smaller of two numbers is four lessthan twice the larger, means:

2 = 2 4 2 = 2 4 22

=22 42

= 2 EQUATION AOn the other hand, to say that the sum of three times the smaller andfour times the larger is 99, means:

3 + 4 = 99 3 + 4 = 99 EQUATION B

3. Solve.Now that we have two working equations now, we plug in equation Ato Equation B:

3 + 4 = 99


1. Twice the larger ofnumbers is four mthan fifteen times smaller. The sum olarger and ten tithe smaller is 37. Ware the numbers?

2.

The larger numbeone less than thricesmaller. Three tthe smaller plus tthe larger is 61. Ware the numbers?

3. Thrice the smalletwo numbers is more than twice smaller. The sumtwice the larger thrice the smaller i

What are the numbe4. Seven times the la

of two numbers less than ten timessmaller. Ten timessmaller plus five tthe larger is 50. Ware the two number

5. Seven times smaller of two numis 3 less than the laSix times the smaltwo more than larger. What are two numbers?


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CAJMP Page 45

3 2 + 4 = 99 3 6 + 4 = 99 3 = 4 = 9 9 + 6

7 = 105 7

7 =

105

7 = 15

From Equation A, the smaller number is given by = 2 = 2

= 15 2 = 13

4. Interpret the answer you got.The problem is asking for the value of the two unknown numbers. We let that to be and .These numbers must be the answer.

Therefore, the larger number is 15 and the smaller is 13.5. Evaluate the result.

We let the larger number be 15 and the smaller to be 13. The problem said that twice thesmaller (that will be 26) is 4 less than twice the larger (that will be 30). So 26 = 30 4. 26 =26 . So far, our values agree with what the problem is describing. Also, thrice the smaller (thatwill be 39) plus 4 times the larger (that will be 60) has a sum of 99. 39 + 60 = 99 . Therefore,our result agrees with the problem. This must be correct.


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CAJMP Page 46

DAY 27 P ROBLEM

A man drives 200km in an early hour drive for 2 hours and returns homeduring rush hours in 3 hours. What is his average speed for the round trip?

SOLUTIONS The problem above is an easier kind of an average motion problem. This is a loteasier than those problems which ask for the rate of the car on the return trip.It is kind of straightforward and needs only a formula to be answered correctly,provided that you are careful enough while solving. Remember that

=

. Total distance refers to the distance it traveled fromthe origin to the destination back to the origin again. On the other hand, thetotal time is the time it took the moving object to reach its destination and backto the origin again. In other words, total distance and total time is the distanceand time for the round trip. Here are the procedures in solving this problem:

1. Identify the given data in the problem.The problem said that the man traveled 200 km when he was about togo to his destination. So 1 = 200 . Note that the distancethat he traveled on his way back is just the same so 2 =

200 . This makes the total distance equal to 400 . The time it tookthe man to drive towards his destination is 2 hours. His return trip took3 hours. This makes the total time 5 .

2. Solve.We simply use the formula for average speed:

=4005


Since the average speed is what is asked from the problem and we got

it from the previous step, this must already be the answer.Therefore, the average speed of the man for his round trip is 80kph.

4. Evaluate the result.In this problem, the best way to check if we got the correct answer is togo over the solutions again and check if all the operations are perfectlyperformed. If yes, then, the result we got must be correct.


1. A truck traveled km from town Atown B for 26 houtraveled back forhours. What isaverage speed for entire trip?

2. A jogger ran 10 khis early stroll fhour and he returhome for 3 hbecause of exhaustWhat is his avespeed for his etrip?

3. A plane traveled Country A to Counwhich are 30 apart, for 6 hoursits way back, it trav

for 4 hours. How iaverage speed of plane for the etrip?

4. A boat cruised 20 mfrom an islandanother for 2 hourtraveled back fohours again. Whathe average speedthe boat?

5. A car traveled 60for 7 hours. It travback the same path5 hours. What isaverage speed of car for the entire tri


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CAJMP Page 47

DAY 28 P ROBLEM

Jan was a lady with a very interesting life. of her life was dedicated to

studying; then she spent two years of her life finding a job, but she found aman instead. One-twentieth of her life she devoted to this man but theybroke up and she spent three years of her life regaining herself and mendingher broken heart. After this span of time, she met Luis and they got along justfine. Luis courted her for years equivalent to 1/25 of Jans life and he becameJans boyfriend for thre e years before they decided to get married. They spent

2 years more than of Jans life waiting for their dearest child. When at last,

it came. The adorable baby attained half her mothers age when she died, andJan wrote the story of her life for 7 years. How old was she when she died?

SOLUTIONS This kind of age proble m usually scare students off. Its long and seems tocontain too many fractions, but actually, this is very easy to solve and verystraightforward. Here are the procedures in solving this problem:

1. Assign variables.Note that we are looking for Jans age wh en she died. This is equivalent

to her total years. So we let to be this value. .

2. Identify the numerical values found in the problem.15 studying, 2 finding a job, 120 first boyfriend,3 mending her broken heart, 125 courting of Luis,3 Luis is Jans boyfriend, 2 + 125 waiting for the child,12 age of the child, 7 Jan wrote the story of her life.Note that all this values equate to the total age of Jan or to .

3. Construct a mathematical equation.

We may ask the question, 15

of what? Then the problem will say that15

, but her total years is equal to , so this will be 15

.

15

+ 2 +1

20+ 3 +

125

+ 3 + 2 +1

25+

12

+ 7 =

4. Solve.


1. On Berthas dia

says that 112

of he

was devoted

childhood, 14

wa

studies, 16

for

marriage life whenfinally had a child,attained half the agher mother then tboth died at the sa

year. How old Bertha when she di

2. Peter spent 120

of h

in his childhood da

of his life practicinprofession,

spending 110

i

medicine school. met Rosalie, to whe spent half hiswith. Rosalie was murdered and Pspent 16 more yearinvestigating his wdeath before he dHow old was when he died?

3. Alessandra spent

her life in

childhood, 16

in stu

and 14

in family

when at last she bochild, who attainedof her age and meaccident. She speyear in curing depression then died eventually. old was Alesswhen she died?


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CAJMP Page 49

DAY 29 P ROBLEM

Anna and Thomas each invested money in a one-year CD at Bagel Bank. Thebank pays 5.15% yearly interest, compounded annually. Thomas invested sixhundred dollars less than three times the amount that Anna invested. BagelBank will pay Anna and Thomas a total of $381.10 in interest for the year.How much money will be in Anna's account after it matures in 1 year?

SOLUTIONS In dealing with investment problems, the key formula is =

( ) . In this problem, the interest is accumulated by twopeople. This makes the formula + = , where A is the first personand B is the second person. Here are the procedures in solving this problem:

1. Identify the given data in the problem.Anna and Thomas invested in the same bank. This means that their rateis the same and is equal to 5.15% compounded annually. Thomasprincipal is not given, and so is Annas. The total interest is $381.10, andthe time is both equal to 1 year.

2.

Assign variables.Since we are not given Thomas and Annas principal, we choose Annasprincipal to be . This makes Thomas principal 3 $600 .

3. Create a working equation.We simply substitute the data we had earlier and the variables weformulated to the formula of total interest:

$381.10 = 0.0515 1 + 3 $600 0.0515 1 $381.10 = 0.0515 + 0.1545 $30.9 $381.10 + $30.9 = 0.0515 + 0.1545

$412 = 0.206

$4120.206

= 0.2060.206

= $2000 4. Interpret the answer you got.

Note that is the principal of Anna, and is equal to $2000. Note thatthis is not what the problem is asking for. It is asking for the amount in


1. Joana and Jeff invested their monea bank which payinterest per annJeff invested $1000than half the amothat Joana investedthe end of the year,bank paid them a tof $980 in interesthe year. How m

money will be in account after matures in 1 year?

2. Bernard and Neil bought mobile phousing a credit which puts 2% intper year. The amof Neils phon$4000 less than tthe cost of Bernphone. After a yearcredit card comsaid that the interest that they going to pay is $How much is Bernphone?

3. An insurance compays 4% simple intper quarter year. Juinvested $5000 than twice the amo

that Pauline invesAfter 4 months, company paid t$1000 in interestthe period. How mmoney will bePaulines ac count it matures in 1 year


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CAJMP Page 50

Annas account after it matured in a year. We first compute for theinterest that it will earn:

= $2000 0.0515 1 = $103 . The amount in Annas account istotal to her principal and the interest that it earned. So it will be equalto $2000 + $103 = $2103 .

Therefore, the amount in Annas account after it matures in 1 year is $2103.

5. Evaluate the result.Let $2000 be the principal of Anna. If this is so, Thomas principal, givenby 3 $600 = 3 $2000 $600 = $6000 $600 = $5400 . Then,we compute the interest that they will earn after a year, in the rate5.15% per annum:

= $2000 0.0515 1 = $103 = $5400 0.0515 1 = $278.1 Their total interest is the sum of their individual interest given by:$103 + $278.1 = $381.1 , which is exactly the total interest given inthe problem. Therefore, our obtained value must be correct.

4. Cedric and Fluer borrowed money fa pawnshop which10% interest month. Fleur borroan amount whic$1800 less than tthe amount that Ceborrowed. Aftermonth, the interest they havepay is $210. How mdid Cedric borrowHow much did borrowed?

5. Victor and Hera invested their mone

a trust bank which 7.5% simple intper year. Hera inve$3000 less than tof the amount Victor invested. Aend of the year, bank paid them a tof $1350 in interesthe year. How mmoney will beVictors account afmatures for a year?


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CAJMP Page 51

DAY 30 P ROBLEM

If a 5m rubber band can be stretched up to 8m, how long can a 10m rubberband be stretched?

SOLUTIONS Proportion and ratio problems are usually easily solved by writing them in thisform: : = : . Because the means are equal to the extremes, = . Hereare the procedures in solving this problem:

1. Write the proportion.Since a 5m rubber band can be stretched up to 8m, we can write 5:8(RATIO A). Note that we are dealing with the same rubber band, so wecan equate the next ratio and equate the two.

2. Assign variables.Let be the length of the 10m stretched rubber band. Now, we canwrite the ratio 10: , and equate it to the first ratio (ratio A).

3. Construct a working equation.5 : 8 = 10 :

Since means are always equal to the extremes, we get: 5 =

10 (8 ) or (5 ) = 802

.4. Solve.

(5 ) = 80 2 (5 )

5=

80 2

5


Since we let be the length of the 10m stretched rubber band, thismay already be the answer.

Therefore, the 10m rubber band can be stretched up to 16m.

6. Evaluate the result.Note that ratios can be written as fractions. So we write 5:8 as 5

8and we

write 10: 16 as 1016

. Because 5:8 = 10:16, we can write them as 58

= 1016

. We cross-multiply,

5 16 = 10 8 80 = 80 . Also, note that from 5 m, we considered a 10 m rubber band.It was multiplied by 2, so we expect 8 m to be multiplied by 2, too. Our answer must be correct.


1. A spring, 7 cm castretched up to 16 How long can a 2spring of the smaterial be stretche

2. If a 3 m rubber can be stretched up7 m, how long can

m rubber band stretched?

3. Louie found out his 5m garter canstretched up to 11How long can a 15the same materialstretched?

4. A 50 cm rubber baa slingshot canstretched up to 60 How long can a 1of the same materiastretched?

5. A 75 cm wireJosephs bow canstretched up to 66 He decided to another one for son. He bought a juversion of the sbow, which wire is50 cm. How long

this wire be stretche


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CAJMP Page 52

DAY 31 P ROBLEM

If the radius of a circle is increased by 10%, what would be the expectedpercent of increase in its circumference?

SOLUTIONS In solving this problem, the only formula that we need to remember is =2 , where is the circumference of the circle and is its radius. What we aregoing to do is we will increase the radius by 10%, then we will subtract it withthe original and see their difference. Here are the procedures in solving thisproblem:

1. Make two circles.We will be constructing two circles, the one with the original radius andthe one, whose radius is increased by 10%.

1 = 2 ; 2 = 2 + 0.10 This makes the second circle equal to 2 = 2 1.1 .

2. Solve.We will subtract the two circles, circle 2 being the subtrahend:

2 1 = 1.1 2 2 = 0.1 2 Note that this is just equal to

1.

So we substitute it: 2 1 = (0.1) 1 . This means that the expectedincrease in the circumference is 0.1 or 10%.

3. Interpret the answer you got.Like what weve said in step 2, the value we got is already the finalanswer.

Therefore, the expected increase in the circumference is 10%.4. Evaluate the result.

In here, we increased the radius by 10% and so the circumference

followed with the same factor. Note that it is not correct to generalize this for all values. Forinstance, if we increase the radius by 10% and we are asked for the percent increase the area, itis not 10% because the formula for the area is quadratic ( 2) , while theformula for circumference is linear. Whenever confused, it is always safe to do all these steps.They may be quite long, but you will have higher chances of getting this right. Also, the best wayto check is to go over the solutions and check if every step was carried out smoothly.


1. If the radius of a cis increased by what would be expected percent increase in circumference?

2. If the diameter ocircle is increase

10%, what wouldthe expected perof increase in circumference?

3. If the sides of a sqare all increased30%, what wouldthe expected perof increase in its ar

4. If the sides of a sqare all increased20%, what wouldthe expected perof increase in perimeter?

5. If the length orectangle is increby 15% and its widincreased by 10%, would be the expepercent of increasits perimeter?


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CAJMP Page 53

DAY 32 P ROBLEM

The time required for an elevator to lift a weight varies directly with theweight of the load and the distance travelled and inversely as the power of the motor. It takes thirty seconds for a 10hp motor to lift 100 lbs through10ft. How much power is required to lift 250 lbs in 1 min through 40ft?

SOLUTIONS This is a combination of variation and work concepts. We only need to recallhow to write variations (inverse and direct) then we can solve right away. Tosay that x varies directly with y, we have = , where is a proportionalityconstant. On the other hand, to say that varies inversely with , we have

= , where is again some proportionality constant. Here are the

procedures in solving this problem:

1. Assign variables.Let be the time, be the weight of the load, be the distance and

be the power of the motor.2. Create a working equation.

varies directly with and , and is inversely proportional to . So

we have: = .

3. Solve.This is a 2-step process. For the first one, we look for . After findingit, we look for .

=

Given = 30 , = 10 , = 100 and = 10 .30 = 100 10

10

30 =100

30100

=100

100

30100

=

=30 100

Similar Problems You Might Wato Work On:

1. Time varies directly the weight and distance and inverselthe power. It takes 70for a 2 hp motor to lifkg through 14 m. much power is requto lift 140 kg in 1through 15 m?

2. varies directly withsquare of and the of and inversely wIf = 4 when = 21 = 3 , whatwhen = 45 = 2 ?

3. varies directly and and inv

with . If = 10= 8, = 5 and

what is= 5, = 7 and

10 ?4. varies directly w

and the square of inversely with the cub

. If = 135, = 53 = 1 , whatwhen = 8, = 4

= 2? 5. The force varies dir

with the masses of ob1 and 2 and invewith the square of distance between thIf the force = 1distance = 9, mass 1kg and mass 2 = 3 kgthe proportionconstant.


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CAJMP Page 55

DAY 33 P ROBLEM

Rationalize the denominator of .

SOLUTIONS The main goal of rationalizing the denominator is to have no radicals in thedenominator. We do it by multiplying both numerators and denominators byit conjugate. For instance, the conjugate of 1 + is 1 . It is often easy todetermine the conjugate of an expression. We simply copy the terms andthen take the opposite sign of the second term, that is if the term only have

two terms. Here are the procedures in solving this problem:

1. Determine the conjugate of the denominator.Like what weve said in the introduction of this problem, we obtainthe conjugate by writing the terms and just reverse the sign of the

second term. The conjugate of ` 2 is therefore 2 + .2. Multiply both the numerator and the denominator by the conjugate.

A student must always remember that we are to multiply both the

numerator and the denominator by the conjugate which is 2 + .In other words, we multiply

72 by

2

+

2 + . Note that we are justmultiplying by 1, the multiplicative identity of multiplication.7

(2 ) 2 +

(2 + ) =7(2 + )

4 + 2 2 =7(2 + )

4 For the multiplication of the numerators, there is no problem since 7 is just a constant. For themultiplication of the denominators, one should always remember that we always use FOIL (first,outer, inner, last) in multiplying polynomials like this.

3. Interpret the answer you got.Note that the answer we got doesnt contain any radicals in the denominator. This is the main

goal of rationalizing. This must be the already be the answer.Therefore, the answer is

( + ).4. Evaluate the result.

The best way to check if the answer is correct is to go over each step and confirm if all theoperations are perfectly performed. If yes, then the result must be correct.


Rationalize the denominators the following:

1. 22 1

2. 1

3. 7 +1 4.

2

2 +1

5. 153 3

6. 173+4

7. 12 +

8. +2

9. +2

10. 32


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CAJMP Page 57

We have two possible solutions. First, we add 10 to 2:

2 + 1 02

=82

= 4

Second, we subtract 10 from 2:

2 102 = 122 = 6 Since we have two answers, we will have to choose between 4 and 6. In the first part of ourdiscussion, Ive mentioned that the condition that is a negative number is important and veryuseful. We are going to let = 6 since 4 is positive. If this is so, + 2 = 6 + 2 = 4.

4. Interpret the answer you got.The problem is actually asking for the two numbers. We let this to be equal to and + 2 ,which we found out to be equal to 6 = 4 .

Therefore, the two numbers are

and

.

5. Evaluate the result.We will work backwards in order to check our answer. Let 6 and 4 be two consecutive evennegative integers. We obtain the product of those two numbers. 6 4 = 24 , which isexactly the value given in the problem. Our result must be correct.


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CAJMP Page 58

DAY 35 P ROBLEM

Two cars started at the same point and travels on different directions. Onetravels at 80 km/hr while the other travelled 100 km/hr. In how manyhours would they be 540 km apart?

SOLUTIONS We only need to remember that = , where is the distance traveled, isthe rate and is the time it took the vehicle to travel the distance. However,analyzing the problem, one must recognize the fact that the distance(540 ) is not done by one car alone. It is the distance in between twocars. We can imagine that a car made 500 and the other one made40 or the first car made 250 and the other made 290 , as long asthe distance was shared. In solving this, we should note that

= 1 1 + 2 ( 2 ) , where the subscriptsdistinguish the first car to another. Here are the procedures in solving thisproblem:

1. Identify the given data in the problem.The problem said that the first car travelled 80 km/hr while the othertravelled for 100km/hr. it also said that the distance between the

two cars is 540 km. This means that 1 =80 , 2 =

100 = 540 . Since both cars left the same

point at the same time, 1 = 2 and we will just denote it by.

2. Construct a working equation.From the equation in the introduction of this problem, we simplyplug in the values we obtained from step 1:

= 1 1 + 2 2

540 = 80 + 100 ( )

We factor out on the right side of the equation:

540 = 80 + 100

3. Solve.From the working equation we obtained earlier, we simply simplify:


1. Jim and Sarah whohiking in wilderdecide to leave their and walk around a lThey start going inopposite directions. hikes at the rate of 3 mper hour. Sarah hikethe rate of 2 miles hour. The perimeter

the lake is 10 miles. long will it be before meet?

2. Two jets leave an airat the same timeopposite directions. first jet is travelinthree hundred seveseven mph and the otat two hundred sevefive mph. How long wtake the jets to be 2,

miles apart?3. Two cyclists start at

same time from oppoends of a course that imiles long. One cyclriding at 14 mph andsecond cyclist is ridin16 mph. How long they begin will they m

4. Chelsea left the WHouse and travtoward the capital at

average speed of km/hr. Jasmine left at same time and travellethe opposite direcwith an average spee65 km/hr. Find number of hours Jasmneeds to travel bethey are 594 km apart.


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CAJMP Page 59

5. Two planes left the airat the same time. domestic plane travelling at a consrate of 150 miles per hThe international plantravelling at an oppodirection at a consrate of 190 miles per hAfter some time, it reported that they already at their respecdestinations which 1700 miles apart. long did it take the plato reach destinations?

540 = 180

We divide both sides by 180 :

540

180=

180

180

540 180

=


Since we are looking for the time that the cars will be 540 apartand

365 problems and solutions (day 70)

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