Math Word Problems and Solutions

BICOL UNIVERSITY

COLLEGE OF ENGINEERING

In partial fulfillment of the requirem requirements ents to be submitted to Engr. Atanacio A. Barajas Jr. Prepared by: KRISTEN GAIL N. MANLAPAZ BS ARCHITECTURE 1C

NEXT CENTURY MATHEMATICS BY FERNANDO ORINES 

2014

1

NEXT CENTURY MATHEMATICS BY FERNANDO ORINES 

2014

1

NEXT CENTURY MATHEMATICS BY FERNANDO ORINES 

2014

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 Problems Involving A Single Person Person Example 1: Five years ago, John’s age ag e was half of the age he will be in 8 years. How old is he now? Solution: Step 1: Let 1: Let x   x  be  be John’s age now. Look at at the question and put the the relevant expressions above it.

Step 2: Write 2: Write out the equation.

Isolate variable Isolate variable x   x  2

NEXT CENTURY MATHEMATICS BY FERNANDO ORINES 

2014

 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 age x + 10 = 3x x- 3x = -10 -2x = -10  x = 5 Example 3: In 20 years, Kayleen will be four times older than she is today. What is her current age? Solution: let x= current age x + 20 = 4x x - 4x = -20 -3x = -20 x = 6 2/3

3

NEXT CENTURY MATHEMATICS BY FERNANDO ORINES 

2014

 Age Problems Involving More Than One Person Example 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 now

age in 5 yrs

John Peter  Alice Step 2: Fill in the table with information given in the question. 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? Let x  be Peter’s age now. Add 5 to get the ages in 5 yrs. age now

age in 5 yrs

John

2 x 

2 x + 5

Peter

 x 

 x + 5

 Alice

 x – 5

 x – 5 + 5

 Write the new relationship in an equation using the ages in 5 yrs. In 5 years, John will be three times as old as Alice. 4

NEXT CENTURY MATHEMATICS BY FERNANDO ORINES 

2014

2 x + 5 = 3( x – 5 + 5) 2 x  + 5 = 3 x  Isolate variable x   x  = 5  Answer: Peter is now 5 years old. Example 5: John’s 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? Solution: Step 1: Set up a table.

age now

age in 2 yrs

John’s father John  Alice Step 2: Fill in the table with information given in the question. John’s 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? Let x  be John’s age now. Add 2 to get the ages in 2 yrs.

John’s father John

age now

age in 2 yrs

5 x   x 

5 x + 2  x + 2

 Alice  Write the new relationship in an equation using the ages in 2 yrs.

5

NEXT CENTURY MATHEMATICS BY FERNANDO ORINES 

2014

In two years time, the sum of their ages will be 58.

 Answer: John is now 8 years old.

6

NEXT CENTURY MATHEMATICS BY FERNANDO ORINES 

2014

7

NEXT CENTURY MATHEMATICS BY FERNANDO ORINES 

2014

There are two main types of average problems commonly encountered in school algebra: Average (Arithmetic Mean and Average 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 numbers Step 2: Using the formula

8

NEXT CENTURY MATHEMATICS BY FERNANDO ORINES 

2014

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

 Average Speed Problems Examples of Average Speed Problems Computation 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 50 miles per hour and for 2 hours at 60 miles per hour. What was his average speed for the whole journey? Solution: Step 1: The formula for distance is Distance = Rate × Time Total distance = 50 × 3 + 60 × 2 = 270 Step 2: Total time = 3 + 2 = 5 Step 3: Using the formula

9

NEXT CENTURY MATHEMATICS BY FERNANDO ORINES 

2014

 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.

10

NEXT CENTURY MATHEMATICS BY FERNANDO ORINES 

2014

11

NEXT CENTURY MATHEMATICS BY FERNANDO ORINES 

2014

Digit word problems are problems that involve individual digits in integers and howdigits are related according to the question. Some problems would involve treating the digits as individual numbers to be related. This would make it similar to an integer problem, except that the integers are between 0 and 9, inclusive.  Example1: The ten’s digit of a number is three times the one’s digit. The sum of the digits in the number is 8. What is the number?  Solution: Step 1: Sentence: The ten’s digit of a number is three times the one’s digit.  Assign variables: Let

 x one’s digit 3 x  ten’s digit

Sentence: The sum of the digits in the number is 8.  x + 3 x = 8 Step 2: Solve the equation  x + 3 x  = 8 Isolate variable x  4 x  = 8  x  = 2 The one’s digit is 2. The ten’s digit is 3 × 2 = 6  Answer: The number is 62.

12

NEXT CENTURY MATHEMATICS BY FERNANDO ORINES 

2014

13

NEXT CENTURY MATHEMATICS BY FERNANDO ORINES 

2014

Distance Problems: Traveling In Different Directions Example1:  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 a rtd  table. r 

t 

d 

 bus car Step 2: Fill in the table with information given in the question. If the bus is traveling at 50 mph and the car is traveling at 55mph, in how many hours will they be 210 miles apart? Let t  = time when they are 210 miles apart. r 

t 

 bus

50

t 

car

55

t 

d 

Step 3: Fill in the values for d  using the formula d = rt  r 

t 

d 

 bus

50

t 

50t 

car

55

t 

55t 

14

NEXT CENTURY MATHEMATICS BY FERNANDO ORINES 

2014

Step 4: Since the total distance is 210, we get the equation: 50t  + 55t  = 210 105t  = 210 Isolate variable t 

 Answer: They will be 210 miles apart in 2 hours.

Distance Problems: Given Total Time Example2: 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 a rtd  table. r 

t 

d 

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 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? Let t  = time to travel to town.

15

NEXT CENTURY MATHEMATICS BY FERNANDO ORINES 

2014

7 – t = time to return from town. r 

t 

d 

Case 1

40

t 

Case 2

30

7 – t 

Step 3: Fill in the values for d  using the formula d = rt  r 

t 

d 

Case 1

40 t 

40t 

Case 2

30 7 – t   30(7 – t )

Step 4: Since the distances traveled in both cases are the same, we get the equation: 40t  = 30(7 – t ) Use distributive property 40t  = 210 – 30t  Isolate variable t  40t  + 30t  = 210 70t  = 210

Step 5: The distance traveled by John to town is 40t  = 120

16

NEXT CENTURY MATHEMATICS BY FERNANDO ORINES 

2014

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

 Wind and Current Word Problems There 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= plane’s speed  y= wind speed Step 1: Set up a rtd  table PLANE

r 

t 

d 

 With

x+y

5

5 (x+y)

 Against

x-y

5.5

5.5 (x-y)

17

NEXT CENTURY MATHEMATICS BY FERNANDO ORINES 

2014

Step 2: 5 (x+y) = 2750

x + y = 550

5 5.5 (x-y) = 2750

x – y = 500

5.5 Step 3: add the two equations to cancel out one variable. (x + y = 550) + (x – y = 500) 2x = 1050 x= 525 So, the Plane’s speed is 525km/h  And the wind is 25km/h Example4: 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= Boat’s speed  y= current

18

NEXT CENTURY MATHEMATICS BY FERNANDO ORINES 

2014

Step 1: Set up a rtd  table BOAT

r 

t 

d 

 With

x+y

 3 / 4

3/4 (x+y)

 Against

x-y

1.5

3/2 (x-y)

Step 2: FIND THE EQUATION

[¾ ( x+y) = 6] 4/3

x+y=8

[3/2 ( x+y) = 6] 2/3

x – y = 4

Step 3: add the two equations to cancel out one variable.

(x+y=8) + (x-y=4) 2x = 12 x=6 Step 4: SUBSTITUTE THE VALUE OF X IN THE EQUATION

x+y=8 6+y=8  y = 2 So, the boat’s speed is 6km/hr and the water is 2km/hr.

19

NEXT CENTURY MATHEMATICS BY FERNANDO ORINES 

2014

Distance Word Problems - Traveling at Different Rates Distance problems are word problems that involve the distance an object will travel at a certain average rate for a given period of  time. The formula for distance problems is: distance = rate × time or d = 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 a rtd  table. r 

t 

d 

Case 1 Case 2

Step 2: Fill in the table with information given in the question.

20

NEXT CENTURY MATHEMATICS BY FERNANDO ORINES 

2014

 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? Let t  = time to make the trip in Case 2. r 

t 

Case 1

50

6

Case 2

45

t 

d 

Step 3: Fill in the values for d  using the formula d = rt  r 

t 

d 

Case 1

50

6

50 × 6 = 300

Case 2

45

t 

45t 

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

Step 5: Beware - the question asked for “how many more minutes 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. 21

NEXT CENTURY MATHEMATICS BY FERNANDO ORINES 

2014

Distance Word Problems - Given the Total Time Distance problems are word problems that involve the distance an object willtravel at a certain average rate for a given period of  time. The formula for distance problems is: distance = rate × time or d = 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.

Example6 : 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 a rtd  table. r 

t 

d 

Case 1 Case 2

Step 2: Fill in the table with information given in the question. 22

NEXT CENTURY MATHEMATICS BY FERNANDO ORINES 

2014

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? Let t  = time to travel to town. 7 – t = time to return from town. r  t  d  Case 1 40 t  Case 2 30 7 – t  Step 3: Fill in the values for d  using the formula d = rt  r 

t 

d 

Case 1

40

t 

40t 

Case 2

30

7 – t  

30(7 – t )

Step 4: Since the distances traveled in both cases are the same, we get the equation: 40t  = 30(7 – t ) Use distributive property 40t  = 210 – 30t  Isolate variable t  40t  + 30t  = 210 70t  = 210

Step 5: The distance traveled by John to town is 40t  = 120

23

NEXT CENTURY MATHEMATICS BY FERNANDO ORINES 

2014

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

24

NEXT CENTURY MATHEMATICS BY FERNANDO ORINES 

2014

25

NEXT CENTURY MATHEMATICS BY FERNANDO ORINES 

2014

Mixture problems are word problems where items or quantities of different values are mixed together.

 Adding to the Solution Example 1: John has 20 ounces of a 20% of salt solution, How much salt should he add to make it a 25% solution? Solution: Step 1: Set up a table for salt. original 

added 

result 

concentration amount

Step 2: Fill in the table with information given in the question. John has 20 ounces of a 20% of salt solution. How much salt should he add to make it a 25% solution? The salt added is 100% salt, which is 1 in decimal. Change all the percent to decimals Let x = amount of salt added. The result would be 20 + x.

concentration amount

original 

added 

result 

0.2

1

0.25

20

 x 

20 + x 

26

NEXT CENTURY MATHEMATICS BY FERNANDO ORINES 

2014

Step 3: Multiply down each column. original 

added 

result 

concentration

0.2

1

0.25

 Amount

20

 x 

20 + x 

Multiply

0.2 × 20

1 × x 

0.25(20 + x )

Step 4: original + added = result 0.2 × 20 + 1 × x = 0.25(20 + x) 4 + x = 5 + 0.25x Isolate variable x x – 0.25x = 5 – 4 0.75x = 1

 Answer: He should add

ounces of salt

27

NEXT CENTURY MATHEMATICS BY FERNANDO ORINES 

2014

Removing From the Solution Example 2: 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 for water. The water is removed from the original. original 

removed 

 Result 

concentration  Amount

Step 2: Fill in the table with information given in the question. John has 20 ounces of a 20% of salt solution. How much water should he evaporate to make it a 30% 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 is 1 in decimal. Change all the percent to decimals. Let x = amount of water evaporated. The result would be 20 – x. original  removed  result  concentration

0.8

1

0.7

amount

20

 x 

20 – x 

28

NEXT CENTURY MATHEMATICS BY FERNANDO ORINES 

2014

Step 3: Multiply down each column. original 

removed  result 

concentration

0.8

1

0.7

amount

20

 x 

20 – x 

multiply

0.8 × 20

1 × x  

0.70(20 – x )

Step 4: Since the water is removed, we need to subtract original – removed = result 0.8 × 20 – 1 × x = 0.70(20 – x) 16 – x = 14 – 0.7x Isolate variable x x – 0.7x = 16 – 14 0.3x = 2

 Answer: He should evaporate 6.67 ounces of water

29

NEXT CENTURY MATHEMATICS BY FERNANDO ORINES 

2014

Replacing the Solution Example 3:  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 for alcohol. The alcohol is replaced i.e. removed and added. original  removed  added  result  concentration amount

Step 2: Fill in the table with information given in the question.  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? Change all the percent to decimals. Let x = amount of alcohol solution replaced. original 

removed  added 

result 

concentration

0.15

0.15

0.8

0.7

amount

10

 x 

 x 

10

30

NEXT CENTURY MATHEMATICS BY FERNANDO ORINES 

2014

Step 3: Multiply down each column. original  removed  added 

result 

concentration 0.15

0.15

0.8

0.7

amount

10

 x 

 x 

10

multiply

0.15 × 10

0.15 × x 

0.8 × x 

0.7 × 10

Step 4: Since the alcohol solution is replaced, we need to subtract and add. original – removed + added = result 0.15 × 10 – 0.15 × x + 0.8 × x = 0.7 × 10 1.5 – 0.15x + 0.8x = 7 Isolate variable x 0.8x – 0.15x = 7 – 1.5 0.65x = 5.5

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

31

NEXT CENTURY MATHEMATICS BY FERNANDO ORINES 

2014

Mixing Quantities Of Different Costs Example 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 different types of chocolate. original 

added 

result 

cost amount Step 2: Fill in the table with information given in the question. 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? Let x = amount of chocolate added. original 

added 

result 

cost

0.9

1.2

1

amount

10

 x 

 x  + 10

original 

added 

result 

cost

0.9

1.2

1

amount

10

 x 

 x  + 10

multiply

0.9 × 10

1.2 × x 

1 × ( x  + 10)

Step 3: Multiply down each column.

32

NEXT CENTURY MATHEMATICS BY FERNANDO ORINES 

2014

Step 4: original + added = result 0.9 × 10 + 1.2 × x = 1 × (x + 10) 9 + 1.2x = x + 10 Isolate variable x 1.2x – x = 10 - 9 0.2x = 1

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

33

NEXT CENTURY MATHEMATICS BY FERNANDO ORINES 

2014

34

NEXT CENTURY MATHEMATICS BY FERNANDO ORINES 

2014

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 Interest

i = prt i represents the interest earned p represents the principal which is the number of dollars invested t represents the time the money is invested which is generally stated in  years or fractions of a year. Formula for Amount

 A = p + i  A  represents 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)1 p = 71,428.57 He must invest $71,429

35

NEXT CENTURY MATHEMATICS BY FERNANDO ORINES 

2014

36

NEXT CENTURY MATHEMATICS BY FERNANDO ORINES 

2014

In this chapter, we will learn how to solve algebra word problems that involve motion. Motion (Distance) Formula Motion problems are based on the formula

d = rt  where d = distance, r = rate and t = 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: Let x = time walked. r 

t 

d 

John

3

 x 

3 x 

Philip

4

 x 

4 x 

3x + 4x = 14 7x = 14 x=2 They will meet in 2 hours.

37

NEXT CENTURY MATHEMATICS BY FERNANDO ORINES 

2014

Example2: In still water, still water, Peter’s boat goes 4 times as fast as the current 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: Let x = rate of the current. r 

t 

d 

down river

4 x  + x   + x 

15 / 5 x 

15

up river

4 x  - x   - x 

15 / 3 x 

15

 ANSWER: The rate of the current is 2 mph. mph.

38

NEXT CENTURY MATHEMATICS BY FERNANDO ORINES 

2014

39

NEXT CENTURY MATHEMATICS BY FERNANDO ORINES 

2014

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: Problems: Value Of A Particular Term The 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 is y. Example1: The first term in a sequence of number is 2. Each even-numbered 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?

40

NEXT CENTURY MATHEMATICS BY FERNANDO ORINES 

2014

Solution: Step 1: Write down the terms until you notice a repetition 2, 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 1 Step 3: The 45th term is the same as the 1st term, which is 2  Answer: 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 by m and then adding n. What is the value of n? Solution: Method 1: The fastest way to solve this would be if you notice that the pattern: 6 × 2 + 1 = 13 13 × 2 + 1 =27  Answer: The value of n is 1.

41

NEXT CENTURY MATHEMATICS BY FERNANDO ORINES 

2014

Method 2: If you were not able to see the pattern then you can come with two equations and then solve for n. 6m + n =13 13m + n = 27

(equation 1) (equation 2)

Use substitution method Isolate n in equation 1 n = 13 – 6m Substitute into equation 2 13m + 13 – 6m = 27 7m = 14 m=2 Substitute m = 2 into equation 1 6(2) + n = 13 n=1  Answer:

n=1

42

NEXT CENTURY MATHEMATICS BY FERNANDO ORINES 

2014

43

NEXT CENTURY MATHEMATICS BY FERNANDO ORINES 

2014

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 Problems and Inversely Proportional Problems.

Directly Proportional Problems The question usually will not tell you that the items are directly 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: If x then y. If x is changed to a then what will be the value of y?

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 out which values to divide and which values to multiply. The following method is helpful: Change the word problem into the form: If x then y. If x is changed to a then what will be the value of y?  which can then be represented as:

44

NEXT CENTURY MATHEMATICS BY FERNANDO ORINES 

2014

For example,  You can think of the sentence: If two pencils cost $1.50, how many pencils can you buy with $9.00? as If $1.50 then two pencils. If $9.00 then 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 seconds Example 2: If of 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:

45

NEXT CENTURY MATHEMATICS BY FERNANDO ORINES 

If

then 2. If 1 then how many? (Whole tank is

2014

)

Step 2: Write the proportional relationship:

 Answer: It took 3.5 minutes Example 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 traveled

miles.

Inversely Proportional Problems Inversely Proportional questions are similar to directly proportional problems, but the difference is that when x increase y will decrease and vice versa - which is the inverse proportion relationship. The most common example of inverse proportion problems would be “the more men on a  job the less time taken for the job to complete”  Again, the technique is to change the proportion problems into the form:

46

NEXT CENTURY MATHEMATICS BY FERNANDO ORINES 

2014

If x then y. If x is changed to a then what will be the value of y? 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 take

hours.

47

NEXT CENTURY MATHEMATICS BY FERNANDO ORINES 

2014

Ratio Word Problems Ratio problems are word problems that use ratios to relate the different items in the question. The main things to be aware about for ratio problems area:   

Change the quantities to the same unit if necessary. Write the items in the ratio as a fraction. Make sure that you have the same items in the numerator and denominator.

Ratio problems: Two-term Ratios Example 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: Step 1: Assign variables : Let x = red sweets Write the items in the ratio as a fraction.

Step 2: Solve the equation Cross Multiply

3 × 120 = 4 × x 360 = 4x

48

NEXT CENTURY MATHEMATICS BY FERNANDO ORINES 

2014

Isolate variable x

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 many more blue marbles than Jane? Solution: Step 1: Sentence: Jane has 20 marbles, all of them either red or blue. Assign variables: Let x = blue marbles for Jane 20 – x = red marbles for Jane We get the ratio from John John has 30 marbles, 18 of which are red and 12 of which are blue.

49

NEXT CENTURY MATHEMATICS BY FERNANDO ORINES 

2014

Direct Variation There are many situations in our daily lives that involve direct variation. For example, a worker may be paid according to the number of hours he worked. The two quantities x (the number of hours worked) and y (the amount paid) are related in such a way that when x changes, y changes proportionately such that the ratio remains a constant. We say that y varies directly with x. Let us represent the constant by k, i.e.

or y = kx where k ≠ 0 If y varies directly as x, this relation is written as y as y varies as x. The sign “ the sign of variation.

x and read

” is read “varies as” and is called

Example3: If y varies directly as x and given y = 9 when x = 5, find:   

the equation connecting x and y the value of y when x = 15 the value of x when y = 6 Solution: a) y

x i.e. y = kx where k is a constant

Substitute x = 5 and y = 9 into the equation:

50

NEXT CENTURY MATHEMATICS BY FERNANDO ORINES 

 y =

2014

x

 b) Substitute x = 15 into the equation

 y =

= 27

c) Substitute y = 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, where k is a constant. Substitute C = 35 and D = 60 into the equation 35 = 60k ⇒k = Therefore, C =

D

Substitute D = 95 into the equation: C=  Answer :

55.42

The cost for 95 km is $55.42 51

NEXT CENTURY MATHEMATICS BY FERNANDO ORINES 

2014

Joint and Combined Variation Word Problems

 Joint variation  is

a variation where a quantity varies directly as 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 and w is the width.

, where A is the area, l is the length

Combined variation is

a variation where a quantity depends on two (or more) other quantities, and varies directly with some of them and varies inversely with 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 length l or the height h, the narrower is the width w. 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. Express s as a joint variation in terms of d and t.

52

NEXT CENTURY MATHEMATICS BY FERNANDO ORINES 

2014

Solution:

 Answer: In other words, the longer the distance or the shorter the time, the faster is the speed.

Inverse Variation Word Problems In general, when two variables x and y are such that xy = k where k is a non-zero constant, we say that y varies inversely with x. In notation, inverse variation is written as

Example7: Suppose that y varies inversely as x and that y = 8 when x = 3. a) Form an equation connecting x and y.  b) Calculate the value of y when x = 10. Solution:

i.e. xy = k where k is a non-zero constant a) Substitute x = 3 and y = 8 into the equation to obtain k 3 × 8 = k ⇒ k = 24 The equation is xy = 24  b) When x = 10, 10 × y = 24 ⇒ y =

53

NEXT CENTURY MATHEMATICS BY FERNANDO ORINES 

2014

Example: Suppose that y varies inversely as x 2 and that y = 10 when x =

.

a) Find the equation connecting x and y .  b) Find the value of y when x = 3. Solution: i.e. yx2 = k a) Substitute x =

and y = 10 into the equation to obtain k

The equation is yx2 =  b) When x = 3,

54

NEXT CENTURY MATHEMATICS BY FERNANDO ORINES 

2014

55

NEXT CENTURY MATHEMATICS BY FERNANDO ORINES 

2014

"Work" Problems: Two Persons The formula for “Work” Problems that involve two persons is

This formula can be extended for more 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: Assign variables: Let x = time to mow lawn together Step 2: Use the formula:

Step 3: Solve the equation The LCM of 40 and 60 is 120 Multiply both sides with 120

56

NEXT CENTURY MATHEMATICS BY FERNANDO ORINES 

2014

 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: Assign variables: Let x = time taken by Peter Step 2: Use the formula:

Step 3: Solve the equation Multiply both sides with 30 x 

57

NEXT CENTURY MATHEMATICS BY FERNANDO ORINES 

 Answer: The time taken for Peter to paint the fence alone is

2014

hours.

“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: Assign variables: Let x = time taken to fill up the tank Step 2: Use the formula: Since pipe C drains the water it is subtracted.

58

NEXT CENTURY MATHEMATICS BY FERNANDO ORINES 

2014

Step 3: Solve the equation The LCM of 3, 4 and 5 is 60 Multiply both sides with 60

 Answer: The time taken to fill the tank is

hours.

59