PCSR REVIEW SERIES WEEK 2: Addition and Subtraction of Fractions

PCSR REVIEW SERIES for October 2016
Week 2: Addition and Subtraction of Fractions

In Week 1 of our PCSR Review Series , we have learned how to get the Least Common Multiple (LCM) and Greatest Common Divisor (GCD) of two or more numbers. The LCM is used to add and subtract dissimilar fractions. To add or subtract fractions, you have to get the LCM of their denominator and convert them to similar fractions.

The GCD is used for converting fractions to lowest terms. In converting fractions to lowest terms, you have to get the GCD of the numerator and denominator of a fraction, and divide both the numerator and the denominator by the GCD.

There are two more skills you need to know: to convert improper fractions to mixed fractions. Most of the time, final answers are required to be in mixed fractions. Links to tutorials on how to do it are included below.

Articles

  1. A Gentle Introduction to Fractions
  2. How to Convert Mixed Fractions to Improper Fractions
  3. How to Convert Improper Fractions to Mixed Fractions
  4. How to Add Fractions
  5. How to Subtract Fractions

Videos

More… 

If you have time and your internet is fast, I suggest that you watch the entire fraction series (21 videos):

Week 1 Review Answers and Solutions

Below are the solutions and answers to the Practice Exercises and Problems for the Week 1 Review on LCM and GCD.

Practice Exercises
I. Find the GCD of each of the following.
a.) 6, 10
Answer: 2

b.) 18, 42
Answer: 6

c.) 12, 48, 60
Answer: 12

d.) 56, 72
Answer: 8

e.) 225, 75
Answer: 75

II. Find the LCM of each of the following.
a.) 3, 4
Answer: 12

b.) 2, 5
Answer: 10

c.) 3, 6, 8
Answer: 24
d.) 3, 4, 5
Answer: 60

e.) 6, 12, 15
Answer: 60

III. Practice Problems Solutions and Answers

1.) 24 (GCD of 3, 4, and 8)

2.) 15 (GCD of 3 and 15)

3.) Solution: GCD of 3, 7, and 21 is 42. They will be seen in the gym on the same day in 42 days. Since June has 30 days, we need an additional 12 days to complete the 42 days. Therefore, they will be seen on the same day on July 12 (of the same year of course).

Answer: July 12

4.) Solution: The LCM of the two sequences is 12 and since we are looking for the tenth common number, we multiply 12 by 10. This gives us 120.
Answer: 120

5.) (a) Solution: LCM of 24 and 30 is 6. That is 6 groups.
(b) Solution: From (a) we can form 6 groups. There are 30 + 24 = 54 students. So in each group, there are 54/6 = 9 members. Since there are 6 groups, we divide each Grade level by 6. That is, 24/6 = 6 Grade 11 and 30/6 = 5 Grade 12 students in each group.
Answer: 4 Grade 11 students and 5 Grade 12 students.

6.) Reduce 42/56 to lowest terms.
Solution: GCD of 42 and 56 is 14. 42 divided by 14 is 3 and 56 divided by 14 is 4. Therefore, the lowest terms is 3/4.
Answer: 3/4

7.) Answer: 24 (LCM of 6 and 8)

8.) Answer: 6 (GCD of 18 and 12)

9.) Solution: The GCD of 21, 35, and 84 is 7. So, the cube has a side length of 7 cm.
Answer: 7 by 7 by 7

10.) Answer: 30 (GCD of 3, 5, and 6).

Week 1 Review: Practice Exercises and Problems

Below are the practice exercises for the Week 1 Review on LCM and GCD. You can read the answers and solutions to these exercises and problems.

Practice Exercises

I. Find the GCD of each of the following.
a.) 6, 10
b.) 18, 42
c.) 12, 48, 60
d.) 56, 72
e.) 225, 75

II. Find the LCM of each of the following.

a.) 3, 4
b.) 2, 5
c.) 3, 6, 8
d.) 3, 4, 5
e.) 6, 12, 15

III. Practice Problems

1.) The fractions 1/3, 1/4 and 1/8 are added. To add them, you need to convert them to similar fractions. What will be the least possible denominator of these similar fractions?

2.) In a disco, the red lights blink every 3 seconds and the blues light blink every 5 seconds. The two lights blink every ___ seconds.

3.) Anna, Karen, and Nina go to the same gym. Anna goes every 2 days, Karen goes every 3 days, and Nina goes every 7 days. On June 1, all of them were seen on the gym. What is the soonest date that they will be seen on the gym on same day?

4.) Consider the following sequences:
Sequence 1: 4, 8, 12, 16, 20, 24, …
Sequence 2: 6, 12, 18, 24, …
Notice that 12 and 24 are the first and second common numbers, respectively, to both sequences.
What is the 10th common number?

5.) In Senior High School athletic meet, there are 24 students from Grade 11 and 30 students from Grade 12. Groups are formed such that in each group should have an equal number students from each Grade level.
(a) If everyone is included, how many of such groups can be formed?
(b) When the largest number of groups is formed, how many students from each Grade level are there in one group?

6.) Reduce 42/56 to lowest terms.

7.) Boxes of chocolates have a height of 6 cm each. Boxes of cookies have a height of 8 cm each. A combination of these boxes will be packed in a large box. What should be the minimum height of the large box so that the smaller boxes would exactly fit?

8.) Square cardboard of the same size are to completely cover a rectangle with dimensions 18 cm by 12 What should be the dimensions of the largest possible squares so that there is no wasted cardboard?

9.) Cubes of the same size are to be placed inside a rectangular box. What is the size of the largest possible cubes that can fit exactly inside the box if its dimensions are 21 cm by 35 cm by 84 cm and no space is to be wasted?

10.) In a flour shop, a cake flour comes in 5-kg packages, a pastry flour comes in 3-kg packages and the bread flour comes in 6 kg packages. If Gina bought the same number of kilograms of these flour, what is the minimum number of kilograms of each she must have bought?

Solving Ratio and Proportion Problems Part 2

In the previous post, we have learned the meaning and notation of ratio. We can write the ratio of 8 girls and 12 boys as 8:12 or as 8/12. However, if we represent this as fraction, we can also reduce the fraction to its lowest terms which is equal to 2/3. Converting to lowest term is dividing the numerator and denominator by the largest possible integer known as the greatest common factor or greatest common divisor. In the example above, the greatest common divisor of 8 and 12 is 4, and 8 divided by 4 is 2, and 12 divided by 4 is 3, so, 8:12 can also be represented as 2:3.

The Meaning of Direct Proportion

Consider the following problem.

A car is traveling at an average speed of 60 kilometers per hour. What is the total distance it traveled after 5 hours?

Solution

We can solve the problem above by simply multiplying 5 hours by 60 kilometers per hour giving us 300 kilometers. We can also answer the problem by simply constructing the table below.

proportion

Notice from the table that if the number of hours is multiplied by 2, then the distance is also multiplied by 2. For example, from 1 hour to 2 hours, the number of hours is multiplied by 2, and the distance is also multiplied by 2, that is 60 × 2 = 120 hours. From 2 hours to 4 hours, the number of hours is also multiplied by 2 and the distance is also multiplied by 2, that is 120 × 2 = 240 hours. If the number of hours is multiplied by 3, the distance is also multiplied by 3. From 1 hour to 3 hours, the number of hours is multiplied by 3 and the distance is also multiplied by 3, that is 60 × 3 = 180 hours.

Suppose we have two quantities and if we multiply one quantity by a number, then the other quantity is also multiplied the same number, then we say that the two quantities are directly proportional. In the example above, time and distance are the two quantities that are directly proportional.

Representing Direct Proportions

We can represent the problem above in ratio. The first ratio is 60 kilometers and 1 hour. The second ratio is 5 hours and an unknown number of kilometers. If we let the unknown number of hours be n, then the ratios are

1 hr :60 km and 5 hrs :n km

Notice that the number of hours is multiplied by 5 (1 hr to 5 hrs), so the distance should also be multiplied by 5. That is, 60 × 5 = 300.

Now that we found the answer to the problem above, let us represent them in ratios as shown.

1:60 and 5:300

Observe from the representation that the product of the outer terms (1 and 300) is equal to the product of the inner terms (60 and 5). The product are both 300. This property is always true in directly proportional quantities: the product of the outside terms (extremes) is equal to the product of the inside terms (means). In the original representation, we had

1:60 = 5:n

Using the relationship between the means and extremes, we can solve for n algebraically. That is,

1 × n = 60 × 5. So, n = 300.

This can also be represented in fraction as 1/60 = 5/n. Cross multiplying, we have n = (60)(5) = 300.

Summary

In a directly proportional relationship, if the ratios are a:b and c:d, then

a: b = c: d

and a × d = b × c.

Practice Problem

Three cubes of sugar is needed to make 1 cups of coffee. How many cubes of sugar is needed to make 20 cups of coffee?

Solution

Let x = number of cubes of sugar needed to make 20 cups of coffee.

3:1 = x:20

The product of the extremes is equal to the product of the means, so solving algebraically, we have

1(x) = (3)(20)
x = 60.

Therefore, we need 60 cubes of sugar for 20 cups of coffee.

In the next post, we are going to have some practice problems on how to solve direct proportion problems.

Solving Ratio and Proportion Problems Part 1

One of the key concepts tested in the Civil Service Exam is ratio and proportion. In this series, we are going to discuss how to solve problems involving ratio and proportion. We first begin below by explaining the meaning and concept of ratio and how to represent it.

Suppose we are cooking, and for every 4 teaspoons of vinegar, we put 3 teaspoons of soy sauce, then we can say the ratio of the volume of vinegar to the volume of soy sauce is “four is to three” and represent it as 4:3. We can also use the fraction 4/3 to represent the ratio above. Now, we discuss more examples about ratio.

Example 1

In a class, there are 24 girls and 18 boys. What is the ratio of (1) the number of girls to the number of boys and (2) the number of boys to the number of girls?

Answer

The number of girls is 24 and the number of boys is 18, so the ratio of the number of girls to the number of boys is 24:18 or 24/18. In contrast, the ratio of the number of boys to the number of girls is 18:24 or 18/24.

Example 2

In a box of colored balls, there are 5 red balls and 8 blue balls. What is the ratio of the number of blue balls to the total number of balls?

Answer

The number of blue balls is 8 and the total number of balls is 5 + 8 = 13. Therefore the ratio of the number of blue balls to the total number of balls is 8:13 or 8/13.

Example 3

Gemma put 2 teaspoons of sugar for every cup of coffee. Represent the ratio of the number of teaspoons of sugar if there are 6 cups of coffee.

Answer

For every cup of coffee, we need 2 teaspoons. Therefore, for 6 cups of coffee, we need 6 times 2 = 12 teaspoons. So, the ratio of the number of teaspoons and 6 cups of coffee is 12:6.

In the three examples above, we have learned how to represent ratio. The ratio A: B means how many times of B is A. For example, the ratio 4:3 means A is 4/3 times of B.

In the next post, we are going to discuss about proportion or equal ratio.

Practice Exercises on Subtraction of Integers

To answer the exercises below, it is assumed that you have already finished reading addition of integers and subtraction of integers.

In subtraction of integers, we have learned two rules:

(1) a – b = a + (-b)
(2) a – (-b) = a + b

We will use these rules in answering the exersises below.

Exercises

1. 2 – 5
2. 18 – ( – 2)
3. 16 – 7
4. -17 – 3
5. -9 – (-3)
6. 0 – (-11)
7. -18 – (-25)
8. -10 – 9
9. 12 – (-9)
10. -6 – 3

Solutions/Answers

1. 2 – 5

Solution 1: 5 is greater than 2. If you subtract two numbers, if the subtrahend is larger than the minuend, the answer will be negative. So, the answer is -3.

Solution 2: From rule 1, a – b = a + (-b), so 2 + 5 = 2 + (-5) = -3
Answer:

2. 18 – ( – 2)

Solution: From rule 2, a – (-b) = a + b, so 18 + 2 = 20.

Answer: 20

3. 16 – 7

Answer: 9

4. -17 – 3

Solution: From rule 1, -17 – 3 = -17 + (- 3) = -20. Recall that in adding two negative numbers, we just add the numbers and then the answer will be negative.

Answer: -20

5. -9 – (-3)

Solution: From rule 2, -9 – (-3) = -9 + 3 = -6.

6. 0 – (-11)

Solution: From rule 2, 0 – (-11) = 0 + 11 = 11.

7. -18 – (-25)

Solution: From rule 2, a –(-b) = a + b. So, -18 + 25 = 7.

8. -10 – 9

Solution: From rule 1, a – b = a + (-b), so -10 + (- 9) = – 19.

9. 12 – (-9)

Solution: From rule 2, a –(-b) = a + b, so 12 + 9 = 21.

10.- 6 – 3

Solution: From rule 1, a – b = a + (-b) = -6 + -3 = -9.

Four Effective Techniques in Adding Integers

We have already discussed addition of integers. In this post, I am going to discuss four different techniques in adding integers.

Adding Numbers with the Same Sign

In adding integers with the same sign, we just add them and then copy the sign. For example, in adding 2 + 8, 2 and 8 are positive integers, so we just add them, and the answer will be positive. So, 2 + 8 = 10. On the other hand, if both integers are negative we also do the same: add them, then copy the sign. For example, -9 + -3 = -12 since both of them are negative integers.

Watch: 24 Taglish Math Videos about Integers in Youtube

The techniques below are for adding integers with different signs. These strategies are important because you can visualize addition even without memorizing the rules.

Techniques in Adding Integers with Different Signs

Technique 1: Using Positive and Negative Chips

You can imagine integers as positive and negative chips. Since +1 + -1 = 0, a pair of positive and negative chips will give a sum of 0. So, 3 means 3 positive chips and -4 means 4 negative chips. Since each pair of positive and negative chip cancels out each other (their sum is 0), then the remaining chips after the pairing will be the answer. So, 3 + (-4) as represented below is -1 since only one negative chip remains.

teachniques in adding integers 1

Another example is -4 + 6 = 2.

adding integers using chips

Watch: How to Add Integers Using Chips in Youtube

Technique 2: Decomposing the Numbers

This strategy uses the fact that a + (-a) = 0. Using this strategy, we can split one of the addends. For example, in 8 + (-5), we split 8 to 3 + 5 so that 5 and -5 will become 0. So,

8 + (-5) = 3 + 5 + (-5)

=3 + (5 + -5)= 3 + 0 = 3.

In -11 + 3, we can split -11 to -8 + -3. So,

-11 + 3 = -8 + (-3 + 3) = -8 + 0 = -8.

Technique 3: Using the Number Line

Integers can also be represented as movement on the number line. A positive integer is a movement to the right of 0 and a negative integer is a movement to the left. Positive 3 and -2 can be represented as shown below.

adding integers using number line 1

So, 3 + (-5) can be represented as a movement of 3 units to the right of 0, then a movement of 5 units to the left. As we can see in the next diagarm, it the movement stopped at -2. So, 3 + (-5) = -2.

adding integers using number line 2

Also, -2 + 3 can be represented as a movement 2 units to the left of 0 and then a movement of 3 units to the right. The movement stopped at 1, so -2 + 3 = 1.

adding integers using number line 3

Watch: How to Add Integers Using the Number Line in Youtube

Technique 4: Grouping Numbers with Similar Signs

In adding more than 2 addends with different signs, group the numbers with the same signs. For example, 9 + (-2) + 4 + (-1), we can do the following:

(1) add the positive integers first: 9 + 4 = 13
(2) add the negative integers (-2 + -1 = -3)
(3): finally, add the two sums: 13 + (-3) = 10.

Try using the techniques above by answering the following and share to us which technique do you like most.

1.) 5 + (-10)
2.) -3 + 8
3.) 12 + – 10 + 7
4.) -10 + -3 + 4

Solving Word Problems by Working Backwards: Summary

In the previous posts, I have shown to you a series on how to solve word problems by working backward.  Although I recommend that you learn this method, it is also very important to learn algebraic methods and others since some problems are difficult to solve by working backward. Below are the posts in this series.

How to Solve Word Problems by Working Backward Part 1 discusses solving number problems by working backward.  Two examples are given.

How to Solve Word Problems by Working Backward Part 2 discusses solving age problems by working backward. Two sample problems are solved in this post.

How to Solve Word Problems by Working Backward Part 3 discusses more complicated number problems. Two sample problems are discussed in this post.

If you are interested to solve more word problems, please visit the Word Problems page. It contains word problems about number, age, motion, work, discount and many more.

 

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