Below are the solutions and answers to the Practice Exercises and Problems for the Week 2 Review on Addition and Subtraction of Fractions.

Practice Exercises 1 Answers
1.) 3/5
2.) 2/5
3.) 4/7
4.) 2
5) 3/5

Practice Exercises 2
Convert the following improper fractions to mixed form.
1.) 3 3/5
2.) 1 5/7
3.) 4 1/2
4.) 12 3/4
5) 10 1/12

Practice Problems Solutions and Answers

1. The LCM of the denominators 2 and 8 is 8. We convert ½ to a fraction whose denominator is 8 in order for the two fractions to be similar. To do this, we divided 8 by 2 and the multiply by 1. The result will be the numerator of the fraction. That is

$\dfrac{8}{2} \times \dfrac{ 1}{8} = \dfrac{4}{8}$.

So, $\dfrac{4}{8} + \dfrac{9}{8} = \dfrac{13}{8}$.

Converting the answer to mixed form, we have $1\dfrac{5}{8}$

2. The LCM of 5 and 4 is 20. After getting the LCM, we convert 3/5 and 1/4 to their equivalent fractions whose denominator is 20.

The equivalent fraction for 3/5 is 12/20.
The equivalent fraction of 1/4 is 5/20.

12/20 + 5/20 = 17/20

3. The LCM of 2, 3 and 4 is 12. After getting the LCM, we convert 1/2, 1/3, and 1/4 to their respective equivalent fractions whose denominator is 12.

The equivalent fraction for 1/2 is 6/12.
The equivalent fraction of 1/3 is 4/12.
The equivalent fraction of 1/4 is 3/12.

6/12 + 4/12 + 3/12 = 13/12

Converting 13/12 to mixed fractions, we get 1 1/12.

4. The LCM of 12, 2 and 3 is 12. After getting the LCM, we convert 5/12, 1/2, and 2/3 to their respective equivalent fractions whose denominator is 12.

The equivalent fraction for 5/12 is still 5/12.
The equivalent fraction of 1/2 is 6/12.
The equivalent fraction of 2/3 is 8/12.

5/12 + 6/12 + 8/12 = 19/12

Converting 19/12 to mixed fractions, we get 1 7/12.

5. The LCM of 4 and 6 is 12. Therefore, we convert 3/4 and 1/6 to their respective equivalent fractions whose denominator is 12.

The equivalent fraction for 3/4 is still 9/12.
The equivalent fraction of 1/6 is 2/12.

9/12 – 2/12 = 7/12

6. The LCM of 15 and 30 is 30. Therefore, we convert 13/15 and 7/30 to their respective equivalent fractions whose denominator is 30.

The equivalent fraction for 13/15 is still 26/30.
The equivalent fraction of 15/30 is 15/30.

26/30 – 7/30 = 19/30

7. In this problem, we can just add the fractions first. We add ¾ and ½ which is equal to 1 ¼ kg. We now add the 4 and 1 which is 5 ¼ kg.

8.  We need to add 1/8 and 1/2.
The LCM of 8 and 2 is 8. Therefore, we convert 1/2 to a fraction whose denominator is 8.

The equivalent fraction of 1/2 is 4/8.

1/8 + 4/8 = 5/8

9. We need to add 1 1/2 and 3/4. We just add the fractions and then add the whole numbers later. We first add ½ and ¾.

The LCM of 2 and 4 is 4. Therefore, we convert 1/2 to a fraction whose denominator is 4.

The equivalent fraction of 1/2 is 3/4.

2/4 + 3/4 = 5/4

Converting 5/4 to mixed fractions, we have 1 ¼.

We add 1 ¼ to 1 from the original given. The answer 2 ¼.

10. We need to add ¼, 1/5, and 3/10.

The LCM of 4, 5 and 10 is 20. Therefore, we convert 1/4, 1/5, and 3/10 to their respective equivalent fractions whose denominator is 20.

The equivalent fraction for 1/4 is still 5/20.
The equivalent fraction of 1/5 is 4/20.
The equivalent fraction of 3/10 is 6/20.

5/20 + 4/20 + 6/20 = 15/20

Changing 15/20 to lowest terms, we have ¾.

## Week 2 Review: Practice Exercises and Problems

Below are the practice exercises for the Week 2 Review on Addition and Subtraction of Fractions. I will post the solutions and answers soon.

Practice Exercises 1

Add/Subtract the following fractions.

1.) 1/5 + 2/5
2.) 1/10 + 3/10
3.) 6/7 – 3/7
4.) 5/6 + 3/6 + 4/6
5) 12/5 – 9/5

Practice Exercises 2
Convert the following improper fractions to mixed form.
1.) 18/5
2.) 12/7
3.) 9/2
4.) 51/4
5) 121/12

Practice Problems

1. 1/2 + 9/8

2. 3/5 + 1/4

3. 1/2 + 1/3 + 1/4

4. 5/12 + 1/2 + 2/3

5. 3/4 – 1/6

6. 13/15 – 7/30

7. A 4 1/2 kg of rice placed inside a container weighing 3/4 kg. What is the total weight of the rice and the container?

8. Alfie bought a cake. He gave 1/8 of the cake to his wife and 1/2 of the cake to his children. What part of the cake was left?
Solution

9. Revie spend 1 1/2 hours studying here homework, 3/4 hours watching the television before sleeping. How much time did she spend studying and watching TV?

10. Ria drinks 1/4 L of milk everyday. Her son drinks 1/5 L of milk everyday. Her husband drinks 3/10 L of milk everyday. How much milk do the three of them drink everyday?

## 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

Videos

More…

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

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

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 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).

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.

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.

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?

## PCSR REVIEW SERIES WEEK 1: LCM and GCD

I have decided to outline a 16-week Philippine Civil Service Review (PCSR) series in mathematics. This way, you will be able to study systematically. In this series, I will post links every week, give exercises, and discuss the solutions of the exercises. Included in the links are articles that I have written and Taglish Youtube videos that I have created.

Let’s start our review in mathematics by learning about GCD and LCM. These two concepts are very important since you will use them in solving problems involving fractions. Fractions appear everywhere in the Civil Service exams. Be sure that you master these concepts before we proceed to our Week 2 review.

LEAST COMMON MULTIPLE

Least Common Multiple (LCM) is used when adding, subtracting, comparing and ordering, fractions. Here are the links with examples.

Articles

GREATEST COMMON DIVISOR
Greatest Common Divisor (GCD) is used in factoring and reducing fractions to lowest terms. GCD is also called Greatest Common Factor (GCF)

Articles

Enjoy!

If you have questions about the articles and the videos just post them below.

## 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.

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?

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?

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.