## Addition of Fractions Exercises – Set 2

1.) $\dfrac{1}{7} + \dfrac{2}{7}$

2.) $\dfrac{1}{12} + \dfrac{5}{12}$

3.) $\dfrac{7}{15} + \dfrac{2}{15} + \dfrac{6}{15}$

4.) $\dfrac{1}{3} + \dfrac{1}{4}$

5.) $\dfrac{2}{9} + \dfrac{1}{3}$

6.) $\dfrac{3}{8} + \dfrac{1}{4}$

7.) $8 \dfrac{2}{3} + \dfrac{3}{2}$

8.) $\dfrac{1}{2} + \dfrac{2}{3} + \dfrac{3}{4}$

9.) $7 \dfrac{2}{7} + 6 \dfrac{3}{14}$

10.) $3 \dfrac{1}{4} + 2 \dfrac{1}{8}$

1.) $\dfrac{3}{7}$

2.) $\dfrac{6}{12} = \dfrac{1}{2}$

3.) $\dfrac{15}{15} = 1$

4.) $\dfrac{7}{12}$

Solution:

## Conversion of Units Part 1: Centimeters, Meters, Kilometers

Converting from one unit to another is one of the skills that you should learn before taking the Civil Service Examinations. It sometimes makes calculations easier and therefore saves time.

In this series, I will discuss the different methods that can be used in converting from one unit to another. I will illustrate several examples such as conversion among centimeters (cm), meters (m) and kilometers (km). Take note that the methods used here can also be used in conversion of other units of measure.

It is important that you memorize some standard conversions such as the following:

1 m = 100 cm

1 km = 1000 m

1 kg = 1000 g

1 ft = 30 cm

1 ft = 12 inches

1 mile = 1.6 km  Continue Reading

## Multiplication and Division of Fractions Exercises – Set 1

Week 4

1. 3/4 x 1/2

2. 2/7 x 7/10

3. 4 1/5 x 2/3

4. 8 x 3/4

5. 1 2/3 x 2 3/4 x 11/12

6. 2/5 ÷ 4/5

7. 3/4 ÷ 8

8. 9 ÷ 3/4

9. 5 1/6 ÷ 3/31

10. 2 5/6 ÷ 4 3/5

1. 3/8 3/4 x 1/2 = 3/8

2. 1/5

Solution

2/7 x 7/10 = 14/70 = 1/5

3. 2 4/5

Solution

4 1/5 x 2/3
= 21/5 x 2/3 = 42/15
= 2 12/15 = 2 4/5

4. 6

Solution

8 x 3/4 = 8/1 x 3/4
= 24/4 = 6

5. 4 29/144

Solution
1 2/3 x 2 3/4 x 11/12
= 5/3 x 11/4 x 11/12
= 605/144
= 4 29/144

6. 1/2

Solution

2/5 x 5/4 = 10/20 = 1/2

7. 3/32

Solution
3/4 ÷ 8
= 3/4 x 1/8
= 3/32

8. 12

Solution

9 ÷ 3/4 = 9/1 x 4/3 = 36/3 = 12

9. 5 1/6 ÷ 31/3
= 31/6 x 3/31
= 961/18 = 53 7/18

10. 8/35

2 5/6 ÷ 4 3/5 = 17/6 ÷ 23/5
= 17/6 x 5/23 = 85/138

You might also like: Multiplication and Division of Fractions Exercises – Set 1

## Subtraction of Fractions Exercises – Set 1

Find the difference of the following:

1.) $\dfrac{8}{11} - \dfrac{3}{11}$

2.) $\dfrac{7}{16} - \dfrac{3}{16}$

3.) $\dfrac{4}{5} - \dfrac{3}{4}$

4.) $7 - \dfrac{4}{9}$

5.) $8 \dfrac{9}{10} - \dfrac{3}{10}$

6.) $6 \dfrac{2}{7} - \dfrac{5}{7}$

7.) $4 \dfrac{1}{3} - \dfrac{1}{6}$

8.) $6 \dfrac{3}{5} - 2 \dfrac{3}{4}$

9.) $11 \dfrac{2}{3} - 7$

10.) $9 \dfrac{2}{5} - \dfrac{11}{9}$

1.) $\dfrac{5}{11}$

2.) $\dfrac{1}{4}$

Solution
$\dfrac{7}{16} - \dfrac{3}{6} = \dfrac{4}{16} = \dfrac{1}{4}$

3.) $\dfrac{1}{20}$

Solution
LCD: 20
$\dfrac{4}{5} - \dfrac{3}{4} = \dfrac{16}{20} - \dfrac{15}{20} = \dfrac{1}{20}$

4.) $6 \dfrac{5}{9}$

Solution

We can decompose 7 into 6 and 9/9.

$6 \dfrac{9}{9} - \dfrac{4}{9} = 6 \dfrac{5}{9}$

5.) $8 \dfrac{3}{5}$

Solution
$8 \dfrac{9}{10} - \dfrac{3}{10} = 8 \dfrac{6}{10} = 8 \dfrac{3}{5}$

6.) $5 \dfrac{4}{7}$

Solution
$5 \dfrac{9}{7} - \dfrac{5}{7} = 5 \dfrac{4}{7}$

7.) $4 \dfrac{1}{6}$

Solution
LCD: 6
$4 \dfrac{1}{3} - \dfrac{1}{6} = 4 \dfrac{2}{6} - \dfrac{1}{6} = 4 \dfrac{1}{6}$

8.) $3 \dfrac{17}{20}$

Solution
LCD: 20
$5 \dfrac{8}{5} - 2 \dfrac{3}{4} = 5 \dfrac{32}{20} - 2 \dfrac{15}{20} = 3 \dfrac{17}{20}$

9.) $4 \dfrac{2}{3}$

Solution
$11 \dfrac{2}{3} - 7 = 4 \dfrac{2}{3}$

10.) $8 \dfrac{8}{45}$

Solution
$9 \dfrac{2}{5} - \dfrac{11}{9} = 8 \dfrac{7}{5} - \dfrac{11}{9}$

LCD: 45
$8 \dfrac{63}{45} - \dfrac{55}{45} = 8 \dfrac{8}{45}$

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## Addition of Fractions Exercises – Set 1

Find the sum of the following.

1. 1/8 + 2/8 + 3/8
2. 2/5 + 1/4
3. 5/8 + 7/12
4. 1/15 + 3/5 + 1/3
5. 4 + 3/5
6. 2 1/3 + 1/2
7. 3 3/4 + 7/10
8. 6 1/5 + 2 7/15 + 1/3
9. 3 + 7 1/8 + 4/5
10. 9 1/2 + 11/3

1. 3/4

2. 13/20
LCD: 20
8/20 + 5/20 = 13/20

3. 1 1/6
LCD: 24
15/24 + 14/24 = 29/24 = 1 5/24

4. 1
LCD: 15
1/15 + 9/15 + 5/15 = 15/15 = 1

5. 4 3/5
4 3/5

6. 2 5/6
LCD: 6
2 2/6 + 3/6 = 2 5/6

7. 4 9/20
LCD: 20
3 15/20 + 14/20
= 3 29/20 = 4 9/20

8. 9
LCD: 15
6 3/15 + 2 7/15 + 5/15
= 8 15/15 = 9

9. 10 37/40
LCD: 40
3 + 7 5/40 + 32/40
= 10 37/40

10. 13 1/6
LCD: 6
9 3/6 + 22/6
= 9 25/6 or 13 1/6

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## LCM and GCD Exercises Set 1

After learning about LCM and GCD in our PCSR Review Guide 1, let’s practice what we have learned by answering the exercises below.

1.) What is the LCM of 6 and 8?

2.) What is the LCM of 3, 4, and 9?

3.) Consider the following sequences:
Sequence 1: 4, 8, 12, 16, 20, 24, 28, …
Sequence 2: 5, 10, 15, 20, 25, 30, …
As we can see, 20 is common to both sequences. What is the 10th common number?

4.) Gina goes to BODY SLIM gym every 3 days, while Sam goes to the same gym every 4 days. If they were in the same gym on a February 12, what’s the nearest date that they will be both in the gym?

5.) In a decoration LED light, the green lights blink every 400 milliseconds, the blue light every 500 milliseconds, and the red light every 750 milliseconds. If the three lights all blinked at the same time, in how many milliseconds will they again blink at the same time?

6.) What is the LCD of 1/3, 1/4, and 1/6?

7.) What is the GCD of 28 and 35?

8.) What is the GCD of 39, 65, and 91?

9.) A rectangular paper with dimensions 12 inches by 18 inches is to be cut into largest possible squares of equal sizes. If no paper is to be wasted, what should be the dimensions of the squares?

Q10.) A rectangular prism piece of wood with dimensions 16 inches by 20 inches by 32 is to be cut into cubes of the same size.

A. If we want to minimize the wasted wood while cutting, what should be the dimensions of the cubes?

B. How many cubes can we make given the above conditions?

1.) 24

2.) 36

3.)  200

4.) LCM of 3 and 4 is 12. Twelve days after February 12 is February 24.

5.) 6000 milliseconds

6.) 12

7.) 7

8.) 13

9.) Side of the square should be GCD of 12 and 18 which is 6 inches.

10.) A. Side of the cube should be GCD of 16, 20, and 32 which is equal to 4 inches.

B. 16/4 × 20/4× 32/4 = 4 × 5 × 8 = 160 cubes

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## The Difference Among LCM, LCD, GCD and GCF

LCM and LCD

LCM means Least Common Multiple. Multiple in this context is used for integers. LCD, on the other hand, means Least Common Denominator. As we know, denominator is used in fractions.

So, in 1/3 and 1/4, should we say, LCM or LCD?

CORRECT STATEMENT: LCM of 3 and 4 (multiple is used for integers).
CORRECT STATEMENT: LCD of 1/3 and 1/4 (denominator is used for fractions).

It is INCORRECT to say LCM of 1/3 and 1/4.

Notice that the LCM of 3 and 4 and the LCD of 1/3 and 1/4 is 12. That’s why most people use them interchangeably.

GCF and GCD

GCF means Greatest Common Factor

A factor of an integer is an integer that can be multiplied with another integer to get the given integer. For example, 3 is a factor of 6 because 3 x 2 = 6.

A divisor of an integer is an integer that divides that integer (without a remainder). For example, 3 is a divisor of 6 because 6/3 = 2, which is an integer.

Basically, factors and divisors have the same meaning when it comes to integers. Factor, however, is the term used in multiplication, while divisor is the term used for division. Here are some examples.

Positive factors of 12 are 1, 2, 4, 6, 12.
Positive divisors of 12 are 1, 2, 4, 6, 12.

GCF and GCD have the same meaning and will give the same values.

## Week 10 Review: Answers and Solutions

After learning how to solve motion problems, let’s answer some exercises and problems. In the solutions, we let d = distance, r = rate, and t = time.

Exercises

1.)   A car travels and average speed of 75 kph. If it traveled for 3.5 hours, what is the total distance traveled?

d = rt
d = (75 kph)(3.5 hrs) =
d = 262.5 km

2.) A bus traveled 4 hours from City A to City B which is 450 kilometers apart. What is its average speed?

d = rt
450 km = (4 hrs)(r)
r = (450 kph)/(4 hrs)
r = 112.5 km

Problem

1.) Two cars left City A at 8:00 am going to City B using the same route. Car 1 traveled at the average speed of 60 kph while Car 2 traveled at an average speed of 50kph. At what time were the two cars 25 kilometers apart?

Let x = time traveled by the two cars
60x – 50x = 25
10x = 25
x = 25/10
x = 2.5 hours

2.5 hours = 2hours and 30 mins
2 hours and 30 minutes after 8:00 am is 10:30 am.

2.) The road distance from Sapiro City to Lireo City is 195 km. Car 1 left Sapiro City going to Lireo City at an average speed of 70kph. Car 2 left City Lireo City going to Sapiro City at an average speed of 60 kph. If both cars left the two cities at the same time and use the same road, after how many hours will the two cars meet?

Car 1
Rate = 70kph
Time = x
Distance = 70x

Car 2
Rate = 60kph
Time = x
Distance = 60x

Total distance = 195kph

70x + 60x = 195
130x = 195
x = 195/130

x = 1.5hrs

3.) A red car left Vigan at 9:00 AM and traveled to Manila at an average speed of 45 kph. After one hour, a white car left the same place for Manila using the same route at an average speed of 60 kph. At what time will the white car overtake the red car?

RED CAR
Rate = 45kph
Time = x+1
Distance = 45(x+1)

WHITE CAR
Rage = 60kph
Time = x
Distance = 60x

60x = 45(x+1)
60x = 45x+45
60x – 45x = 45
15x = 45
x = 45/15
x = 3hours

3 hours after 10:00am is 1:00 p.m.

Note: We add 3 hours to 10:00 am because the second car left at 10:00 am.

4.) Two cars started from the same point, at 12nn, traveling to opposite directions at 50 and 60 kph, respectively. What is the distance between them at exactly 3:30 PM?

CAR 1
Rate = 50kph
Time = 3.5 hrs
Distance = (50 kph)(3.5 hrs) = 175

CAR 2
Rate = 60kph
Time = 3.5 hrs
Distance = (60 kph)( 3.5 hrs) = 210

What is the distance between them at exactly 3:30 PM?
175 km + 210 km = 385 km

5.) Two cars from the same point traveling to opposite directions at 75 and 85 kph, respectively. After how many hours will they be 240 kilometers apart?

75x + 85x = 240
160x = 240
240/160 = x
x = 1.5 hours

6.) A blue car leaves City A for City B at exactly 8:00 AM traveling at average speed of 55 kph. A gray car leaves City B for City A at the same time traveling at an average speed of 45 kph. The distance between the two cities is 75 kilometers.

If the two cars use the same route, what time will they pass each other?

let x be the time
d = r x t

sum of the distance traveled by 2 cars is equal to 75km
so

55x + 45x = 75
100x = 75
x = 75/100 or 0.75 hours
0.75 hours (45 mins)

They will pass each other 45 mins after 8:00am so the answer is 8:45 am.