How to Calculate Faster using Cancellation Part 1

Cancellation is one of the great techniques in making calculations faster. This technique is used in simplifying fractions, rational expressions, and equations in Algebra. In this post, we are going to learn some of the cancellation techniques that are usually not in schools but can be a helpful strategy in taking examinations like the Civil Service Exam.

1. Getting the Lowest Terms of a Fraction

Cancellation can be used to simplify fractions in order to convert them to lowest terms. In general, in order to simplify fractions, we have to get the greatest common denominator of the numerator and the denominator; however, making use of cancellation several times until the fraction is in lowest terms is also a good strategy especially for large numbers .

cancellation

 

In the first example above, cancellation is used to simplify 6/9 to 2/3 by dividing both the numerator and denominator by 3.  In the second example, cancellation was used twice: first, 24/32 is divided by 4 to obtain 6/8, and then was divided by 2 to obtain 3/4.

2. Multiplying Fractions

Cancellation can also be used to simplify multiplication of fractions. You can cancel any pair of number where one is on the numerator and the other is on the denominator.

cancellation

In the example above, 4/9 is multiplied by 3/16.

(1) We can cancel out 3 and 9 by dividing both of them by 3. We get 1 in the numerator and 3 in the denominator.

(2) We can also cancel out 4 and 16 by dividing both of them by 4. This gives us 1 and 4 respectively.

cancellation

Click image to enlarge

In the last example, we have

\dfrac{7}{15} \times \dfrac{9}{8} \times \dfrac{4}{21}.

(1)  9 and 15 are cancelled by dividing both of them by 3. This results to 3 and 5, respectively.
(2) 4 and 8 are cancelled out by dividing both of them by 4. This results to 1 and 2, respectively.
(3) 7 and 21 are cancelled out by dividing both of them by 7. This results to 1 and 3, respectively.
(4) 3 and 3 are cancelled out by dividing both of them by 3. The result is 1.

This results to the simplified fraction to

\dfrac{1}{5} \times \dfrac{1}{2} \times \dfrac{1}{1} = \dfrac{1}{10}.

The examples above are discussed in the following video. The language used is mixed Tagalog and English.

In the next post, we will discuss more about cancellation.

3 Helpful Strategies in Comparing Fractions

There are questions in Civil Service Examinations that may require you to compare fractions or even arrange them in order. In this post, I am going to teach you three strategies in comparing fractions.

Strategy 1: Cross Multiplication

Which is greater, 5/7 or 8/11?

If only two fractions are compared, the easiest way is to cross multiply. However, take note of the following:

1.) You multiply the denominator of the fraction to the numerator of the other fraction.
2.) Place the product above the numerator.

Screen Shot 2015-10-08 at 11.02.23 PM

The larger product is the larger fraction. As shown in the example above, 56 is larger than 55, therefore, 8/11 is larger than 5/7.

Strategy 2: Converting to Similar Fractions

Sample Question: Which is the largest: 13/16, 5/8, 3/4?

We can get the least common denominators of these fractions. Now, the LCM of these denominators is 16. So, we convert everything to fractions whose denominators is 16.

To convert 5/8 to something over 16, we divide 16 by 8 then multiply by 5 which gives us 10. So, 5/8 is equal to 10/16.

To convert ¾ to 16, we divide 16 by 4, then multiply by 3. This gives us 12.

So, we have converted all fractions to fractions whose denominator is 16.

We have 13/16, 10/16, and 12/16. Obviously, the largest is 13/16. Note that using this strategy does not only tell us which is the largest. In fact, we can order the fractions from smallest to largest or vice versa.

Strategy 3: Converting to Decimals

Which is larger: 2/5, 3/4, or 7/10.

We can convert them to decimal by manually dividing the numerator by the denominator (watch video above). The equivalent of 2/5 = 0.4, 3/4 is 0.75 and 7/10 = 70.

The strategies above can be used effectively by looking at the fractions. If two fractions are compared, use Strategy 1. If the numerators are not very large, you can use strategy 2 or 3.

How to Solve Simple Interest Problems Part 2

This is the second part of the Solving Simple Interest Problems Series. In the previous post, we have discussed the basics of simple interest problems. We have learned that the simple interest (I) is equal to the product of the amount of money invested or the principal (P), the percentage of interest or the rate (R) and the time (T). Therefore, we can use the following formula:

I = PRT.

In this post, we are going to discuss more problems particularly interests that are not yearly and finding unknowns other than interest.

Example 3

Dr. Lopez invested his Php120,000 in a bank that gives 2% interest every quarter. What is the interest of his money if he is to invest it for 1 years?

Solution and Explanation

Notice that the interest is applied quarterly and not every year. Quarterly means every three months and therefore it will be applied four times a year since there are four quarters every year. So in this case, the time is 4. So,

P = Php120,000
R = 2%
T = 4
I = ?

Note that the rate percent must be converted to decimal by dividing it by 100. So 2% equals 0.02. Now, using the formula, we have

I = PRT
I = (Php120,000)(.02)(4)
I = Php9600.00

So, the interest of the money for 1 year is Php9600.

Example 4

Danica invested here money amounting to Php150,000 in a bank that offers a 5% simple interest every year. She went abroad and never made any deposit or withdrawal in her account. After coming back, she immediately checked her account and found out that her money got an interest of Php37,500. How many years was the money invested?

Solution and Explanation

In this problem, interest is given and time is unknown. Assigning the values we have

I = Php37,500
P = Php150,000
R = 5%
T = ?.

Using the formula, we have

I = PRT

Converting 5 percent to decimal and substituting, we have

37,500 = (150,000)(.05)(T)
37,500 = 7500(T).

Dividing both sides by 7500, we have

5 = T.

That means that the money was invested for 5 years.

In the next post, we will be solving simple interest problems whose uknowns are rate and principal.

How to Solve Simple Interest Problems Part 1

This is the first part of the Solving Simple Interest Problem Series for the Civil Service Examination.

Simple interest problems are usually included in many examinations such as the Civil Service Exams. It is important that you practice solving these types of problems in order to increase your chance of passing the exams.

Before solving simple interest problems, let us familiarize ourselves with the terms used in simple interest problems. These are the money invested which is called the principal, the rate of interest which is the percent and the interest which is the income or return of investment, and time. Time may vary depending on the investment. It can range from months to years.

Example 1

Mr. Reyes invested Php50,000 at an interest rate of 3% per year.

a.) Identify the principal and rate of interest.
b.) Calculate the interest earned after 1 year.

Solution

For (a)
The money invested or principal is Php50,000, the interest rate is 3%, and the time is 1 year.

For (b)
We want to calculate 3% of Php50,000. To multiply, we must convert 3 percent to decimal which is equal to 0.03.

interest = Php50,000 × 0.03
interest = Php1500

This means that for a year, the money earned Php1500.

Example 2

Ms. Gutierrez invested Php60,000 at a simple interest of 4% per year for 4 years.

a.) Identify the principal, rate of interest, and time.
b.) How much money will Ms. Gutierrez have after four years?

Solution

For (a),
The principal or money invested is Php60,000.
The rate of interest is 4%.
The time is 4 years.

For (b),
We need to calculate 4% of Php60,000. Just like above, we must first convert 4 percent to decimal which is equal to 0.04.

Now,
interest (1 year) = P60,000 × 0.04 = 2,400
That is the interest for 1 year. To be able to calculate the interest for four years, we have

interest (4 years) = 2,400 × 4 = 9600.

So, the money Ms. Gutierrez will have by the end of four years is the Principal which is 60,000 and the interest for 4 years which is 9600. So, in total, her money will be 69,600.

***

From the two problems above, we can see the interest (I) is the product of the principal (P), the rate (R), and the time(T). Therefore, we can have the formula.

I = P × R × T

or simply

I = PRT.

In the next part of this series, we will be solving more simple interest problems.

The Solving Mixture Problems Series

The Solving Mixture Problems Series is a series of tutorials that explain how to solve mixture problems. Mixture problems can be classified into two, those involved with percents and the other one involved with prices.

How to Solve Mixture Problem Part 1 discusses the basics of base and percentage. This is a preparation of the mixture problems involving percents.

How to Solve Mixture Problem Part 2 discusses the most basic of the mixture problems. A detailed solution is discussed about the following problem.

How many liters of 80% alcohol solution must be added to 60 liters of 40% alcohol solution to produce a 50% alcohol solution.

How to Solve Mixture Problem Part 3 discusses another mixture problems involving percents. A detailed solution is discussed about the following problem.

How many liters of pure water must be added to 15 liters of a 20% salt solution to make a 5% salt solution?

How to Solve Mixture Problem Part 4 discusses one more mixture problems involving percents. A detailed solution is discussed about the following problem.

A chemist creates a mixture with 5% boric acid and combined it with another mixture containing 40% boric acid to obtain a 800 ml of mixture with 12% boric acid. How much of each mixture did he use?

How to Solve Mixture Problem Part 5 discusses the basics of mixture problem involving prices. A detailed solution is discussed about the following problem.

A seller mixes 20 kilograms of candy worth 80 pesos per kilogram to candies worth 50 pesos per kilogram. He sold at 60 pesos per kilogram. After selling all the candies he discovered that he had no gain or loss. How much of the 50-pesos per kilogram candies did he use?

How to Solve Mixture Problem Part 6 discusses one more mixture problem involving prices. A detailed solution is discussed about the following problem.

A bakery owner wants to box assorted chocolates. Each box is to be made up of packs of black chocolates worth $10 per pack and packs of ordinary chocolates worth $7 each pack. How many packs of each kind should he use to make 15 packs which he can sell for $8 per pack?

How to Solve Mixture Problems Part 6

This is the 6th part and last part of the Solving Mixture Problems Series. In the previous 4 parts, we have learned how to solve mixture problems involving percent and in part 5, we have learned how to solve problems involving percents. In this post, we solve another problem involving percent.

A bakery owner wants to box assorted chocolates. Each box is to be made up of packs of black chocolates worth $10 per pack and packs of ordinary chocolates worth $7 each pack. How many packs of each kind should he use to make 15 packs which he can sell for $8 per pack?

Solution

Let x = number of $10 packs
15 – x = number of $7 packs

Multiplying the cost per pack and the number of packs we have

($10)(x) = total cost of $10 packs
($7)(15 – x) = total cost of the $7 packs
($8)(15) = total cost of all the chocolates

Now, we know that

total cost of $10 packs + total cost of the $7 packs = total cost of all the chocolates.

Substituting the values, we have

($10)(x) + ($7)(15 – x) = ($8)(15).

Eliminating the dollar sign and solving for x, we have

(10)(x) + (7)(15 – x) = (8)(15)
10x + 105 – 7x = 120
3x + 105 = 120
3x = 120 – 105
3x = 15
x = 5.
This means that we need 5 packs of $10 and 10 packs of $7 chocolates.

Check:
($10)(5) + ($7)(10) = ($8)(15)
$50 + $70 = $120
$120 = $120
Therefore, we are correct.

How to Solve Mixture Problems Part 5

In the previous posts, we have learned how to solve mixture problems involving percentages and liquid mixture problems. In this post, we are going to solve mixture problems involving prices. Although these two types of problems are different, they are very similar when you set up the equation. Below is our first problem.

Problem

A seller mixes 20 kilograms of candy worth 80 pesos per kilogram to candies worth 50 pesos per kilogram. He sold at 60 pesos per kilogram.

candy solve mixture problems

After selling all the candies he discovered that he had no gain or loss. How much of the 50-pesos per kilogram candies did he use?

Solution and Explanation

Let x = number of kilograms of candy worth 50-pesos per kilogram.

The total price of 20 kilograms of candy at 80 pesos per kilogram is (20 kg)(80 pesos/kg) = 1600 pesos.

The total price of x kilograms of candy at 50 pesos per kilogram is (x kg)(50 pesos/kg) pesos.

When we add these 2, the number of kilograms of candy is x + 20 (can you see why?) and it is sold at 60 pesos. So, its total price is (x + 20)(60 pesos/kg).

Using these facts, we have the following equation:

total price of 80pesos/kg candy + total price of 50pesos/kg candies = total price of 60pesos/kg candies.

Substituting the expressions above, we have

(20 kg)(80 pesos/kg) + (x kg)(50 pesos/kg) = (x + 20 kg)(60 pesos/kg)

1600 + 50x = 60(x + 20)
1600 + 50x = 60x + 1200
1600 – 1200 = 60x – 50x
400 = 10x
40 = x.

Therefore, he used 40 kilograms of candy worth 50 pesos per kilogram.

Check:

1600 + 50(40) = 60(40 + 20)
1600 + 2000 = 60(60)
3600 = 3600.

This means that we are correct.

How to Solve Mixture Problems Part 4

This is the fifth post of the Solving Mixture Problems Series on PH Civil Service Review. In this post, we are going to solve a problem in which only the total amount of mixture is given.

Problem

A chemist creates a mixture with 5% boric acid and combined it with another mixture containing 40% boric acid to obtain a 800 ml of mixture with 12% boric acid. How much of each mixture did he use?

Solution and Explanation

Let

x = mixture with 5% boric acid

800 – x = mixture with 40% boric acid.

Note that if these two mixtures are combined, we will produce a mixture with 800 ml of solution with 12% boric acid. As we have learned in the previous tutorials, the amount of boric acid in the first mixture added to the amount of boric acid in the second mixture is equal to the amount of boric acid in the combined mixtures. That is,

(5%)(x) + (40%)(800-x) = (12%)(800).

Take note that x, 800 – x, and 800 are amount of the mixture and if these are multiplied by the percentage of boric acid, then we will get the exact amount of pure boric acid. 

Converting percent to decimals, we have

(0.05)(x) + (0.4)(800-x) = (0.12)(800)

0.05x + 320 – 0.4x = 96.

Simplifying, we have

-0.35x = -224

x= 640.

That means that we need 640 ml of mixture with 5% boric acid and 800 – 640 = 160 ml of mixture with 40% boric acid.

Check:

Amount of boric acid in mixture with 5% boric acid: (640 ml)(0.05) = 32ml
Amount of boric acid in mixture with 40% boric acid: (160 ml)(0.4) = 64ml
Amount of boric acid in mixture with 12% boric acid: (800 ml)(0.12) = 96ml

As we can see, 32 ml + 64 ml = 96 ml which means that we are correct.

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