How to Solve Investment Problems Part 3

This is the third part of the Solving Investment Problems Series. In this part, we discuss solve invest problem which is very similar to the second part. We discuss an investment at two different interests.

Problem

A government employee invested a part of Php60000 in bonds at 6% yearly interest and the remaining part in stocks at a 5% yearly interest. The annual interest in stocks is Php850 less than the annual interest in bonds. How much was invested at each rate?

Solution and Explanation

Let x = amount invested on bonds (6% yearly interest)
60,000 – x = amount invested on stocks (5% yearly interest).

The interest in bonds is 6% per year times the amount invested or

(0.06)(x)

when the percentage is converted to decimals.

In addition, the interest in stocks is 5% per year times the amount invested or

(0.05)(60000 – x)

when the percentage is converted to decimals.

Now, the interest in stocks is 850 less than the interest in bonds which means that if we subtract 850 from the interest in bonds they will be equal. That is

interest in bonds – 850 = interest in stocks.

Substituting the expressions of each, we have

(0.06)(x) – 850 = (0.05)(60000 – x).

Simplifying, we have

0.06x – 850 = 3000 – 0.05x.

To eliminate the decimal numbers, we multiply everything by 100. The equation becomes

6x – 85000 = 300000 – 5x

We simplify by adding 85000 and 5x to both sides. The equation becomes.

11x = 385000.

Dividing both sides by 11, we have
x = 35000.

So, 35000 was invested in bonds and 60000-35000 = 25000 was invested in stocks.

Check:

35000 × 0.06 = 2100
25000 × 0.05 = 1250

Indeed, the amount invested in bonds is 2100 – 1200 = 850 less than the interest in stocks.

How to Solve Investment Problems Part 2

This is the second part of the Solving Investment Problems Series. In the first part, we discussed in detail the solution of a problem at two different rates of interest. In this post, we discuss another problem.

Problem

Mr. Reyes invested a part of Php70000 at 3% yearly interest and the remaining part at a 5% yearly interest. The annual interest on the 3% investment is Php100 more than the annual interest on the 5% investment. How much was invested at each rate?

Solution

If we let x be the amount invested at 3%, then, 70000 – x is the amount invested at 5%. The yearly interest is the product of the rates and the amount invested so,

(3%)(x) = yearly interest of the amount invested at 3%
(5%)(70000 – x) = yearly interest of the amount invested at 5%

Now, the annual interest at 3% is 100 more than the annual interest at 5%. This means that if we add 100 to the yearly interest at 5%, the interests will be equal. That is,

(3%)(x) = (5%)(70000 – x) + 100.

Next, we convert percent to decimal by dividing the percentage by 100. So,

(0.03)(x) = (0.05)(70000 – x) + 100.

Simplifying, we have

0.03x = 3500 – 0.05x + 100
0.03x = 3600 – 0.05x
0.03x + 0.05x = 3600
0.08x = 3600.

Tip: You can calculate better by eliminating the decimal. You can do this by multiplying both sides by 100.

Dividing both sides by 0.08, we have

x = 45000.

This means that 45000 is invested at 3% yearly interest.

Now, the remaining amount is 70000 – 45000 = 25000.

This means that 25000 is invested at 5%.

Check:

Yearly interest of 45000 at 3% interest = (45000 x 0.03) = 1350

Yearly interest of 25000 at 5% interest = (25000 x 0.05) = 1250

As we can see, the interest at 45000 at 3% interest is 100 more than the interest of 25000 at 5% interest.

Therefore, we are correct.

How to Solve Investment Problems Part 1

In the Simple Interest Problems Series, we have learned how to calculate the interest, rate, or time of the principal invested. In that series, the principal is invested to a single bank or company. In this series, we will learn how to calculate the interest of money invested at different companies (different rates). Before starting with our first example, familiarize yourself with the following terms:

principal – the amount of money invested
rate of interest – the percent of interest yearly or any period of time (e.g. monthly, quarterly)
interest – income or return of investment

Hence, if Php10000 is invested at a bank with 3% interest is 300, Php10000 is the principal, 3% is the rate of interest, and 300 is the interest.

Example 1

Mr. Molina invested Php100,000.00. A part of it was invested in a bank at 4% yearly interest and another part of it at a credit cooperative at 7% yearly interest. How much investment he made in each if his yearly income from the two investments is Php5950.00?

Scratch Work

If we let x be the money invested at a bank, then 100000 – x is the amount invested at the credit cooperative. To calculate for the interest, we must apply the percent of each interest at each amount. That is

(4%)(x) = yearly interest from at the bank
(7%)(100000 – x) = yearly interest from the credit cooperative

If we add the interest, it will amount to Php5950. So, here’s our solution.

Solution

Let x = amount of money invested at the bank
100000 – x = amount of money invested at the credit cooperative

Total Interest = Interest From the Bank + Interest from the Credit Cooperative
5950 = (4%)(x) + (7%)(100000 – x)

We need to convert percent to decimals in order to multiply. We do this by dividing the percentage by 100. So, 4% = 0.04 and 7% = 0.07. Substituting to the previous equation, we have

5950 = (0.04)(x) + (0.07)(100000 – x)
5950 = 0.04x + 7000 – 0.07x
5950 – 7000 = -0.03x
-1050 = -0.03x.

We can eliminate the decimals by multiplying by 100.

-105000 = -3x
Dividing both sides by -3, we have

35000 = x.

That means that Mr. Molina invested Php35000 in the bank and Php100000 – Php35000 = Php65000 in the credit cooperative.

Check
(4%)(35000) + (7%)(65000) = (.04)(35000) + (.07)(65000)
= 1400 + 4550 = 5950

As we can see, our interest from the two investments is Php5950.00

The Solving Simple Interest Problems Series

The Solving Simple Interest Problems is a series of tutorials on how to solve simple interest problems. In solving interest problems you need to know the following terms:

the principal (P) is the money invested, the rate (R) of interest is the percentage of interest (that is the number with percent sign), the time (T) and the interest (I) is the earnings or return of investment. The interest is the product of the principal, the rate, and the time,  or I = PRT is explained in the first part of the series.

How to Solve Simple Interest Problems Part 1 discusses the basics of simple interest problems and the terms used in such problems. Two examples worked examples are solved in which interests are both unknowns.

How to Solve Simple Interest Problems Part 2 is a continuation of the simple interest discussion. One problem is an example of a rate of interest which is not given yearly and the other one is an investment made for more than a year.

How to Solve Simple Interest Problems Part 3 is a discussion of simple interest problems where the unknowns are the principal and the rate.

I will be posting exercises and problems with solutions about this type of problems soon, so keep posted.

 

How to Solve Simple Interest Problems Part 3

This is the third and last part of the Solving Simple Interest Problems Series. In the first and second part of this series, we have learned how to solve simple interest problems. In this post, we continue with two more problems, one with the rate missing and the other one with the principal missing.

Example 5

Mr. Wong deposited $15,000 in a bank for a certain rate of interest per year. After two years, the interest was $900. What was the rate of interest in percent?

Solution and Explanation

We have the following given:

Principal (P) = $15,000
Interest (I) = 900
Time = 2 (years)
Rate (R) = ?

As we have learned in the previous posts, simple interest is the product of the principal, the rate, and time or

I = PRT.

Substituting the given we have

900 = (15000)(R)(2)
900 = 30000R

Dividing both sides by 30000, we have

R = 900/30000
R = 0.03

The rate is 0.03 which is we need to convert to percent by multiplying it by 100. Therefore, the rate is 3%.

Example 6

Mrs. Lansangan invested a certain amount of money in a bank that gives 4% interest per year. She got an interest of Php2400 after 3 years.

Solution and Explanation

Given
I = 2400
R = 4%
T = 3 years

Using the simple interest formula mentioned above, we have

I = PRT
2400 = (P)(4%)(3)

Converting 4 percent to decimal, we have

2400 = (P)(0.04)(3)
2400 = 0.12P

Dividing both sides by 0.12, we have

20000 = P

Therefore, Mrs. Lansangan invested Php20000.

That’s it for our series on how to solve simple interest problems. I hope you have learned well. Good luck for the exams.

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?

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