About the topic compound interest problems

Compound interest problems play a major role in quantitative aptitude test. It is bit difficult to score marks in competitive exams without knowing the shortcuts related to compound interest  problems.  We might have already learned this topic in our school. Even though we have been already taught this topic in our higher  classes in school, we need to learn some more short cuts which are being used to solve the problems in the above topic.

The only thing we have to do is, we need to apply the appropriate short cut and solve the problems in a limited time. This limited time will be one minute or less than one minute in most of the competitive exams.

Points to remember

Compound Interest:
                  Here interest will be calculated on principal at the end of every term. The interest earned at the end of each term will be added to the principal of that term and taken it as new principal for the next term. So, principal will be different in each term.
Difference between Simple Interest and Compound Interest:
                  In simple interest, principal remains constant. But in compound interest principal will be different for each term.
Formula to find accumulated value in compound interest:
                                         A = P(1+i)n
A = Accumulated value (final value of an investment)
P = Principal (initial value of an investment)
How to find the value of "i":
                  To find the value of "i", first convert the rate of interest in to decimal form. Then divide the decimal value by the number of conversion periods per year.
                  For example,in an investment rate of interest =15% and compounded quarterly. First we write 15% as decimal form.That is 0.15. Now we have to divide this value by 4 (because compounded quarterly). Finally i = 0.15/4 = 0.0375
How to find the value of "n":
n = (no. of years) X (no. of conversion periods per year)
For example, in an investment no.of years = 2 and compounded quarterly.The value of n = 2X4 =8.

Why is this topic needed?

Not only to answer questions being asked in competitive exams, and also in our regular life, we need the knowledge of simple interest in lending and borrowing money. Students who are preparing to improve their aptitude skills and those who are preparing for this type of competitive test must prepare this topic in order to have better score. Because, today it is bit difficult to score marks in competitive exams without knowing how to solve compound interest problems. Whether a person is going to write placement exam to get placed or a student is going to write a competitive exam in order to get admission in university, they must be knowing shortcuts related to compound interest problems. This is the reason for why people must study this topic.

What is the advantage of this topic?

As we mentioned in the above paragraph, a person who wants to get placed in a company and a students who wants to get admission in university for higher studies must write competitive exams like placement test and entrance exam. To meet the above requirement, it is very important to score more marks in the above mentioned competitive exams. To score more marks, they have to prepare this topic. Preparing this topic would definitely improve their marks in the above exams. Preparing this topic is not difficult task. We are just going to remember the stuff that we have already learned in our lower classes

How can students know compound interest shortcuts?

Students will learn, how and when they have to apply shortcuts to solve the problems which are related to compound interest. Apart from the regular shortcuts, students can learn some additional tricks in this topic simple interest problems. Already we are much clear with the four basic operations which we often use in math. They are addition, subtraction, multiplication and division. Even though we are much clear with these four basic operations, we have to be knowing some more stuff to solve the problems which are being asked from the topic compound interest in competitive exams. The stuff which I have mentioned above is nothing but the tricks and shortcuts which need to solve compound interest problems in a very short time. 

Shortcuts we use to solve the problems

Short cut is nothing but the easiest way to solve problems related to compound interest. In competitive exams, we will have very limited time to solve each problem. Then only we will be able to attend all the questions. If we do problems in competitive exams in perfect manner with all the steps, it will definitely take much time and we may not able to attend the other questions. So we need some other way in which the problems can be solved in a very short time. The way we need to solve the problem quickly is called as shortcut.

Here, we are going to have some compound interest problems such that how shortcuts can be used. You can check your answer online and see step by step solution.

1. A sum of money placed at compound interest doubles itself in 3 years. In how many years will it amount to four times itself?

          (A) 5 years                   (B) 6 years
          (C) 7 years                   (D) 8 years

jQuery UI Dialog functionality
Let "P" be the amount invested initially.

From the given information, P becomes 2P in 3 years.
Since the investment is in compound interest, in the 4th year the principal will be 2P

And 2P becomes 4P (it doubles itself) in the next 3 years.
Therefore, at the end of 6 years accumulated value will be 4P

Hence, the amount deposited will amount to 4 times itself in 6 years.

2. A man borrows a certain sum of money and pays back in two years in two installments. If compound interest is reckoned at 5% per year compounded annually and he pays back annually $ 441, what sum did he borrow?

          (A) $ 820                   (B) $ 825
          (C) $ 830                   (D) $ 835

jQuery UI Dialog functionality
Let "x" be the amount borrowed which becomes $ 441 in one year.
Let "y" be the amount borrowed which becomes $ 441 in two years.

We know the formula in compound interest. That is
     A = P(1+i)n ===> P = A/(1+i)n

P = x,A = 441,i = 0.05/1 = 0.05 (Compounded annually, So divided by 1)
Therefore, x = 441/(1+0.05)1= 441/1.05 = 420
x = 420

P = y,A = 441,i = 0.05/1 = 0.05 (Compounded annually, So divided by 1)
Therefore, y = 441/(1+0.05)2= 441/1.1025 = 400
y = 400

Total = x+y = 420+400 = 820

Hence, the total money borrowed is $ 820

3. The compound interest and simple interest on a certain sum for 2 years is $ 1230 and $ 1200 respectively. The rate of interest is same for both compound interest and simple interest and it is compounded annually. What is the principle?

          (A) $ 11,000              (B) $ 12,000
          (C) $ 13,000              (D) $ 14,000

jQuery UI Dialog functionality
Simple interest for two years = 1200 and interest for one year = 600
So, C.I for 1st year is 600 and for 2nd year is 630.(Since it is compounded annually, S.I and C.I for 1st year would be same)

When we compare the C.I for 1st year and 2nd year, it is clear that the interest earned in 2nd year is 30 more than the first year.

Because, interest 600 earned in 1st year earned this 30 in 2nd year.

It can be considered as if it were simple interest for one year. That is principle = 600, interest = 30
I = Pit ===> 30 = 600i(1) ===> i = 0.05 ===>rate of interest = 5%

In the given problem, simple interest earned in two years = 1200.
I = Pit ===> 1200 = PX0.05X2 ===> P = 1200/0.1 = 12000

Hence, the principal is $ 12,000

4. Mr. David borrowed $ 15,000 at 12% per year compounded annually. He repaid $ 7000 at the end of 1st year. What amount should he pay at the end of second year to completely discharge the load?

          (A) $ 10676                   (B) $ 10776
          (C) $ 10876                   (D) $ 10976

jQuery UI Dialog functionality
The formula for accumulated value in C.I is A = P(1+i)n
To find the accumulated value for the first year, plug P = 15000, i = 0.12, n = 1 in the above formula

A = 15000(1+0.12)1 = 15000(1.12) = 16800
Amount paid at the end of 1st year = 7000
Balance to be repaid = 16800-7000 = 9800
This 9800 is going to be the principal for the 2 nd year.
Now we need the accumulated value for the principal 9800 in one year. (That is at the end of 2nd year)

A = 9800(1+.012)1 = 9800X1.12 = 10976

Hence to completely discharge the loan, at the end of 2nd year, he has to pay $ 10,976

5. There is 60% increase in an amount in 6 years at simple interest. What will be the compound interest of Rs. 12,000 after 3 years at the same rate?

          (A) $ 3672                   (B) $ 3772
          (C) $ 3872                   (D) $ 3972

jQuery UI Accordion - Default functionality
Let the principal in simple interest be $ 100.
Since there is 60% increase, simple interest = 60
We already know the formula for S.I. That is I = Pit
Here I = 60, P = 100, t = 6
60 = 100i(6) ===> i = 60/600 = 0.1 = 10%
Therefore, rate of interest is 10% or i = 0.1

In C.I, formula for accumulated value, A = P(1+i)n
Here P =12000,i = 0.1/1 = 0.1,n = 3X1 = 3 (compounded annually)
A = 12000(1+0.1)3 = 12000(1.331)
A = 15972
C.I = A - P = 15972 - 12000 = 3972

Hence the compound interest after 3 years at the same rate of interest is $ 3972.

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