About the topic simple interest problems

Simple 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 problems in simple interest.  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

Simple Interest:
                   Simple interest is the interest computed on the principal for the entire period of borrowing. It is calculated on the outstanding principal balance and not on interest previously earned.It means no interest paid on interest earned during the term of loan.
Formulas related to simple interest:

                                                 I = Pit

                                                 A = P + I
                                                 A = P + Pit
                                                 A = P(1 + it)

                                                 I = A - P
A = Accumulated value (final value of an investment)
P = Principal (initial value of an investment)
i = Annual interest rate in decimal
I = Amount of interest
t = Time in years

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 shortcuts to related to simple interest. 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 simple 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 simple interest shortcuts?

Students will learn, how and when they have to apply shortcuts to solve the problems which are related to simple 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 simple interest in competitive exams. The stuff which I have mentioned above is nothing but the tricks and shortcuts which need to solve simple 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 simple 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 simple interest problems such that how shortcuts can be used. You can check your answer online and see step by step solution.

1. In simple interest, a sum of money amounts to $ 6200 in 2 years and $ 7400 in 3 years. The principal is  

    (A)$ 3600, 20.58%       (B)$ 3700, 25.58%
    (C)$ 3800, 30.58%       (D)$ 3900, 35.58%

jQuery UI Dialog functionality
Pincipal becomes $ 6200 in 2 years and $7400 in 3 years.
From this information, it is very clear that interest earned in the third year = 7400 - 6200 = 1200
In the given investment, interest for one year = 1200
In the first two years, interest earned = 2X1200 = 2400
In the first two years, amount = 6200

Therefore P = A-I = 6200 - 2400
P = 3800

I = Pit
Here,P=3800, t=2, I =2400 (for two years)
2400 = 3800iX2
2400 = 7600i ===> i = 0.3158

Rate of interest = 0.3158X100 = 31.58%

Hence the principal and rate of interest are $ 3800 and 31.58%

2. In simple interest, a sum of money doubles itself in 10 years. The number of years it would triple itself is 

            (A)19 years                     (B)20 years
            (C)21 years                     (D)22 years

jQuery UI Dialog functionality
Let "P" be the principal
We know A = P + I ----(1)
In 10 years, P becomes 2P (from the given information)
That is A = 2P ----(2)

From (1) & (2), we have P + I = 2P
I = 2P - P ===> I = P
Therefore in 10 years, interest earned (I) = P (invested amount)

In 20 years, interest earned (I) = 2P
In 20 years, A = P + I = P + 2P
A = 3P (triple of investment P)

Hence, time taken for the principal would triple itself is 20 years.

3. Mr. Abraham  invested an amount of $ 13,900 divided in two different schemes A and B at the simple interest rate of 14% p.a. and 11% p.a. respectively. If the total amount of simple interest earned in 2 years be $ 3508, what was the amount invested in Scheme B?

            (A)$ 6100                     (B)$ 6200
            (C)$ 6300                     (D)$ 6400

jQuery UI Dialog functionality
Let "x" be the amount invested in scheme B.
Then the amount invested in scheme A = 13900-x

Interest in scheme (A)+ Interest in scheme (B) = 3508

(13900-x)X0.14X2 + (x)X0.11X2 = 3508
(13900-x)X0.28 + 0.22x = 3508
3892 - 0.28x + 0.22x = 3508
0.06x = 384 ===> x = 6400

Hence, the amount invested in scheme B is $ 6400

4. Lily took a loan of $ 1200 with simple interest for as many years as the rate of interest. If she paid $ 432 as interest at the end of the loan period, what was the rate of interest?

                (A) 6%                                 (B) 7%
                (C) 8%                                 (D) 9%

jQuery UI Dialog functionality
Let "x" be the rate of interest.
Then, the number of years = x (from the given information, rate of interest and number of years are same)

I = PNR/100
Here, I = $432,P = $1200, R = x, N = x
So, 432 = (1200.x.x)/100 ===> x2 = 36
Therefore x = 6

Hence, the rate of interest is 6%

5. A lent $ 5000 to B for 2 years and $ 3000 to C for 4 years on simple interest at the same rate of interest and received $ 2200 in all from both of them as interest. The rate of interest per year is:

                (A) 8%                                 (B) 9%
                (C) 10%                              (D) 11%

jQuery UI Accordion - Default functionality
Let "x" be the rate of interest

Interest from B + Interest from C = 2200
(5000.x.2)/100 + (3000.x.4)/100 = 2200
10000x + 12000x = 220000
22000x = 220000 ===> x = 10

Hence, the rate of interest is 10%

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