7th Class Mathematics Comparing Quantities Compound Interest

     Compound Interest                

 

In this case we observe that the interest paid by borrower is same for every year.                

There are different methods to calculate the interest in case of bank transaction loan etc. In these methods, interest is calculated either quarterly, half yearly or yearly, what so ever may be the case. i.e. The agreement between the lender and borrower on the principle. The amount after that first fixed interval of time will be principle for second interval of time, the amount after second interval of time will be the principle for the third interval and so on. The interest paid by borrower under the above conditions is called compound interest.

 

      Abbreviation used in Interest

•  Principle as P              
•  Time as T
•   Rate as R or r
•   Simple interest as S.I.
•   Compound interest as C.I
•   Amount as A  

 

               

 

Find the compound interest on a sum of $ 1500 for two years at the rate opf 5% per annum compounded annually.

Solution:

The following steps are followed

Step 1: Principle for the first year = $ 1500, Time (T) =1 year R = 5 % p.a.                

\[S.I.=\frac{P\times R\times T}{100}=\frac{1500\times 5\times 1}{100}\$75\]                

Amount after 1st year \[=\$1500+\$75=\$1575\]                                

Step 2: Principle for second year = $ 1575, Time = 1 year, R = 5 %                

\[S.I.=\frac{P\times R\times T}{100}=\frac{1575\times 5\times 1}{100}\$78.75\]                

The amount after second year \[=\$1575+\$\] \[78.75=\$1653.75\]                

Therefore, the compound interest \[=A-P\]                

\[=\$1653.75-\$1500=\$153.75\]                                

Form the above discussion we conclude that

• The principle for simple interest is same whereas it is different in the case of compound interest for every fixed interval of time.
• The interest for simple interest is same whereas it is different in the case of compound interest for every fixed interval of time.  

 

 

               

Find the compound interest on $15,000 for 2 year at 8 % per annum.                

(a) Rs 2496                                         

(b) Rs 2393                

(c) Rs 2293                                          

(d) Rs 2593                

(e) None of these                  

 

Answer: (a)                                

Explanation                                

Step 1: Principle for 1st year =$ 15,000, time (T) = 1 years, R = 8 % p.a.                

\[S.I.=\frac{P\times R\times T}{100}=\frac{15000\times 8\times 1}{100}\$1200\]                                                                       

The amount after 1st year =$15,000 +$ 1,200 = $16,200                                

Step 2: Principle for the second year = Amount after 1st year =$ 16,200                

Time (T) = 1 year, R = 8 % P.a.                

\[S.I.=\frac{P\times R\times T}{100}=\frac{15000\times 8\times 1}{100}\$1296\]                

Amount after 2nd year =$16200 +$ 1296 = $17496                

Compound interest (C.I.) =Amount - Principle                

=$ 17496-$15000 =$ 2496  

 

 

       Calculating Compound Interest by using Formula                

Apart from calculating compound interest by compounding the interest annually which has been described earlier), we can calculate it directly by using formula.                                

Formula is as follows:                

If principle is P, rate of interest is R % per annum compounded annually and time is T then the amount A is:                

\[A=P{{\left( 1+\frac{R}{100} \right)}^{T}}\]and \[C.I.=A-P\]                

\[\Rightarrow C.I.=P{{\left( 1+\frac{R}{100} \right)}^{T}}\]\[-P\Rightarrow C.I.=P\left\{ {{\left( 1+\frac{R}{100} \right)}^{T}}-1 \right\}\]  

 

      Some Special Cases                

Suppose principle = P, Time = t years and rate of interest = R %.

• The amount will be when interest compounded annually: \[A=P{{\left( 1+\frac{R}{100} \right)}^{t}}\]
• The amount will be when interest compounded half yearly: \[A=P{{\left( 1+\frac{R/2}{100} \right)}^{2t}}\]

In this case time will be doubled and rate of interest will become half .

• The amount will be when interest compounded quarterly: \[A=P{{\left( 1+\frac{R/4}{100} \right)}^{4t}}\]

In this case the time will be four times and rate will become one fourth.

• The amount will be when the rate of interest is different in the different interval of time \[A=P(1+\frac{{{R}_{1}}}{100})(1+\frac{{{R}_{2}}}{100})(1+\frac{{{R}_{3}}}{100})...........\]  

 

 

               

Find the compound interest on Rs 12,000 for 3 years at 20% per annum.                

(a) Rs 8076                                         

(b) Rs 8736                

(c) Rs 3524                                          

(d) Rs 3250                

(e) None of these                  

 

Answer: (b)                  

Explanation                

First of all calculate the compound interest on Rs. 100 for 3 years at the rate of 20% per annum.                

Principal for the first year = Rs 100                

Interest for the first year \[=Rs\frac{100\times 20\times 1}{100}=Rs20\]                

\[\therefore \]Amount at the end of first year = Rs 100 +Rs 20 = Rs 120                

Now principal for the second year = Rs 120.                

Interest for the second year \[=Rs\frac{120\times 20\times 1}{100}=Rs24\]                

\[\therefore \]Amount at the end of second year = Rs 120 + Rs Rs 24 =Rs144                

Now principal for the third year = Rs 144                

Interest for the third year \[=Rs\frac{144\times 20\times 1}{100}=Rs28.80\]                

\[\therefore \]Amount at the end of third year = Rs 144 + Rs 28.80 = Rs 172.80                

Now amount on Rs 100 = Rs 172.80                

Amount on \[Rs1=Rs\frac{172.80}{100}\]                

Hence, amount on \[Rs12,000=Rs\frac{172.80}{100}\times 12,000=20,736\]                

Compound interest = Amount - Principal = Rs 20,736 - Rs12, 000 = Rs 8.736  

 

 

 

 

A sum of Rs 1300 is lend into two parts so that interest on the 1st 3 years at 5 % may be equal to the interest at the second part for 4 year at 6 %. The 1st sum is:                                                          

(a) Rs 800                                            

(b) Rs 600                        

(c) Rs 500                                           

(d) Rs 400                

(e) None of these                                

 

Answer: (a)                

Explanation                

Let the 1st  sum be Rs. 100. Interest on Rs 100 for 3 year at 5 % is Rs 15. As the interest on the second sum is also the same as that or the 1st sum.    

The second sum \[=\frac{15\times 100}{4\times 6}=\frac{125}{2}.\] Hence, the ratio of the sums                

\[=100:\frac{125}{2}=200:125=8:5\]so, the 1st sum \[=\frac{8}{13}\times 1300=Rs800.\]                

 

 

Find the compound interest on Rs 33,280 for 3 years, if the rate of interest is \[\text{12}\frac{1}{2}%\]% p.a.                

(a) Rs 14,102                                      

(b) Rs 14,105                

(c) Rs 15,207                                      

(d) Rs 16,209                

(e) None of these                                

 

Answer: (b)                

Explanation

Here, \[P=33,280,R=12\frac{1}{2}=\frac{25}{2},n=3\]                

We know that\[A=P{{\left( 1+\frac{R}{100} \right)}^{n}}\]                

               

\[=Rs=33,280{{\left( \frac{9}{8} \right)}^{3}}=Rs33,280\frac{9}{8}\times \frac{9}{8}\times \frac{9}{8}=Rs47,385\]                

\[\therefore \]Compound interest = Amount - Principal = 47,385 - 33,280 = Rs 14,105      

 

 

 

 

• If principle is P, rate of interest is R % per annum compounded annually and time is T then                

(i)  \[A=P{{\left( 1+\frac{R}{100} \right)}^{T}}\]                

(ii) \[C.I.=A-P\]                

(iii) \[C.I=P\left\{ {{\left( 1+\frac{R}{100} \right)}^{T}}-1 \right\}\]

• When the rate of interest are different interval of time then, \[A=P(1+\frac{{{R}_{1}}}{100})(1+\frac{{{R}_{2}}}{100})(1+\frac{{{R}_{3}}}{100}).........\]  

 

 

 

 

• \[{{2682440}^{4}}+{{15365639}^{4}}+{{18796760}^{4}}={{20615673}^{4}}\]
• \[111,111,111\times 111,111,111\text{ }\]\[=12,345,678,987,654,321\]
• There are just four numbers (after 1) which are the sums of the cubes of the digits:
• \[153={{1}^{3}}+{{5}^{3}}+{{3}^{3}}\]
• \[370={{3}^{3}}+{{7}^{3}}+{{0}^{3}}\]
• \[371={{3}^{3}}+{{7}^{3}}+{{1}^{3}}\] and
• \[407={{4}^{3}}+{{0}^{3}}+{{7}^{3}}\]