"> Word Problem 13 | Job Exam Rare Mathematics (JEMrare)

Word Problem 13

Problem 13:  A saving bank pays interest at 4% per annum and interest being calculated every 6 months. A sum of Rs 100 was deposited on January 1, 1973. After one year the amount will be;

a. Rs 104          b. Rs 104.40             c. Rs 104.04                 d. Rs 102.02

Solution:

simple arithmetic questions and answers

P(principal or initial amount) = Rs 100

R(rate) = 4% pa

Interest paid =every six months

T(term) =1 year

A (amount after the term) =?

This problem can be solved using compound interest formula !

The formula is ;

A=P (1+\frac{R}{100})^{n}

Here n is the number of terms or transactions which will be 2 in one year, so ;

n=2

R=4/2=2% for six months term each

Putting values in the formula;

 \Rightarrow A=P (1+\frac{2}{100})^{2}

 \Rightarrow A=100 (\frac{102}{100})^{2}

 \Rightarrow A=100 (\frac{51}{50})^{2}

 \Rightarrow A=\frac{100\times51\times51}{50\times50})

 \Rightarrow A=\frac{2601}{25}

 \Rightarrow A=Rs 104.04

So, the option c is correct

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