
The difference between C.I. and S.I. on a certain sum of money at 10% per annum for 3 years is Rs.620. find the principle if it is known that the interest is compounded annually.
Answer
446.1k+ views
Hint: Here in this problem we are given with difference in C.I. and S.I. and the rate of interest with period. Now we are going to use the formulas directly in order to find the principal value.
Formulas used:
1. Simple interest = \[\dfrac{{PRT}}{{100}}\]
2. Compound interest = \[P\left( {{{\left( {1 + \dfrac{R}{{100}}} \right)}^T} - 1} \right)\]
Step by step solution:
Given that,
Time period T=3years
Rate R=10%
Now the difference in C.I. and S.I. is given by
\[ \Rightarrow P\left( {{{\left( {1 + \dfrac{R}{{100}}} \right)}^T} - 1} \right) - \dfrac{{PRT}}{{100}}\]
Putting the values of T and R
\[ \Rightarrow P\left( {{{\left( {1 + \dfrac{{10}}{{100}}} \right)}^3} - 1} \right) - \dfrac{{P \times 10 \times 3}}{{100}}\]
Using the algebraic identity \[{\left( {a + b} \right)^3} = {a^3} + {b^3} + 3{a^2}b + 3a{b^2}\] we get
\[ \Rightarrow P\left( {1 + {{\left( {\dfrac{{10}}{{100}}} \right)}^3} + 3 \times \dfrac{{10}}{{100}} + 3 \times {{\left( {\dfrac{{10}}{{100}}} \right)}^2} - 1} \right) - \dfrac{{P \times 10 \times 3}}{{100}}\]
Cancelling 1 from above expansion
\[ \Rightarrow P\left( {{{\left( {\dfrac{{10}}{{100}}} \right)}^3} + \dfrac{{30}}{{100}} + 3 \times {{\left( {\dfrac{{10}}{{100}}} \right)}^2}} \right) - \dfrac{{P \times 10 \times 3}}{{100}}\]
Multiplying with P,
\[ \Rightarrow P{\left( {\dfrac{{10}}{{100}}} \right)^3} + \dfrac{{30P}}{{100}} + 3P{\left( {\dfrac{{10}}{{100}}} \right)^2} - \dfrac{{30P}}{{100}}\]
Cancelling second and last term we get,
\[ \Rightarrow P{\left( {\dfrac{{10}}{{100}}} \right)^3} + 3P{\left( {\dfrac{{10}}{{100}}} \right)^2}\]
\[ \Rightarrow P{\left( {\dfrac{{10}}{{100}}} \right)^2}\left( {\dfrac{1}{{10}} + 3} \right)\]
Taking LCM of second bracket
\[ \Rightarrow P \times \dfrac{1}{{100}} \times \dfrac{{31}}{{10}}\]
\[ \Rightarrow P \times \dfrac{{31}}{{1000}}\]
But this difference is already given to us as Rs. 620
\[ \Rightarrow P \times \dfrac{{31}}{{1000}} = 620\]
\[ \Rightarrow P = 620 \times \dfrac{{1000}}{{31}}\]
Cancelling by 31 we get
\[ \Rightarrow P = 20 \times 1000\]
\[ \Rightarrow P = 20000\]
This is the principle amount \[ \Rightarrow P = 20000\]
Note:
In this problem instead of using or expanding the bracket we can directly use the formula if 3 years of compound interest are given \[\dfrac{{{{\Pr }^2}}}{{{{100}^3}}}\left( {300 + r} \right)\]. Otherwise we have solved it with traditional methods.
Formulas used:
1. Simple interest = \[\dfrac{{PRT}}{{100}}\]
2. Compound interest = \[P\left( {{{\left( {1 + \dfrac{R}{{100}}} \right)}^T} - 1} \right)\]
Step by step solution:
Given that,
Time period T=3years
Rate R=10%
Now the difference in C.I. and S.I. is given by
\[ \Rightarrow P\left( {{{\left( {1 + \dfrac{R}{{100}}} \right)}^T} - 1} \right) - \dfrac{{PRT}}{{100}}\]
Putting the values of T and R
\[ \Rightarrow P\left( {{{\left( {1 + \dfrac{{10}}{{100}}} \right)}^3} - 1} \right) - \dfrac{{P \times 10 \times 3}}{{100}}\]
Using the algebraic identity \[{\left( {a + b} \right)^3} = {a^3} + {b^3} + 3{a^2}b + 3a{b^2}\] we get
\[ \Rightarrow P\left( {1 + {{\left( {\dfrac{{10}}{{100}}} \right)}^3} + 3 \times \dfrac{{10}}{{100}} + 3 \times {{\left( {\dfrac{{10}}{{100}}} \right)}^2} - 1} \right) - \dfrac{{P \times 10 \times 3}}{{100}}\]
Cancelling 1 from above expansion
\[ \Rightarrow P\left( {{{\left( {\dfrac{{10}}{{100}}} \right)}^3} + \dfrac{{30}}{{100}} + 3 \times {{\left( {\dfrac{{10}}{{100}}} \right)}^2}} \right) - \dfrac{{P \times 10 \times 3}}{{100}}\]
Multiplying with P,
\[ \Rightarrow P{\left( {\dfrac{{10}}{{100}}} \right)^3} + \dfrac{{30P}}{{100}} + 3P{\left( {\dfrac{{10}}{{100}}} \right)^2} - \dfrac{{30P}}{{100}}\]
Cancelling second and last term we get,
\[ \Rightarrow P{\left( {\dfrac{{10}}{{100}}} \right)^3} + 3P{\left( {\dfrac{{10}}{{100}}} \right)^2}\]
\[ \Rightarrow P{\left( {\dfrac{{10}}{{100}}} \right)^2}\left( {\dfrac{1}{{10}} + 3} \right)\]
Taking LCM of second bracket
\[ \Rightarrow P \times \dfrac{1}{{100}} \times \dfrac{{31}}{{10}}\]
\[ \Rightarrow P \times \dfrac{{31}}{{1000}}\]
But this difference is already given to us as Rs. 620
\[ \Rightarrow P \times \dfrac{{31}}{{1000}} = 620\]
\[ \Rightarrow P = 620 \times \dfrac{{1000}}{{31}}\]
Cancelling by 31 we get
\[ \Rightarrow P = 20 \times 1000\]
\[ \Rightarrow P = 20000\]
This is the principle amount \[ \Rightarrow P = 20000\]
Note:
In this problem instead of using or expanding the bracket we can directly use the formula if 3 years of compound interest are given \[\dfrac{{{{\Pr }^2}}}{{{{100}^3}}}\left( {300 + r} \right)\]. Otherwise we have solved it with traditional methods.
Recently Updated Pages
What percentage of the area in India is covered by class 10 social science CBSE

The area of a 6m wide road outside a garden in all class 10 maths CBSE

What is the electric flux through a cube of side 1 class 10 physics CBSE

If one root of x2 x k 0 maybe the square of the other class 10 maths CBSE

The radius and height of a cylinder are in the ratio class 10 maths CBSE

An almirah is sold for 5400 Rs after allowing a discount class 10 maths CBSE

Trending doubts
In Indian rupees 1 trillion is equal to how many c class 8 maths CBSE

Which king started the organization of the Kumbh fair class 8 social science CBSE

The first matrimonial alliance with the Rajput was class 8 social science CBSE

Application to your principal for the character ce class 8 english CBSE

When will we use have had and had had in the sente class 8 english CBSE

Distinguish between SouthWest and NorthEast monsoo class 8 social science CBSE
