Answer
385.5k+ 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
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Why Are Noble Gases NonReactive class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let X and Y be the sets of all positive divisors of class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
At which age domestication of animals started A Neolithic class 11 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which are the Top 10 Largest Countries of the World?
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give 10 examples for herbs , shrubs , climbers , creepers
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Write a letter to the principal requesting him to grant class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Change the following sentences into negative and interrogative class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)