Question

Divide \[{\rm{Rs}}1600\] into two parts in such a way that simple interest on the first part at \[10\% \] for 2 years is equal to the interest on the other part at \[8\% \] for \[1\dfrac{1}{2}\] years.

Answer 1

Hint:
Here, we will assume the first part to be \[x\] and subtract it from the given amount to get the second part. We will then find the simple interest for both the parts for the given period and rate using the formula of simple Interest. Then, we will equate both of the obtained interests to find the value of \[x\] and hence, both the required parts.

Formula Used:
\[S.I = \dfrac{{P.R.T}}{{100}}\], where, \[S.I\] is the Simple Interest, \[P\] is the Principal, \[R\] is the rate of interest per annum and \[T\] is the time period.

Complete Step by Step Solution:
Let the total principal Rs 1600 be divided into two parts such that:
First part, i.e. the first Principal, \[{P_1} = x\]
Hence, the second part or the second Principal, \[{P_2} = \left( {1600 - x} \right)\]
Now, according to the question, for the first part:
Rate of interest\[ = {R_1}\% = 10\% \]
Time, \[{T_1} = 2\] years
And, for the second part:
Rate of interest\[ = {R_2}\% = 8\% \]
Time, \[{T_2} = 1\dfrac{1}{2} = \dfrac{3}{2}\] years
Now we will find the interest using the simple interest formula \[S.I = \dfrac{{P \cdot R \cdot T}}{{100}}\].
Therefore, the Simple Interest for the first part will be \[S.{I_1} = \dfrac{{{P_1} \cdot {R_1} \cdot {T_1}}}{{100}}\].
Now, substituting the values, \[{T_1} = 2\] years , \[{R_1} = 10\] and \[{P_1} = x\] in the above equation, we get,
\[S.I{._1} = \dfrac{{\left( x \right) \cdot \left( {10} \right) \cdot \left( 2 \right)}}{{100}}\]……………………………..\[\left( 1 \right)\]
Similarly, the Simple Interest for the second part will be \[S.I{._2} = \dfrac{{{P_2} \cdot {R_2} \cdot {T_2}}}{{100}}\].
Now, substituting the values, \[{T_2} = 1\dfrac{1}{2} = \dfrac{3}{2}\] years, \[{R_2} = 8\] and \[{P_2} = \left( {1600 - x} \right)\] in the above equation, we get,
\[S.I{._2} = \dfrac{{\left( {1600 - x} \right) \cdot \left( 8 \right) \cdot \left( 3 \right)}}{{100\left( 2 \right)}}\]…………………………….\[\left( 2 \right)\]
Now, it is given that the simple interest on the first part at \[10\% \] for 2 years is equal to the interest on the other part at \[8\% \] for \[1\dfrac{1}{2}\]years.
Hence, equating \[\left( 1 \right)\] and \[\left( 2 \right)\], we get,
\[\dfrac{{\left( x \right) \cdot \left( {10} \right) \cdot \left( 2 \right)}}{{100}} = \dfrac{{\left( {1600 - x} \right) \cdot \left( 8 \right) \cdot \left( 3 \right)}}{{100\left( 2 \right)}}\]
\[ \Rightarrow \left( x \right) \cdot \left( {10} \right) \cdot \left( 2 \right) = \left( {1600 - x} \right) \cdot \left( 4 \right) \cdot \left( 3 \right)\]
Now, multiplying the brackets,
\[ \Rightarrow 20x = \left( {1600 - x} \right) \cdot \left( {12} \right)\]
\[ \Rightarrow 20x = 19200 - 12x\]
Adding \[12x\] on both sides, we get
\[ \Rightarrow 32x = 19200\]
Dividing both sides by 32, we get
\[ \Rightarrow x = 600\]
Therefore, the first part or the first Principal, \[{P_1} = x = {\rm{Rs}}600\]
And, the second part or the second Principal, \[{P_2} = \left( {1600 - 600} \right) = {\rm{Rs}}1000\]
Therefore, \[{\rm{Rs}}1600\] is divided into two parts, i.e. \[{\rm{Rs}}600\] and \[{\rm{Rs}}1000\] such that the simple interest on the first part at \[10\% \] for 2 years is equal to the interest on the other part at \[8\% \] for \[1\dfrac{1}{2}\] years.

Hence, this is the required answer.

Note:
In this question we have used the formula of Simple Interest. Simple Interest is the interest earned on the Principal or the amount of loan. There is another type of interest, which is the Compound Interest. Compound Interest is calculated both on the Principal as well as on the accumulated interest of the previous year. Hence, this is also known as ‘interest on interest’.