Question

A person invested some amount at the rate of $12\%$ simple interest and some other amount at the rate of $10\%$ simple interest. He received yearly interest of rupees 130. But if he had interchanged the amount received, he would have received Rs. 4 more as interest. How much did he invest at different rates?

Answer 1

Hint: In this particular question use the concept that assumes any different variable be the principal amount a person invested at the rate of $12\%$ and $10\%$ respectively, then use the formula of simple interest to calculate the interest which is given as, $S.I = \dfrac{{p.r.t}}{{100}}$ where symbols have their usual meaning so use these concepts to reach the solution of the question.

Complete step-by-step solution
Given data:
A person invested some amount at the rate of $12\%$ simple interest and some other amount at the rate of $10\%$ simple interest.
Let’s he invested Rs x as principal at $12\%$ simple interest and Rs y as principal at $10\%$ simple interest.
Now as we know that the formula of simple interest to calculate the interest is given as, $S.I = \dfrac{{p.r.t}}{{100}}$, where p = principal amount, r = rate of interest and t = time in years.
Now it is given that he receives the yearly interest of Rs. 130.
So the time of investment = 1 year.
Now when the rate of interest is $12\%$.
So ${P_1}$ = x Rs, ${r_1}$ = $12\%$, and ${t_1} = 1$ year.
So the simple interest is, ${\left( {S.I} \right)_1} = \dfrac{{{p_1}.{r_1}.{t_1}}}{{100}} = \dfrac{{x.12.1}}{{100}} = \dfrac{{12x}}{{100}} = \dfrac{{3x}}{{25}}$
Now when the rate of interest is $10\%$.
So ${P_2}$ = y Rs, ${r_2}$ = 10%, and ${t_2} = 1$ year.
So the simple interest is, \[{\left( {S.I} \right)_2} = \dfrac{{{p_2}.{r_2}.{t_2}}}{{100}} = \dfrac{{y.10.1}}{{100}} = \dfrac{{10y}}{{100}} = \dfrac{y}{{10}}\]
Now the sum of interest is, ${\left( {S.I} \right)_1} + {\left( {S.I} \right)_2} = 130$
$ \Rightarrow \dfrac{{3x}}{{25}} + \dfrac{y}{{10}} = 130$
\[ \Rightarrow \dfrac{{6x + 5y}}{{50}} = 130\]
\[ \Rightarrow 6x + 5y = 6500\]............... (1)
Now when he interchanges the amount he receives Rs 4 more as interest.
So the total interest he receives = $130 + 4 = Rs. 134$
$ \Rightarrow {\left( {S.I} \right)_1} + {\left( {S.I} \right)_2} = 134$
Now the interchanges the amount i.e. Rs y is invested at the rate of 12% and Rs. x invested at a rate of 10%.
$ \Rightarrow {\left( {S.I} \right)_1} = \dfrac{{y.12.1}}{{100}} = \dfrac{{3y}}{{25}}$
And
$ \Rightarrow {\left( {S.I} \right)_2} = \dfrac{{x.10.1}}{{100}} = \dfrac{x}{{10}}$
$ \Rightarrow \dfrac{{3y}}{{25}} + \dfrac{x}{{10}} = 134$
$ \Rightarrow \dfrac{{6y + 5x}}{{50}} = 134$
$ \Rightarrow 6y + 5x = 6700$................. (2)
Now from equation (1), we have,
$ \Rightarrow x = \dfrac{{6500 - 5y}}{6}$................. (3)
Now substitute this value in equation (2) we have,
$ \Rightarrow 6y + 5\left( {\dfrac{{6500 - 5y}}{6}} \right) = 6700$
Now simplify we have,
$ \Rightarrow 36y + \left( {32500 - 25y} \right) = 6700\left( 6 \right)$
$ \Rightarrow 11y = 40200 - 32500 = 7700$
$ \Rightarrow y = \dfrac{{7700}}{{11}} = 700$ Rs.
Now substitute this value in equation (3) we have,
$ \Rightarrow x = \dfrac{{6500 - 5\left( {700} \right)}}{6} = \dfrac{{6500 - 3500}}{6} = \dfrac{{3000}}{6} = 500$ Rs.
So he invested 500 Rs at the rate of $12\%$ and 700 Rs at the rate of $10\%$ initially. So this is the required answer.

Note: Whenever we face such types of questions the key concept we have to remember is that there are lots of methods to solve a pair of linear equations such as substitution, elimination, graphical, and cross multiplication method here we use substitution method, in elimination method multiply first equation with the coefficient of x of the second equation and multiply second equation with the coefficient of x of the first equation and subtract them so that variable x cancels out only variable remain is variable y so solve it and substitute it in any linear equation we will get the required answer.