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 Rs.130. But if he had interchanged the amount invested, he would have received Rs.4 more as interest. How much did he invest at different rates?

Answer 1

Hint: Before attempting this question, one should have prior knowledge about the concept of Simple interest also remember to use the formula of simple interest i.e. $S.I = \dfrac{{P \times R \times T}}{{100}}$, use this information to approach the solution.

Complete step-by-step answer:
Case first:
According to the given information it is given that when a person invests some amount at rate of 12% simple interest and other amount at rate of 10% simple interest so he receives Rs. 130 yearly invest
So, it is given that yearly interest = Rs. 130
Now let, X be the amount person invested at rate of 12% and Y be the amount person invested at rate of 10%
So, the given equation will be simple interest on X amount at rate of 12% + simple interest on Y amount at rate of 10% = Rs. 130
As we know that formula of simple interest is given as $S.I = \dfrac{{P \times R \times T}}{{100}}$Where, P = Principle value, R = Rate of interest and T = Time
Now for first case substituting the values in the given equation using the formula of simple interest we get
$\dfrac{{12 \times X \times 1}}{{100}} + \dfrac{{10 \times Y \times 1}}{{100}} = 130$
$ \Rightarrow $$12X + 10Y = 13000$ (equation 1)
Case second:
Now it is said that when the person interchanged the invested amount, he received Rs. 4 more than before
So, the equation will be Simple interest on X amount at rate of 10% + Simple interest on Y amount at rate of 12% = Rs. (130 + 4)
Now using the simple interest formula substituting the values in the above equation we get
$\dfrac{{10 \times X \times 1}}{{100}} + \dfrac{{12 \times Y \times 1}}{{100}} = 130 + 4$
$ \Rightarrow $$10X + 12Y = 13400$ (equation 2)
Now multiplying equation 1 by 12 and equation 2 by 10 we get
$12\left( {12X + 10Y = 13000} \right) = 144X + 120Y = 156000$ and $10\left( {10X + 12Y = 13400} \right) = 100X + 120Y = 134000$
Now subtracting resulting equation 1 by equation 2 we get
$144X + 120Y - 100X - 120Y = 156000 - 134000$
$ \Rightarrow $$44X = 22000$
$ \Rightarrow $$X = \dfrac{{22000}}{{44}}$
$ \Rightarrow $X = Rs. 500
Now substituting the value of X in the equation 2 we get
$10\left( {500} \right) + 12Y = 13400$
$ \Rightarrow $$12Y = 13400 - 5000$
$ \Rightarrow $$12Y = 8400$
$ \Rightarrow $$Y = \dfrac{{8400}}{{12}}$
$ \Rightarrow $Y = Rs. 700
Therefore, He invested Rs.500 at 12% and Rs.700 at 10%

Note: In these types of questions always remember to carefully read the given information in the question and identify the given values as in the above solution we had principle value, interest rate and time so we directly formed the equation as it was given that person received Rs. 130 yearly in first case and then we formed the second equation where it was given that the amount of investment have interchanged so we interchanged the rate of simple interest in equation and the yearly interest person received which was 4 more than before after obtaining both the equations we directly multiplied both the equation with some number in order to make coefficient of the same variable in the both the equation equal and hence we got our invested amount.