Question

A man invested 10,000 rupees, split into two schemes, at annual rates of interest 8 % and 9%. After one year he got 875 rupees as interest from both. How much did he invest in each?

Answer 1

Hint: According to the question 10,000 rupees are invested at different annual rates. So, assume the invested amount at the annual rate of interest 8 % is $x$ and the remaining amount out of 10,000 rupees are invested at the rate of 9%.
Now, determine the Simple interest for the principal amount $x$ at the rate of 8 %. Again, find the Simple interest for the principal amount $10000 - x$ at the rate 9 %. Add both the Simple interest and equate to the interest of one year that is 875 rupees. And, solve the equation for $x$.
The Formula for Simple interest is , $SI = \dfrac{{P \times R \times T}}{{100}}$
Where P is the Principal amount, R is the rate of interest and T is the time.
After substituting the value of $x$ into $10000 - x$ we get the remaining amount invested at the rate of 9%.

Complete step-by-step solution:
Given is the 10,000 rupees are invested at the different rates.
The time for the amount is invested in one year.
Suppose $x$ rupees are invested at the rate of 8% and the remaining amount is invested at the rate of 9% is,
$10000 - x$
We now find the simple interest at both rates of interest.
The Simple interest can be found by,
$SI = \dfrac{{P \times R \times T}}{{100}}$
P=Principle Amount
R=Rate of Interest
T=Time
First, determine the simple interest of one year for the rate of interest 8%,
Here, the principal amount is $P = x$, rate of interest is $R = 8$%, and time $T = 1$ year.
Substitute all the values into the formula, $SI = \dfrac{{P \times R \times T}}{{100}}$
$SI = \dfrac{{x \times 8 \times 1}}{{100}}$
$SI = \dfrac{{8x}}{{100}} \ldots \ldots (1)$
Now, determine the simple interest of one year for the rate of interest 9%,
Consider the principal amount is $P = 10000 - x$, rate of interest is $R = 9$%, and time $T = 1$ year.
Substitute all the values into the formula, $SI = \dfrac{{P \times R \times T}}{{100}}$
$ \Rightarrow SI = \dfrac{{(10000 - x) \times 9 \times 1}}{{100}}$
$SI = \dfrac{{9(10000 - x)}}{{100}} \ldots \ldots (2)$
Now, the combined interest at the rate of interests 8% and 9% is obtained by the addition of the individual simple interest. So, add the equation $(1)$ and the equation $(2)$ we get,
$ \Rightarrow SI = \dfrac{{8x}}{{100}} + \dfrac{{9(10000 - x)}}{{100}} \ldots \ldots (3)$
After one year the combined interest found from both the schemes is 875 rupees . $\therefore $ substitute \[SI = 875\] into the equation $(3)$ .
$875 = \dfrac{{8x}}{{100}} + \dfrac{{9(10000 - x)}}{{100}}$
Solve the equation to find the value of $x$.
$ \Rightarrow 875 = \dfrac{{8x + 9(10000 - x)}}{{100}}$
Simplify the bracket in the right-hand side of the equation and cross multiply,
\[ \Rightarrow 87500 = 8x + 90000 - 9x\]
\[ \Rightarrow 87500 = 90000 - x\]
Simplify further calculation of the equation,
\[ \Rightarrow x = 90000 - 87500\]
\[ \Rightarrow x = 2500\]
The amount that is invested at a rate of 8% is $2500$rupees.
The amount that is invested at the rate 9% is $10000 - x$
Put \[x = 2500\] into $10000 - x$ we get,
$10000 - 2500 = 7500$
The amount that is invested at the rate 9% is 7500 rupees.

The amount that is invested at the rate of 8% is $2500$ rupees and The amount that is invested at the rate of 9% is 7500 rupees.

Note: The important steps to be remembered are to find the simple interest individual at both rates of interest and add both the interest that is equal to the one year interest obtained as 875 rupees. Sometimes students got confused with the formulas for finding simple interest and compound interest. Here are the formulas are given below,
Simple Interest:
$SI = \dfrac{{P \times R \times T}}{{100}}$
P=Principle Amount
R=Rate of Interest
T=Time
Compound interest:
$A = P{\left( {1 + \dfrac{r}{n}} \right)^{nt}}$
A=Amount
P=Principle Amount
R=Rate of Interest
T=Time
N=Number of times interest is compounded per unit time