 QUESTION

# Peter invested an amount of Rs. 12,000 at the rate of 10 % p.a. simple interest and another amount at the rate of 20 % p.a. simple interest. The total interest earned at the end of one year on the total amount invested became 14 % p.a. Find the total amount invested.(a). Rs. 20,000(b). Rs. 22,000(c). Rs. 24,000(d). Rs. 25,000

Hint: The formula for simple interest is $S.I. = \dfrac{{P \times N \times R}}{{100}}$, where S.I. is the simple interest, P is the principal amount, R is the rate p.a. and N is the number of years. Use this formula to find the interests for the invested amounts at the rate of 10 % p.a and 20 % p.a. respectively and equate it to the interest gained from the total amount at the rate of 14 % p.a.

It is given that Peter invested an amount of Rs. 12,000 at the rate of 10
% annually for one year.
We know the formula for simple interest S.I. earned in N years for the principal amount P at the rate of R p.a. is given as follows:
$S.I. = \dfrac{{P \times N \times R}}{{100}}............(1)$
Hence, the interest earned from Rs. 12,000 at the rate of 10 % in one year is given as follows:
${I_1} = \dfrac{{12,000 \times 1 \times 10}}{{100}}$
${I_1} = Rs.1,200...........(2)$
Hence, the interest is Rs. 1,200.
Let another amount invested be Rs. X. Then, the interest earned at the end of one year at the rate of 20 % p.a. is given by equation (1) as follows:
${I_2} = \dfrac{{X \times 1 \times 20}}{{100}}$
${I_2} = \dfrac{X}{5}...........(3)$
The total interest earned is given at the rate of 14 % p.a. of the total amount invested.
The total amount invested is Rs. (12,000 + X).
Using equation (1), we have:
${I_{total}} = \dfrac{{(12,000 + X) \times 1 \times 14}}{{100}}$
We know that the total interest earned is the sum of ${I_1}$ and ${I_2}$.
${I_1} + {I_2} = \dfrac{{(12,000 + X) \times 1 \times 14}}{{100}}.........(4)$
Using equation (2) and equation (3) in equation (4), we have:
$1,200 + \dfrac{X}{5} = \dfrac{{(12,000 + X) \times 1 \times 14}}{{100}}$
Separating the terms in the bracket of the right-hand side of the equation, we get:
$1,200 + \dfrac{X}{5} = \dfrac{{12,000 \times 1 \times 14}}{{100}} + \dfrac{{X \times 1 \times 14}}{{100}}$
Simplifying, we get:
$1,200 + \dfrac{X}{5} = 1680 + \dfrac{{7X}}{{50}}$
Taking all X terms to the left-hand side of the equation, we get:
$\dfrac{X}{5} - \dfrac{{7X}}{{50}} = 1680 - 1,200$
Solving for X, we get:
$\dfrac{{10X - 7X}}{{50}} = 1680 - 1,200$
$\dfrac{{3X}}{{50}} = 480$
$X = \dfrac{{480 \times 50}}{3}$
$X = Rs.8000$
Hence, the total amount invested is Rs. (12,000 + 8,000) which is Rs. 20,000.

Note: There are two unknowns, the interest earned from the second amount and the second amount. So you need two equations relating them to solve for them. One comes from the simple interest formula for the second amount and the other from the simple interest formula for the total amount.