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# Peter invested an amount of Rs.12000 at a rate of 10% p.a. simple interest and another amount at a 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.20,000B. Rs.22,000C. Rs.24,000D. Rs.25,000

Last updated date: 13th Jun 2024
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Hint: The given question is a simple interest problem so, the formula for the simple interest is given by, $S.I = \dfrac{{P \times T \times R}}{{100}}$, where S.I is the simple interest, P is the principal, T is the number of years invested, and R is the rate of interest per annum, now using the formula and substituting the value given values in the given formula we can solve the given question.

Complete step-by-step solution:
Given Peter invested an amount of Rs.12000 at a rate of 10% p.a. simple interest and another amount at a 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, now we have to find the total amount invested.
We know that the formula for the simple interest is given by,
$S.I = \dfrac{{P \times T \times R}}{{100}} - - - - (1)$,
From the first investment, Peter invested an amount of Rs.12000 at a rate of 10% p.a. simple interest,
Here from the given question,
$P = Rs.12000$,
$T = 1$, and
$R = 10$.
Now substituting the values in the simple interest formula we get,
$S.{I_1} = \dfrac{{12000 \times 1 \times 10}}{{100}}$,
Now simplifying the above equation we get,
$S.{I_1} = \dfrac{{120000}}{{100}}$,
Now dividing the term on R.H.S, we get,
$S.{I_1} = Rs.1200 - - - - (2)$,
Hence the interest is Rs.1200.
Let another amount that is invested be Rs.$x$, then the interest earned at the end of the one year at the rate of 20% p.a. is given by using the formula of simple interest, we get,
$S.{I_2} = \dfrac{{{P_2} \times {T_2} \times {R_2}}}{{100}}$,
Now from the given data,
${P_2} = Rs.x$ ,
${T_2} = 1$, and
${R_2} = 20$,
Now substituting these values in the formula , we get,
$S.{I_2} = \dfrac{{x \times 1 \times 20}}{{100}}$,
Now simplifying the above we get,
$S.{I_2} = \dfrac{x}{5} - - - - - (3)$,
Now from the above given data the total interest earned at the end of one year is given at a rate of 14% of the total amount invested.
$\therefore$The total amount invested is$Rs.\left( {12000 + x} \right)$.
Now using equation (1) we get,
${I_{total}} = \dfrac{{{P_{total}} \times {I_{total}} \times {R_{total}}}}{{100}}$
Here from the given data, ${P_{total}} = Rs.\left( {12000 + x} \right)$,
${T_{total}} = 1$,
${R_{total}} = 14\%$,
Now substituting these values we get,
${I_{total}} = \dfrac{{\left( {12000 + x} \right) \times 1 \times 14}}{{100}}$,
Now we know that the total interest earned is the sum of two interests, i.e.,
${I_{total}} = S.{I_1} + S.{I_2}$,
Now substituting the values we got in the above equation, we get,
$\Rightarrow 1200 + \dfrac{x}{5} = \dfrac{{\left( {12000 + x} \right) \times 1 \times 14}}{{100}}$,
Now simplifying the right hand side by releasing the brackets we get,
$\Rightarrow 1200 + \dfrac{x}{5} = \dfrac{{12000 \times 1 \times 14}}{{100}} + \dfrac{{x \times 1 \times 14}}{{100}}$,
Now calculating and simplifying we get,
$\Rightarrow 1200 + \dfrac{x}{5} = 1680 + \dfrac{{7x}}{{50}}$,
Taking $x$terms to one side and constants to other side we get,
$\Rightarrow \dfrac{x}{5} - \dfrac{{7x}}{5} = 1680 - 1500$,
Now taking L.C.M on the L.H.S and solving , we get,
$\Rightarrow \dfrac{{10x - 7x}}{{50}} = 1680 - 1200$,
Now further simplifying we get,
$\Rightarrow \dfrac{{3x}}{{50}} = 480$,
Now taking all constant terms to one side we get,
$\Rightarrow x = \dfrac{{480 \times 50}}{3}$
Now further simplifying we get,
$x = 8000$,
The second investment is Rs.8000,
$\therefore$The total amount invested is Rs (12000+8000), which is Rs.20,000.

Option A is the correct answer.

Note: In these types of questions there will be two unknown values, the interest earned from the second amount and the amount, so here we acquire two equations which are needed to be solved to get the required values, these can be done by using the simple interest formula.