# Mr. Thomas invested an amount of $Rs.13,900$ divided into two different schemes ${\text{A}}$

and ${\text{B}}$ at the simple interest rate of $14\% p.a$ and $11\% p.a.$respectively. If the total

amount of simple interest earned in 2 years be $Rs.3508,$what was the amount invested

in scheme ${\text{B}}$?

$

{\text{A}}{\text{. Rs}}{\text{.6400}} \\

{\text{B}}{\text{. Rs}}{\text{.6500}} \\

{\text{C}}{\text{. Rs}}{\text{.7200}} \\

{\text{D}}{\text{. Rs}}{\text{.7500}} \\

{\text{E}}{\text{. none of these}} \\

$

Last updated date: 20th Mar 2023

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Answer

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306.6k+ views

Hint: You have to find amount invested in scheme ${\text{B}}$ so assume it $x$ and use simple interest formula and what is given in question and you will find amount invested in scheme ${\text{B}}$.

For scheme ${\text{A}} \to {\text{14\% p}}{\text{.a}}$ and for scheme ${\text{B}} \to {\text{11\% p}}{\text{.a}}$

Let amount $x$ has been invested in scheme ${\text{B}}$.

So amount invested in scheme ${\text{A = 13900 - }}x$

Amount earned from simple interest in scheme ${\text{A = }}\dfrac{{\left( {13900 - x} \right) \times 14 \times 2}}{{100}}$$\left( {\because s.i = \dfrac{{p \times r \times t}}{{100}}} \right)$

Amount earned from simple interest in scheme ${\text{B = }}\dfrac{{x \times 11 \times 2}}{{100}}$

As given in question total amount earned from both scheme in two year is ${\text{R}}{\text{s}}{\text{.3508}}$

$\therefore \dfrac{{\left( {13900 - x} \right) \times 14 \times 2}}{{100}} + \dfrac{{x \times 11 \times 2}}{{100}} = 3508$

$ \Rightarrow \dfrac{{13900 \times 28 - 28x + 22x}}{{100}} = 3508$

$ \Rightarrow - 6x + 389200 = 350800$

$ \Rightarrow - 6x = - 38400 \Rightarrow x = 6400$

Hence option ${\text{A }}$is the correct option.

Note: Whenever you get this type of question the key concept of solving is if you have a principal rate and time then just apply the simple interest formula and get the answer. In this question only you have to notice the principal and use what is given in the question.

For scheme ${\text{A}} \to {\text{14\% p}}{\text{.a}}$ and for scheme ${\text{B}} \to {\text{11\% p}}{\text{.a}}$

Let amount $x$ has been invested in scheme ${\text{B}}$.

So amount invested in scheme ${\text{A = 13900 - }}x$

Amount earned from simple interest in scheme ${\text{A = }}\dfrac{{\left( {13900 - x} \right) \times 14 \times 2}}{{100}}$$\left( {\because s.i = \dfrac{{p \times r \times t}}{{100}}} \right)$

Amount earned from simple interest in scheme ${\text{B = }}\dfrac{{x \times 11 \times 2}}{{100}}$

As given in question total amount earned from both scheme in two year is ${\text{R}}{\text{s}}{\text{.3508}}$

$\therefore \dfrac{{\left( {13900 - x} \right) \times 14 \times 2}}{{100}} + \dfrac{{x \times 11 \times 2}}{{100}} = 3508$

$ \Rightarrow \dfrac{{13900 \times 28 - 28x + 22x}}{{100}} = 3508$

$ \Rightarrow - 6x + 389200 = 350800$

$ \Rightarrow - 6x = - 38400 \Rightarrow x = 6400$

Hence option ${\text{A }}$is the correct option.

Note: Whenever you get this type of question the key concept of solving is if you have a principal rate and time then just apply the simple interest formula and get the answer. In this question only you have to notice the principal and use what is given in the question.

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