# 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}} \\

$

Answer

Verified

363k+ 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.

Last updated date: 26th Sep 2023

â€¢

Total views: 363k

â€¢

Views today: 11.63k

Recently Updated Pages

What do you mean by public facilities

Paragraph on Friendship

Slogan on Noise Pollution

Disadvantages of Advertising

Prepare a Pocket Guide on First Aid for your School

10 Slogans on Save the Tiger

Trending doubts

How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Difference Between Plant Cell and Animal Cell

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

The equation xxx + 2 is satisfied when x is equal to class 10 maths CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

Drive an expression for the electric field due to an class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

What is the past tense of read class 10 english CBSE