Question

Mr T invested an amount of Rs13900 divided in two different schemes A and B at the simple interest rate of 14/ and 11/ p.a. respectively if the total amount of simple interest earned in 2 years is Rs3508 what was the amount invested initially in both the scheme​

Answer 1

Answer:-

Given:

Principle (P) = Rs. 13900

Time (T) = 2 years

Rate of interest (R) = 14 % & 11% p.a

Sum of simple interest of both the schemes = Rs. 3508

Let us assume that,

Principle of scheme A be x.

So, Principle of scheme B = Rs. (13900 - x)

[ since , total principle is Rs. 13900 ]

We know that,

Simple interest (SI) = PTR/100

So,

S.I of scheme - A :

⟶ SI (A) = (x)(2)(14)/100

⟶ SI (A) = Rs. 7x/25

SI of scheme - B :

⟶ SI (B) = (13900 - x)(2)(11)/100

⟶ SI (B) = 11(13900 - x)/50

⟶ SI (B) = 152900 - 11x / 50

Now,

SI(A) + SI(B) = 3508

⟶ 7x/25 + 152900 - 11x/50 = 3508

⟶ [ 2(7x) + 152900 - 11x ] / 50 = 3508

⟶ 14x + 152900 - 11x = 3508 × 50

⟶ 3x = 175400 - 152900

⟶ 3x = 22500

⟶ x = 22500/3

⟶ x = Rs. 7500

∴

• Amount invested in scheme A = x = Rs. 7500.

• Amount invested in scheme B = 13900 - x = 13900 - 7500 = Rs. 6400

Answer 2

Question :-

Mr T invested an amount of Rs13900 divided in two different schemes A and B at the simple interest rate of 14/ and 11/ p.a. respectively if the total amount of simple interest earned in 2 years is Rs3508 what was the amount invested initially in both the scheme.

Answer :-

$\large\sf\underline\red{Given:}$

• Principal (P) = Rs. 13900

• Rate of interest (R) = 14 % and 11%

• Time (T) = 2 years

• Total amount of simple interest = Rs.3508

$\large\sf\underline\red{To\:find:}$

what was the amount invested initially in both the scheme ?

$\large\sf\underline\red{Solution:}$

Let the investment in scheme A be Rs. x

and the investment in scheme B b Rs. (13900-x)

We know that,

$\boxed{\bf{\purple{SI=\dfrac{PRT}{100}}}}$

simple interest for Rs. x in 2 yeras at 14% p.a :-

$\sf{\implies \dfrac{x×14×2}{100}=\dfrac{28}{100}}$

simple interest for Rs.(13900-x) in 2 years at 11% p.a :-

$\sf{\implies \dfrac{(13900-x)×11×2}{100}=\dfrac{22(13900-x)}{100}}$

Total interest = Rs. 3508

ATQ,

$\sf{\implies \dfrac{28x}{100}+\dfrac{22(13900-x)}{100}=3508}$

$\sf{\implies 28x +305800 - 22x = 350800}$

$\sf{\implies 6x = 4500}$

$\sf{\implies x =\dfrac{4500}{6}}$

$\sf{\implies x = 7500}$

Therefore,

The amount invested in scheme A = Rs.7500 and

Investment in scheme B = Rs. (13900-7500)= Rs. 6400

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