Question

(3) The simple interest on a certain sum for 7 years at 11% per annum is 400 more than
the simple interest on the same sum for 5 years at 14% per annum. Find the sum.
a. 5728.29
b. 5714.29
c. 5716.29
d. 5722.29​

Answer 1

Given :

$\sf\:1^{st}\:case\::$

• Time = 7 yrs

• Rate = 11 %

$\sf\:2^{nd}\:case\::$

• Time = 5 yrs

• Rate = 14 %

➷ The question also says that in both the cases the Principal is same .

➷ It is also said that the Simple Interest earned in First case is 400 more than the Simple Interest earned in Second case.

‎

‎

To find :

• The principal on the both cases

‎

‎

Concept :

In order to solve this question we would frame up an equation. Equation needs to be framed between the Simple Interest earned on the first case followed by the Simple interest earned on the Second case. So firstly we would calculate the SI earned on both the cases. Then after equating both we would get our required sum or Principal.

Hope am clear let's solve !~

‎

‎

Formula :

$\sf\color{navy}{\maltese}$ $\tt\:SI~=~\frac{P \times R \times T}{100}$

‎

Here :

‎

$\sharp$ Simple Interest ( SI ) $\twoheadrightarrow$ Additional amount of money which is received at the end of fixed time for a certain sum of money deposited at a fixed rate .

‎

$\sharp$ Principal ( P ) $\twoheadrightarrow$ The amount of money deposited for a fixed interval of time in an organization or is taken as a loan.

‎

$\sharp$ Rate ( R ) $\twoheadrightarrow$ Rate is simply a percentage in which an interest is to be determined.

‎

‎

$\sharp$ Time ( T ) $\twoheadrightarrow$ The period for which the money is to be deposited.

‎

‎

Solution :

Let the Principal for both the cases be Rs P.

Calculating the Simple Interest for the first case.

• Time = 7 yrs and Rate = 11 %

$\tt\therefore\:SI~=~\frac{P \times 11 \times 7}{100}$

$\tt\implies\:SI~=~\frac{P \times 77}{100}$

$\small{\mathfrak{\implies\:SI~=~\frac{77P}{100}}}$

‎

Calculating the Simple Interest for the second case.

• Time = 5 yrs and Rate = 14 %

$\tt\therefore\:SI~=~\frac{P \times 14 \times 5}{100}$

$\tt\implies\:SI~=~\frac{P \times 70}{100}$

$\small{\mathfrak{\implies\:SI~=~\frac{70P}{100}}}$

‎

According to the question :

Simple interest earned on First case = 400 + Simple Interest earned on the second case

Substituting the values we got :

$\tt\implies\:\frac{77P}{100}=400+\frac{70P}{100}$

LCM of 1 and 100 in RHS = 100

$\tt\implies\:\frac{77P}{100}=\frac{(100 \times 400) + (1 \times 70P)}{100}$

$\tt\implies\:\frac{77P}{100}=\frac{40000+70P}{100}$

Cross multiplying

$\tt\implies\:100 \times 77P = 100(40000+70P)$

$\tt\implies\:7700P = 4000000+7000P$

Transposing +7000P from RHS to LHS it becomes -7000P

$\tt\implies\:7700P - 7000P = 4000000$

$\tt\implies\:700P=4000000$

Transposing 700P to RHS it goes to the denominator

$\tt\implies\:P=\frac{4000000}{700}$

Reducing the fraction to the lower terms

$\tt\implies\:P=\frac{40000\cancel{00}}{7\cancel{00}}$

$\tt\implies\:P=\frac{40000}{7}$

$\small{\underline{\boxed{\mathrm\red{\implies\:P=Rs\:5714.29}}}}$ ★

‎

_______________

$\dag\:\underline{\sf So\:the\:required\:sum\:of\:money\:is\:Rs\:5714.29}$

‎

Therefore $\sf\color{blue}{option~b}$ i.e $\sf\color{purple}{Rs~5714.29}$ is correct.

‎

Answer 2

$Given :</p><p></p><p>[tex]\sf\:1^{st}\:case\::$

Time = 7 yrs

Rate = 11 %

$\sf\:2^{nd}\:case\::$

Time = 5 yrs

Rate = 14 %

➷ The question also says that in both the cases the Principal is same .

➷ It is also said that the Simple Interest earned in First case is 400 more than the Simple Interest earned in Second case.

‎

‎

To find :

The principal on the both cases

‎

‎

Concept :

In order to solve this question we would frame up an equation. Equation needs to be framed between the Simple Interest earned on the first case followed by the Simple interest earned on the Second case. So firstly we would calculate the SI earned on both the cases. Then after equating both we would get our required sum or Principal.

Hope am clear let's solve !~

‎

‎

Formula :

$\sf\color{navy}{\maltese}$ $\tt\:SI~=~\frac{P \times R \times T}{100}$

‎

Here :

‎

$\sharp$ Simple Interest ( SI ) $\twoheadrightarrow$ Additional amount of money which is received at the end of fixed time for a certain sum of money deposited at a fixed rate .

‎

$\sharp$ Principal ( P ) $\twoheadrightarrow$ The amount of money deposited for a fixed interval of time in an organization or is taken as a loan.

‎

$\sharp$ Rate ( R ) $\twoheadrightarrow$ Rate is simply a percentage in which an interest is to be determined.

‎

‎

$\sharp$ Time ( T ) $\twoheadrightarrow$ The period for which the money is to be deposited.

‎

‎

Solution :

Let the Principal for both the cases be Rs P.

Calculating the Simple Interest for the first case.

Time = 7 yrs and Rate = 11 %

$\tt\therefore\:SI~=~\frac{P \times 11 \times 7}{100}$

$\tt\implies\:SI~=~\frac{P \times 77}{100}$

$\small{\mathfrak{\implies\:SI~=~\frac{77P}{100}}}$

‎

Calculating the Simple Interest for the second case.

Time = 5 yrs and Rate = 14 %

$\tt\therefore\:SI~=~\frac{P \times 14 \times 5}{100}$

$\tt\implies\:SI~=~\frac{P \times 70}{100}$

$\small{\mathfrak{\implies\:SI~=~\frac{70P}{100}}}$

‎

According to the question :

Simple interest earned on First case = 400 + Simple Interest earned on the second case

Substituting the values we got :

$\tt\implies\:\frac{77P}{100}=400+\frac{70P}{100}$

LCM of 1 and 100 in RHS = 100

$\tt\implies\:\frac{77P}{100}=\frac{(100 \times 400) + (1 \times 70P)}{100}$

$\tt\implies\:\frac{77P}{100}=\frac{40000+70P}{100}$

Cross multiplying

$\tt\implies\:100 \times 77P = 100(40000+70P)$

$\tt\implies\:7700P = 4000000+7000P$

Transposing +7000P from RHS to LHS it becomes -7000P

$\tt\implies\:7700P - 7000P = 4000000$

$\tt\implies\:700P=4000000$

Transposing 700P to RHS it goes to the denominator

$\tt\implies\:P=\frac{4000000}{700}$

Reducing the fraction to the lower terms

$\tt\implies\:P=\frac{40000\cancel{00}}{7\cancel{00}}$

$\tt\implies\:P=\frac{40000}{7}$

$\small{\underline{\boxed{\mathrm\red{\implies\:P=Rs\:5714.29}}}}$ ★

‎

_______________

$\dag\:\underline{\sf So\:the\:required\:sum\:of\:money\:is\:Rs\:5714.29}$

‎

Therefore $\sf\color{blue}{option~b}$ i.e $\sf\color{purple}{Rs~5714.29}$ is correct.

‎[/tex]