7676\frac{7}{6} of itself in 3 years at a certain rate of simple interest. The rate of interest per annum is ?" /> 7676\frac{7}{6} of itself in 3 years at a certain rate of simple interest. The rate of interest per annum is ?" /> 7676\frac{7}{6} of itself in 3 years at a certain rate of simple interest. The rate of interest per annum is ?" />

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A sum of money becomes $\frac{7}{6}$ $\frac{7}{6}$ of itself in 3 years at a certain rate of simple interest. The rate of interest per annum is ?

1 $5 \frac{5}{9}$ $5\frac{5}{9}$ %
2 $6 \frac{5}{9}$ $6\frac{5}{9}$ %
318 %
425 %

Answer:- 1

Explanation:-

Solution:

\begin{aligned} Let principal   =  6P \\ Hence Amount \\ =  6P \times \frac{7}{6} = 7 P \\ ∴ SI  =  7P - 6P =  P \\ Time  =  3 years \\ SI = \frac{P \times T \times R}{100} \\ \Rightarrow P = \frac{6 P \times R \times 3}{100} \\ \Rightarrow R = \frac{100}{18} \\ = \frac{50}{9} \\ = 5 \frac{5}{9} % \end{aligned}

$\eqalign{ & {\text{Let principal = 6P}} \cr & {\text{Hence Amount}} \cr & {\text{ = 6P}} \times \frac{7}{6} = 7{\text{P}} \cr & \therefore {\text{SI = 7P}} - {\text{6P = P}} \cr & {\text{Time = 3 years}} \cr & \boxed{{\text{SI = }}\frac{{{\text{P}} \times {\text{T}} \times {\text{R}}}}{{100}}} \cr & \Rightarrow {\text{P = }}\frac{{6{\text{P}} \times {\text{R}} \times {\text{3}}}}{{100}} \cr & \Rightarrow {\text{R = }}\frac{{100}}{{18}} \cr & \,\,\,\,\,\,\,\,\,\,\,\, = \frac{{50}}{9} \cr & \,\,\,\,\,\,\,\,\,\,\,\, = 5\frac{5}{9}\% \cr}$

Alternate:
Note: In such type of questions to save your valuable time try to think like that.

\begin{matrix} \begin{matrix} Amount & Principal \end{matrix} \\ \underset{+ 1}{\underset{⏟}{7 6}} \end{matrix}

$\begin{gathered} \begin{array}{*{20}{c}} {{\text{Amount}}}&{{\text{Principal}}} \end{array} \hfill \\ {\text{ }}\underbrace {7\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{6}}}_{ + 1} \hfill \\ \end{gathered}$

\begin{aligned} Required rate % \\ = \frac{1}{6} \times \frac{100}{3} \\ = 5 \frac{5}{9} % \end{aligned}

$\eqalign{ & {\text{Required rate % }} \cr & {\text{ = }}\frac{1}{6} \times \frac{{100}}{3} \cr & = 5\frac{5}{9}\% \cr}$

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of itself in 3 years at a certain rate of simple interest. The rate of interest per annum is ?", "text": "A sum of money becomes

\frac{7}{6}

$\frac{7}{6}$ of itself in 3 years at a certain rate of simple interest. The rate of interest per annum is ?", "dateCreated": "7/24/2019 10:09:12 AM", "author": { "@type": "Person", "name": "Nitin Sir" }, "answerCount": "4", "acceptedAnswer": { "@type": "Answer", "text": "

Solution:

\begin{aligned} Let principal   =  6P \\ Hence Amount \\ =  6P \times \frac{7}{6} = 7 P \\ ∴ SI  =  7P - 6P =  P \\ Time  =  3 years \\ SI = \frac{P \times T \times R}{100} \\ \Rightarrow P = \frac{6 P \times R \times 3}{100} \\ \Rightarrow R = \frac{100}{18} \\ = \frac{50}{9} \\ = 5 \frac{5}{9} % \end{aligned}

$\eqalign{ & {\text{Let principal = 6P}} \cr & {\text{Hence Amount}} \cr & {\text{ = 6P}} \times \frac{7}{6} = 7{\text{P}} \cr & \therefore {\text{SI = 7P}} - {\text{6P = P}} \cr & {\text{Time = 3 years}} \cr & \boxed{{\text{SI = }}\frac{{{\text{P}} \times {\text{T}} \times {\text{R}}}}{{100}}} \cr & \Rightarrow {\text{P = }}\frac{{6{\text{P}} \times {\text{R}} \times {\text{3}}}}{{100}} \cr & \Rightarrow {\text{R = }}\frac{{100}}{{18}} \cr & \,\,\,\,\,\,\,\,\,\,\,\, = \frac{{50}}{9} \cr & \,\,\,\,\,\,\,\,\,\,\,\, = 5\frac{5}{9}\% \cr}$

Alternate:
Note: In such type of questions to save your valuable time try to think like that.

\begin{matrix} \begin{matrix} Amount & Principal \end{matrix} \\ \underset{+ 1}{\underset{⏟}{7 6}} \end{matrix}

$\begin{gathered} \begin{array}{*{20}{c}} {{\text{Amount}}}&{{\text{Principal}}} \end{array} \hfill \\ {\text{ }}\underbrace {7\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{6}}}_{ + 1} \hfill \\ \end{gathered}$

\begin{aligned} Required rate % \\ = \frac{1}{6} \times \frac{100}{3} \\ = 5 \frac{5}{9} % \end{aligned}

$\eqalign{ & {\text{Required rate % }} \cr & {\text{ = }}\frac{1}{6} \times \frac{{100}}{3} \cr & = 5\frac{5}{9}\% \cr}$ ", "dateCreated": "7/24/2019 10:09:12 AM", "author": { "@type": "Person", "name": "Nitin Sir" } }, "suggestedAnswer": { "@type": "Answer", "text": "

Solution:

\begin{aligned} Let principal   =  6P \\ Hence Amount \\ =  6P \times \frac{7}{6} = 7 P \\ ∴ SI  =  7P - 6P =  P \\ Time  =  3 years \\ SI = \frac{P \times T \times R}{100} \\ \Rightarrow P = \frac{6 P \times R \times 3}{100} \\ \Rightarrow R = \frac{100}{18} \\ = \frac{50}{9} \\ = 5 \frac{5}{9} % \end{aligned}

$\eqalign{ & {\text{Let principal = 6P}} \cr & {\text{Hence Amount}} \cr & {\text{ = 6P}} \times \frac{7}{6} = 7{\text{P}} \cr & \therefore {\text{SI = 7P}} - {\text{6P = P}} \cr & {\text{Time = 3 years}} \cr & \boxed{{\text{SI = }}\frac{{{\text{P}} \times {\text{T}} \times {\text{R}}}}{{100}}} \cr & \Rightarrow {\text{P = }}\frac{{6{\text{P}} \times {\text{R}} \times {\text{3}}}}{{100}} \cr & \Rightarrow {\text{R = }}\frac{{100}}{{18}} \cr & \,\,\,\,\,\,\,\,\,\,\,\, = \frac{{50}}{9} \cr & \,\,\,\,\,\,\,\,\,\,\,\, = 5\frac{5}{9}\% \cr}$

Alternate:
Note: In such type of questions to save your valuable time try to think like that.

\begin{matrix} \begin{matrix} Amount & Principal \end{matrix} \\ \underset{+ 1}{\underset{⏟}{7 6}} \end{matrix}

$\begin{gathered} \begin{array}{*{20}{c}} {{\text{Amount}}}&{{\text{Principal}}} \end{array} \hfill \\ {\text{ }}\underbrace {7\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{6}}}_{ + 1} \hfill \\ \end{gathered}$

\begin{aligned} Required rate % \\ = \frac{1}{6} \times \frac{100}{3} \\ = 5 \frac{5}{9} % \end{aligned}

$\eqalign{ & {\text{Required rate % }} \cr & {\text{ = }}\frac{1}{6} \times \frac{{100}}{3} \cr & = 5\frac{5}{9}\% \cr}$ ", "dateCreated": "7/24/2019 10:09:12 AM" } }

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