Question

31. At what rate of compound interest, a sum amounts
to Rs. 672 in 2 years and to Rs. 714 in 3 years
when the interest is payable annually?
I want I maths

Answer 1

❏ Formula used:-

❏ Used ForMuLaS:-

➔ For Compound interest

$\sf\longrightarrow\boxed{ P_c=P(1+\frac{r}{100}){}^{n\times t}}$

$\sf\longrightarrow\boxed{I_c=P[(1+\frac{r}{100}){}^{n\times t}-1]}$

➔ For Simple interest

$\sf\longrightarrow\boxed{I_s=\frac{Prt}{100}}$

$\sf\longrightarrow\boxed{P_s=(P+\frac{Prt}{100})}$

Where ➔ • $P_c$= Total amount after

compound interest .

• $P_s$= Total amount after

Simple interest .

• P= principal amount.

• t= time.

• r= rate of interest.

• n= No. of interest cycle per year .

Ex:-

➝n=1 (when compounded annually)

➝n=2 (when compounded half yearly)

➝n=4 (when compounded quarterly)

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❏ Question:-

@ At what rate of compound interest, a sum amounts to Rs. 672 in 2 years and to Rs. 714 in 3 years when the interest is payable annually?

❏ Solution:-

❚❚❚ Given:

• Time( $</strong><strong>t_</strong><strong>1</strong><strong>$ )= 2 yrs

• $P_</strong><strong>{</strong><strong>c_</strong><strong>1</strong><strong>}</strong><strong>$ = 672Rs

• Time( $t_</strong><strong>2</strong><strong>$ )= 3 years

• $P_{c_</strong><strong>2</strong><strong>}$ = 714 Rs

• n=1 [∵ compounded only]

❚❚❚To Find:

• Principal=?= P (let)
• interest rate=?= r (let)

❍ From the 1st part :-

• Time( $t_1$ )= 2 years
• $P_{c_1}$ = 672Rs
• principal= P
• interest rate=r %

$\sf\longrightarrow P_{c_1}=P(1+\frac{r}{100}){}^{n\times t_1}$

$\sf\longrightarrow 672=P(1+\frac{r}{100}){}^{1\times 2}$

$\sf\longrightarrow \boxed{672=P(1+\frac{r}{100}){}^{2}}............(i)$

❍ From the 2nd part :-

• Time( $t_</strong><strong>2</strong><strong>$ )= 3 years
• $P_{c_</strong><strong>2</strong><strong>}$ = 714 Rs
• principal= P
• interest rate=r %

$\sf\longrightarrow P_{c_2}=P(1+\frac{r}{100}){}^{n\times t_2}$

$\sf\longrightarrow 714=P(1+\frac{r}{100}){}^{1\times 3}$

$\sf\longrightarrow 714=P(1+\frac{r}{100}){}^{2+1}$

$\sf\longrightarrow 714=[P(1+\frac{r}{100}){}^{2}]\times(1+\frac{r}{100})$

[putting the value from equation (i)]

$\sf\longrightarrow 714=672\times(1+\frac{r}{100})$

$\sf\longrightarrow \frac{714}{672}=\frac{100+r}{100}$

$\sf\longrightarrow \frac{714}{672}\times100=100+r$

$\sf\longrightarrow 1.0625\times100=100+r$

$\sf\longrightarrow 106.25=100+r$

$\sf\longrightarrow 106.25-100=r$

$\sf\longrightarrow\boxed{ \large{\red{r=6.25\:\%}}}$

Now, putting the value r=6.25 in equation (i);

$\sf\longrightarrow 672=P(1+\frac{6.25}{100}){}^{2}$

$\sf\longrightarrow 672=P(\frac{106.25}{100}){}^{2}$

$\sf\longrightarrow 672\times(\frac{100}{106.25}){}^{2}=P$

$\sf\longrightarrow\boxed{\large{\red{ P=595.27\:Rs}}}$

∴ Interest rate per annum is =6.25 %.

∴ The principal amount was = 595.27 Rs.

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$\underline{ \huge\mathfrak{hope \: this \: helps \: you}}$

#answerwithquality & #BAL

Answer 2

❏ Some ForMuLaS:-

➔ For Compound interest

$\sf\longrightarrow\boxed{ P_c=P(1+\frac{r}{100}){}^{n\times t}}$

$\sf\longrightarrow\boxed{I_c=P[(1+\frac{r}{100}){}^{n\times t}-1]}$

➔ For Simple interest

$\sf\longrightarrow\boxed{I_s=\frac{Prt}{100}}$

$\sf\longrightarrow\boxed{P_s=(P+\frac{Prt}{100})}$

Where ➔ • $P_c$= Total amount after

compound interest .

• $P_s$= Total amount after

Simple interest .

• P= principal amount.

• t= time.

• r= rate of interest.

• n= No. of interest cycle per year .

Ex:-

➝n=1 (when compounded annually)

➝n=2 (when compounded half yearly)

➝n=4 (when compounded quarterly)

━━━━━━━━━━━━━━━━━━━━━━━

❏ Solution:-

❚❚❚ Given:

• Time( $t_1$ )= 2 yrs

• $P_{c_1}$ = 672Rs

• Time( $t_2$ )= 3 years

• $P_{c_2}$ = 714 Rs

• n=1 [∵ compounded only]

❚❚❚To Find:

Principal=?= P (let)

interest rate=?= r (let)

❍ From the 1st part :-

Time( $t_1$ )= 2 years

$P_{c_1}$ = 672Rs

principal= P

interest rate=r %

$\sf\longrightarrow P_{c_1}=P(1+\frac{r}{100}){}^{n\times t_1}$

$\sf\longrightarrow 672=P(1+\frac{r}{100}){}^{1\times 2}$

$\sf\longrightarrow \boxed{672=P(1+\frac{r}{100}){}^{2}}............(i)$

❍ From the 2nd part :-

Time( $t_2$ )= 3 years

$P_{c_2}$ = 714 Rs

principal= P

interest rate=r %

$\sf\longrightarrow P_{c_2}=P(1+\frac{r}{100}){}^{n\times t_2}$

$\sf\longrightarrow 714=P(1+\frac{r}{100}){}^{1\times 3}$

$\sf\longrightarrow 714=P(1+\frac{r}{100}){}^{2+1}$

$\sf\longrightarrow 714=[P(1+\frac{r}{100}){}^{2}]\times(1+\frac{r}{100})$

[putting the value from equation (i)]

$\sf\longrightarrow 714=672\times(1+\frac{r}{100})$

$\sf\longrightarrow \frac{714}{672}=\frac{100+r}{100}$

$\sf\longrightarrow \frac{714}{672}\times100=100+r$

$\sf\longrightarrow 1.0625\times100=100+r$

$\sf\longrightarrow 106.25=100+r$

$\sf\longrightarrow 106.25-100=r$

$\sf\longrightarrow\boxed{ \large{\red{r=6.25\:\%}}}$

Now, putting the value r=6.25 in equation (i);

$\sf\longrightarrow 672=P(1+\frac{6.25}{100}){}^{2}$

$\sf\longrightarrow 672=P(\frac{106.25}{100}){}^{2}$

$\sf\longrightarrow 672\times(\frac{100}{106.25}){}^{2}=P$

$\sf\longrightarrow\boxed{\large{\red{ P=595.27\:Rs}}}$

∴ Interest rate per annum is =6.25 %.

∴ The principal amount was = 595.27 Rs.

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$\underline{ \huge\mathbb{hope \: this \: helps \: you}}$