Question

Rs 4800 become 7500 at a certain rate of interest compounded annually in 2 years 8 months. then in how many years rs 19200 will amounts to 37500 at the same rate of interest compounded annually?

Answer 1

Part 1) :-

• P = 4800
• A = 7500
• T = 2 years , 8 months = 2(w) + (2/3)(F) Years.
• Let Rate = R% .

we know That,

☛ Amount = P [ 1 + (R/100) ]^W * [ 1 + {(F*R)/100} ]

Putting values we get :-

→ 7500 = 4800[ 1 + (R/100)]²[ 1 + (R/150)]

→ (75/48) = [ (100+R)²/10000][(150 + R)/150]

→ (25/16) = [(10000 + R² + 200R)(150+R)]/[10000*150]

→ 25 = [1500000 + 150R² + 30000R + 10000R + R³ + 200R²] / [ 93750 ]

→ 25*93750 = R³ + 350R² + 40000R + 1500000

→ R³ + 350R² + 40000R - 843750 = 0

Solve This cubic Polynomial , find Value of R.

____________

Part 2)

→ (37500 /19200) = (1 + R/100)^T

→ 1.953125 = (1+R/100)^T

Multiply by log both sides

→ log(1.953125) = T[log(1+R/100)]

→ T = [ log(1.953125) ] / [log(1+R/100)]

Put value of R here now, You will get Your Answer...

Answer 2

Answer:

Given :-

Rs 4800 become Rs 7500 at a certain rate of interest compound annually in 2 years 8 months.

Rs 19200 will amount to Rs 37500 at the same rate of interest compound annually.

To Find :-

How many years Rs 19200 will amounts to Rs 37500 at the same rate of interest compounded annually.

Formula Used :-

$\longmapsto \sf\boxed{\bold{\pink{Amount =\: P\bigg(1 + \dfrac{r}{100}\bigg)^{n} \bigg(1 + \dfrac{r}{100}\bigg)^{F}}}}$

where,

P = Principal

r = Rate of Interest

n = Time

F = Fraction Time

Solution :-

${\small{\bold{\purple{\underline{\bigstar\: In\: the\: {1}^{{st}}\: case\: :-}}}}}$

$\mapsto$ Rs 4800 become Rs 7500 at a certain rate of interest compound annually in 2 years 8 months.

Given :

Time = 2 years 8 months

Then,

$\implies \sf Time =\: 2\: years + 8\: months$

2 years = 12 + 12 = 24

$\implies \sf Time =\: 24\: months + 8\: months$

$\implies \sf\bold{\green{Time =\: 32\: months}}$

Again, convert it into years we get,

$\implies \sf \dfrac{\cancel{32}}{\cancel{12}}$

$\implies\sf\dfrac{\cancel{16}}{\cancel{6}}$

$\implies \sf\dfrac{\cancel{8}}{\cancel{3}}$

$\implies \sf\bold{\green{2\dfrac{2}{3}}}$

Now, we can say that :

Time (n) = 2

Fraction Time (F) = ⅔

According to the question by using the formula we get,

Given :

Principal = Rs 4800

Amount = Rs 7500

Time = 2

Fraction Time = ⅔

Then,

$\leadsto\sf 7500 =\: 4800\bigg(1 + \dfrac{r}{100}\bigg)^{2} \bigg(1 + \dfrac{r}{100}\bigg)^{\frac{2}{3}}$

$\leadsto \sf \dfrac{75\cancel{00}}{48\cancel{00}} =\: \bigg(1 + \dfrac{r}{100}\bigg)^{2 + \frac{2}{3}}$

$\leadsto \sf \dfrac{75}{48} =\: \bigg(1 + \dfrac{r}{100}\bigg)^{\frac{8}{3}}$

$\leadsto \sf {\bigg(\dfrac{5}{4}\bigg)}^{2} =\: \bigg(1 + \dfrac{r}{100}\bigg)^{\frac{8}{3}}$

$\leadsto \sf \bigg(\dfrac{5}{4}\bigg)^{\cancel{2} \times \frac{3}{\cancel{8}}} =\: \bigg(1 + \dfrac{r}{100}\bigg)$

$\leadsto \sf \bold{\green{\bigg(\dfrac{5}{4}\bigg)^{\frac{3}{4}} =\: \bigg(1 + \dfrac{r}{100}\bigg)\: ----\: (Equation\: No\: 1)}}$

${\small{\bold{\purple{\underline{\bigstar\: In\: the\: {2}^{{nd}}\: case\: }}}}}$

$\mapsto$ Rs 19200 will amounts to Rs 37500 at the same rate of interest compound annually.

Given :

Principal = Rs 19200

Amount = Rs 37500

As we know that,

$\longmapsto \sf\boxed{\bold{\pink{A =\: P\bigg(1 + \dfrac{r}{100}\bigg)^{n}}}}$

Then,

$\leadsto \sf 37500 =\: 19200\bigg(1 + \dfrac{r}{100}\bigg)^{n}$

$\leadsto \sf \dfrac{374\cancel{00}}{192\cancel{00}} =\: \bigg(1 + \dfrac{r}{100}\bigg)^{n}$

$\leadsto \sf \bigg(\dfrac{375}{192}\bigg) =\: \bigg(1 + \dfrac{r}{100}\bigg)^{n}$

$\leadsto \sf {\bigg(\dfrac{5}{4}\bigg)}^{3} =\: \bigg(1 + \dfrac{r}{100}\bigg)^{n}$

$\leadsto\sf\green{\bold{{\bigg(\dfrac{5}{4}\bigg)}^{\frac{3}{n}} =\: \bigg(1 + \dfrac{r}{100}\bigg)\: ----\: (Equation\: No\: 2)}}$

Now, by putting the equation no 1 in the equation no 2 we get,

$\leadsto \sf \bigg(\dfrac{5}{4}\bigg)^{\frac{3}{4}} =\: \bigg(\dfrac{5}{4}\bigg)^{\frac{3}{n}}$

$\leadsto \sf {\cancel{\bigg(\dfrac{5}{4}\bigg)}}^{\frac{3}{n}} =\: {\cancel{\bigg(\dfrac{5}{4}\bigg)}}^{\frac{3}{4}}$

$\leadsto \sf \dfrac{3}{n} =\: \dfrac{3}{4}$

$\leadsto \sf\bold{\red{n =\: 4\: years}}$

$\therefore$ The time is 4 years.