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

$\sf\small\underline\green{Given:-}$

$\sf{\implies Condition\:_{(as\:per\: first)}}$

$\sf{\implies P=Rs.4800\:\:A=Rs.7500}$

$\sf{\implies T= 2\: years\:and\: 8\: months}$

$\sf\small\underline\green{To\: Find:-}$

$\sf{\implies Time\:_{(as\:per\:2nd\: condition)}=?}$

$\sf\small\underline\green{Solution:-}$

• To calculate time here by applying formula and setting up equation as per given condition:-]

$\sf\small\underline\green{Formula\: Used:-}$

$\tt\red{\implies A=P\bigg[1+\dfrac{r}{100}\bigg]^n\bigg[1+\dfrac{r}{100}\bigg]^F}$

• Here P=4800 A=7500 N=2 F=2/3

$\tt{\implies 7500=4800\bigg[1+\dfrac{r}{100}\bigg]^2\bigg[1+\dfrac{r}{100}\bigg]^\frac{2}{3}}$

$\tt{\implies \dfrac{7500}{4800}=\bigg[1+\dfrac{r}{100}\bigg]^2\bigg[1+\dfrac{r}{100}\bigg]^\frac{2}{3}}$

$\tt{\implies \dfrac{7500}{4800}=\bigg[1+\dfrac{r}{100}\bigg]^{2+\frac{2}{3}}}$

$\tt{\implies \dfrac{75}{48}=\bigg[1+\dfrac{r}{100}\bigg]^\frac{6+2}{3}}$

$\tt{\implies \dfrac{75}{48}=\bigg[1+\dfrac{r}{100}\bigg]^\frac{8}{3}}$

$\tt{\implies \bigg[\dfrac{5}{4}\bigg]^2=\bigg[1+\dfrac{r}{100}\bigg]^\frac{8}{3}}$

$\tt{\implies \bigg[\dfrac{5}{4}\bigg]^{2\times\frac{3}{8}}=\bigg[1+\dfrac{r}{100}\bigg]}$

$\tt{\implies \bigg[\dfrac{5}{4}\bigg]^\frac{3}{4}=\bigg[1+\dfrac{r}{100}\bigg]-----(i)}$

$\sf{\implies Condition\:_{(as\:per\:2nd)}}$

• Here P=19200 A=37500 n=?

$\tt{\implies A=P\bigg[1+\dfrac{r}{100}\bigg]^n}$

$\tt{\implies 37500=19200\bigg[1+\dfrac{r}{100}\bigg]^n}$

$\tt{\implies \dfrac{375}{192}=\bigg[1+\dfrac{r}{100}\bigg]^n}$

$\tt{\implies \dfrac{5^3}{4^3}=\bigg[1+\dfrac{r}{100}\bigg]^n}$

$\tt{\implies \bigg[\dfrac{5}{4}\bigg]^\frac{3}{n}=\bigg[1+\dfrac{r}{100}\bigg]-----(ii)}$

• Putting the value eq (i) in (ii) here :-]

$\tt{\implies \bigg[\dfrac{5}{4}\bigg]^\frac{3}{4}=\bigg[1+\dfrac{r}{100}\bigg]}$

$\tt{\implies \bigg[\dfrac{5}{4}\bigg]^\frac{3}{4}=\bigg[\dfrac{5}{4}\bigg]^\frac{3}{n}}$

$\tt{\implies \dfrac{3}{4}=\dfrac{3}{n}}$

$\tt{\implies n=4}$

$\sf\large{Hence,}$

$\sf{\implies Time\:_{(as\:per\:2nd\: condition)}=4\: years}$

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.