[tex] \large \tt{Question : - }[/tex]
★ A sum of money doubles itself at compounded interest in 15 yr. In how many years will it becomes eight times?
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[tex] \large \tt{Question : - }[/tex]
★ A sum of money doubles itself at compounded interest in 15 yr. In how many years will it becomes eight times?
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[tex]let \: the \: sum \: be \: x \: and \: ci = 2x \\ \\ we \: know \: that \\ ci = p(1 + \frac{r}{100}) r \\ 2x \: = x(1 + \frac{r}{100} )15 \\ (1 + \frac{r}{100} )15 = 2 \: \: \: (i) \\ \\ let \: m \: be \: number \: of \: years \\ therefore \\ x(1 + \frac{r}{100})m = 8x \\ (1 + \frac{r}{100} )m = {2}^{2} \\ (1 + \frac{r}{100})m =((1 + \frac{r}{100)}15 ) 3 \\ (1 + \frac{r}{100})m = (1 + \frac{r}{100} )45 \\ on \: comparing \: both \: the \: powers \\ we \: get \: m = 45 \: years \\ hence \: this \: is \: the \: answer [/tex]
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Answer:
According to the question:
[tex] \pink{ \underline{ \underline {\sf{1st \: Case : }}}} [/tex]
[tex] \longrightarrow{ \sf{A = P \left( 1 + \frac{R}{100} \right)^{n} }}[/tex]
Putting their values:
[tex] \longmapsto{ \sf{2P = P \left( 1 + \frac{R}{100} \right)^{{ \bold{15}}} }}[/tex]
[tex] \longmapsto{ \sf{2{ \cancel{P}} = { \cancel{P}} \left( 1 + \frac{R}{100} \right)^{15} }}[/tex]
[tex] { \longmapsto{ \sf\left( 1 + \frac{R}{100} \right)^{15} = { \boxed{ \green{2}}}} \: \quad \: - - - (1)} [/tex]
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Again, According to question:
[tex] \pink{ \underline{ \underline {\sf{2nd \: Case : }}}} [/tex]
[tex]\longrightarrow{ \sf{A = P \left( 1 + \frac{R}{100} \right)^{n} }}[/tex]
Again, Putting their values:
[tex]\longmapsto{ \sf{8P = P \left( 1 + \frac{R}{100} \right)^{n} }}[/tex]
[tex]\longmapsto{ \sf{8{ \cancel{P}} = { \cancel{P}} \left( 1 + \frac{R}{100} \right)^{n} }}[/tex]
[tex]{ \longmapsto{ \sf\left( 1 + \frac{R}{100} \right)^{n} = { \boxed{{8 = {2}^{3} }}}}}[/tex]
[tex]{ \longmapsto{ \sf{ \left[ \left(1 + \frac{R}{100} \right)^{15} \right] ^{3} }}}[/tex]
Multiplying 15 and 3 using this formula:
[tex]{\boxed{ \large{ \sf{( {a}^{m})^{n} = {a}^{mn}}}}}[/tex]
[tex] { \longmapsto{ \sf{\left(1+ \frac{R}{100} \right)^{n} = \left(1+ \frac{R}{100} \right)^{45}}}}[/tex]
Hence, In [tex]{ \boxed{ \purple{ \bold{ \large{45 \: years}}}}}[/tex] it will become eight times.