Question

In how many years will a sum double itself at 8% per annum simple interest

Answer 1

Step-by-step explanation:

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Answer 2

$\huge{ \underline{ \overline{ \mid{ \mathfrak{ \purple{ \: \: SOLUTION \: \: } \mid}}}}}$

$\huge{\underline{ \mathfrak{ \pink{ \: \: Method \: \: \: 1 \: : \mapsto \: }}}}$

$\underline{ \mathfrak{ \: Given, \: }} \\ \\ \text{ \pink{A sum of money become double of itself}} \\ \text{ \pink{ at 8\% simple interest.}} \\ \\ \underline{ \mathfrak{ \:Suppose, \: }} \\ \\ \text{ \red{Rate of interest, r = 8\% }} \\ \\\text{ \red{ Principal = P }} \\ \\ \text{ \red{Amount, A = 2P}} \\ \\ \sf{ \blue{\therefore \: Simple \: \: Interest = Amount - Principal }} \\ \\ \sf{ \blue{\implies S.I. = 2P - P }} \\ \\ \: \: \: \: \: \: \: \: \: \sf{ \blue{\therefore \: S.I. = P}} \\ \\ \: \: \: \: \text{ \red{No. of years = n}}$

$\underline{ \mathfrak{ \: \: We \: \: know \: \: that, \: \: }} \\ \\ \: \: \: \: \: \: \: \: \: \: \: \tt{S.I. = \frac{P \times 8 \times n}{100}} \\ \\ \tt{ \implies \:P = \frac{P \times 8 \times n}{100} } \\ \\ \tt{ \implies \: 100 \times P = P \times 8 \times n }\\ \\ \tt{ \implies \:P \times 8 \times n = 100 \times P} \\ \\ \tt{ \implies \:8\times n = \frac{100 \times \cancel{P}}{ \cancel{P}} } \\ \\ \tt{ \implies \:8 \times n = 100} \\ \\ \tt{ \implies \: n = \frac{100}{10} }\\ \\ \: \: \: \: \: \: \: \tt{ \therefore \: \: n = 12.5}$

$\: \: \: \: \sf{ \therefore \: \: \red{No. \: \: of \: \: years , n = \underline{ \: 12 \: years \: \: and \: \: 6 \: \: months.\: }}}$

$\underline{ \underline{ \: \: \: \: \: \: \: \: \: \: \: \: \: \:\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:\: \: \: \: \: \: \: \: \: \: \: \: \:\: \: \: \: \: \: \: \: \: \: \: \: \: }}$

$\huge{\underline{ \mathfrak{ \pink{ \: \: Method \: \: \: 2 \: : \mapsto \: }}}}$

$\underline{ \mathfrak{ \: Given, \: }} \\ \\ \text{ \pink{A sum of money become double of itself}} \\ \text{ \pink{ in 8\% under simple interest.}} \\ \\ \underline{ \mathfrak{ \:Suppose, \: }} \\ \\ \text{ \red{Rate of interest , r = 8\%}} \\ \\ \: \: \: \: \text{ \red{No. of years = n }} \\ \\ \text{ \red{ Principal , P =Rs. 100 }} \\ \\ \text{ \red{Amount, A = Rs. 200}} \\ \\ \sf{ \blue{\therefore Simple \: \: Interest = Amount - Principal }} \\ \\ \sf{ \blue{\implies S.I. = Rs.(200- 100 )}} \\ \\ \: \: \: \: \: \: \: \: \: \sf{ \blue{\therefore \: \: S.I. = Rs. 100}}$

$\underline{ \mathfrak{ \: \: We \: \: know \: \: that, \: \: }} \\ \\ \: \: \: \: \: \: \: \: \: \: \: \tt{S.I. = \frac{P \times r \times n}{100}} \\ \\ \tt{ \implies \:100 = \frac{ \cancel{100} \times 8 \times n}{ \cancel{100}} } \\ \\ \tt{ \implies \:100 = 8 \times n } \\ \\ \tt{ \implies \: n = \frac{100}{8} }\\ \\ \: \: \: \: \tt{ \therefore \: \: n = 12.5}$

$\: \: \: \: \sf{ \therefore \: \: \red{No. \: \: of \: \: years , n = \underline{ \: 12 \: years \: \: and \: \: 6 \: \: months.}}}$