Question

The rate per cent per annum simple interest at which a sum will be double of itself in 8 years is

Answer 1

Let the principal be P
So, amount=2P
Time=8 years
Rate of interest=R%

ATQ:-
Amount=Simple Interest+Principal

$= > 2p = \frac{p \times r \times t}{100} + p$

$= > 2p = \frac{p \times r \times 8}{100} + p$

$= > 2p - p = \frac{8pr}{100}$

$= > p = \frac{8pr}{100}$

After cutting P (Principal) from both the sides, we get:-

$= > 1 = \frac{8r}{100}$

$= > 1 \times 100 = 8r$

$= > \frac{1 \times 100}{8} = r$

$= > \frac{25}{2} = r$

$= > 12.5 = r$

Hence, the rate of interest at which a sum will double itself in 8 years is 12.5%

Answer 2

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

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

$\underline{ \mathfrak{ \: Given, \: }} \\ \\ \text{ \pink{A sum of money become double of itself}} \\ \text{ \pink{ in 8 years under simple interest.}} \\ \\ \underline{ \mathfrak{ \:Suppose, \: }} \\ \\ \text{ \red{Rate of interest = r }} \\ \\\text{ \red{ Principal = P }} \\ \\ \text{ \red{Amount, A = 2P}} \\ \\ \sf{ \blue{\therefore \: Simple \: \: Interest = Amount - Principal }} \\ \\ \sf{ \blue{\implies S.I. = 2P - P }} \\ \\ \: \: \: \: \: \: \: \: \: \sf{ \blue{\therefore \: S.I. = P}} \\ \\ \: \: \: \: \text{ \red{Time, n = 8 years.}}$

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

$\: \: \: \: \sf{ \therefore \: \: \red{Rate \: \: of \: \: interest, r = \underline{ \: 12.5\% \: }}}$

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

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

$\underline{ \mathfrak{ \: Given, \: }} \\ \\ \text{ \pink{A sum of money become double of itself}} \\ \text{ \pink{ i 8 years under simple interest.}} \\ \\ \underline{ \mathfrak{ \:Suppose, \: }} \\ \\ \text{ \red{Rate of interest = r }} \\ \\ \: \: \: \: \text{ \red{Time, n = 8 years.}} \\ \\ \text{ \red{ Principal , P =Rs. 100 }} \\ \\ \text{ \red{Amount, A = Rs. 200}} \\ \\ \sf{ \blue{\therefore Simple \: \: Interest = Amount - Principal }} \\ \\ \sf{ \blue{\implies S.I. = Rs.(200- 100 )}} \\ \\ \: \: \: \: \: \: \: \: \: \sf{ \blue{\therefore \: \: S.I. = Rs. 100}}$

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

$\: \: \: \: \sf{ \therefore \: \: \red{Rate \: \: of \: \: interest, r = \underline{ \: 12.5\% \: }}}$