Question

at what rate per cent per annum simple interst will a sum be double of itself in 8 years

Answer 1

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

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

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

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

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

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

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

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

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

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

Answer 2

$\fbox \purple {Answer:-}$

Let the s be ₹ x.

Then, amount = ₹ 2x.

∴ S.I. = Amount - Principal = ₹ (2x - x) = ₹ x.

Now, P = ₹ x, S.I. = ₹ x and T = 8 years.

Let the required rate be R% per annum. Then

$R = \frac {100\ ×\ S.I.}{P\ ×\ T}$ $= \frac {100\ ×\ x}{x\ ×\ 8}$ % p.a = 25/2 % p.a. = 12 1/2 % p.a.

Hence, the required rate is 12 1/2 % per annum.

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