Question

In how much time will a sum of money double itself at 12% per annum simple interest.​

Answer 1

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

$\underline{ \bold{ \: \: Given \: :\mapsto }} \\ \\ \text{Rate \: of \: interest, r = 12\% } \\ \\ \text{Suppose,} \\ \\ \: \: \: \: \: \: \: \: \text{Principal = P} \\ \\ \bold{ \therefore \: Amount,A = 2P} \\ \\ \: \: \: \: \: \: \: \: \text{Time = n} \\ \\ \\ \text{Interest \: is \: calculated \: in \: Simple \: Interest.} \\ \\ \bold{ \therefore \: S.I. = Amount - Principal} \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \bold{= 2P - P} \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \bold{= P} \\ \\ \\ </p><p>$

$\underline{ \text{ \: We know that, \: }} \\ \\ \bold{S.I. = \frac{P \times r \times n}{100} } \\ \\ \bold{ \Rightarrow \: P = \frac{P \times 12 \times n}{100} } \\ \\ \bold{\Rightarrow \: 100 \times \: P = P \times 12 \times n} \\ \\ \bold{\Rightarrow \: \frac{100 \times \cancel{P}}{ \cancel{P}} = 12 \times n } \\ \\ \bold{\Rightarrow \: \frac{100}{12} = n } \\ \\ \bold {\therefore \: n = 8.33} \\ \\ \\ \mathsf{ \therefore \:Required \: \: number \: \: of \: \: years = 8.33 \: years}$

Answer 2

Answer:

We know that SI = PRT/100

We know that SI = A - P.

A - P = PRT/100

2400 - 1200 = 1200 * 8 * T/100

1200 = 96T

T = 1200/96

  = 12.5 years.

Step-by-step explanation:

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