Question

in how many years will a sum double itself at the rate of 10% simple interest per annum?

Answer 1

Answer:

Step-by-step explanation:

Let the initial sum be p.

Time = t years

r =10%.

Given,

Amount= 2p after t years.

SI= Amount - Principal = 2p-p = p

SI= prt/100

p=p*10*t/100

t=10 years.

Answer 2

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

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

$\underline{ \mathfrak{ \: Given, \: }} \\ \\ \text{ \pink{A sum of money become double of itself}} \\ \text{ \pink{ at 10\% simple interest.}} \\ \\ \underline{ \mathfrak{ \:Suppose, \: }} \\ \\ \text{ \red{Rate of interest, r = 10\% }} \\ \\\text{ \red{ Principal = P }} \\ \\ \text{ \red{Amount, A = 2P}} \\ \\ \sf{ \blue{\therefore \: Simple \: \: Interest = Amount - Principal }} \\ \\ \sf{ \blue{\implies S.I. = 2P - P }} \\ \\ \: \: \: \: \: \: \: \: \: \sf{ \blue{\therefore \: S.I. = P}} \\ \\ \: \: \: \: \text{ \red{No. of years = n}}$

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

$\: \: \: \: \sf{ \therefore \: \: \red{No. \: \: of \: \: years , n = \underline{ \: 10 \: years.\: }}}$

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

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

$\underline{ \mathfrak{ \: Given, \: }} \\ \\ \text{ \pink{A sum of money become double of itself}} \\ \text{ \pink{ in 10\% under simple interest.}} \\ \\ \underline{ \mathfrak{ \:Suppose, \: }} \\ \\ \text{ \red{Rate of interest , r = 10\%}} \\ \\ \: \: \: \: \text{ \red{No. of years = n }} \\ \\ \text{ \red{ Principal , P =Rs. 100 }} \\ \\ \text{ \red{Amount, A = Rs. 200}} \\ \\ \sf{ \blue{\therefore Simple \: \: Interest = Amount - Principal }} \\ \\ \sf{ \blue{\implies S.I. = Rs.(200- 100 )}} \\ \\ \: \: \: \: \: \: \: \: \: \sf{ \blue{\therefore \: \: S.I. = Rs. 100}}$

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

$\: \: \: \: \sf{ \therefore \: \: \red{No. \: \: of \: \: years , n = \underline{ \: 10 \: years \: }}}$