Question

a sum of money doubles itself at simple interest in 15 years in how many years will it become eight times​

Answer 1

Answer:

45 years

Explanation:

Say initial amount = 100

Final amount = 800

100 >>>>>>>> 200 >>>>>>>>> 400 >>>>>>>>>> 800

........... 15 ............ 15 ................. 15

Total = 15*3 = 45

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Answer 2

If a sum of money doubles in 15 years, at same rate of interest at simple interest, it will become eight times of itself.

Explanation:

• We know that the amount can be calculated using the formula:  $Amount (A) = Principal (P) + Simple Interest (SI)$  which can also be written as $A=[P+(PNR/100)]^{n}$
• Here A is the Amount at the end of the given time period, P is the Principal , R is the rate of interest and t is the Time period. In the first 15 years amount becomes twice of the Principal P, which can be framed as     $2P=P[1+(R/100)]^{15}$ cancelling P on both sides we get, $2=[1+(R/100)]^{15}$ (Equation 1)
• In eight years we get $8=[1+(R/100)]^{t}$ (Equation 2). Now cubing the first equation, we get $8=[1+(R/100)]^{45}$ (Equation 3). Comparing equations 2 and 3 we get, Time period to be 45 years.  

To learn more:

What is the derivation of Simple Interest formula ?​

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Simple interest formula and compound interest formula

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