Multiple choice

A sum of money at 20% rate of compound interest per annum becomes more than 100 times in n years. What is the least value of n? (Use log10 2 = 0.301, log10 3 = 0.477)

  1. 23

  2. 24

  3. 25

  4. 26
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

A = P(1 + r/100)^n. 100P < P(1.2)^n => 100 < 1.2^n. n * log(1.2) < log(100). n * (log 12 - log 10) < 2. n * (log(3*4) - 1) < 2. n * (log 3 + 2*log 2 - 1) < 2. n * (0.477 + 0.602 - 1) < 2. n * (0.079) < 2. n > 2/0.079 = 25.31. The least integer n is 26.

AI explanation

Using the compound interest formula, we need the sum to become more than 100 times itself, so P times (1.2) raised to the power of n is greater than 100P. Dividing both sides by P and taking the base-10 logarithm of both sides gives n times log(1.2) is greater than log(100), where log(1.2) equals log(12 divided by 10), which is log(4 times 3) minus 1, or (2 times 0.301 plus 0.477) minus 1, equaling 0.079. Since n times 0.079 must be greater than 2, n must be greater than 2 divided by 0.079, which is 25.316; therefore, the least integer value for n is 26.

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