Multiple choice

A father divides his property between his two sons A and B. A invests the amount at compound interest of $8\%$ p.a. B invests the amount at 10% p.a simple interest. At the end of $2$ years , the interest received by B is Rs $1336$ more than the interest received by A. Find A 's share in the father's father's property of Rs $25, 000$.

  1. Rs $12,000$
  2. Rs $13,000$
  3. Rs $10,000$
  4. Rs $12,500$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let A's share be x, B's be 25000-x. Interest for A = x(1.08^2 - 1) = 0.1664x. Interest for B = (25000-x) * 0.1 * 2 = 5000 - 0.2x. Given (5000 - 0.2x) - 0.1664x = 1336 => 3664 = 0.3664x => x = 10000.

AI explanation

Let A's share be x, making B's share 25000 minus x. A's compound interest for two years at 8% is found using the formula CI = P((1 + R/100)^n - 1), which gives x multiplied by 0.1664. B's simple interest is (25000 - x) multiplied by 10 multiplied by 2 divided by 100, yielding 5000 - 0.2x. The equation for the difference in interest is (5000 - 0.2x) - 0.1664x = 1336, which simplifies to 0.3664x = 3664, resulting in A's share being Rs. 10,000.

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