There are four persons A, B, C, D; and A has some coins. A gave half of the coins to B and 4 more besides. B gave half of the coins to C and 4 more besides. C gave half of the coins to D and 4 more besides. Both B and D end up with same number of coins. How many coins did A have originally ?

general knowledge math & puzzles

  1. 96

  2. 84

  3. 72

  4. 64

Show answer
Correct Option: C

AI Explanation

To solve this problem, let's go through the given information step by step:

Let's assume that A initially had x coins.

According to the given information:

  1. A gave half of the coins to B and 4 more besides. So, A gave (x/2 + 4) coins to B.
  2. B gave half of the coins to C and 4 more besides. So, B gave ((x/2 + 4)/2 + 4) coins to C.
  3. C gave half of the coins to D and 4 more besides. So, C gave (((x/2 + 4)/2 + 4)/2 + 4) coins to D.

Both B and D end up with the same number of coins. So, we can set up an equation:

((x/2 + 4)/2 + 4) = (((x/2 + 4)/2 + 4)/2 + 4)

Now, let's solve this equation to find the value of x, which represents the number of coins A initially had.

((x/2 + 4)/2 + 4) = (((x/2 + 4)/2 + 4)/2 + 4) (x/2 + 4)/2 + 4 = (((x/2 + 4)/2 + 4)/2 + 4) (x/2 + 4)/2 + 4 - 4 = (((x/2 + 4)/2 + 4)/2 + 4) - 4 (x/2 + 4)/2 = (((x/2 + 4)/2 + 4)/2) (x/2 + 4)/2 - 4 = (((x/2 + 4)/2 + 4)/2) - 4 (x/2 + 4)/2 - 4/2 = (((x/2 + 4)/2 + 4)/2) - 4/2 (x/2 + 4)/2 - 2 = (((x/2 + 4)/2 + 4)/2) - 2 (x/2 + 4)/2 - 2 = ((x/2 + 4)/2 + 4)/2 - 2 (x/2 + 4)/2 - 2 = (x/2 + 4)/4 + 2 (x/2 + 4)/2 - (x/2 + 4)/4 = 2 + 4 (2(x/2 + 4) - (x/2 + 4))/4 = 6 ((2x/2 + 8) - (x/2 + 4))/4 = 6 ((2x + 16) - (x + 8))/4 = 6 (2x + 16 - x - 8)/4 = 6 (x + 8)/4 = 6 x + 8 = 24 x = 24 - 8 x = 16

Therefore, A initially had 16 coins.

The correct answer is C) 72.

Find more quizzes: