Question 271715: A thief steals a number of rare plants from a nursery. On his way out, the thief meets three security guards, one after another. To each security guard the thief is forced to give one-half the plants that he still has, plus two more. Finally, the thief leaves the nursery with no plants. How many plants were originally stolen?
Found 2 solutions by solver91311, CharlesG2:
Answer by solver91311(24713)   (Show Source):
You can put this solution on YOUR website!
He gave the last guard everything, which was half of what he had plus 2. So 2 had to be the other half of what he had -- hence he had 4 when he met the last guard. The guard before then got 2 more than half, so half must have been 6, and he had 12 after his encounter with the first guard. 12 plus 2 is 14, times 2 is 28 which is the number he had at the start.
John
Answer by CharlesG2(834)  (Show Source):
You can put this solution on YOUR website! A thief steals a number of rare plants from a nursery. On his way out, the thief meets three security guards, one after another. To each security guard the thief is forced to give one-half the plants that he still has, plus two more. Finally, the thief leaves the nursery with no plants. How many plants were originally stolen?
x plants stolen
x - (1/2 * x + 2) 1st guard
x/2 - 2 after 1st guard
x/2 - 2 - (1/2 * (x/2 - 2) + 2) 2nd guard
x/2 - 2 - (x/4 - 1 + 2)
x/2 - 2 - x/4 - 1
2x/4 - x/4 - 3
x/4 - 3 after 2nd guard
x/4 - 3 - (1/2 * (x/4 - 3) + 2) 3rd guard
x/4 - 3 - (x/8 - 3/2 + 2)
2x/8 - 3 - (x/8 + 1/2)
2x/8 - 3 - x/8 - 1/2
x/8 - 7/2 after 3rd guard
x/8 - 7/2 = 0
x/8 = 7/2
x = 7/2 * 8
x = 56/2 = 28 plants originally stolen
check: 28
28 - 16 = 12 after 1st guard
12 - 8 = 4 after second guard
4 - 4 = 0 after 3rd guard
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