Question

help with a math problem

Answer 1

2. Late one night a burglar somehow got into one of the vaults in Bank of America and started out with a big sack of money. No one really knows how much he stole. On his way out, he was stopped by one of the guards, who caught him "holding the bag" so to speak. Fortunately for him, the burglar was able to talk his way out of trouble by offering the guard half the money he had taken with a bonus of $2,000 thrown in. Just as he was walking away, praising his good luck at having gotten free, he was stopped by a second guard. He took the same bribe, half of all the money he had left, with $2,000 thrown in, to get by the second guard. Just as he was about to leave, you guessed it, he was stopped by yet a third guard who let him go only after receiving half of all the burglar had left, with $2,000 thrown in.

By the time the burglar left Bank of America he had mixed emotions. After all, he did leave with $9,000 more than he had when he arrived and he escaped a free man. But as he thought of all the money he had left behind with the guards, he wept.

Answer 2

work backwards
before he met the third guard he had \(9000+9000+2000=20,000\)

Answer 3

@satellite73 how come you added $9000 twice? and why would you only add $2000 once?

Answer 4

@kirbykirby help please

Answer 5

@imranmeah91 help please

Answer 6

Working backwards...
Assume the burglar had x dollars before he met the third guard.
He gave half the money to the third guard and an extra $2000.
So AFTER meeting the third guard, the burglar has: x - x/2 - 2000 = 9000
x/2 = 11,000
x = 22,000

Do a similar procedure with the second guard.
Assume the burglar had y dollars before meeting the second guard.
He gave half the money to the second guard and an extra $2000.
So AFTER meeting the second guard, the burglar has: y - y/2 - 2000 = 22,000
solve for y.

do the same with the first guard.

Answer 7

PS: I am assuming the question is how much money was in the sack originally before he met the three guards. Otherwise there is nothing to do because there is no question.

Answer 8

@aum im sorry im still lost, can you explain it

Answer 9

Assume the burglar had "x" dollars BEFORE meeting guard #3.
How much does he give to guard #3? Half what he has, which is x/2. So the burglar has x - x/2 = x/2 left. He gives an additional $2000 to guard #3. So the burglar has x/2 - 2000 dollars. We are told the burglar had $9000 after he met guard #3.
So x/2 - 2000 = 9000
solve for x.

Answer 10

and you get $22.00 correct?

Answer 11

$22,000 (twenty two thousand, not twenty two dollars)

Answer 12

oh yeah my bad

Answer 13

So do the same thing with guard #2. We know the burglar had $22,000 AFTER meeting guard #2. Assume the burglar had "y" dollars BEFORE meeting guard #2. Use the same method as above to find y.

Answer 14

so i'll have to do y/2-2000=22,000

Answer 15

yes. solve for y.

Answer 16

@aum ok and i get $48000

Answer 17

Yes. Do the same thing with guard #1.
AFTER guard #1 the burglar has $48,000.
Assume the burglar had "z" dollars BEFORE meeting guard #1.
He gives half to guard #1 and an additional $2000. So how much will the burglar have left?

Answer 18

and i did the same thing with the last guard and i get $100,000 is that correct? @aum

Answer 19

Correct.

Answer 20

ah okay so that means that will be the final answer correct?

Answer 21

If you wish to verify, you can now work forward.
Out of 100,000 he gives 50,000 to guard #1 and an additional 2,000. So he has 48,000 left. Next he gives 24,000 to guard #2 and an additional 2,000. So he has 22,000 left. Next he gives 11,000 to guard #3 and an additional 2,000. So he has 9,000 left. And that is what the problem states.

So the sack originally had $100,000.
In your original posting there is no question asked and so I cannot tell for sure if this is the answer they are expecting.

Answer 22

@aum thank you so much for you help! i really appreciate it, especially since other people weren't really helping! but thank you, it means alot!

Answer 23

Glad to be able to assist.