Thread: double false position

Hello everyone,

Could someone please help me with this double false position problem?

A buisness man doubled his money and he spent $12. He then left and doubled his money in another city and spent $12. He returned and doubled his money, and spent $12. At the end of the 3 trips and after the expenses he has $9 left. How much did he have at the begining?

I know the answer is 11.625

Do you see where I went wrong?

the algebraic way:
(2x-12)+(2x-12)+(2x-12)=9
6x=45
x=7.5

Double false position:

Guess $100

($100)2-12=188
($100)2-12=188
($100)2-12=188
________________
564

564-9=555

Guess #2 $50

(50)2-12=88
(50)2-12=88
(50)2-12=88
__________
264

264-9=255


100 555
50 255
cross multiply: (100)(255)=25500
(50)(555)=27750

Take the difference:
27750-25500=2250
2250/difference in error=
2250/555-255 =
7.5

Thank you very much

2(2(2x-12)-12)-12=9

He's still the same man when he goes to another city. Plug the result of one city into the equation of the next city.

Thank you very much

Could you please show me how to do it with double false position?

Thank you

Could someone please show me how to do this problem with double false position?

Thank you