1. ## Another word problem

hi everyone,

Some of the balls in bag A are divided into two bags B and C. If five are put in bag B and a third of the remainder are put in bag C, twelve remain.
Find the number of balls originally in bag A.

first attempt was

$\displaystyle 5+\frac{x}{3}=12$

which is wrong, then

$\displaystyle 5+\frac{1}{3}(x-5)=12$

which is still wrong

2. Lets say we have x balls in bag A

Take 5 balls to bag B

remainder = (x-5)

Balls taken to Bag C is

$\displaystyle \frac{1}{3}(x-5)$

Now remainder = (x-5) - (x-5)/3

This remainder is = 12

2(x-5)/3 = 12

x =36/2 + 5

x = 18+5

x= 23

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