
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
$\displaystyle 5+\frac{1}{3}(x5)=12$
which is still wrong
the answer is 23. but can someone please show me. thanks

Lets say we have x balls in bag A
Take 5 balls to bag B
remainder = (x5)
Balls taken to Bag C is
$\displaystyle \frac{1}{3}(x5)$
Now remainder = (x5)  (x5)/3
This remainder is = 12
2(x5)/3 = 12
x =36/2 + 5
x = 18+5
x= 23

Adarsh

adarsh, you did it again...
thanks. (Clapping)