Thread: 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



which is wrong, then



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 = (x-5)

Balls taken to Bag C is




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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Adarsh

adarsh, you did it again...
thanks.