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
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