Hi Rishab2409,Your final answer and almost all of your reasoning are spot on. You correctly saw that Bag B's whites must be a multiple of
4, then tested the splits of
30, landed on Bag B =
12 or
24, leaving Bag A =
18 or
6 whites, giving
6 or
2 red in Bag A - and only
6 is a listed choice. That's exactly the right path.
There's one constraint you glossed over, though, and it's worth tightening up.
The missing half of Bag A's constraintYou required Bag A's whites to be a multiple of
3 (from red:white = 1:3). But Bag A also has the
white:blue = 2:3 ratio. That means white must be even too - so Bag A's whites must actually be a multiple of
6, not just
3.
Combine the ratios: red:white:blue =
2:6:9. So white comes in blocks of
6.
Here you got lucky: both survivors (
18 and
6) happen to be multiples of
6, so your "multiple of 3" filter gave the same answers. But had a candidate like
12 or
24 needed checking against the blue ratio, using only the multiple-of-
3 rule could have let a bad value slip through. So the fully rigorous version is:
- Bag A whites: multiples of
6 -
6,
12,
18,
24- Bag B whites: multiples of
4 -
4,
8,
12,
16,
20,
24- Pairs summing to
30: (
6,
24) and (
18,
12)
- Red in A = white/
3 -
2 or
6Same destination, but now every constraint is doing its job.
Quick habit to lock in: whenever two ratios share a term (white, here), fuse them into one chain first -
2:6:9 -
before you start testing numbers. That way you carry every divisibility rule at once and never rely on luck.
Answer: DRishab2409
Kindly confirm my logic - I saw that for Red to white the ratio in Bag A is 1:3 and Bag B is 1:4. Bag B will contain white ball in multiple of 4, but balance white balls should be a multiple of 3 for Bag A, which helps us with either 12 or 24 white balls in Bag B. The balance being 18 or 6 white balls for Bag A, bringing out the fact that Bag A has either 6 or 2 red balls. And upon going through the options, we have only 6 red balls as the answer.