just suppose that you have a total of 60 hats.
no. of brow hats = 1/4*60 = 15
of these hats, 4/5*15 = 12 were sold, so remaining brown hats = 3

no. of hats sold = 2/3*60 = 40
unsold hats = 20

ANS = 3/20

Suppose he bought n hats, n/4 brown and 3n/4 not brown.

He sold 2n/3 of the hats in total ,(4/5)(n/4)=n/5 of which were brown


Remaining, then, are n/4-n/5=n/20 brown hats of a total of n/3 unsold hats.

Thus the fraction of the unsold hats that are brown is n/20 divided by n/3 =3/20

C

Alternatively, suppose there are 60 hats bought (LCM(3,4,5)). 20 were unsold, and of the 15 brown hats, 3 were unsold

i went a more difficult way to solve it,

if x = the total number of hats, then x/4 are brown hats and 3x/4 are black hats

if you sell 2/3 of the hats, and since you don't know how many of those are black or brown luckily they give you the statement that 4/5 of the brown hats were sold off in that

so 2/3x = (x/4)*(4x/5) + Y(3x/4)

I let Y be the fraction of Black hats sold, and as you can see, I included the 3x/4 which is the amount of black hats originally.

2/3*x - 1/5*x = Y(3x/4)
7x/15 = Y(3x/4)

you end up with Y = 28/60 or 14/30, or in other words, the amount of black hats sold from the original 3/4.

but earlier we also figured out that 1/5*x is the amount of brown hats sold from the original 1/4. So i guess the tricky part is realizing you need to find out what is remaining after being sold. that would equal the original amount - the sold amount.

1/4*x - 1/5*x = the remaining brown hats = 1/20*x
3/4*x - 14/30*x = the remaining black hats = 45/60*x - 28/60*x = 17/60*x

the question asks for the ratio of remaining brown to total, so divide remaining brown / total

(1/20*x) / (1/20*x + 17/60*x)
x(3/60 + 17/60)
= 1/20*x / (x*20/60)

= 1/20 * 60/20 = 1/20 * 3 = 3/20

We know 1/4 of the hats are brown
acc. to given conditions 4/5th of brown hats are sold
so 1/5 are unsold
i.e. 1/5*(no. of brown hats are unsold)
Let total no. of hats be H
no.of brown hats remaining=1/4*(total no. of hats)=1/4H
acc. to given conditions,
1/3 hats remain
hence no. of hats remaining=1/3H
there required fraction=(1/20*H)/(1/3*H)=3/20
3/20

let T be the total and B be the brown ones.

(1/4)T = B

(2/3)T were sold, of which (4/5)T were brown.
left over brown hats(unsold brown hats) = (1/4)T - (4/5)T = T/20

(1/3)T were unsold.

the fraction of unsold hats that were brown = (T/20)/ ((1/3)T) = 3/20

Lets solve it by picking a smart number, say the milliner bought total 60 hats:-



Correct answer is option C
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---------Brown---------Not Brown------
Sold-----12---------------28------------40
Unsold---3----------------17------------20
----------15---------------45------------60

3/20
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I started this out by finding the LCM of the 3, 4, and 5, which is 60.

So let's say 60 total hats.

1/4 of them were brown = 15 brown hats.

He sold 2/3 of the hats, which would be 40 total, but 4/5 brown hats were sold off along with the 40, so 4/5 of 15 is 12 brown hats gone.

Left remaining are 20 hats, with 3 of them being brown.

3/20

Given: A milliner bought a job lot of hats 1/4 of which were brown. The milliner sold 2/3 of the hats including 4/5 of the brown hats.
Asked: What fraction of the unsold hats were brown?

Brown hats: 1/4
Other hats: 3/4
Brown hats sold: 4/5*1/4 = 1/5

Unsold hats: 1/3
Unsold brown hats :1/4 - 1/5 = 1/20

The fraction of the unsold hats were brown = (1/20)/(1/3) = 3/20

IMO C
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