A milliner bought a job lot of hats 1/4 of which were brown.

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

---------Brown---------Not Brown------
Sold-----12---------------28------------40
Unsold---3----------------17------------20
----------15---------------45------------60

3/20

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

STRATEGY CHECK: One option here is to let x = the total number of hats, and then calculate each value in terms of x.
Another option is to assign a "nice" to the total number of hats (i.e., a number that works well with all of the fractions)
Let's take the second approach.

The least common multiple of the three denominators (4, 3 and 5) is 60
So, let's say there are 60 hats in the job lot.

A milliner bought a job lot of (60) hats 1/4 of which were brown.
1/4 of 60 = 15, which means there were 15 brown hats in the job lot.

The milliner sold 2/3 of the hats including 4/5 of the brown hats.
2/3 of 60 = 40, which means the milliner sold a total of 40 hats, and 20 hats remained unsold
4/5 of 15 = 12, which means the milliner sold a total of 12 brown hats, and 3 brown hats remained unsold

What fraction of the unsold hats were brown?
3 of the 20 unsold hats were brown.
Answer = 3/20 = C

how to chose a number according to the question ?

When picking numbers, we need to choose a number with which the calculations required for the question become easier. Often, it means getting an integer value for the fractions or percentages mentioned in the question. For example, for this question, choosing the least common multiple of each denominator mentioned (4, 3, and 5), which is 60, ensures that at each step of the question we get an integer value of hats, which simplifies the calculations.

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Hope it helps.