pushpitkc wrote:

2. If he needs to withdraw 20 markers to have 8 of a color,

2 cases are possible such that 8 markers of either color can be taken out.

Case 1: Red = 9, Blue = 8

Case 2: Red = 8, Blue = 9

We cannot come to a unique value for the number of packets containing Blue markers(Insufficient)

What if Josh selects 7 black, 7 blue, 6 red? In that case, he will not have 8 markers of any one color.

Here's what I did...

40 packets

23 of them have black markers

(# of packets with red markers) + (# of packets with blue markers) = 17

We're asked to find the # of packets with blue markers.

Statement 1

(# of packets with blue markers) < (# of packets with red markers)

We can have (7 blue, 10 red), or (6 blue, 11 red).

Insufficient

Statement 2

This means that when Josh selects 19 packets, he is not guaranteed to have 8 packets of any one color. In the worst case scenario, Josh will select 7 packets of one color, 7 packets of another, and 5 packets of another color (total selected = 19 packets). The 20th packet he selects MUST give him 8 packets of one color, thus one of the colors must have 5 packets. So we either have (5 red, 12 blue) or (12 red, 5 blue).

Insufficient

Combine Statements 1 & 2

Per Stmt 1: red < blue

Per Stmt 2: (5 red, 12 blue) or (12 red, 5 blue)

Combined, we get 5 red, 12 blue

Answer: C