I12-25

Question

Shyam placed a total of 20 red and blue balls in an empty bag. Ramesh removed 10 balls without revealing their colors. Despite this, Shyam was able to correctly deduce that at least 3 of them must have been red.
 
Based on this information, select for  x  the option that  cannot  be the number of Red balls Shyam originally placed in the bag, and select for  y  the maximum number of blue balls he could have placed. Make only two selections, one in each column.

Bunuel · Answer

Official Solution:

Since Shyam correctly states that  at least  3 of the 10 removed balls were red, then  at most  7 of the removed balls were blue. This, in turn, means there could not have been more than 7 blue balls in the bag either. If there had been more than 7 blue balls in the bag, it would have been possible to remove 10 balls with more than 7 blue, and respectively fewer than 3 red, which contradicts Shyam's statement.
 
Thus, the maximum number of blue balls initially in the bag was 7, meaning the minimum number of red balls was 13. This makes having only 7 red balls impossible.

Correct answer:

x  "7"

y  "7"

Bunuel · Answer

I have revised the question, solution, and formatting by adding more details to enhance clarity.

tthuy · Answer

I did not quite understand the solution. i still don't get it. Why the minimum number of red balls was 13? Shouldn't it be 10? As the total number of balls the man removed was 10.

Bunuel · Answer

The 13 refers to the  minimum total number of red balls in the bag , not the number of red balls removed. The number of balls removed is still 10. The reason 13 is the minimum is because if there were fewer, say 12 (meaning 12 red and 8 blue), then it would have been possible to pick  8 blue and 2 red , contradicting the fact that at least  3 red were removed .

Dereno · Answer

Bunuel  
After so much thought and brain storming I am seeking ur help once again to clarify my doubt.

Case 1: max blue balls can be equal to 7. So anything you pick more than 7, in this case will land us within 🔴 red ball. Hence the max value of y is 7. Blue balls more than 7 i.e., 8,9 will contradict shyam’s deduction of at least 3.

Case 2: which cannot be the value of red balls in the bag “X”

At least 3 red balls are there in the bag. How did u arrive at the threshold value of 13. Could you please help me out.

Bunuel · Answer

The threshold value of  13 red balls  comes from the fact that the total number of balls in the bag was  20 , and the maximum possible number of  blue balls was 7 .

Here’s why:

Since  at least 3 of the 10 removed balls were red , that means at most  7 of the removed balls were blue . 
 This also means that there could not have been  more than 7 blue balls in the bag initially . 
 If there had been more than 7 blue balls (say, 8 or more), then it would have been possible to remove 10 balls in a way that fewer than 3 were red, contradicting Shyam's certainty. 
 Since the total number of balls was  20 , having at most  7 blue balls  means the minimum number of red balls must have been  20 - 7 = 13 . 
 Therefore, having  7 red balls is impossible  because that would imply  13 blue balls , which contradicts the reasoning above.

Dereno · Answer

Thanks ☺️  Bunuel  for ur time and effort. I missed the initial count of 20.

agrasan · Answer

I like the solution - it’s helpful.

Adit_ · Answer

This problem uses the Pigeonhole Principle IMO
You prolong a negative outcome as much as u can till a point where you have to pick red. 
If there were 7 red balls, then 13 blue balls and he could pick 10 balls from the blue section and have 0 red balls leading to contradiction

So basically you need 3 more red balls when compared to equilibrium or
13:7 red to blue ratio so that worst case u pick all blue (7+3 is always red)

Is this approach valid for all similar cases?  Bunuel

Also IMO this is a 655-705 problem at the very minimum

RJJ · Answer

I did not quite understand the solution. but what if we take 2 cases 
in the first case the remaining 10 balls are red and the second case has 10 blue balls as remaining

Bunuel · Answer

Your two “remaining 10” cases don’t matter. Shyam’s claim is about the 10 removed: he can guarantee at least 3 red only if it’s impossible for the removed 10 to have 8 blue (and only 2 red). That is possible unless the bag has at most 7 blue total. You can check alternative solutions here: https://gmatclub.com/forum/shyam-placed ... 40950.html

ManmeetSinghSaggi · Answer

I don’t quite agree with the solution. for X, 18 can also be the answer because if there are minimum 3 red balls, 18 cannot be the maximum blue balls out of 20 balls. Therefore, max blue balls can be 17. I agree even 7 cant be the max but 18 also cant be. Please clarify.

Bunuel · Answer

That's wrong. 18 red is not “impossible.” If the bag has 18 red (so only 2 blue), then any 10-ball removal must include at least 8 red, so Shyam can still safely say “at least 3 red.”

Your “max blue can be 17” mixes up the condition. The condition is that Shyam could deduce at least 3 red no matter which 10 were removed. If the bag had 17 blue (only 3 red), Ramesh could remove 10 blues, so Shyam could not be certain. That is why y is 7, not 17.

x	y
		7
		14
		15
		16
		17
		18

x	y
		7
		14
		15
		16
		17
		18

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