Shyam placed red and blue balls in an empty bag. Ramesh removed 10

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 · Accepted 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"

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Attachment:

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Suyash1331 · Answer

since there are at least 3 red balls even after removing 10 balls, this means that there are more than 10 balls. hence x = 7

maximum blue balls can be 7 as well because after removing 10 balls he says that at least 3 balls are red, which means that after removing all the blue balls 3 red will remain if 10 are total balls.
if y > 7 then we cannot say that minimum 3 red balls are still there. its a common sense question

AbhiS101 · Answer

Maximum No of blue balls should be 7, even if Ramesh has taken all the blue balls i.e 7 it will guarantee atleast 3 red balls
As per questions at least 3 balls should be red, i believe maximum could be 10 so can't be less than 10

IMO 7 & 7

technie0903 · Answer

Did not get how we deduced the value of x = 7. How were we able to reach this conclusion : "meaning the minimum number of red balls was 13"? GMAT-Club-Forum-zsdd2am7.png

Bunuel · Answer

The minimum number of red balls was 13 because there could not have been more than 7 blue balls in the bag. If there were more than 7 blue balls, it would have been possible to remove 10 balls with more than 7 blue, leaving fewer than 3 red, which contradicts Shyam’s certainty that at least 3 red balls were removed. GMAT-Club-Forum-0l589u37.png

Rishm · Answer

Why can't there be 17 blue balls in total he placed, considering he removed 3 red balls in those 10 balls, meaning 7 more blue balls were there as we minimize this and max the blue balls and add the remaining 10 balls as 10 hence 7+10?

Bunuel · Answer

Because Shyam’s claim must hold for  all possible  ways of removing 10 balls, not just the case where exactly 3 red are removed.

If there were 17 blue balls, it would be possible to remove 10 balls with fewer than 3 red. That possibility alone contradicts Shyam’s certainty, so 17 blue cannot work.

pappal · Answer

given R+B=20 and 10 balls are removed randomly but the case is that at least 03 R balls were removed.
1.what can't be no. of R balls placed? 2. what is the maximum no. of B balls placed?
let if R=7 so B=13 then there is a possibility that all the removed balls are blue that violates the given condition.
so  R can't be 7  is the correct choice.
now for the satisfaction of the removal condition, at least 10 red balls should have been placed so as to facilitate removal of red balls in number from 3 to 10,
which restricts removal of blue balls in number from 7 to 0 respectively.
however if there are initially 08 blue balls placed and if all are removed then only 02 red balls can be removed, which again violates the condition.
hence  we can restrict the maximum no. of B balls to 07.

x	y
		7
		14
		15
		16
		17
		18

Shyam placed red and blue balls in an empty bag. Ramesh removed 10

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Shyam placed red and blue balls in an empty bag. Ramesh removed 10

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