In a bag, there are 55 balls of different colors such that there is 1

Question

In a bag, there are 55 balls of different colors such that there is 1 ball of the first color, 2 balls of the second color, 3 balls of the third color, and so on. If each ball is of only one color, what is the minimum number of balls that must be removed, without looking, from the bag to ensure that balls of 5 different colors are removed?

(A) 5
(B) 15
(C) 25
(D) 35
(E) 45

(A) 5 
(B) 15 
(C) 25 
(D) 35 
(E) 45

Bunuel · Answer

First let's find out how many colors of balls there are in the bag. We can use the formula for the sum of the first n positive integers, which is 1 + 2 + 3 + ... + n = n(n+1)/2. By solving the equation n(n+1)/2 = 55, we get n = 10, indicating that there are 10 different colors of balls in the bag: 1 + 2 + 3 + ... + 10 = 55.

To determine the minimum number of balls that must be removed from the bag to  guarantee  that 5 different colored balls are removed, we should assume the worst-case scenario, which would be when we remove all 10 balls of one color, then all 9 balls of another color, then all 8 balls of another color, and then all 7 balls of another color. At this point we'd have 10 + 9 + 8 + 7 = 34 balls of 4 different colors. After removing these 34 balls, we will have removed all the balls of 4 different colors, and the next ball we remove will necessarily be of a fifth different color. Therefore, the minimum number of balls that we need to remove to ensure that we remove balls of 5 different colors is 34 + 1 = 35.

Answer: D.

Check other  Worst Case Scenario Questions  from our  Special Questions Directory  to understand the concept better.

Hope it helps.

Rabab36 · Answer

Bunuel  sir can you pls explain

if we consider the scenario given in q

there are 10 different colors

1 +2+ 3+ 4+ 5+ 6+ 7+ 8+ 9+10=55

minimum no of balls to be removed of 5 different colors

why cannot be 1 ,2,3,4,5
which is total 15 balls

Bunuel · Answer

The question asks for the minimum number of balls that need to be removed to ensure with 100% certainty that 5 different colored balls are selected.

While it's possible to get lucky and select 5 balls of different colors with just 5 picks, this isn't a guaranteed outcome. Similarly, selecting five balls of different colors with 15 picks isn't a guaranteed outcome either. For example, you could pick 10 balls of one color, and 5 balls of another color and you'll be left with only 2 different colors.

Therefore, to guarantee the selection of five balls of different colors, the worst-case scenario must be considered, which is the scenario where the desired outcome is guaranteed with 100% certainty even with the least favorable picking stream. By considering the worst-case scenario, you get the minimum number of balls required to guarantee the desired outcome with 100% certainty.

Hope it helps.

Archit3110 · Answer

total balls are 55
total colors possible can be determined by
n*(n+1) = 55*2
n= 10
so as to get minimum number of balls which are to be removed we have to take cases of maximum balls which are removed 
10 of same color
9 of same color
8 of same color
7 of same color
this makes total 34 
now any other ball taken out will be of different color and will increase total balls count to 35
so option D is correct

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In a bag, there are 55 balls of different colors such that there is 1

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