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by chaitanya.bhansali » Thu Aug 07, 2014 10:08 am
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
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by [email protected] » Thu Aug 07, 2014 10:40 am
Hi chaitanya.bhansali,

This is more of a logic question than anything else. We need to be clear on the facts given to us and find the "worst-case" scenario (what number of balls would GUARANTEE that 5 different colors were picked?).

With a total of 55 balls and the progression described (1, 2, 3, etc.), you can determine that the number of each color would be....

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

So we have 10 different colors.

As we pull balls from the bag, the only way to GUARANTEE that 5 different colors are chosen is to figure out the worst-case possibility (the maximum you COULD pick before you found the 5th color)....

If we picked the 10 balls of the same same color, then we'd only have 1 color, then....
If we picked the 9 balls of the same same color, then we'd only have 2 colors, then...
If we picked the 8 balls of the same same color, then we'd only have 3 colors, then...
If we picked the 7 balls of the same same color, then we'd only have 4 colors, then....
We'd need just 1 more ball, of any color, to end up with 5 colors...

10 + 9 + 8 + 7 + 1 = 35

We would need to pull 35 balls to GUARANTEE that we ended up with 5 different colors.

Final Answer: D

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by shrivats » Thu Aug 07, 2014 9:29 pm
In all there are 55 balls. 1 ball of colour1, 2 balls of colour2 and so on..

To absolutely ensure we take out 5 colours, we must ensured that colours that have the maximum representation are takn out first.

10+9+8+7=34 contain 4 colours. So if we just draw out one more ball, we can be sure it'll be a different colour, so 34+1 = 35.
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by GMATinsight » Fri Aug 08, 2014 6:29 am
chaitanya.bhansali wrote: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
Step 1: In order to calculate the total number of Colors of the Balls we must consider

Total Number of Balls = 55 = 1+2+3+4+5+6+7+8+9+10

i.e. There are 10 Colors (Because 1 ball of 1st color, 2 balls of 2nd color,...etc.)

Now Take the WORST POSSIBLE SCENARIOS

i.e After picking 10 balls there are chances that you haven't picked any ball of second color as there are a maximum of 10 balls of 1 Color

Similarly, After picking 19 balls there are chances that you haven't picked any ball of Third color as there are a maximum of 19 (10 of 10th color and 9 of 9th color) balls of 1 Color


Similarly even if you pick (10+9+8+7 = 34 Balls) then there is possibility that you haven't picked a ball of 5 Color as the there are chances that all balls are of 10th, 9th, 8th and 7th Color

BUT If you pick 1 more ball i.e.(10+9+8+7+'1' = 35 Balls) then 35th ball will be of 5th different color and will make sure that we have picked at-least 1 ball of 5 different colors

Answer: Option D
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