QR question

[topdent1]

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Grace has 16 jellybeans in her pocket. She has 8 red ones, 4 green ones, and 4 blue ones. What is the minimum number of jellybeans she must take out of her pocket to ensure that she has one of each color?

I did 7+3+3=13, which happens to be the correct answer, but I don't think I did it right. In fact, I don't even know what this question is asking.

 

[PooyaH]

Grace has 16 jellybeans in her pocket. She has 8 red ones, 4 green ones, and 4 blue ones. What is the minimum number of jellybeans she must take out of her pocket to ensure that she has one of each color?

I did 7+3+3=13, which happens to be the correct answer, but I don't think I did it right. In fact, I don't even know what this question is asking.

hmm I'm not sure how to explain this one, but just think of it this way: if you have 4 blue ones and 4 green ones after you have taken 13 jellybeans out of your pocket, there's no way you could still have 4 blue ones or 4 green ones in your pocket since there are only 3 left!!! If you still had 4 jellybeans left in your pocket, then there was a possiblity for them to be all blue ones or all green ones, but since there are only 4 left you must have one of each color in your hands!

Hope it helps although I don't really know if I explained it well

 

[TheGreenWave]

Grace has 16 jellybeans in her pocket. She has 8 red ones, 4 green ones, and 4 blue ones. What is the minimum number of jellybeans she must take out of her pocket to ensure that she has one of each color?

I did 7+3+3=13, which happens to be the correct answer, but I don't think I did it right. In fact, I don't even know what this question is asking.

Im pretty sure thats not the best logic. Try mine:

Since it says what number do you need to take to ENSURE you will have at least one of each color, you have to pick the minimum amount of beans that will eliminate any possibility of you not having one of each. There are tons of different possibilities (for instance you could pick three and each could be a different color), but we want the worst one.

Say you kept picking and only got RED (8) and GREEN (4) . That means you could potentially have picked 12 of the 16 beans and not gotten any BLUE ones. So the only way to make sure you got one of each is to get ONE more than the other two combined. So 8 + 4 + 1* = 13

NOTE if there had been 8 red, 4 green, and 5 blue you would add 8 + 5 + 1.

 

[Streetwolf]

No that isn't the right way, you got lucky (topdent1). For instance if you had 8 blue, 8 red, and 2 green, your method would get you 7+7+1 = 15 but the answer here would be 17.

You need to start with what you have the most of and work your way down. The poster above me got it.

The first 8 you pick could all be red. The next 4 could all be blue. Only with the 13th one can you say FOR SURE that you picked all 3 colors.