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A bowl contains equal numbers of red, orange, green, blue, a

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A bowl contains equal numbers of red, orange, green, blue, a  [#permalink]

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New post 02 May 2013, 12:38
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A
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Question Stats:

76% (02:31) correct 24% (03:04) wrong based on 248 sessions

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A bowl contains equal numbers of red, orange, green, blue, and yellow candies. Kaz eats all of the green candies and half of the orange ones. Next, he eats half of the remaining pieces of each color. Finally, he eats red and yellow candies in equal proportions until the total number of remaining candies of all colors equals 15% of the original number. What percent of the red candies remain?

A. 0%
B. 5%
C. 10%
D.15%
E. 20%
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Re: A bowl contains equal numbers of red, orange, green, blue,  [#permalink]

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New post 02 May 2013, 16:14
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4
See image attached.
Remianing RED candies = 0
hence 0%

Ans A
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Re: A bowl contains equal numbers of red, orange, green, blue,  [#permalink]

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New post 02 May 2013, 12:47
R | O | G | B | Y - colors
10 | 10 | 10 | 10 | 10 - initial situation
10 | 5 | X | 10 | 10 - eats all G and half O
5 | 2.5 | X | 5 | 5 - eats half of the remaining

Now we know that remain 15% of 50 => \(50*15%=7.5\) candies. We have to "eat" R and Y till the remaining candies will be 7.5.

We have to eat ALL R and Y.

0 | 2.5 | X | 5 | 0 = 7.5 remains

\(%\)of \(R\) = \(0%\)
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Re: A bowl contains equal numbers of red, orange, green, blue, a  [#permalink]

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New post 19 Nov 2013, 15:49
Did almost the same as Zarrolou did. I just used numbers that sum 100, I think this is easier to work with % using numbers that sum 100.

R | O | G | B | Y - colors
20 | 20 | 20 | 20 | 20 - initial situation
20 | 10 | 0 | 20 | 20 - eats all G and half O
10 | 5 | 0 | 10 | 10 - eats half of the remaining
0 | 5 | X | 10 | 0 - eats R and Y candies in equal proportions until the total number of remaining candies of all colors equals 15%

Ans A
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Re: A bowl contains equal numbers of red, orange, green, blue, a  [#permalink]

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New post 23 Jun 2014, 04:04
Pick smart numbers here. 5 different colors of candy: Let's 100 Total, hence 20 pieces each in original. In the end we want to have 15 % of original total, so 15 candies.

20 R + 20 O + 20 G + 20 B + 20 Y = TOTAL (original=100)
First step: Eats all green and half of orange:
20 R + 10 O + 0 G + 20 B + 20 Y = Total (new=70)
Second step: Eats half of the remaining:
10 R + 5 O + 0 G + 10 B + 10 Y = Total (new = 35)
third step: eats equal of R and Y so that 15 % of the ORIGINAL TOTAL remain (=15 candies)
Hence he eats all red and all yellow candies:
5 O + 10 B = 15 Total

Hence we get Answer A: 0 % of red candies remain!
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Re: A bowl contains equal numbers of red, orange, green, blue, a  [#permalink]

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New post 24 Apr 2019, 05:41
ratinarace wrote:
A bowl contains equal numbers of red, orange, green, blue, and yellow candies. Kaz eats all of the green candies and half of the orange ones. Next, he eats half of the remaining pieces of each color. Finally, he eats red and yellow candies in equal proportions until the total number of remaining candies of all colors equals 15% of the original number. What percent of the red candies remain?

A. 0%
B. 5%
C. 10%
D.15%
E. 20%


Given : Equal number of candies : R + O + G + B + Y = 100%

Weightage : 20 + 20 + 20 + 20 + 20 = 100


Condition 1 : Kaz eats all of the green candies and half of the orange ones

Remaining candies : 20 + (20 - 10) + (20 - 20) + 20 + 20

Remaining candies : 20 + 10 + 0 + 20 + 20 = 70

Condition 2 : He eats half of the remaining pieces of each color.

Remaining candies : (20 - 10) + (10 - 5) + (20 - 10) + (20 - 10)

Remaining candies : 10 + 5 + 10 + 10 = 35

Condition 3 : He eats red and yellow candies in equal proportions until the total number of remaining candies of all colors equals 15% of the original number.

Original number of candies = 100

15% of 100 = 15 (Number of candies that must not be eaten)

This means, he can eat 20 candies from 35 remaining candies

Now he can eat Red and Yellow candies in equal proportion, means

Remaining Red = 10 : Remaining Yellow = 10

He has to eat 10 of the remaining red candies and 10 yellow candies.

Remaining red candies = 0%

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Re: A bowl contains equal numbers of red, orange, green, blue, a  [#permalink]

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New post 25 Apr 2019, 18:55
ratinarace wrote:
A bowl contains equal numbers of red, orange, green, blue, and yellow candies. Kaz eats all of the green candies and half of the orange ones. Next, he eats half of the remaining pieces of each color. Finally, he eats red and yellow candies in equal proportions until the total number of remaining candies of all colors equals 15% of the original number. What percent of the red candies remain?

A. 0%
B. 5%
C. 10%
D.15%
E. 20%


Let’s let the initial number of candies of each color be 12 (and thus the total number of candies is 60). So he eats all 12 green candies and 6 orange candies at first. Next he eats 3 orange, 6 red, 6 blue and 6 yellow candies. Thus far, we have

60 - (12 + 6 + 3 + 6 x 3) = 60 - 39 = 21

candies left. At this point, there are 3 orange, 6 red, 6 blue and 6 yellow candies left. Since 15% of 60 is 9, he must eat 12 more candies to get to 9. However, since he eats an equal number of of red and yellow candies, he must eat 6 red and 6 yellow candies to have 9 candies left. Therefore, there will be no more red candies left.

Answer: A
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Re: A bowl contains equal numbers of red, orange, green, blue, a   [#permalink] 25 Apr 2019, 18:55
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