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Danny has boxes colored either red or blue. In each blue box [#permalink]

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29 May 2012, 07:17

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56% (06:52) correct
44% (01:54) wrong based on 172 sessions

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Danny has boxes colored either red or blue. In each blue box there is a fixed number of blueberries. In each red box there is a fixed number of strawberries. If Danny disposed of one blue box for one additional red box, the total number of berries (of both kinds) would increase by 25, and the difference between the total number of strawberries and the total number of blueberries would increase by 95. Each blue box contains how many blueberries?

Re: Danny has boxes colored either red or blue. In each blue box [#permalink]

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29 May 2012, 10:40

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Let b be the number of blue boxes and r the number of red boxes. Let x be the number of blueberries and y the number of strawberries in each blue and red box respectively.

If Danny disposed of one blue box for one additional red box, the total number of berries (of both kinds) would increase by 25 => (b-1)x + (r+1)y = bx + ry + 25 => y - x = 25 [Eqn 1]

If Danny disposed of one blue box for one additional red box, the difference between the total number of strawberries and the total number of blueberries would increase by 95 => (r+1)y - (b-1)x = ry - bx + 95 => y + x = 95 [Eqn 2]

Re: Danny has boxes colored either red or blue. In each blue box [#permalink]

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29 May 2012, 10:48

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Here is another way to solve this question: Replacing one blue box with a red box causes an increase of 25 berries overall => There are 25 more strawberries in a red box than blueberries in a blue box

Replacing one blue box with a red box causes the difference between the total number of strawberries and blueberries to increase by 95 => If there are x blueberries in a box, then we have taken away x blueberries, and replaced them by x+25 strawberries (using the statement above). By doing this, we have caused the difference between the remaining blueberries and strawberries to increase by x (blueberries removed) + x+25 (strawberries added) => x + x + 25 = 95 => x = 35 or choice A
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Re: Danny has boxes colored either red or blue. In each blue box [#permalink]

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01 Jun 2012, 02:10

GyanOne wrote:

Let b be the number of blue boxes and r the number of red boxes. Let x be the number of blueberries and y the number of strawberries in each blue and red box respectively.

If Danny disposed of one blue box for one additional red box, the total number of berries (of both kinds) would increase by 25 => (b-1)x + (r+1)y = bx + ry + 25 => y - x = 25 [Eqn 1]

If Danny disposed of one blue box for one additional red box, the difference between the total number of strawberries and the total number of blueberries would increase by 95 => (r+1)y - (b-1)x = ry - bx + 95 => y + x = 95 [Eqn 2]

Re: Danny has boxes colored either red or blue. In each blue box [#permalink]

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02 Oct 2012, 07:25

GyanOne wrote:

Here is another way to solve this question: Replacing one blue box with a red box causes an increase of 25 berries overall => There are 25 more strawberries in a red box than blueberries in a blue box

Replacing one blue box with a red box causes the difference between the total number of strawberries and blueberries to increase by 95 => If there are x blueberries in a box, then we have taken away x blueberries, and replaced them by x+25 strawberries (using the statement above). By doing this, we have caused the difference between the remaining blueberries and strawberries to increase by x (blueberries removed) + x+25 (strawberries added) => x + x + 25 = 95 => x = 35 or choice A

Re: Danny has boxes colored either red or blue. In each blue box [#permalink]

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02 Oct 2012, 11:57

@Gyanone - This statement is confusing "Replacing one blue box with a red box causes an increase of 25 berries overall => There are 25 more strawberries in a red box than blueberries in a blue box"

The problem only states that blueberries are fixed in the blue box but they still exist in the redbox. The increase in 25 is a combination of blueberries and redberries. It is not 25 redberries more.

Re: Danny has boxes colored either red or blue. In each blue box [#permalink]

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03 Oct 2012, 00:36

vsprakash2003 wrote:

@Gyanone - This statement is confusing "Replacing one blue box with a red box causes an increase of 25 berries overall => There are 25 more strawberries in a red box than blueberries in a blue box"

The problem only states that blueberries are fixed in the blue box but they still exist in the redbox. The increase in 25 is a combination of blueberries and redberries. It is not 25 redberries more.

by replacing one blue box for one red box...increase it by 25..that means..red box contains more strawberries thats it increase by 25.. rather assume if both wud b same..how cud it b increased by 25 by replacing one blue box for red..

thats y we assume from the statement that red contains 25 more straw berries.
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Re: Danny has boxes colored either red or blue. In each blue box [#permalink]

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21 Oct 2013, 10:37

Joy111 wrote:

Danny has boxes colored either red or blue. In each blue box there is a fixed number of blueberries. In each red box there is a fixed number of strawberries. If Danny disposed of one blue box for one additional red box, the total number of berries (of both kinds) would increase by 25, and the difference between the total number of strawberries and the total number of blueberries would increase by 95. Each blue box contains how many blueberries?

A. 35 B. 40 C. 45 D. 50 E. 60

I solved it with reverse PI process.

Say

R | B | 70| 45

Let both boxes are 1 each. After I remove blue box, no of berries - 140 (25 more than initial sum) Now the diff as per qs should be 95. Let's see, The Initial diff = (70-45=25) Final diff = (140) Diff - 140-25 = 115. We need less diff, so,

R | B | 60| 35

Let both boxes are 1 each. After I remove blue box, no of berries - 120 (25 more than initial sum) Now the diff as per qs should be 95. Let's see,

Initial diff = (60-45=25) Final diff = (120-0= 120) Diff - 120-25 = 95. This matches the question stem, so, it's my answer.

Re: Danny has boxes colored either red or blue. In each blue box [#permalink]

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18 Nov 2013, 15:54

Oh this is an interesting one!

The new box brought in 25 [more] Straw berries. The difference between the two increased 95 more [meaning that Straw Berries > Blue Berries, and not the other way around] Now to explain the Increase of 95; 25 came from the exchange of boxes, and we are left with 70 more straw berries.

{Now if you remember how elections work, Candidate A has 24 votes, Candidate B has 18 votes, but if three people turn from A to B than A will have 21 votes and B will have 21 votes. So in actuality Candidate A has only three votes lead (but whose effect is double or equal to Six votes).}

Okay back to the 70 straw berries, all we need to do is to divide the 70 by 2 (to cover up for the double effect) we get 35 or Answer choice A.

Re: Danny has boxes colored either red or blue. In each blue box [#permalink]

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28 Nov 2014, 17:27

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Re: Danny has boxes colored either red or blue. In each blue box [#permalink]

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04 Feb 2016, 02:43

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