Danny has boxes colored either red or blue. In each blue box

It's a great shortcut. Thanx!

awesome solution: +1

@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.

Let R and B be the number of strawberries and bluberries in red and blue boxes.
Let x and y be the number of Red and Blue boxes.

Equation1:
\((x+1)R + (y-1)B = xR + yB + 25\)
\(xR + R + yB - B = xR + yB + 25\)
\(R-B=25\)

Equation2:
\((x+1)R - (y-1)B = xR - yB + 95\)
\(xR + R - yB + B = xR - yB + 95\)
\(R+B=95\)

Combine two equations
\(B+B+25=95\)
\(B=35\)

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.

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.

Its given that if Danny disposed of one blue box for one additional red box, the total number of berries (of both kinds) would increase by 25 .

Let's say that no of Strawberries in a box be S and no of blueberries in a box be B. So, from the above statement we can say that
S - B = 25

It's also given that the difference between the total number of strawberries and the total number of blueberries would increase by 95. So, we can conclude that

S + B = 95

In case if you don't understand the above conclusion, please have a look at this.

Let's say the total no of strawberries in all boxes together be Stotal and the total no of blueberries in all boxes together be Btotal .
​
The difference in number of Strawberries and Blueberries = Stotal - Btotal = D.

After Danny disposed of one blue box for one additional red box,
The total no of strawberries in all boxes together = Stotal + S
The total no of blueberries in all boxes together = Btotal - B

​Their difference have increased by 95. That means,

Stotal + S - (Btotal - B)= Stotal + S - Btotal + B
​
= Stotal - Btotal + S + B
= D + (S + B )
​
Since D is their previous difference and its given that difference has been increased by 95 , we can conclude that S + B = 95
​
​Now we have two conclusions,
S - B = 25​
S + B = 95 ​​​
​
On solving these 2 equations,
2B = 70
B = 35
So, the no of blueberries in a box is 35.
​
Thanks,
Clifin J Francis
GMAT SME

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