How many blueberries

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How many blueberries

by binaras » Sat Mar 12, 2016 12:23 am
Hi

Need help answering the attached.

Thanks
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by GMATGuruNY » Sat Mar 12, 2016 4:06 am
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?
35
40
45
50
60
Let S = the number of strawberries in each box of strawberries and B = the number of blueberries in each box of blueberries.
Let Danny start with ONE BOX OF STRAWBERRIES and ONE BOX OF BLUEBERRIES.
Thus:
Original total number of berries = S+B.
Original difference between the total number of strawberries and the total number of blueberries = S-B.

Danny disposed of one blue box for one additional red box.
After the blue box is exchanged for a second red box, the result is TWO BOXES of strawberries and NO BLUEBERRIES.
Thus:
New total number of berries = 2S + 0B = 2S.
New difference between the total number of strawberries and the total number of blueberries = 2S - 0B = 2S.

The total number of berries (of both kinds) would increase by 25.
Since the new total is 25 more than the original total, we get:
2S = (S+B) + 25
S = B+25.

The difference between the total number of strawberries and the total number of blueberries would increase by 95.
Since the new difference is 95 more than the original difference, we get:
2S = (S-B) + 95
S = -B+95

Since S = B+25 and S =-B+95, the two expressions in red are EQUAL:
B+25 = -B+95
2B = 70
B = 35.

The correct answer is A.
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by Brent@GMATPrepNow » Sat Mar 12, 2016 1:15 pm
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?
35
40
45
50
60
Let B = number of blueberries PER BOX
Let S = number of strawberries PER BOX

If Danny disposed of one blue box for one additional red box, the TOTAL number of berries (of both kinds) would increase by 25
So, S - B = 25
Here's why. Let's pretend there is only ONE blue box and ONE red box. So, the TOTAL number of berries = B + S
If we remove a blue box and add a red box, the TOTAL number of berries = 2S
We're told that the TOTAL DECREASES by 25. So, we can write: B + S + 25 = 2S
Rearrange to get: S - B = 25

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
So, S + B = 95
Here's why.
If we REMOVE a blue box, the difference is INCREASED by B
If we ADD a red box, the difference is INCREASED by S
So, the net effect here is increasing the difference by S + B
In other words, S + B = 95

We now have the system:
S + B = 95
S - B = 25
SUBTRACT the bottom equation from the top equation to get: 2B = 70
Solve to get [spoiler]B = 35[/spoiler]

Answer: A

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by sandipgumtya » Sun Mar 13, 2016 9:58 pm
Hi Mitch,
The method I followed took me nearly 4 min.Can u pl suggest a quicker method or make it easier to understand?


Sandip.

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by GMATGuruNY » Mon Mar 14, 2016 6:19 am
sandipgumtya wrote:Hi Mitch,
The method I followed took me nearly 4 min.Can u pl suggest a quicker method or make it easier to understand?


Sandip.
An alternate approach is to PLUG IN THE ANSWERS, which represent the number of blueberries in each blue box.

Answer choice D: 50 blueberries per box
If Danny disposed of one blue box for one additional red box, the total number of berries (of both kinds) would increase by 25.
In other words, subtracting one box of 50 blueberries and adding in one box of S strawberries is equal to an increase of 25:
-50 + S = 25
S = 75 strawberries per box.

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.
Let Danny start with 2 boxes of strawberries and 2 boxes of blueberries.
Original difference = (total strawberries) - (total blueberries) = (75+75) - (50+50) = 50.
After one box of blueberries is exchanged for one box of strawberries, new difference = (75+75+75) - 50 = 175.
(new difference) - (original difference) = 175-50 = 125.
The increase in the difference is too great.
Eliminate D.

Answer choice B: 40 blueberries per box
If Danny disposed of one blue box for one additional red box, the total number of berries (of both kinds) would increase by 25.
In other words, subtracting one box of 40 blueberries and adding in one box of S strawberries is equal to an increase of 25:
-40 + S = 25
S = 65 strawberries per box.

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.
Let Danny start with 2 boxes of strawberries and 2 boxes of blueberries.
Original difference = (total strawberries) - (total blueberries) = (65+65) - (40+40) = 50.
After one box of blueberries is exchanged for one box of strawberries, new difference = (65+65+65) - 40 = 155.
(new difference) - (original difference) = 155-50 = 105.
The increase in the difference is still too great.
Eliminate B.

Notice the pattern:
In D, (new difference) - (original difference) = 125.
In B, (new difference) - (original difference) = 105.
The smaller the answer choice, the smaller the value of (new difference) - (original difference).

Implication:
To yield (new difference) - (original difference) = 95, the correct answer must be SMALLER than answer choice B.

The correct answer is A.

Answer choice A: 35 blueberries per box
If Danny disposed of one blue box for one additional red box, the total number of berries (of both kinds) would increase by 25.
In other words, subtracting one box of 35 blueberries and adding in one box of S strawberries is equal to an increase of 25:
-35 + S = 25
S = 60 strawberries per box.

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.
Let Danny start with 2 boxes of strawberries and 2 boxes of blueberries.
Original difference = (total strawberries) - (total blueberries) = (60+60) - (35+35) = 50.
After one box of blueberries is exchanged for one box of strawberries, new difference = (60+60+60) - 35 = 145.
(new difference) - (original difference) = 145-50 = 95.
Success!
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by Matt@VeritasPrep » Thu Mar 17, 2016 10:29 pm
We could also work with a couple equations.

Suppose we have x red boxes, each of which has s strawberries, and y blue boxes, each of which has b blueberries.

Before the change, we have (x + y) boxes and xs + by berries.

After the change, we have (x + y) boxes and (x+1)s + (y - 1)b, or xs + by + s - b berries.

We're told that

Strawberries in a box - Blueberries in a box = 25, or

s - b = 25

and

(New Strawberries - New Blueberries) - (Old Strawberries - Old Blueberries) = 95, or

((x + 1)*s - (y - 1)*b) - (xs - yb) = 95

The second equation simplifies to

s + b = 95

so we're set!

s - b = 25
s + b = 95

Adding these together, we have 2s = 120, from which we find s = 60 and b = 35.

Neat Q, surprisingly tricky to set up.