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by rajatvmittal » Thu Jun 13, 2013 10:18 am
A grocery store bought some mangoes at a rate of 5 for a dollar. They were separated into two stacks, one of which was sold at a rate of 3 for a dollar and the other at a rate of 6 for a dollar. What was the ratio of the number of mangoes in the two stacks if the store broke even after having sold all of its mangoes?

1:4
1:5
2:3
1:2
2:5

Correct Answer A
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by Brent@GMATPrepNow » Thu Jun 13, 2013 11:43 am
rajatvmittal wrote:A grocery store bought some mangoes at a rate of 5 for a dollar. They were separated into two stacks, one of which was sold at a rate of 3 for a dollar and the other at a rate of 6 for a dollar. What was the ratio of the number of mangoes in the two stacks if the store broke even after having sold all of its mangoes?

1:4
1:5
2:3
1:2
2:5

Correct Answer A
NOTE:
3 mangoes for $1.00 is the same as 100/3 cents each
6 mangoes for $1.00 is the same as 100/6 cents each
5 mangoes for $1.00 is the same as 100/5 cents each (aka 20 cents)


Let E = # of mangoes sold for 100/3 cents each (expensive)
Let C = # of mangoes sold for 100/6 cents each (cheap)

Altogether, a total of C+E mangoes were bought and sold.
At 20 cents apiece, the total price that the store paid for the mangoes = 20(C + E)

As far as revenue goes . . .
At 100/3 cents apiece, the revenue from the expensive mangoes = (100/3)E
At 100/6 cents apiece, the revenue from the cheap mangoes = (100/6)C
So, total revenue = (100/3)E + (100/6)C

If the store broke even, it must be the case that . . .
20(C + E) = (100/3)E + (100/6)C
Multiply both sides by 6: 120(C + E) = 200E + 100C
Expand: 120C + 120E = 200E + 100C
Rearrange: 20C = 80E
Divide both sides by 20: C = 4E
Divide both sides by E: C/E = 4
Flip both side: E/C = 1/4 = A

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by GMATGuruNY » Thu Jun 13, 2013 4:02 pm
rajatvmittal wrote:A grocery store bought some mangoes at a rate of 5 for a dollar. They were separated into two stacks, one of which was sold at a rate of 3 for a dollar and the other at a rate of 6 for a dollar. What was the ratio of the number of mangoes in the two stacks if the store broke even after having sold all of its mangoes?

1:4
1:5
2:3
1:2
2:5

Correct Answer A
Let C = the cost of each mango, H = the higher selling price, and L = the lower selling price.

Since the mangos are purchased at 5 per dollar, C = 1/5 of a dollar.
Since the higher selling price is 3 per dollar, H = 1/3 of a dollar.
Since the lower selling price is 6 per dollar, L = 1/6 of a dollar.

To break even when the mangos are sold, the average revenue per mango must be equal to C = 1/5 of a dollar.
This is a MIXTURE problem.
A selling price of 1/3 of a dollar (H) is MIXED with a selling price of 1/6 of a dollar (L) to yield an AVERAGE revenue of 1/5 of a dollar (C).
Use ALLIGATION -- a great way to handle mixture problems.

Step 1: Put the fractions over a COMMON DENOMINATOR.
C = 1/5 = 6/30.
H = 1/3 = 10/30.
L = 1/6 = 5/30.

Step 2: Plot the 3 NUMERATORS on a number line, with the numerators for the selling prices (10 and 5) on the ends and numerator for the average revenue (6) in the middle.
H 10-----------6-----------5 S

Step 3: Calculate the distances between the numerators.
H 10-----4-----6-----1-----5 S

Step 4: Determine the ratio of H to L in the mixture.
The required ratio of H to L is equal to the RECIPROCAL of the distances in red.
H:L = 1:4.

The correct answer is A.

For two other problems that I solved with alligation, check here:

https://www.beatthegmat.com/ratios-fract ... tml#484583
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by kul512 » Thu Jun 13, 2013 7:51 pm
Rate of buyig=1/5

Rate of selling in loss=1/6
Rate of selling in Profit= 1/3

loss of one mango=1/5-1/6= 1/30

profit on one mango= 1/3-1/5= 2/15=4/30

So to break even (means Loss= Profit) ratio of the mangoes must be in ratio of the profit and loss=
1/30:4/30=1:4
Sometimes there is very fine line between right and wrong: perspective.
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by Brent@GMATPrepNow » Fri Jun 14, 2013 4:33 am
kul512 wrote:Rate of buyig=1/5

Rate of selling in loss=1/6
Rate of selling in Profit= 1/3

loss of one mango=1/5-1/6= 1/30

profit on one mango= 1/3-1/5= 2/15=4/30

So to break even (means Loss= Profit) ratio of the mangoes must be in ratio of the profit and loss=
1/30:4/30=1:4
Great solution, kul512. I hadn't considered that approach.

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by jainpiyushjain » Sat Jun 29, 2013 5:30 am
I am getting the answer 1:2 by applying the weighted average. Dear experts can you please let me know what is the flaw in this approach.

Thank you
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by Brent@GMATPrepNow » Sat Jun 29, 2013 5:51 am
jainpiyushjain wrote:I am getting the answer 1:2 by applying the weighted average. Dear experts can you please let me know what is the flaw in this approach.

Thank you
It would help if you provided the steps you performed to reach this answer.

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by chandni170 » Fri Jul 05, 2013 1:21 am
Hi,

I used the weighted average approach as well and got 1:2. Not sure why this is wrong.

Here's my solution:

Let the number of mangoes= M. Therefore the total cost is 5M.

The number of mangoes (M) are distributed into two stacks: Say A and B. (M= A+B)

A is sold for $3 and B is sold for $6. Therefore total sales = 3A+6B

From this, we've got two equations:
1) Break-even equation--- 5M = 3A+6B
2) Initial equation---- M = A+B

Solving the two equations we get: A= M/3 and B= 2M/3... Hence, A:B= 1:2

Experts, kindly correct my approach. (I'd appreciate a quick response as my exam's quite close!)

Thanks :)
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by chandni170 » Fri Jul 05, 2013 1:50 am
Right after my post, I realized my silly error. I read the question wrong!

Corrected solution:

Let the total number of mangoes= M.
5 mangoes---> $1 (20 cents per mango)

Let the two stacks be A and B.
For A: 3 mangoes---> $1 (100/3 cents per mango)
For B: 6 mangoes---> $1 (100/6 cents per mango)

From here on it's the same as my previous solution...
The two equations are:-
1) M= A+B
2) 20M= (100/3)A + (100/6)B

Solving this equation we get A:B = 1:4.

Cheers
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by Sukant » Thu Dec 12, 2013 10:33 pm
Dear experts,
I still am confused with the different solutions provided.
I am getting the answer 1:2 simply by using the alligation method. Also, if we take an example, say I have 30 mangoes.
The total cost price of the mangoes= 5X30= $150
If I sell the mangoes at a ratio of 1:2, the net S.P will be= 10 X 3 + 20 X 6 = 30 + 120 = $150
Thus, it can be seen that using these ratios, the store is having a breakeven as total CP= total SP
Please tell me where I am wrong.
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by Sukant » Thu Dec 12, 2013 10:39 pm
I realized my mistake quickly after submitting the reply.
I was doing a major blunder. The question said 5 mangoes for $1 and not 1 mango for $5. Same for Selling prices also.
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