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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NOTE: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
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
Cheers,
Brent
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Let C = the cost of each mango, H = the higher selling price, and L = the lower selling price.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
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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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
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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Great solution, kul512. I hadn't considered that approach.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
Cheers,
Brent
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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
Thank you
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It would help if you provided the steps you performed to reach this answer.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
Cheers,
Brent
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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
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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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
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
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.
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.