A grocery store bought some mangoes at a rate of 5

Lets say he bought mangoes worth '$z'
so total no. of mangoes = 5z

he made a stash of 3 mangoes to be sold at $x,
total no. of mangoes in this stash = 3x

and a second stash of 6 mangoes to be sold at $y
total no. of mangoes in this stash = 6y

so we get two equations
z = x+ y &
5z = 3x+ 6y

solving these two
we get x/y = 1/2 (WHICH IS PRICE)
for no. of mangoes i.e. 3x/6y = x/2y = 1/4
OA : A

nice explanation. When equating (3) and(4)

1/3 x + 1/6 y = 1/5 . (x+y)
(2y+x)/6xy = 1/(5x+5y)
5(2y+x)(x+y)=6xy
10y^2 + 5x^2 + 9xy = 0

i got stuck here

Hello buddy.

I think you put wrong equation.

Assume we have total mangoes = T
Pack 1 has x mangoes
Pack 2 has y mangoes

=> T = x + y
Cost = T/5 = (x +y)/5 [you put 5/(x+y)]
Revenue pack 1 = x/3 [you put 3/x]
Revenue pack 2 = y/6 [you put 6/y]
=> (x+y)/5 = x/3 +y/6
=> 4x = y
=> Ratio 1:4

A is correct.

Please explain whats the issue with my approach. Break even means no profit no loss.

3 mangoes bought at 5$ per mango = 15$ cost price.
1 mango sold at 3$ per mango = 3 $
2 Mangoes sold at 6$ per mango = 12 $

Ratio should be 1:2

Please re-read the question.

A grocery store bought some mangoes at a rate of 5 for a dollar --> 5 mangoes for a dollar --> the cost price of a mango = 1/5 dollars.

One stack was sold at a rate of 3 for a dollar (3 mangoes for 1 dollar) --> the selling price of a mango from the first stack = 1/3 dollars.

Another stack was sold at a rate of 6 for a dollar (6 mangoes for 1 dollar) --> he selling price of a mango from the second stack = 1/6 dollars

Please re-read the Question
it says he bought 5 mangoes for $1
and sold at 3 mangoes for $1
sold at 6 mangoes for $1

Each mango from the first stack sold generates (1/3-1/5) = 2/15 profit
Each mango from the second stack sold generates (1/5-1/6) = 1/30 loss

Since 2/15 / (1/30) = 4, to break even, we need to sell 1 mango from the first stack for each 4 sold from the second stack.

Answer choice A.

1/3*x+1/6*y=1/5(x+y)
x/y=1/4

To avoid fractions, assume there were 45 mangoes. So the store bought them for 45/5 = 9 dollars.

To break even, the selling price should be $9 too. 45 mangoes can be split into 1:4, 2:3 or 1:2. So let's try these.

45 split in the ratio 1:4 gives 9 and 36.

9 mangoes split into 3 mangoes each will give $3.
36 mangoes split into 6 mangoes each will give $6.
They add up to $9 so we have hit the right answer.

Answer (A)

Resolved it through primes
some mangoes at a rate of 5 for a dollar - so 5 is the known prime factor
one stack was sold at a rate of 3 for a dollar - so 3 is the known prime factor
and the other at a rate of 6 for a dollar - so 2 and 3 are prime factors

total (avoiding duplicating 3) - 5 * 2 * 3 = 30
so if 30 mangoes were bought then the dollar amount is 6

how we can divide 30 in 2 stacks by keeping the same total dollar amount
2*3 = 2$ and 6 * 4 = 4$

Then ration between first stack and the second one: 6 mangoes / 24 mangoes = 1:4

Let initial number of mangoes = 60
cost of 60 mangoes = 60/5 = 12
stack 1 = x mangoes, stack 2 = y mangoes
x + y = 60 ----------------(1)
store broke even, 12 = (x/3) + (y/6) ------(2)
solving two equations x = 12, y = 48, ratio of x:y = 1:4 -> (A)

5 mangoes @ 1/5$, x mangoes @ 1/3$ and y mangoes @ 1/6$. Also, x + y=5.
Therefore, 5*1/5= x*1/3 + y*1/6 to breakeven.
Hence, x=3-y/2 & x+y=5, thus, 5-y=3-y/2, y=4. x=1.

B.

let x=number of 3 for a dollar mangoes sold
y=number of 6 for a dollar mangoes sold
x/3+y/6=(x+y)/5
x/y=1/4
x:y=1:4
A

Bunuel's explanation is clear. Thank you Bunuel.

If I may add, I think the trickiest part of this problem is the " reading comprehension " bit of understanding the question.

In practice mode, to be honest I spent 3 minutes reading and re-reading the question stem to no avail.

To be specific, the part that threw me off was the first sentence " A grocery store bought some mangoes at a rate of 5 for a dollar ". It did not occur to me on my first read that this is a prompt for " COGS of 1 mango is 0.2 USD ".

To benefit the GMATclub community, I would suggest reading Bunuel's explanation and at least my way of studying this for exam, is to build a mental framework should this type of question occur in the GMAT exam.

If we use just plain logic, my framework for the exam is:
Step 1: identify input - process - output -> get COGS (0.2 USD per mango) and ASP (0.333 USD per mango for the first basket and 0.167 USD per mango for the second basket)
Step 2: identify unit profit and/or unit loss for all basket -> (the first basket yields us (0.333 - 0.2 = 0.133 USD per mango; the second basket we lose 0.033 USD per mango)
Step 3: see, we if we sell 4 mangoes from the second basket, it annihilates any profit we make from selling 1 mango from the first basket.

Thus, the answer is A.

Apologies for the simplistic language. At least for my brain, this is digestible. Hope this post is useful.

The question is essentially a disguised weighted averages/mixture type of question.


The Data Points = ($) per Mango sold

“Weighted Average” or the Break-Even Cost Point = $1 dollar per 5 mangoes

And the “Weighting” for each Data Point is the number of mangoes sold at each price


Price A: $1 dollar / per 3 mangoes ———Sold A Mangoes and broke even


Price B: $1 dollar / per 6 mangoes——- Sold B mangoes and broke even


The Total mangoes purchased = (A + B)

And the Cost (i.e., “weighted average”) is = $1 / per 5 mangoes

The following equation would describe the break-even point and the Ratio of Mangoes Sold at the two different price structure is given by (A/B) ———>

$1/3(A) + $1/5(B) = $1/6(A + B)

20A + 10B = 12A + 12B

8A = 2B

A/B = 2/8 = 1/4

Answer (A)
1 : 4

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With these problems, I find converting to cents helps.

Sale Price: \(\frac{100}{5}\) (100 cents=1 dollar per 5)

Stack 1 (x) Sale Price:\(\frac{100}{3}\)(1 dollar per 3)
Profit X:\(\frac{100}{3}-\frac{100}{5}=\frac{200}{15}\)

Stack 2 (y) Sale Price: \(\frac{100}{6}\)(1 dollar per 6)
Profit Y: \(\frac{100}{6}-\frac{100}{5}=-\frac{100}{30}\)


Now, we are told that the store breaks even, implying a 0 profit level.
Profit Y+Profit X=0
\(-\frac{100}{30}y+\frac{200}{15}x=0\)
\(-\frac{100}{30}y+\frac{400}{30}x=0\)
\(\frac{400}{30}x=\frac{100}{30}y\)
Which solves to y/x=4/1, which is not a choice but the equivalent (x/y=1/4) is. Note that since the question only asks for the ratio of mangos, and doesn't specify the type of ratio, both types work. (So if it had asked, for say the ratio of stack 1 to stack 2, 4:1 would be incorrect)
A.

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?

A. 1:4
B. 1:5
C. 2:3
D. 1:2
E. 2:5

Buying price 5 $/ pc ; stacking into 2 different rate i.e., $ 3 & $ 6.
Mango stacking at price $3 would require 2 times than stacking at $6.
For break even distribution, equation shall follow = Total Buying / Stacking
For $ 3 stacking equation = (5 x 3) / 2 = 7.5
And for $ 6 stacking equation = (5 x 6) / 1 = 30
Therefore; $ 3 : $ 6 = 7.5:30 = 1:4.