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GMAT Club Olympics 2025 (Day 6): A department store received

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07 Jul 2025, 11:36

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1) we dont know about electronics ns

2) we don't know about fragile NS

Combined
E=4/5T
F=2/3T
NF=1/3T

If all non fragile were electronics we d still have 46% as fragile which is >1/2 of 2/3 T
4/5T-1/3T= 7/15T
Suff
Ans C

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07 Jul 2025, 11:44

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Bunuel

A department store received a shipment of two types of boxes on Monday: fragile and non-fragile. Did more than half of the fragile boxes contain electronics?

(1) Two-thirds of the boxes in the shipment are fragile.
(2) Four-fifths of the boxes in the shipment contain electronics.

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Given:
Shipment of two types of boxes: Fragile & Non-Fragile

Question: Did more than half of the fragile boxes contain electronics?

Let the total shipment = 30 (LCM of 2, 3 & 5)

Statement (1): 2/3*30 = 20 = Fragile boxes; Non Fragile = 10 boxes.

Since we do not know the number of boxes that contain electronics, statement (1) alone is Insufficient.

Statement (2): 4/5*30 = 24 = boxes that contain electronics. However, we do not know the number of boxes that are fragile, hence statement (2) alone is Insufficient.

(C): Both statements together:

Fragile boxes = 20 & Non Fragile boxes = 10
Boxes containing electronics = 24.

Now, even if all the 10 non fragile boxes contain electronics, the minimum number of fragile boxes that contain electronics is 14 & 14>10 [1/2*20]

Hence statement (1) & (2) together are Sufficient.

Answer (C)

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07 Jul 2025, 11:47

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We need to determine out of all fragile boxes, do >50% contain electronics?

Statement (1): Two‐thirds of the boxes are fragile
This alone tells us nothing about electronics boxes. So it’s Insufficient.

Statement (2): Four‐fifths of all boxes contain electronics
This also doesn't tell about fragile/non fragile. Insufficient.

Both 1 & 2,
Let total boxes be 15. (LCM of 3 & 5)

2/3 of 15 => 10 are fragile.

4/5 of 15 => 12 are electronics

So out of 15, 10 are fragile and 12 contains electronics.
Therefore, we can say Yes >50% contain electronics.

C) Both 1 & 2 together are sufficient.

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07 Jul 2025, 11:50

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Looking at the answer choices, it's clear that we need information regarding fragile non-fragile split as well as something about the composition of electronics in both the categories. There individual statements are therefore insufficient. On combining them, I assume that the total number of boxes are 15 (LCM of 3 and 5 based on the options).

the F and NF split is 10 & 5. The E & NE split is 12 & 3. I try to minimize electronics in NF box by taking 5 electronics there and am left with 10 electronics in fragile box. At the minimum my ratio is 10/12 here therefore it will have be more than 0.5 - the answer is C

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07 Jul 2025, 11:56

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Total Boxes: T
No of Fragile boxes (F), Non Fragile (NF), Electronics (E), Non Electronics (NE)
Fragile box contain electronics : FE, Doesn't contain electronics: FE'
NF box contain lectronics : NFE, Doesn't contain electronics: NFE'

Question: FE > F/2

A. $\frac{2}{3}$ T = F (no information on Electronics in boxes) INSUFFICIENT
B. $\frac{4}{5}$ T = E (no info on fragile boxes) INSUFFICIENT

Combine, $\frac{2}{3}$ T = F , $\frac{4}{5}$ T = E

Lets check extreme case, all NF have Electronics => NFE= $\frac{1}{3}$ T, No. of E left = ($\frac{4}{5}$ - $\frac{1}{3}$)
= $\frac{7}{15}$ T => they all must be in Fragile so FE =$\frac{7}{15}$ T > $\frac{1}{3}$ T ($\frac{F}{2}$) YES

If we take any other case it will definitely be greater than F/2 so It is sufficient

Bunuel

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07 Jul 2025, 12:09

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For problems like this one it can come in handy to use a set matrix to order the data. Because the statements are clearly insufficient alone i will move on to show only both statements together on the set matrix.

	Electronics	Non-Electronics	Total
Fragile			2/3
Non-Fragile			1/3
Total	4/5	1/5	1

Using the set matrix in which we see the information that both statements provide us together doesn't help us to determine if more than half of the fragile boxes contain electronics, because we have to distribute 4/5 on two cells one for fragile and one for non-fragile. We know that the non-fragile cell could be at max 1/3 and at min 2/15 so we could determine a case in which the fragile boxes with electronics are 4/5-1/3=(12-5)/15=7/15 which would be below 50 percent and a case in which it would be 10/15 if we apply the minimum value of non-fragile (notice that in this case the value of fragile boxes with electronics would be more than half). Answer E.

Regards,
Lucas

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07 Jul 2025, 12:30

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Bunuel

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The departmental store received two types of boxes as shipments on Monday - Fragile and Non Fragile.

Did electronics _fragile / total fragile > 50% ?

Since the statements contain fractions as 2/3 and 4/5 , let’s assume the shipment received on Monday as 30.

Statement 1:

(1) Two-thirds of the boxes in the shipment are fragile.

(2/3)*30 = 20 boxes are fragile. Which means the remaining 10 boxes are non fragile.

No information about electronics equipment. Hence, Insufficient

Statement 2:

(2) Four-fifths of the boxes in the shipment contain electronics.

(4/5)* 30 contain electronics = 24 boxes contain electronics.

No categorisation of fragile and non fragile boxes are made. Hence, Insufficient

Combining Statements 1 and Statements 2, we get

Out of 30 boxes - 20 are fragile and 10 are non fragile.

24 contain electronics and the rest may be others.

Even in worst case scenario, if we put all the 10 non fragile boxes as electronics. We are left with 24-10 = 14 electronic boxes , which can be accommodated into fragile boxes.

Out of 20 fragile, 14 are electronics and rest (6) boxes are others.

14/20 > 50%

Hence, sufficient

option C

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07 Jul 2025, 12:36

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total boxes = t

statement A - fragile boxes = 2/3t | non fragile = 1/3t
but we don't have any idea about the quantity of electronic items in any of the boxes insufficient.

statement B - boxes containing electronics = 4/5 t

we have no clarity about the number of fragile and non-fragile shipments
not fragile boxes could be 4/5 t or greater

combining statement A and B

fragile boxes = 2/3 t = 10/15 t | Non-fragile boxes = 5/15t
electronic items in = 4/5 t = 12/15 t

minimum value : let say all the non-fragile boxes are consisting electronics
5/15t non-fragile boxes consisting electronics
Remaining boxes with electronics = 7/15t
Total fragile boxes = 10/15 t

let say 7/15t of these boxes consisting electronics > half of the total fragile boxes

Whether true or false we can conclude that the Statement A and Statement B together is sufficient but alone insufficient.

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07 Jul 2025, 12:52

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Bunuel

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Statement 1 and Statement 2 are not sufficient individually. In both the statements one parameter that we need for evaluation is missing.

In statement 1, we are not given which or how many boxes contain electronics and in Statement 2, we are not given information on the fragile box.

Combined
+
Combined the minimum value for Fragile + Electronics = 4/5 - Max (Not Fragile + Electronics)

Max (Not Fragile + Electronics) = 1/3

minimum value for Fragile + Electronics = 4/5 - 1/3 = approx 0.44

This value is greater than 0.33

Hence, the statements combined can yield an answer.

Option C

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07 Jul 2025, 13:10

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Let's assume total boxes = 90

From statement (1):Two-thirds of the boxes in the shipment are fragile.
Fragile boxes = 60
Non Fragile = 30

From statement (2):Four-fifths of the boxes in the shipment contain electronics.
Electronics = 72
Non Electronic = 18

Now put all 30 non-fragile boxes as electronics.
That leaves: 72-30= 42 Electronics that must be in fragile boxes

Now fragile boxes = 60
Electronics in fragile boxes = 42
Percentage of Fragile boxes which contains Electronic items = 42/60 = 70%

So more than half of the fragile boxes contains electronics by combining Statement I and II

Option C

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07 Jul 2025, 13:12

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	Fragile	non fragile	Total
Electronics	A	B
Non electronics	C	D
Total			1

If we assign letters for each fraction, like the table above, the question asks if A>C.

Statement 1:
This statement says $A+C=\frac{2}{3}$. This is not enough information to evaluate whether A>C or not. So, this statement is not sufficient.

Statement 2:
This statement says that $A+B=\frac{4}{5}$; this statement is not enough to figure out any relationship between A and C. Not sufficient.

Both statements together:
If we consider both statements together, we can see that C is being limited to $\frac{1}{5}$, but A+C equals to $\frac{2}{3}$, which is way more than double of $\frac{1}{5}$.
So A is probably more than C.
To prove that, we can use inequalities:

$C<=\frac{1}{5}$, and $A+C=\frac{2}{3}$
So: $\frac{2}{3}-A<=\frac{1}{5}$, then : $A=>\frac{7}{15}$
We can conclude that A is certainly greater than C.

	Fragile	non fragile	Total
Electronics	A	B	4/5
Non electronics	C	D	1/5
Total	2/3	1/3	1

The answer is C.

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07 Jul 2025, 14:25

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Fragile (F) and Non Fragile (NF)
Did more than 0.5 * F contain electronics ?

Statement 1:
2/3rds are Fragile. Not sufficient.

Statement 2:
4/5ths have electronics. Not sufficient.

Combining:
Suppose total = 15
Fragile = 10; Non fragile = 5
Electronics = 12
Suppose from 12, all non fragile had electronics. 7 are left and 7/10 is greater than 50%.

Hence C is the answer.

Bunuel

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07 Jul 2025, 14:57

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Bunuel

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1) Alone insufficient, as we dont know anything about the quantity of the boxes, that contain electronics.
2) Alone insufficient, as we dont know anything about the quantity of the boxes, that contain fragile electronics.

Together sufficient.
We know, that we need at least 2/3 of all boxes to contain electronics to fulfill the statement.
Because 1/3 at max does have no electronics and we need 1/3 to agree or disagree with the statement, that half of the boxes contain electronics.

Knowing, that more than 2/3 contain electronics, we can agree with the statement.

Therefore, the answer is C) Both together are sufficient.

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07 Jul 2025, 14:59

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Topic(s)- Overlapping Sets, Max/Min
Strategy- Max/Min
Variable(s)- Fragile = "F"; Non-Fragile = "nF"; Electronics = "E"; Non-Electronics = "nE"; Total Shipment = "T"; Unknown category = "?"
Grid Notation- Blank cell = "-"

0. Pre-Work- Setup a 4-column, 4-row grid
1st Column: "-", "F", "nF", "Sum"
1st Row: "-", "E", "nE", "Sum"
Rephrase the Question: Is (FE)min/(Sum(F)) ?> (1/2)

(1) (2/3)(T) = Sum(F)
This implies (1/3)(T) do NOT contain fragile boxes: Sum(nF) = (1/3)(T)
Sum(F) = (F?) + (Fn?)
At this point, we have no idea what other categories could be involved.
[Eliminate A, D]

(2) (4/5)(T) = Sum(E)
This implies (1/5)(T) do NOT contain electronics: Sum(nE) = (1/5)(T)
What is FE?
[Eliminate B]

(3) = (1)+(2): Is (FE)min/Sum(F) ?> 1/2
Evaluate (FE)min scenario
1. (FE)min + (EnF)max + (FnE)max + (nEnF)min = T
i) Scan last row and last column for absolute minimum value: Sum(nE) = (1/5)T
(FnE)max + (nEnF)min = (1/5)T
(1/5)T + (0)T = (1/5)T -> (FnE)max = (1/5)T
ii) Sum(F) = (2/3)T = (FnE)max + (FE)min
(FE)min = (2/3)T - (1/5)T = (7/15)T
2. (FE)min / Sum(F) ?> 1/2
(7/15)T / (2/3)T = (21/30) > 1/2 [YES]

Answer: C

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07 Jul 2025, 15:40

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The correct answer is answer choice (E) neither statement is sufficient

If we take a look at the information provided in the question stem, all we know is that our goal is to determine if more than half of the fragile boxes contain electronics. We do not know if ALL electronics are fragile or if a certain proportion are fragile vs non fragile. We also don't know if non electronic items can also be fragile and non fragile.

Statement 1 - this gives us more insight into the proportion that is fragile but again we don't know about the electronics breakdown enough to answer the question. This statement is not sufficient.

Statement 2 - this gives us insight into the electonics proportion vs non electronics but since we don't know enough about fragile vs non fragile, this statement is also not sufficient.

If we try to use both statements together, we still do not know what proportions are both fragile and electronic, so we cannot answer the question, making the answer choice E.

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Bunuel

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So, we need to find can (Number of electronic in fragile) > amount of fragile box/2

To do that, from equation 1, we can understand that 2/3 boxes are fragile which means 1/3 are non fragile.

But, this is not helping us to tell anything which can help us know number of electronics in fragile so not sufficinet.

Similalrly, in second equation we can see that 4/5 of shipment was electronic. okey this is a good info but are wwe sure with just this we can also find hoow mnay were fragile ? We dont know that how the distrobution happened so this is also not sufficient.

But, if we consider both equation 1 and 2, we get that if their are 15 boxes, 10 would be fragile and 5 non fragile. Simillarly out of those 15, 12 would be electonics (from second equation). so in worst case we can imagine let's say all the non fragile shipment were electronic so we will left with just 7 electronics to be put as fragile right. (12-5)

But, then 7 out of 10 shipment in fragile will become electronics. which means 7/10 is more than 50% which is what we want to get... So, (C) both equations together are sufficient but alone does not sufficient.

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C. Together

(1) by itself says nothing about electronics
(2) by itself says nothing about box types

Together:
Half fragile is 1/3. That is more than the 1/5 no electronic containing boxes. SUFFICIENT

Bunuel

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07 Jul 2025, 18:38

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A department store received a shipment of two types of boxes on Monday: fragile and non-fragile. Did more than half of the fragile boxes contain electronics?

Let's assume a total shipment of T=15 (considering both options)

F + NF = 15; (Fragile + Non Fragile)

E + O = 15; (Electronics + Others)

FE > F/2 ?

(1) Two-thirds of the boxes in the shipment are fragile.

F = 2 * 15/3 = 10;

Non Fragile = 5;

FE > F/2 ?

From here, we don't get anything about FE, so alone it's not sufficient.

(2) Four-fifths of the boxes in the shipment contain electronics.

E = 4*15/5 = 12; O = 3;

FE > F/2 ?

From here, there could be many combinations; we can't say anything. So, along with it's not sufficient.

But with both, we knew F = 10; & O = 3; so Possible combinations 3(o) + 7(e) or 2(o) + 8(e) or 10(e), like that each combination would provide a definitive answer for the comparison.

FE > F/2 : This would be false.

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07 Jul 2025, 19:27

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Bunuel

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For better understanding and clarity, Lets consider the total items are 30x.

F, NF means Fragilee and Non Fragile items.
E, NE means Electronics and Non Eletronics.

Now let's draw a matrix for more clarity,

	F	NF	Total
E			24x
NE			6x
Total	20x	10x	30x

With just stmt 1 or stmt 2 we have info about only one row or cloumn but we need the details about EF. Hence we need to use both statements.

If you look the diagram the F can take a total of 20x, To make EF minimum, NE F has to b maximum. So NE F can take 6x max. Making the min value of EF as 14x.

This is more than 50% of the F always. Hence, using stmt 1 and stmt 2, we can always find the values.

Hence IMO C

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07 Jul 2025, 19:54

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E and F > F/2?

(1)
No information about boxes in the shipment that contains electronics.

INSUFFICIENT

(2)
No information about boxes in the shipment that are fragile.

INSUFFICIENT

(1)+(2)

Total boxes = 15

Two-thirds of the boxes in the shipment are fragile (10) -> One-third of the boxes in the shipment are non-fragile (5).
Four-fifths of the boxes in the shipment contain electronics (12) -> One-fifth of the boxes in the shipment contain non-electronics (3).