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

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bart08241192

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07 Jul 2025, 09:15

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Condition One:
Only the proportion of fragile boxes, which is 2/3 of the total number of boxes, is known.
No information is available regarding the distribution of electronic products.
It is impossible to determine whether electronic products exceed half of the fragile category.
(1) Insufficient

Condition Two:
The total proportion of boxes containing electronic products is known to be 4/5.
However, the distribution of fragile boxes within this category is unknown.
It is not possible to ascertain how many fragile boxes contain electronic products.
(2) Insufficient

Condition One + Condition Two:
Firstly, let's analyze the distribution.

For example,
It is known that the proportion of electronic products in the total is 12/15.
The proportion of fragile items in the total is 10/15.
In other words, as long as the distribution of electronic products within the fragile category exceeds 5/15,
the condition is satisfied.

Let's consider the extreme distributions:
10/15 electronic products within fragile boxes satisfy the distribution.
2/15 electronic products + 3/15 non-electronic products also satisfy the 1/3 distribution of non-fragile items.
10/15 > 5/15 satisfies the condition.

If the non-fragile category has the maximum distribution of electronic products:
5/15 electronic products are in non-fragile boxes.
The remaining 7/15 are distributed within fragile boxes + 3/15 non-electronic products within fragile boxes, which also satisfies the distribution.
7/15 > 5/15 satisfies the condition.

Conditions are sufficient. Answer: C

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07 Jul 2025, 09:16

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Assume total boxes to be 150.
Statement 1 - 2/3 of the boxes is Fragile. Therefore, 100 boxes is fragile and 50 boxes non-fragile. There is no information on electronics. Hence, insufficient.

Statement 2 - 4/5 of the boxes contain electronics.
Therefore, 4/5*150 = 120 boxes contain electronics.
Again, insufficient since, we don't know anything about fragile boxes from this statement alone.

When you combine the two statements:
Fragile is 100
Even if all the 30 non-electronic are considered to be part of fragile boxes, fragile electronics = 70 which is more than half.

Hence, C.

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07 Jul 2025, 09:26

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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?

	Fragile	Not fragile	Total
Electronics	x
Not electronics
Total	F

Is x > F/2 ?

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

	Fragile	Not fragile	Total
Electronics	x
Not electronics
Total	2/3	1/3	1

Since there is no information about shipments containing electronics.
NOT SUFFICIENT

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

	Fragile	Not fragile	Total
Electronics	x		4/5
Not electronics			1/5
Total			1

Since there is no information about shipments that are fragile.
NOT SUFFICIENT

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

	Fragile	Not fragile	Total
Electronics	x	1/3 + 7/15 - x = 4/5 - x	4/5
Not electronics	2/3-x	1/5+x-2/3 = x - 7/15	1/5
Total	2/3	1/3	1

x>=0
4/5 - x >= 0; x<=4/5
x - 7/15>=0 ; x>=7/15
2/3-x>=0; x<=2/3

Combing all above,
0<7/15 <= x <= 2/3 < 4/5

F/2 = 1/3
x > 7/15 > 5/15 = 1/3
SUFFICIENT

IMO C

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07 Jul 2025, 09:27

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Solution:

Let's assume the total number of boxes is 300.

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

	Fragile	Non-Fragile	Total
Electronics	More than 1/2 contain electronics
Non-Electronics
	200	100	300

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

	Fragile	Non-Fragile	Total
Electronics	More than 1/2 contain electronics		240
Non-Eectronics			60
			300

By combining both statements, we won't get an answer

	Fragile	Non-fragile	Total
Electronics	More than 1/2 contain electronics		240
Non-Electronics			60
	200	100	300

Hence, Option E

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07 Jul 2025, 09:32

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Each fact by itself leaves open too many possibilities about how the electronics‐boxes overlap with the fragile ones:

Statement (1) tells us only that 2/3 of all boxes are fragile, but says nothing about how many of those fragile boxes contain electronics so you can’t tell whether it’s more or less than half.
Statement (2) tells us only that 4/5 of all boxes contain electronics, but again leaves the fragile/non‐fragile breakdown of those electronics‐boxes unknown.

However, if you know both together that 2/3 of the boxes are fragile and 4/5 of the boxes contain electronics then even in the worst case the overlap of fragile boxes and electronics‐boxes must exceed half of the fragile ones. By the principle that the minimum overlap of two groups is (sum of their proportions - 1), you get
minimum overlap=2/3+4/5−1=7/15
of the total boxes. Since the fragile boxes themselves make up 2/3=10/15 of the total, the fraction of fragile boxes that contain electronics is at least
(7/15)/(10/15 )= 7/10 = 70%
which is definitely more than half. Thus you need both statements together to be sure.

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07 Jul 2025, 09:39

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Let F denote the number of fragile boxes, NF denote the number of non-fragile boxes, and E denote the number of boxes that contain electronics.
Let the total number of boxes in the shipment be T = 30.

(1) F $= \frac{2}{3}*30 = 20$
This does not tell us anything about E so options A and D are out.

(2) E $= \frac{4}{5}*30 = 24$
This does not tell us anything about F so option B is out.

(1) and (2)
F $= 20$ and E $= 24$
NF $=$ T $-$ F $= 30-20 = 10$
Out of E $= 24$, maximum NF can be 10
=> minimum F $= 24-10 = 14$
=> $\frac{14}{20} > \frac{1}{2}$

Sufficient. The answer is C.

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07 Jul 2025, 09:42

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The first option doesn’t mention about how many boxes of electronics, however second option provides sufficient details to come to a conclusion. Hence point C, both options taken together are sufficient .

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07 Jul 2025, 09:43

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The question asks whether more than half of the fragile boxes contained electronics. Statement one says that two-thirds of the boxes are fragile, but it doesn't tell us how many of those fragile boxes contained electronics, so it's not sufficient. Statement two tells us that four-fifths of all boxes contain electronics, but it doesn't tell us how those electronics are distributed between fragile and non-fragile boxes, so it also isn't sufficient.

Now, on combining both statements, let assume there are 15 boxes.
From 1st statement, 10 fragile and 5 non fragile boxes
From 2nd statement, 12 boxes has electronics

Electronics can be arranged in-
1. Max Fragile: 2 non fragile + 10 fragile
2. Min Fragile: 5 non fragile + 7 fragile

In both cases, more than 50% of fragile box has electronics.

So both statements together is sufficient. ANSWER: C

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07 Jul 2025, 09:51

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Context: Total(T) = F + E - FE + Neither

Question: is FE/F >0.5 ?

Statement 1:
F = (2/3)T
We have insufficient info to calculate total
Hence insufficient

Statement 2:
E = (4/5)T
We have insufficient info to calculate total
Hence insufficient

Combining both 1 and 2:
We still have insufficient info regarding boxes which are fragile and contain electronic
Hence E is correct answer

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07 Jul 2025, 09:52

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Let:

T = Total number of boxes in the shipment

F = Number of fragile boxes

NF = Number of non-fragile boxes (T=F+NF)

FE = Number of fragile boxes containing electronics

NFE = Number of non-fragile boxes containing electronics

E = Total number of boxes containing electronics (E=FE+NFE)

The question is: Is FE>0.5F?

lets check Statement (1): Two-thirds of the boxes in the shipment are fragile.
This translates to F=(2/3)T.
From this, we also know NF=T−F=T−(2/3)T=(1/3)T.
This statement tells us the proportion of fragile boxes in the shipment but gives no information about whether these boxes contain electronics.
Therefore, Statement (1) alone is not sufficient.

lets check Statement (2): Four-fifths of the boxes in the shipment contain electronics.
This translates to E=(4/5)T.
This statement tells us the total proportion of boxes with electronics but gives no information about how these electronics are distributed between fragile and non-fragile boxes.
Therefore, Statement (2) alone is not sufficient.

lets combine Statements (1) and (2):
From (1): F=(2/3)T
From (2): E=(4/5)T

We want to know if FE>0.5F.
Substitute F=(2/3)T into the inequality:
Is FE>0.5×(2/3)T?
Is FE>(1/3)T?

We know that FE is the number of fragile boxes that contain electronics.
We also know that E=FE+NFE.
The total number of boxes with electronics is E=(4/5)T.
The maximum possible value for FE is F (if all fragile boxes contain electronics).
The minimum possible value for FE is E−NF (if all non-fragile boxes contain electronics first, then the remaining electronics must be in fragile boxes).

Let's use the proportions.
Total boxes = T
Fragile boxes F= 2/3 T
Non-fragile boxes NF= 1/3 T
Total electronics boxes E= 4/5T

We want to know if the proportion of fragile boxes that contain electronics, relative to the total fragile boxes, is greater than 0.5.

Let's find the possible range for FE.
We know that 0≤FE≤F.
And we know that 0≤NFE≤NF.

Also, FE=E−NFE.
So, FE= 4/5T−NFE.

Since NFE can be at most NF=1/3T
Minimum FE= 4/5T−1/3T =7/15T.
So, FE≥7/15T

Since FE can be at most F=2/3T
Maximum FE=2/3T

So, we know that 7/15T≤FE≤2/3T

Now let's check the condition FE>0.5F.
We know F=2/3T
So, 0.5F=0.5×2/3T=1/3T

The question is: Is FE>1/3T?

We found that the minimum possible value for FE is 7/15T.
Since 7/15T>5/15T.
Since the minimum possible number of fragile boxes containing electronics (7/15T) is already greater than half the total number of fragile boxes (1/3T), it means that in all possible scenarios consistent with both statements, more than half of the fragile boxes must contain electronics.

Therefore, both statements combined are sufficient.

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07 Jul 2025, 10:03

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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?

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

Let's say we have 300 boxes in the shipment.

F = Fragile boxes
FN = Non-fragile boxes
E = Boxes with electronics
EN = Boxes with not electronics

We need to find: (E and F) > 0.5F

Statement 1:

Two-thirds of the boxes in the shipment are fragile.

We have,
F = 200
FN = 100

No info on electronics, so insufficient.

Statement 2:

Four-fifths of the boxes in the shipment contain electronics.

We have,
E = 240
EN = 60

No info on fragile boxes, so insufficient.

Statements 1 and 2 together:

F = 200
FN = 100
E = 240
EN = 60

We know,

(E and F) + (EN and F) = F ----> (E and F) + (EN and F) = 200 [eq 1]

and, (EN and F) + (EN and FN) = EN ----> (EN and F) + (EN and FN) = 60 [eq 2]

and, (E and F) + (E and FN) = E ----> (E and F) + (E and FN) = 240 [eq 3]

Max value of (EN +F) from eq 2, is 60; putting that in eq 1, we get (E and F) = 140 ---> This is greater than 0.5F (0.5*200 = 100)
Min value of (EN +F) from eq 2, is 0; putting that in eq 1, we get (E and F) = 200 ---> This is greater than 0.5F (0.5*200 = 100)

Hence, sufficient.

Answer C.

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07 Jul 2025, 10:11

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statement I (insufficient because we don't have total number of boxes)
statement II also insufficient
let's combine
total boxes= 15 (LCM of 3 & 5)
Fragile boxes=10
electronics boxes= 12
so yes we can find the value hence ans. is C here.

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07 Jul 2025, 10:15

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Statement (1):
Two-thirds of the boxes in the shipment are fragile.
So, F = (2/3)T
This tells us the number of fragile boxes, but gives no information about how many of them contain electronics. Not sufficient.

Statement (2):
Four-fifths of the boxes in the shipment contain electronics.
So, E = (4/5)T
This tells us the total number of boxes with electronics, but nothing about how these are distributed between fragile and non-fragile boxes. Not sufficient.

Combine (1) and (2):

F = (2/3)T
E = (4/5)T

But we still don’t know how many of the electronics boxes are fragile. For example, all electronics boxes could be in fragile boxes, or some in non-fragile boxes. The overlap is unknown.

Answer E

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07 Jul 2025, 10:20

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Let the number of boxes be 15 for ease of calculation(LCM of 3, 5).

Statement 1: Two-thirds of the boxes in the shipment are fragile.
No. of fragile boxes = 2/3 x 15 = 10.
No information about electronics.

Statement 2: Four-fifths of the boxes in the shipment contain electronics.
No. of boxes with electronics = 4/5 x 15 = 12.
No information about the fragile boxes.

Both Statements:
Now we know the no. of fragile boxes=10 and the no. of boxes with electronics=12.
Even in the extreme that the 3 boxes that have electronics are not fragile boxes, 7 out of 100 fragile boxes will still have electronics.
It can be inferred that whatever the distribution of electronics in the boxes is, at least half of the fragile boxes will contain electronics.

Answer: C) Both statements together are sufficient.

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07 Jul 2025, 10:27

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Total no of boxes = 15

(1) Fragile = 2/3+15 = 10 boxes
Non-fragile = 5 boxes. Not sufficient

(2) no of boxes with electronics = 4/5*15= 12. Not sufficient

(3) Even if all 5 non-fragile boxes contained electronics, definitely more than half of the fragile boxes contain electronics. Answer is (c)

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07 Jul 2025, 10:30

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(1) Two-thirds of the boxes in the shipment are fragile.
F = 2/3 and NF = 1/3, but we don't know the fraction of electronic vs non-electronic

Not Sufficient

(2) Four-fifths of the boxes in the shipment contain electronics.
E = 4/5 and NE = 1/5, but we don't know the fraction of Fragile vs Non-Fragile

Not Sufficient

Combined F = 2/3, NF = 1/3
E = 4/5, NE = 1/5

Let's say all the electronics in the non - fragile, then max fragile capacity is 1/3, the rest will have Fragile Electronic = 4/5- 1/3 = 7/15
total F = 2/3 then Fraction = (7/15)/(2/3) = 7/10 > 1/2

Let's say all the electronics the fragile, then out of 4/5, max 2/3 is the capacity, then 2/3, in this case it's 100%

So in both extreme case Fragile electronic fraction will always cross 1/2.

Ans. - C

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07 Jul 2025, 10:40

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Ans: E
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?
so we need a relationship between electronics and fragile boxes or non-fragile one

(1) Two-thirds of the boxes in the shipment are fragile.
2/3 of the boxes are fragile = 66.66% are fragile
Nothing is given about the Electronics (not sufficient)

(2) Four-fifths of the boxes in the shipment contain electronics.
4/5 of the boxes are electronics = 80% are electronics
Nothing is given about the Electronics (not sufficient)

Together, we know that 66.66% of the total is fragile, and 80% of the total is electronics
But we still do not know the segregation between Fragile and non-Fragile.
For example, what if only 10% of the boxes are fragile, and all the electronics are in the remaining (90%) non-fragile boxes.
Or what if all of the 10% of fragile boxes have electronics, and rest 70% is in the non-fragile boxes?

[Not Sufficient]

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07 Jul 2025, 10:42

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Let F= Fragile boxes, N = Non-fragile boxes, T = Total boxes, E = Boxes with electronics, X = Fragile boxes with electronics.

Our question can be rephrased as: "Is X > 1/2*F?" or "If X/F > 1/2?".

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.

F = 2/3*T

Is this sufficient? No, it is not. We know the F amount, but we do not know the E nor the relationship between F and E.

Eliminate answer choices A and D.

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

E = 4/5T

Is this sufficient? No, it is not. We know the E amount, but we do not know the F amount nor the relantionship between F and E (we could have all Electronics boxes in Non-fragile boxes, for example).

Eliminate answer choices B.

(1) plus (2)

F = (2/3)*T
E = (4/5)*T

LCM ( 3 , 5 ) = 15

F = (10/15)*T
E = (12/15)*T

N = T - F = (15/15)*T - (10/15)*T = (5/15)*T.

In the case with the lowest proportion of X, we will have all the N cointaining Electronics and only the remaining boxes with Electronic being F:
Lowest X:
Lowest possible B = E - N = (12/15)*T - (5/15)*T = (7/15)*T.
(7/15)*T > (1/2)*F?
(1/2)*F = (1/2)*(2/3)*T = (1/3)*T = (5/15)*T

(7/15)*T > (5/15)*T
Since any other configuration of Electronics distribution will bring a X even grater, we can answer that Yes, statements (1) and (2) together are sufficient to answer.

Answer: C

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07 Jul 2025, 10:49

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We will solve this by Double set matrix. Suppose the total items in the shipment are 15 (LCM of 3 & 5)

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

	Fragile	Non-Fragile	Total
Electronics
Non-Electronics
Total	10	5	15

No info about electronics. Insufficient

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

	Fragile	Non-Fragile	Total
Electronics			12
Non-electro			3
Total			15

No info about fragile/not fragile. Insufficient

Combining both info

	Fragile	Non-Fragile	Total
Electronics.			12
Non-electro			3
Total	10	5	15

We need Fragile electronics. From the 10 fragile boxes, all 10 could contain electronics, making it a yes. Explained the case below

	Fragile	Not-Fragile	Total
Electro	10	2	12
Non-electro	0	3	3
Total	10	5	15

Or all 3 non-electronics could be fragile, giving us fragile electronics as 7, making it a no. Explained the case below:

	Fragile	Not-Fragile	Total
Electro	7	5	12
Non-electro	3	0	3
Total	10	5	15

Hence we have both yes/no case when we combine the info. Ans E.

sabareeshmc1405

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07 Jul 2025, 11:12

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Bunuel

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Let’s say there are 30 boxes in total (just picking a simple number).

S1:

Two-thirds are fragile → that means 20 fragile boxes
The rest (10) are non-fragile
No not sufficient individual

S2:

Four-fifths of all boxes have electronics → that means 24 boxes have electronics
The other 6 boxes do not have electronics
No not sufficient individual S1+ S2:

Now, let’s try to put as many electronics boxes as possible into the non-fragile group (which has only 10 boxes).

At most, we can fit 10 electronics boxes there.

That leaves:

24−10=14 electronics boxes

So, at least 14 of the 20 fragile boxes must have electronics.

That’s more than half (half of 20 is 10).

So yes — more than half of the fragile boxes have electronics.

Answer: C. The two statements together are enough to answer the question