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

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kurruhee

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07 Jul 2025, 20:26

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Statement (1)

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

We have no information about the quantum of electronics. So A and D are out.

Statement (2)

We have no information about split of fragile and non-fragile boxes. So B is out.

Statement (1) and (2)

If all non-fragile boxes turn out to have electronics, then we won't get 2/3rd fragile electronics.

If only half of all non-fragile boxes have electronics, we can get 2/3rd fragile electronics.

Unable to conclude. Therefore, C is out. E is the right answer.

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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07 Jul 2025, 20:28

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The correct answer is C.

This is because, A does not tell us about amount of shipments which could contain electronics.

B does not tell us about what fraction of the shipments could be fragile and how exactly does it get divided among fragile and non-fragile, i.e. there could be zero fragile electronic shipments too.

As a result, D also gets eliminated.

Now, we have two choices, C and E.

C is correct because when we combine, even if entirety of non-fragile items, i.e. 1/3 or 33% of the entire shipment is electronic, we have 47% of the total shipment which is electronic and fragile. When we take 47/66, i.e. remaining electronic/fragile items in the shipment, we get ≈ 71% which is a definite answer to this Yes/No question.

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Question analysis: We are dealing with 4 types of contents: Fragile boxes "F", non-fragile "F'", Eletronic "E" and non-eletronics "E'"

Question: The number of fragile boxes that contains Eletronics "FE" > F/2 (F = FE + FE')

I- F = 2B/3 -> F' = B/3 (B = all Boxes) -> Not Suf

II- FE + F'E = 4B/5 -> Only 1/5 of boxes does not contains eletronics. By it self, this alternative does not answe, because we don't know the proportion of fragile boxes (e.g, we can have 50 boxes, 40 of which are non-fragile and all eletronics are on those) -> Not Suf

Together: F = 2B/3 = 10B/15 | E = 4B/5 = 12B/15 -> Minimum fragile boxes with eletronics will happen when all non-fragile boxes (5B/15) contains eletronics. So, at least 7B/15 fragile boxes will contain eletronics, more than 50% -> Suf (answer is "Yes") -> C
--------------------
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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07 Jul 2025, 21:00

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Let total boxes=t
Fragile boxes, f=2/3t
Electronics boxes= 4/5t
Fragile electronics boxes = x

To find: x/f>1/2

Statement (1) says two-thirds of the boxes are fragile, so we know the fragile box count but not how many contain electronics.

Statement (2) says four-fifths of all boxes contain electronics, but it doesn’t specify how many are fragile.

Combining them, we know total fragile boxes and total electronics boxes, but not how many electronics boxes are fragile versus non-fragile.

Without knowing this distribution, we can’t tell if more than half of the fragile boxes contain electronics—both yes and no remain possible.

Thus, neither statement alone nor both statements together are sufficient.

Answer: E

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07 Jul 2025, 21:19

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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	Non-fragile	Total
Electronics	x
Others	y
Total	a		1

is $x>\frac{a}{2}?$

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

	Fragile	Non-fragile	Total
Electronics	x
Others	y
Total	$\frac{2}{3}$	$\frac{1}{3}$	1

$y$ is unknown

Not Sufficient

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

	Fragile	Non-fragile	Total
Electronics	x		$\frac{4}{5}$
Others	y		$\frac{1}{5}$
Total	a		1

$x$ is unknown

Not sufficient

Statements Combined:

	Fragile	Non-fragile	Total
Electronics	x		$\frac{4}{5}$
Others	y		$\frac{1}{5}$
Total	$\frac{2}{3}$	$\frac{1}{3}$	1

$x$ is unknown

But let's consider few scenarios. Let's take total boxes be 15

	Fragile	Non-fragile	Total
Electronics	x		12
Others	y		3
Total	10	5	15

$y\leq{3}$ & $x+y=10$

is $x>5?$

even when there are 3 fragile boxes containing non-electronic components, $x+3=10$, $x=7$

even when there are no fragile boxes containing non-electronic components, $x+0=10$, $x=10$

$7\leq{x}\leq{10}$

Sufficient

Answer: C

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07 Jul 2025, 21:21

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Two type of boxes are present 1.Fragile, 2. non fragile. we dont know how many electronics number are present in each boxes.

Let us assume total no of boxes as 30
option 1: 2/3 boxes are fragile. i.e, 20 boxes are fragile. It doesn't tell anything about the Electronics.

option 2: total 4/5 boxes contain electronics i.e, 24 boxes but it doesn't say anything about fragile and non fragile.

Both statements combined= 20 boxes are fragile and 10 boxes are non fragile and 24 boxes are electronics.
So even if 10 boxes from electronics belong to non fragile the rest 14 boxes are greater than half of the fragile items.

So both statements combined is sufficient

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07 Jul 2025, 21:44

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

Statement 1:
Two-thirds of the boxes in the shipment are fragile.
Means, Fragile = 2/3 = 0.66%
No mention of Electronics,
So, not sufficient.

Statement 2:
Four-fifths of the boxes in the shipment contain electronics.
Means, Electronics = 4/5 = 80%
No mention of Fragile,
So, not sufficient.

Now, combining both statements info, we have,
(i) Fragile = 66%
(ii) Non- fragile = 33%
(iii) Electronics = 80%
(iv) Non-Electronics = 20 %

Minimum value of Fragile electronics = 80% - 33% = 47% > half of fragile (i.e. 66%/2 = 33%)
Maximum value of Fragile electronics = 66 % > half of fragile (i.e. 66%/2 = 33%)

Together, the statements are sufficient to answer the question.

Answer: C

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From 1st statement we get that 2/3 are fragile means 1/3 are non fragile.
But this doesnt tell about electronics so moving on.
2nd statement tells us that 4/5th of total boxes contain electronics so 1/5th will not contain electronics.
We want fragile and electronics so taking common denominator we get fragile are 10/15 are fragile and 12/15 contain electronics.
Even if we consider non fragile is only 5/15 so 7/15 should be atleast in the fragile one out of 12/15. which is greater than half.
Both statements are sufficient to answer.

Bunuel

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lets take sweet number 15, lcm( 3, 5);

1) n_frag = 2/3 = 10
2) n_e = 4/5 = 12

individually they are insufficient.

cross tabulate
. n_frag n_nonfrag
n_e a b = (a+b) = 12
n_other c d = (c+d) = 3
(a+c)=10 (b+d) =5 ; total = 15

now
if a = 10; then b = 2 => a's fraction = 2/5 ; more than half of frag contained electronic

if b = 5; then a = 7 => a's fraction = 7/10; more than half of frag contained electronic;

Hence combined they provide a certain answer.

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[*]Total boxes = T
[*]Fragile boxes = F
[*]Electronics boxes = E
[*]Fragile boxes with electronics = X

We have:
From the statements:
(1) F = (2/3)T
(2) E = (4/5)T

- Fragile boxes: F = (2/3)T
- Non-fragile boxes: T - F = (1/3)T
- Total electronics boxes: E = (4/5)T

Since only 1/3 of boxes are non-fragile, but 4/5 contain electronics, the non-fragile boxes can't hold all the electronics (1/3 < 4/5).

Therefore, some fragile boxes must contain electronics. In fact, more than half the fragile boxes must contain electronics to make the math work.

Answer: C - Both statements together are sufficient.

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Bunuel

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Types: Fragile(F) and Non-Fragile (NF)
Did more than half of the F boxes contain electronics?
Let total boxes be 60.

Statement 1.
2/3 are F
= 40 F and 20 NF
But we don't know how many boxes contain electronics in it.
Not Sufficient.

Statement 2.
4/5 are electronics.
4/5*60 =48 boxes contain electronics.
We don't know how many are fragile and non-fragile.
Not Sufficient.

Statement 1 and 2:
48 boxes have electronics.
40 are F and 20 are NF.
This means even if all 20 NF boxes contain electronics, we are still left with 28 boxes of electronics which will have to be classified under F.
The statement: "Did more than half of the F boxes contain electronics?"
Would be a yes, if F and Electronics >21.
This would be true, as there would have to be at least 28 boxes which contain electronics and are fragile.
Sufficient.

ANSWER: C

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08 Jul 2025, 00:33

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Correct Answer: Yes, more than half of the fragile boxes contain electronics.
This can be find out using both the statements.

What we need to find out ?
Did more than half of the fragile boxes contain electronics?
Let's check one by one:

(1) Two-thirds of the boxes in the shipment are fragile.
This includes how may boxes are fragile, here we don't know any information about electronics fragile boxes.

(2) Four-fifths of the boxes in the shipment contain electronics.
If we take this option only, than we don't know about boxes divided among fragile and non fragile.

3) If we take both option together we can assume,
Fragile boxes= 2/3
Assume here total boxes= 30
Fragile= 20
Non fragile= 10
Using statement II,
Electronic contain boxes= 30*4/5= 24
Now if all the boxes of non fragile contains electronic goods, we have at least 14 boxes which contains electronics goods and this must be fragile.
So here we can conclude that from 20 fragile boxes 14 box contains electronic and 6 contains other. So more than half of the fragile boxes contain electronics.

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08 Jul 2025, 00:36

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LET, Fragile=F , Non-fragile=NF,E=electronics, FE = electronics in fragile box, T=F+NF

(1) Two-thirds of the boxes in the shipment are fragile.
F = (2/3)T
INSUFFICIENT, as no information about electronics
(2) Four-fifths of the boxes in the shipment contain electronics.
E = (4/5)T
INSUFFICIENT, as no information about fragility.
Now , combining both the statements we have,
F = (2/3)T then make N=1T/3
E = (4/5)T
Let all the non-fragile boxes have electronics, this makes the number of boxes left to be in fragile boxes: 4T/5-T/3=7T/15 (worst case scenario)
Now is EF>F/2?
EF in this case should be greater than F/2=(2T/3)/2=T/3=5T/15
7T/15 is greater than 5/15, so, EF>F/2
IMO:C

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C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.

(1) Two-thirds of the boxes in the shipment are fragile. --> We don't know anything about either the fraction or actual number about how many have electronics, Not sufficient

(2) Four-fifths of the boxes in the shipment contain electronics. --> We don't know anything about either the fraction or actual number about how many are fragile/nonfragile, Not sufficient

Let's assume total boxes as x
combining both we know,

Fragile = 2x/3
Electronics = 4x/5

To find the least intersection(fragile boxes containing electronics) between these two, picture this (representative, not to scale).

<----------------->2x/3 (Fragile)
<--------------------------> x
<-------------------->4x/5 (Electronics)
<---->x/5 (not Electronics)
<----------->assume y = least intersection

y = 2x/3 - x/5 = 7x/15

To answer the question,
(7x/15)/(4x/5) = 7/12

Since 7/12 is greater than 1/2, we know that even in the worst case more than half of the fragile boxes contain electronics. Hence, 1 and 2 combined are sufficient.

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lets create a table based on the information from both the statements:

	fragile	non fragile	total
electronics			4/5
non electronics			1/5
total	2/3	1/3	1

based on statement 1, we only know about fraction of fragile boxes which is not enough to know the answer
based on statement 2, we only know about fraction of boxes which contain electronics so we don't know the answer

combining both we get the table, still we don't know if the number of fragile boxes containing electronics is more than half

so both statements are not sufficient

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the answer is E

Explanation
1) States Fragile - but we don't know how many electronics- NOT SUFFICEINT
2) States Electronics- don't know about fragile and non fragile- NOT SUFFICEINT

Now 1 +2
we still do not have any data on how may electronics are interlinked to hel;p us find some solution to the question.
SO NOT SUFFICEINT

Therefore answer is E.

Bunuel

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Lets suppose there are 30 boxes in total.
We can make a matrix based on the type of boxes: Fragile(F) & Non-Fragile(NF) and their contents, Electronics(E), and Not Electronics (NE).
to find : if total Fragile boxes are T, is E&F >0.5T ?

	F	NF
E	>0.5T?
NE
	T		30

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

Now we know T = 2/3*30 = 20

	F	NF
E	>10?
NE
	20		30

But we still do not know anything about number of boxes with electronics and if E&F is more than 10.
Statement 1 is insufficient

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

From statement 2:

	F	NF
E	>0.5T?		24
NE			6
	T		30

But we do not know anything about number of fragile boxes (T)
Statement 2 is insufficient

Statement 1 and 2:

	F	NF
E	>10? Min= 14		24
NE	Max = 6		6
	20	10	30

Now we know total Fragile boxes = 20, and total boxes with electronics = 24.
To find the least number of boxes with E&F, we can maximise NE&F= 6
Then E&F= 20 - NE&F = 14
So at minimum. E&F will be 14 which is more than 10(1/2 of total fragile boxes)

so 1 and 2 together are sufficient

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	Fragile	~~Fragile~~	Total
Electronic			4/5
~~Electronic~~			1/5
Total	2/3	1/3	T

For individual statement, we can use test cases and determine that each statement alone won't be sufficient. However, when we combine both statements, illustrated from the double matrix:
- Even at the maximum 1/5 of the total as Fragile & non-electronics boxes, we will 2/3- 1/5 = 7/15 of the total as Fragile & Electronics. Hence, 7/15 : 2/3 = 70% or always more than 50% of the fragile boxes contain electronics
- Now, obviously with the minimum case, we will have even more percentage of fragile boxes contain electronics. Therefore, both statement is sufficient: Choice (C)

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08 Jul 2025, 02:32

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Bunuel

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Categorize boxes as

Fragile + Electronics --> a
Fragile + Non-Electronics --> b
Non-Fragile + Electronics --> c
Non-Fragile + Non-Electronics --> d
Total --> T

=> a + b + c + d = T

To verify whether
a > b

(1) Two-thirds of the boxes in the shipment are fragile.
a + b = 2T/3

a + b + c + d = T
=> 2T/3 + c + d = T
=> c + d = T/3

Insufficient

(2) Four-fifths of the boxes in the shipment contain electronics.
a + c = 4T/5

a + b + c + d = T
=> 4T/5 + b + d = T
=> b + d = T/5

Insufficient

Combining both sentences we get
a + b = 2T/3
c + d = T/3

a + c = 4T/5
b + d = T/5

Now maximum value of c = T/3. Using that we get lowest value of a
=> a + T/3 = 4T/5
=> a = 12T/15 - 5T/15 = 7T/15

Now a + b = 2T/3
=> b = 2T/3 - 7T/15 = 3T/15

if lowest value of 'a' is greater than 'b' then a > b
Sufficient

Option C

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Assume there's total of 15 shipments (LCM of 3 and 5)

Statement 1
Two thirds of the shipments are fragile. So there's 10 fragile and 5 non fragile shipments. Question becomes do more than 5 of these contain electronics. No information about electronics. Hence, insufficient.

Statement 2
Four fifths of the shipments contain electronics. So there's 12 shipments with electronics. No information about distribution between fragile and non fragile. Hence, insufficient.

Together
If there's 12 shipments with electronics, and only 5 non fragile shipments, max non fragile shipments containing electronics is 5. Meaning, minimum 7 fragile shipments carry electronics. As 7 is greater than half of 12, we have an answer to the question. If not all non fragile shipments carry electronics, the number of fragile shipments carrying electronics further increases.

Together, the statements are sufficient. Option C.

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