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Information given:
- Shipment has two types of boxes: fragile and non-fragile
- Question: Did more than half of the fragile boxes contain electronics?

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

Solution:
- Statement 1: Two-thirds of the boxes in the shipment are fragile
- 2/3 of all boxes are fragile, but we do not know anything about how many fragile boxes have electronics vs. not
- We can't answer whether more than half of the fragile boxes contain electronics
- Insufficient

- Statement 2: Four-fifths of the boxes contain electronics
- 4/5 of all boxes contain electronics, but we don't know how the electronics boxes are split, fragile or non-fragile
- Insufficient

- Statement 1 and 2 combined
- We can use numbers to test this scenario
- If we take total boxes = 15
- Statement 1 tells us 10 boxes are fragile, and 5 are non-fragile
- Statement 2 tells us 12 boxes contain electronics, and 3 do not contain electronics
- If we maximize the amount of non-fragile boxes with electronics (5), that still leaves 7 electronics for fragile boxes, which is more than half
- Therefore, combining the statements forces more than half of fragile boxes to contain electronics
- Sufficient

Answer: C, both statements together are sufficient, but neither alone is sufficient
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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The question asks whether the number of fragile boxes containing electronics is more than half the total number of fragile boxes. Let T be the total number of boxes, F be the number of fragile boxes, and F_E be the number of fragile boxes containing electronics. The question is whether F_E > (1/2)F. This is a Yes/No question that requires a definitive answer.

Statement (1) alone is insufficient. It states that F = (2/3)T. This provides the total number of fragile boxes but gives no information about how many of them contain electronics. The number of fragile boxes with electronics, F_E, could be zero or it could be F, allowing for both "No" and "Yes" answers to the question.

Statement (2) alone is insufficient. It states that the total number of boxes with electronics is (4/5)T, but it provides no information about the number of fragile boxes, F. Without knowing F, the threshold for the question, (1/2)F, cannot be determined.

Combining both statements provides sufficient information. Let the total number of boxes, T, be 15 (a common multiple of the denominators 3 and 5). From statement (1), the number of fragile boxes is F = (2/3) * 15 = 10. From statement (2), the total number of boxes with electronics is E = (4/5) * 15 = 12. The question becomes: Is F_E > (1/2) * 10, or is F_E > 5? The minimum possible overlap between the fragile and electronics groups is given by the formula: Minimum Overlap = Group1 + Group2 - Total. In this case, the minimum value for F_E is 10 + 12 - 15 = 7. Since the minimum number of fragile boxes containing electronics is 7, the number must always be greater than 5. This provides a definitive "Yes" answer.

Because the statements together are sufficient to answer the question, but neither statement alone is sufficient, the correct answer is (C).
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We have boxes that can be either fragile or non-fragile, and they either contain electronics or don't.
The question asks: Do more than half of the fragile boxes contain electronics?
Statement 1: Two-thirds of all boxes are fragile.
This only tells us how many fragile boxes there are, but nothing about electronics. Not enough information.
Statement 2: Four-fifths of all boxes contain electronics.
This only tells us how many boxes have electronics, but nothing about which ones are fragile. Not enough information.
Using both statements together:
Let's say we have 15 total boxes (a convenient number that works with our fractions):
* 10 boxes are fragile (2/3 of 15)
* 12 boxes contain electronics (4/5 of 15)
At most, 5 boxes are non-fragile (15 - 10). Even if ALL these non-fragile boxes contain electronics, that's only 5 boxes with electronics that aren't fragile.
This means at least 7 fragile boxes must contain electronics (12 total electronics boxes - 5 non-fragile boxes).
Since 7 out of 10 fragile boxes contain electronics, that's more than half.
Therefore, both statements together are sufficient to answer the question (answer C).
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Let's take the Total number of boxes = 30
So Fragile (F) + Non-Fragile (NF) = 30

Now,
Statement 1 :
Two-thirds of the boxes in the shipment are fragile.
Thus, F = 2/3*30 = 20
So NF = 30-F = 10.

But since Statement 1 does not tell us anything about Electronics,
Therefore Not Sufficient.

Statement 2 :
Total Electronics (E) = 4/5*30 = 24
But if we look at statement 2 individually, we can't say about the number of Fragile boxes.
Therefore Not Sufficient.

Both Together,
We have F=20,
NF = 10,
E = 24
Now even if all of NF are Electronics boxes , we still have 24-10 =14 Electronics boxes.
Thus we will have at least 14 Fragile Electronics boxes.
And 14/24 >1/2
i.e. more than half of the fragile boxes will always contain electronics.

Thus, Answer is C.
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solve using 2x2 matric

A department store received a shipment of two types of boxes on Monday: fragile and non-fragile.

----- F----NF-----total
E----- a ----- -------
NE----- ----- -------
Total--- x ---- ------
determine value of a wrt to x for target Did more than half of the fragile boxes contain electronics?

#1
Two-thirds of the boxes in the shipment are fragile.
----- F----NF-----total
E----- ----- -------
NE----- ----- -------
Total--- 2x/3 ----x/3 ------x

insufficient as no info about Electronics

#2

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


----- F----NF-----total
E----- ----- ------- 4x/5
NE----- ----- ------- x/5
Total--- ---- ------ x

insufficient as fragile info not known
from 1 &2

----- F----NF-----total
E----- ----- ------- 12
NE----- ----- ------- 3
Total--- 10 ---- 5 ------ 15

if we maximize value of Electronics in fragile we get
----- F----NF-----total
E----- 7----- 5 ------- 12
NE----- 3----- 0 ------- 3
Total--- 10 ---- 5 ------ 15

true to target

if we minimize value of electronics in fragile we get

----- F----NF-----total
E----- 10-----2 ------- 12
NE----- 0----- 3 ------- 3
Total--- 10 ---- 5 ------ 15

true to target

OPTION C is correct
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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First lets be clear on the exact information provided,
--> only 2 boxes - fragile and non-fragile
---> Electronics are present in some boxes, electronics may be both fragile or non-fragile

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

(1) Two-thirds of the boxes in the shipment are fragile.
This states that 66.67 % of boxes are fragile. We still don't know the share of electronics inside these boxes. Insufficient

(2) Four-fifths of the boxes in the shipment contain electronics.
This states that 80% of the boxes have electronics. Here we don't know share of fragile and non-fragile boxes. Let say if non-fragile boxes are 10%, then there could be any possibility for electronics present inside these boxes. Insufficient

Considering option 1 and 2 together,

66.67% boxes are fragile and 80% boxes contains electronics.

Lets find try to find out the least probability of electronics in fragile shipment.
For this, maximize the electronics present in non-fragile boxes which would be 33.33%. Hence remaining electronics = 80-33.33 = 46.67% will be definitely in non-fragile box and 46.67% is more than half of 66.67% fragile boxes. If this condition satisfies for minimum possible share of electronics in fragile boxes, the same will hold true for any other combination.


Hence correct answer is C.
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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.


 


This question was provided by GMAT Club
for the GMAT Club Olympics Competition

Win over $30,000 in prizes such as Courses, Tests, Private Tutoring, and more

 

Given: two categories of boxes: fragile and non-fragile

asked: How many fragile contain electronics

Statement 1: 2/3 of total = Fragile.
No mention of electronics. So not sufficient.

Statement 2: 4/5 of total = electronics
No detail about Fragile boxes. So not sufficient.

Combined Statements 1 and 2:
Total number of Fragile and electronics are separately mentioned with no detail of both or neither. So the formula Total = Fragile + Electronics - Both + neither cannot be applied so Option E Answer
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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.
F=2/3 , no information about electronics
Not sufficient

(2) Four-fifths of the boxes in the shipment contain electronics.
E=4/5, no information about fragile boxes
Insufficient

(1)&(2) together
Let, total number of boxes=15
Fragile boxes=2/3 *15=10............. Non Fragile boxes=5
Electronic = 4/5 *15=12............... Box that doesn’t contain electronic =3

Even if all 5 non fragile boxes contain electronic items. We have still 12-5=7 electronic item left that have to be put in fragile boxes and number of remaining electronic item is more than halftge number of fragile boxes.
Sufficient

C
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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.


 


This question was provided by GMAT Club
for the GMAT Club Olympics Competition

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2/3 is fragile ,4/5 is electronic therefore 2/3*4/5=8/15 hence it's more than half
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Let total be x
Fragile all be F
Non Fragile all be N
Fragile Electronics be FE
Fragile No Electronics be FN
Non-fragile Electronics be NE
Non-fragile No Electronics be NN

F = FE+FN
E = FE+NE

Is FE> 0.5*F?

S1
F = FE+FN = 2x/3
We do not know the split of FE and FN
Insufficient

S2
E = FE+NE = 4x/5
We do not know the split of FE and NE
Insufficient

Combined,
Not F = x-2x/3 = x/3
Not E = x-4x/5 = x/5

FN max can be x/5, which means FE min can be 2x/3 - x/5 = 7x/15
FE/F = 7x/15 / 2x/3 = 7/10 = 0.7, which is more than 0.5
Sufficient

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.


 


This question was provided by GMAT Club
for the GMAT Club Olympics Competition

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Total

FNFTotal
E24
NE66
total 2010

To minimize the Electronics and Fragile count, we need to maximize the NE fragile count --> Making it 6
in that case E/F = 14/20
and this can only go up but won't come down.

both statements needed to 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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At first C might look like the answer, but E is the answer.

Each option by itself is not sufficient.
When we take both, we get fragile=2/3 of total=10/15 of total boxes
electronics=4/5 of total=12/15 of total boxes
If you take the total boxes as 15,it might look sufficient to answer that half of the fragile boxes contain electronics. Because this will always be yes in the worst-case scenario.But if the total boxes are 60,100. Then things change. Then it's not sufficient to answer Did more than half of the fragile boxes contain electronics?.Because there will be cases where we can find fewer than half fragile boxes containing electronics.

Hence the answer is E.
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.


 


This question was provided by GMAT Club
for the GMAT Club Olympics Competition

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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.


 


This question was provided by GMAT Club
for the GMAT Club Olympics Competition

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Total number of box should be multiple of 3 and 5.

Assume number of boxes = 15

1) We don't know the number of boxes containing electronics. Hence, the statement alone is not sufficient.

2) We don't know the number of boxes that are fragile. Hence the statement alone is not sufficient.

Combined

Number of fragile boxes = 2/3 * 15 = 10
Number of non fragile box = 5

Number of boxes that contain electronics = 4/5*15 = 12

Now if all of the 5 boxes that are non fragile contain electronics, 7 of 10 fragile boxes must have electronics. Hence, we have a "Yes" for the question asked.

Option C
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What we know:

Two shipments. Two types of boxes in each shipment
Types: Fragile, not fragile.

Ask:

Did more than half the fragile boxes contain electronics?

1) 2/3 the boxes in the shipment are fragile - We do not know if fragile means it has electronics. so this alone is not sufficient

2) 4/5 of the boxes in the shipment contain electronics. - Alone this is not sufficient as we do not know how many of those electronics are in fragile labeled boxes.

Combined)
we know that 2/3 or 10/15 of the boxes are fragile
we know that 4/5 or 12/15 of the boxes contain Electronic.
I am going to double that one more time to work with even numbers. 20/30 fragile. 24/30 electronics.
in a worst case 100% of the not electronics (N/E) are in fragile boxes. That would put 6 N/E inside the 20/30 fragile boxes. leaving us 14/30. That means that there is a scenario where the less than half the fragile boxes contain electronics and a scenario where more than half them do. Not sufficient.

Answer: E - Not sufficient alone or combined.
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Okay lets solve using the Table method we need to find if Fragile electronics > 0.5 Fragile
lets assume total items = 300

F = fragile
NF = non fragile
E = elctronics
NE = non-elctronics

(1) Two-thirds of the boxes in the shipment are fragile. =>
F = 2/3 * 300 = 100 that means NF= 1/3*300 = 100. but this does not give us any info about amount of electronic items in the F and NF. so Not Sufficient
(2) Four-fifths of the boxes in the shipment contain electronics. =>
E = 4/5 * 300 = 240 that mean NE = 60. but this does not give us info about Fragile items and Electronics in them so Not Suffcient

Combine 1 and 2 =>

ENFTotal
Fpq200
NFrs100
Total24060300

Now we need to find if p > 0.5 * 200
is p > 100
Now if we see p = 240 - r and r can never be greater than 100 when r(max)=100 so p(min) = 140 which is always > 0.5 * 200 = 100, so this is Suffcient

Hence Ans C: Both together sufficient.
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Total 90
1. Fragile=60.. Electronix box can be more or less than 30...NOT SUFFICIENT
2. Total electronix=72..fragile Electronix can be ascertained as fragile is not known..NOT SUFFICIENT

Together.. If all the 30 non-fragile are electronix box, then the remaining 42 has be to fragile electronix. so >50%

Ans C
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(1) - No mention of no. of boxes containing electronics. Not sufficient
(2) - we don’t know how many are fragile or whether the electronics are in fragile or non-fragile boxes - Not sufficient.

Combine (1) and (2):
Let’s assume total boxes = 15
From (1):

Fragile = (2/3) * 15 = 10
Non-fragile = 5

Electronics boxes = (4/5) * 15 = 12
Non-Electronics boxes = 3

Let’s suppose worst case: all 3 Non-Electronics are inside Fragile Boxes
Even then, we would have 10 - 3 = 7(more than half) Fragile Boxes that contain Electronics

Sufficient.

(C)

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.


 


This question was provided by GMAT Club
for the GMAT Club Olympics Competition

Win over $30,000 in prizes such as Courses, Tests, Private Tutoring, and more

 

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