Last visit was: 19 Nov 2025, 02:40 It is currently 19 Nov 2025, 02:40
Close
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Close
Request Expert Reply
Confirm Cancel
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 19 Nov 2025
Posts: 105,379
Own Kudos:
Given Kudos: 99,977
Products:
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 105,379
Kudos: 778,174
 [52]
3
Kudos
Add Kudos
49
Bookmarks
Bookmark this Post
Most Helpful Reply
User avatar
BrentGMATPrepNow
User avatar
Major Poster
Joined: 12 Sep 2015
Last visit: 31 Oct 2025
Posts: 6,739
Own Kudos:
35,337
 [10]
Given Kudos: 799
Location: Canada
Expert
Expert reply
Posts: 6,739
Kudos: 35,337
 [10]
6
Kudos
Add Kudos
4
Bookmarks
Bookmark this Post
User avatar
KarishmaB
Joined: 16 Oct 2010
Last visit: 18 Nov 2025
Posts: 16,267
Own Kudos:
76,987
 [7]
Given Kudos: 482
Location: Pune, India
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 16,267
Kudos: 76,987
 [7]
6
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
General Discussion
User avatar
rajatchopra1994
Joined: 16 Feb 2015
Last visit: 22 Jun 2024
Posts: 1,056
Own Kudos:
1,252
 [7]
Given Kudos: 30
Location: United States
Posts: 1,056
Kudos: 1,252
 [7]
5
Kudos
Add Kudos
2
Bookmarks
Bookmark this Post
Explanation:

75% lost an arm; 85% lost a leg
So, 60% lost an arm or leg. (85 + 75 - 100)

70% lost an eye; 80% lost an ear
So, 50% lost an eye or ear. (80 + 70 - 100)

Minimum value of x who lost all 4 limbs are = 60-50
= 10%

IMO-A
avatar
Pankilpatel
Joined: 27 Sep 2020
Last visit: 01 Apr 2021
Posts: 1
Own Kudos:
2
 [2]
Posts: 1
Kudos: 2
 [2]
2
Kudos
Add Kudos
Bookmarks
Bookmark this Post
in this case, why did you take eye and ear together and arm and leg together?

if we take ear and arm together and eye and leg together and follow your method, we will get zero. but that shouldn't be the case right.
User avatar
Fdambro294
Joined: 10 Jul 2019
Last visit: 20 Aug 2025
Posts: 1,350
Own Kudos:
741
 [3]
Given Kudos: 1,656
Posts: 1,350
Kudos: 741
 [3]
2
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
The way I learned how to analyze these kinds of 4 Overlapping Set Problems was from a famous poster (might have been the exact person who posted this question).



Set up a Line Graph from 0 through 100

then, try to Spread Out and Limit the Overlap as much as possible

There exists 100 Unique Elements (100%):


0.........10..........20.....25.....30.........40..........50............60............70.........80........90.......100

[----------------------------------70 = eye----------------------------------]

[------------------------------50 - ear---------------].............................[--------------30 - ear-------]

[-----25-arm------------]...................................[----------------------50 - arm-----------------------]


.................................[----------------------------------------75-Leg-------------------------------------]


Now, Given the Last (and Highest) Set of Losing a Leg = 85:

we could place the Set from the 25th Marker to the 100th Marker (as shown above) and we would AT MOST have only 3 Overlapping Sets at any 1 place on the Line Graph.

However, this will only cover 75 of the 85. We still have to account for 10 more.


Assuming that there does not exist a "Neither/Nor" Group:

No matter where we put the last 10 on the Graph Above, we will have 4 Overlapping Sets.


-A- 10% is the Minimum that X% can be

No matter how you try to spread out the 4 Sets among the 100%, you will always have AT MINIMUM 10% Overlap among all 4 Sets.
User avatar
Nups1324
Joined: 05 Jan 2020
Last visit: 12 Sep 2023
Posts: 105
Own Kudos:
Given Kudos: 353
Posts: 105
Kudos: 64
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Fdambro294
The way I learned how to analyze these kinds of 4 Overlapping Set Problems was from a famous poster (might have been the exact person who posted this question).



Set up a Line Graph from 0 through 100

then, try to Spread Out and Limit the Overlap as much as possible

There exists 100 Unique Elements (100%):


0.........10..........20.....25.....30.........40..........50............60............70.........80........90.......100

[----------------------------------70 = eye----------------------------------]

[------------------------------50 - ear---------------].............................[--------------30 - ear-------]

[-----25-arm------------]...................................[----------------------50 - arm-----------------------]


.................................[----------------------------------------75-Leg-------------------------------------]


Now, Given the Last (and Highest) Set of Losing a Leg = 85:

we could place the Set from the 25th Marker to the 100th Marker (as shown above) and we would AT MOST have only 3 Overlapping Sets at any 1 place on the Line Graph.

However, this will only cover 75 of the 85. We still have to account for 10 more.


Assuming that there does not exist a "Neither/Nor" Group:

No matter where we put the last 10 on the Graph Above, we will have 4 Overlapping Sets.


-A- 10% is the Minimum that X% can be

No matter how you try to spread out the 4 Sets among the 100%, you will always have AT MINIMUM 10% Overlap among all 4 Sets.

Hi,

I tried a lot to understand your explanation but I couldn't. If you could explain it in detail or in some other way then it'll be great.

Tagging others just in case
Bunuel chetan2u GMATinsight IanStewart ScottTargetTestPrep yashikaaggarwal VeritasKarishma BrentGMATPrepNow

Thank you :)

Posted from my mobile device
User avatar
chetan2u
User avatar
GMAT Expert
Joined: 02 Aug 2009
Last visit: 15 Nov 2025
Posts: 11,238
Own Kudos:
43,700
 [5]
Given Kudos: 335
Status:Math and DI Expert
Location: India
Concentration: Human Resources, General Management
GMAT Focus 1: 735 Q90 V89 DI81
Products:
Expert
Expert reply
GMAT Focus 1: 735 Q90 V89 DI81
Posts: 11,238
Kudos: 43,700
 [5]
3
Kudos
Add Kudos
2
Bookmarks
Bookmark this Post
Bunuel
In a battle, 70% of the combatants lost one eye, 80% an ear, 75% an arm, 85% a leg, x% lost all the four limbs. What is the minimum value of x?

A. 10
B. 12
C. 15
D. 20
E. 25


Are You Up For the Challenge: 700 Level Questions


I believe the four limbs here mean the four organs given - eye, ear, leg and arm.
For ease of understanding, let us take eye, ear, arm and leg as a, b, c and d respectively.
Now if we want to minimise the value of x, which is people with all the four injuries, then distribute the injuries one after another in a way that it fills up 100% and then only moves to a person having second disease.

So 70% lost an eye. Next 80% lost an ear.
So the 100-70 or 30 are filled up for ear alone and then the remaining 80-30 of ear injuries are given to those who had already lost an eye.
At the end of this distribution, we have 50% with both injuries, 20% with just eye and 30% with just ear.

Next 75% lost an arm.
So we will fill up those with just one injury first. => So 20% now have injury in eye and arm, while 30% have in eye and arm. But we still have 75-20-30 or 25 arm injuries to fill up, who will go to those with eye and ear.
In the end of this distribution we have 25% with 3 injuries and remaining 75% with 2 injuries.

Next 85% lost a leg.
First we fill up those with two injuries, that is 75%, so 85-75 or 10% are still left and we have everyone with 3 injuries. Thus, the renaming 10% will have to be given to those people who already have 3 injuries.

So x has to be 10.

If you understand the concept of distribution above, the method is straightforward
Add all of them => 70+80+75+85=310%
So 100% have three injuries, leading to 300%. So remaining 10% will make 10% of people to have all 4 injuries.

If you were looking for maximum value of x, we first give a set of people all the injuries, so the minimum of 4 will become the maximum value of x. Here 70 is the least of 70, 80, 75 and 85. So Max value of x is 70.
User avatar
Gokul20
User avatar
Current Student
Joined: 16 Oct 2019
Last visit: 30 May 2023
Posts: 121
Own Kudos:
38
 [4]
Given Kudos: 896
Location: India
GMAT 1: 710 Q49 V36 (Online)
GMAT 2: 720 Q50 V38
GMAT 3: 730 Q50 V38
GPA: 4
GMAT 3: 730 Q50 V38
Posts: 121
Kudos: 38
 [4]
2
Kudos
Add Kudos
2
Bookmarks
Bookmark this Post
The question is confusing in what it actually wants. If by sheer meaning I go with limbs it essentially means arms and legs.
If that is the case,
75% arms and 85% legs -> Which gives a 60% minimum overlap. This should be the answer. But I don't find this in the options.
So, instead the question is asking what is the minimum overlap of all the four organs - eyes, ear, arm, legs (not limbs)
In that case, the overlap would be
100 -70 = 30(eyes)
100 - 80 = 20(ear)
100 - 75 = 25(arm)
100 - 85 = 15(legs)
Minimum overlap = 100 - (30 + 20 + 25 + 15) = 10.

Option (A) is the Answer
User avatar
Fdambro294
Joined: 10 Jul 2019
Last visit: 20 Aug 2025
Posts: 1,350
Own Kudos:
Given Kudos: 1,656
Posts: 1,350
Kudos: 741
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Great way to explain it VeritasKarishma

I tried to do it with a Line Graph, but it didn’t come across as clearly as I hoped it would.

VeritasKarishma
Bunuel
In a battle, 70% of the combatants lost one eye, 80% an ear, 75% an arm, 85% a leg, x% lost all the four limbs. What is the minimum value of x?

A. 10
B. 12
C. 15
D. 20
E. 25


Are You Up For the Challenge: 700 Level Questions

All 4 limbs means both arms and both legs. The intent of the question is to be say all four - an eye, an ear, an arm and a leg.

You want to minimise the overlap of all 4. You can do it orally. Consider this:

Of 100 people, 80 lost an ear and 85 a leg. I want to minimise overlap. Of 100, 85 people lost a leg. Let's make the other 15 lose an ear. But there are still 65 people who also lost an ear. So there must be an overlap in case of 65 people. Let's colour them green.
Now, 75 lost an arm too. I want to minimise overlap with greens (65). So I make other 35 lose an arm. But 40 more people still lost an arm so they will overlap with greens. Now we have 40 with all 3 losses. Let's colour them blue.
Now, 70 lost an arm too. I want to minimise overlap with blues (40). So I make other 60 lose an eye. But 10 more still lost an eye so they will overlap with blues. Hence, 10 will have the overlap of all 4 conditions.

Answer (A)

Posted from my mobile device
User avatar
ScottTargetTestPrep
User avatar
Target Test Prep Representative
Joined: 14 Oct 2015
Last visit: 18 Nov 2025
Posts: 21,712
Own Kudos:
Given Kudos: 300
Status:Founder & CEO
Affiliations: Target Test Prep
Location: United States (CA)
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 21,712
Kudos: 26,995
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Nups1324
Fdambro294
The way I learned how to analyze these kinds of 4 Overlapping Set Problems was from a famous poster (might have been the exact person who posted this question).



Set up a Line Graph from 0 through 100

then, try to Spread Out and Limit the Overlap as much as possible

There exists 100 Unique Elements (100%):


0.........10..........20.....25.....30.........40..........50............60............70.........80........90.......100

[----------------------------------70 = eye----------------------------------]

[------------------------------50 - ear---------------].............................[--------------30 - ear-------]

[-----25-arm------------]...................................[----------------------50 - arm-----------------------]


.................................[----------------------------------------75-Leg-------------------------------------]


Now, Given the Last (and Highest) Set of Losing a Leg = 85:

we could place the Set from the 25th Marker to the 100th Marker (as shown above) and we would AT MOST have only 3 Overlapping Sets at any 1 place on the Line Graph.

However, this will only cover 75 of the 85. We still have to account for 10 more.


Assuming that there does not exist a "Neither/Nor" Group:

No matter where we put the last 10 on the Graph Above, we will have 4 Overlapping Sets.


-A- 10% is the Minimum that X% can be

No matter how you try to spread out the 4 Sets among the 100%, you will always have AT MINIMUM 10% Overlap among all 4 Sets.

Hi,

I tried a lot to understand your explanation but I couldn't. If you could explain it in detail or in some other way then it'll be great.

Tagging others just in case
Bunuel chetan2u GMATinsight IanStewart ScottTargetTestPrep yashikaaggarwal VeritasKarishma BrentGMATPrepNow

Thank you :)

Posted from my mobile device

Let's assume there were 100 combatants. The given information implies the following: 30 combatants did NOT loose an eye, 20 combatants did NOT loose an ear, 25 combatants did NOT loose an arm and 15 combatants did NOT loose a leg. If a combatant lost all four, then the combatant will not belong to any of these groups. Since we want the minimum value for the "all four", we must maximize the number of compatants who belong to either one of these groups. The number of combatants who belong to either one of these groups is maximized when there are no overlaps between any two of these groups, in which case the number of combatants who belong to either one of these groups (i.e. the number of combatants who either kept both eyes, both ears, both arms or both legs) will be 30 + 20 + 25 + 15 = 90. Thus, the minimum value of "all four" is 100 - 90 = 10.
User avatar
Ganeshmantri
Joined: 30 Aug 2020
Last visit: 15 Aug 2022
Posts: 23
Own Kudos:
11
 [2]
Given Kudos: 95
Location: India
Concentration: Healthcare, General Management
GRE 1: Q168 V157
GPA: 3.36
GRE 1: Q168 V157
Posts: 23
Kudos: 11
 [2]
1
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
Bunuel
In a battle, 70% of the combatants lost one eye, 80% an ear, 75% an arm, 85% a leg, x% lost all the four limbs. What is the minimum value of x?

A. 10
B. 12
C. 15
D. 20
E. 25


Are You Up For the Challenge: 700 Level Questions

I can suggest a simple aproach for such type of questions.

You must move step by step. Now since we have to minimize the overlap, i.e. "x", we will approach as follows

1. 70% lost eye, 80% lost ear - so there will be ATLEAST 50% who lost both (since 70% + 80% = 150% --> min 50% overlap. There can be all the 70% eyeless in the 80% earless.. but since we have to "minimize the overlap", we shall consider 50%)

2. ATLEAST 50% (without eye and ear), and 75% lost an arm - so to minimize overlap here as well, there will be ATLEAST 25% who lost an eye, a ear and an arm

3. ATLEAST 25% (without eye, ear and arm), and 85% without leg - here again, to minimize we can consider 25 out of 100 ppl are without eye, ear and arm; 85 out of 100 are without leg. So atleast 10% will be there without eye, ear and arm AND without a leg

And thats the answer.
User avatar
bhupendersingh27
Joined: 02 Jan 2020
Last visit: 23 Dec 2020
Posts: 26
Own Kudos:
37
 [1]
Given Kudos: 45
Posts: 26
Kudos: 37
 [1]
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
I find line graph easy,

|____(eye70)______________ (blank 30) |
|____(ear 80)_______________ (Blank 20)|
|____(arm 75)_____________ (blank 25) |

last we have leg 85%, to distribute evenly, out of these 85%, first we need to fill blank in above three line

balance after filling blank will be minimum numbers who lost all four limbs
so 85 - (30+20+25) = 10%
User avatar
Fdambro294
Joined: 10 Jul 2019
Last visit: 20 Aug 2025
Posts: 1,350
Own Kudos:
741
 [1]
Given Kudos: 1,656
Posts: 1,350
Kudos: 741
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Looking back at my ridiculous answer months ago, wow 😳

Let the 100% be 100 people.

We want to minimize the number of people who get all 4 of the injuries. So we don’t want any one person to have none. All 100 people have at least 1 injury

There are: 70 + 80 + 75 + 85 = 310 recorded injuries.


(1st) since every one of the 100 ppl have at least 1 injury, give them 1 injury.

Right now:
100 ppl = have exactly 1 injury
And
210 recorded injuries still need to be distributed.


(2)to minimize the number of people who have all 4, we want to maximize the number of people who get exactly 3 injuries.

210 recorded injuries are left

If we give another + 2 injuries to the 100 people, they will all have exactly 3 injuries.

210 - (2) (100 ppl) = 10 recorded injuries remain

Right now:
We have all 100 people = have exactly 3 injuries
And
We still have 10 recorded injuries to hand out

We have to hand those 10 recorded injuries out to 10 people (nobody can have more than 4)


10 ppl will have exactly 4

90 ppl will have exactly 3

(A)10

Posted from my mobile device
avatar
Jaya6
Joined: 09 Jan 2021
Last visit: 21 Jun 2022
Posts: 68
Own Kudos:
Given Kudos: 142
Location: India
Schools: ISB '23 (S)
GPA: 3.2
Schools: ISB '23 (S)
Posts: 68
Kudos: 14
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
In a battle, 70% of the combatants lost one eye, 80% an ear, 75% an arm, 85% a leg, x% lost all the four limbs. What is the minimum value of x?

A. 10
B. 12
C. 15
D. 20
E. 25


Are You Up For the Challenge: 700 Level Questions



I used a very simple approach:


Assume total count to be 100:
If 70% lost an eye then 30% are fit.
If 80% lost an ear then 20% are fit.
If 75% lost an arm then 25% are fit.
If 85% lost a leg then 15% are fit.

Now out of 100, 90% of them are fit which means percent of people who injured all four limbs are 10.
User avatar
umeshv
Joined: 19 Nov 2024
Last visit: 27 Aug 2025
Posts: 16
Own Kudos:
Given Kudos: 6
Location: India
Posts: 16
Kudos: 6
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Use Inclusion-Exclusion Principle.

We know:

If the sum of all people who suffered each individual injury exceeds the total population multiple times, then some people must have suffered multiple injuries. That overlap is what reduces the total count.

So we calculate the minimum number who must have all 4 injuries:

Add up all percentages:
70+80+75+85=310

So — 310 "injury instances" among 100 people
Each person can be counted up to 4 times
So the minimum number of people who must have all 4 injuries is when the overlaps are maximized, i.e., injury count is packed into fewest people.

Let’s suppose:

People who lost only 1, 2, or 3 limbs account for as much as possible.

Then, those who lost all 4 must make up the rest.

Let’s define this formula from PIE logic:

Minimum number of people with all 4 injuries = (Sum of individual injuries) - 3 × Total people

Because each person can have max 4 injuries, but anything above 3 injuries overlaps into a "4th injury" group.

x=70+80+75+85−3×100=310−300= 10


Correct Answer: (A) 10
Moderators:
Math Expert
105379 posts
Tuck School Moderator
805 posts