Problem Solving

In a village of 100 households, 75 have at least one DVD, 80 have at least one cell phone, and 55 have at least one MP3. If x and y are respectively the greatest and lowest possible number of households that have all three of these devices, x - y is:

A. 65
B. 55
C. 45
D. 35
E. 25

What's the fastest way to solve this?

Thanks.

IMO B.

The greatest value x can be is 55, because it is the least of all 3 values. you can do a quick check by doing (75 - 55) + (85 - 55) + (55-55) + 55 = 100

you can assume here that none of all three villagers had only 2 items.

Considering all of the villagers have at most two products in common and no one has all three products common, we can set up the equation like:

210 - 4D - 4C - 4M + (2M+2D+2C) + 0 = 100

110= 2(C+D+M) = 55

all three villages have atleast 55 of two products in common while they have 0 of all 3 products common with each other.

55-0 = 55

IMO I figured out y in the long approach...maybe someone has a faster solution.

What is the OA?

IMO E)45. (after help from another forum)

But my method is kinda crude..

IMO B. On similar lines just a short explanation being. the maximum possible with all three will be 75+80+55-100=110=x

The minimum possible no. with all three stuff is 55 (since its mentioned atleast 55 own and its the smallest no) . therefore, y=55

hence x-y=110-55=55.

what the OA?

Naruto wrote:IMO B. On similar lines just a short explanation being. the maximum possible with all three will be 75+80+55-100=110=x

The minimum possible no. with all three stuff is 55 (since its mentioned atleast 55 own and its the smallest no) . therefore, y=55

hence x-y=110-55=55.

what the OA?

I think my answer 25 is right..I am working on how to do it properly. Right now I have just substituted the values.

Naruto the flaw in your method is:
x cannot be equal to 110.
there are only 100 households

ssmiles08 wrote:IMO B.

The greatest value x can be is 55, because it is the least of all 3 values. you can do a quick check by doing (75 - 55) + (85 - 55) + (55-55) + 55 = 100

you can assume here that none of all three villagers had only 2 items.

Considering all of the villagers have at most two products in common and no one has all three products common, we can set up the equation like:

210 - 4D - 4C - 4M + (2M+2D+2C) + 0 = 100

110= 2(C+D+M) = 55

all three villages have atleast 55 of two products in common while they have 0 of all 3 products common with each other.

55-0 = 55

IMO I figured out y in the long approach...maybe someone has a faster solution.

What is the OA?

ssmiles:

flaw in your method:

if you say number of people with all three is 0.
then number of people with exactly one of those should be 45.

because
100=(number of people with exactly one item)+(number of people with exactly two items)+(number of people with exactly three items)

(number of people with exactly three items)=0
(number of people with exactly two items)=55
so (number of people with exactly one item)=45
from the pic shown:
75-(x+y)+80-(y+z)+55-(x+z)=45
or 210-2(x+y+z)=45
or,x+y+z=165/2------ wrong

tohellandback wrote:

Naruto wrote:IMO B. On similar lines just a short explanation being. the maximum possible with all three will be 75+80+55-100=110=x

The minimum possible no. with all three stuff is 55 (since its mentioned atleast 55 own and its the smallest no) . therefore, y=55

hence x-y=110-55=55.

what the OA?

I think my answer 25 is right..I am working on how to do it properly. Right now I have just substituted the values.

Naruto the flaw in your method is:
x cannot be equal to 110.
there are only 100 households

You are correct, i made a mistake it cant be 110, but then now to think of it, Maximum no. of ppl who can own all three is 100 and the minimum is 55. so the answer is 100-55=45, i.e C.
Doesnt that seem right?

OK for some reason I am so computer illiterate that I cannot make a venn-diagram on my computer! arghh! Anyway Here is the Long approach I took. I am still getting 55. I still don't get what I am missing.

Let us take scenario 1. (Lowest possible of all 3) = y

All 3 = y (lets say y = 0)

Only 2
D+C
D+M
M+C

Only 1
Only D = 75 - [2D + C + M]
Only C = 80 - [2C + D + M]
Only M = 55 - [2M + D + C]

so total = 100

100 = [(75 - 2D - C - M) + (80 - 2C - D - M) + (55 - 2M - D - C)] + [2D + 2M+ 2C] + [0]

210 - 2(D + M + C) = 100

110/2 = (D+M+C) = 55

If you reverse check it: (75-55) + (80-55) + (55-55) + 55 + 0 = 100


Scenario 2: Highest possible (x)

Let all only 2's = 0
D+C
D+M
M+C

the only thing remaining is x; so everything needs to be subracted by x.

75 - x = Only D
80 - x = Only C
55 - x = Only M

so 100 = [(75 - x) + (80 - x) + (55 - x)] + [0] + [x]

100 = 210 - 3x + x

110 = 2x

x = 55

So in entirety; x = 55, y = 0, So 55 - 0 = 55

Hope I did not mess up somewhere!

Naruto wrote:

tohellandback wrote:

Naruto wrote:IMO B. On similar lines just a short explanation being. the maximum possible with all three will be 75+80+55-100=110=x

The minimum possible no. with all three stuff is 55 (since its mentioned atleast 55 own and its the smallest no) . therefore, y=55

hence x-y=110-55=55.

what the OA?

I think my answer 25 is right..I am working on how to do it properly. Right now I have just substituted the values.

Naruto the flaw in your method is:
x cannot be equal to 110.
there are only 100 households

You are correct, i made a mistake it cant be 110, but then now to think of it, Maximum no. of ppl who can own all three is 100 and the minimum is 55. so the answer is 100-55=45, i.e C.
Doesnt that seem right?

naruto maximum will be 55.
we need to find the minimum.I will post my method soon

Oks,
please post the OA. Its one of those toughest ones I guess. Ones we know he OA, we might find a way

Maximum can obviously be 55.....all those who have mp3 players can have everything else....

Minimum can be 10...suppose houses are numbered 1 to 100..suppose first 75 have dvd player and last 80 have cell phone...this way 45 houses dont have either cell phone or dvd..now divide 55 mp3 in this..so onlt 10 houses will have all three...

so 55-10=45..
i got it from another forum(sorry). and this is a very cool reasoning method.. without calculations
Naruto you answer is right but values wrong

tohellandback,

I think you are right. Minimum should be 10, I was thinking too much into the problem..

IMO 45

OA is C - 45

tohellandback, why is minimum 10?? i'm not sure i follow...

i get the maximum part. it is the minimum what i struggle with.

thanks!!!

oks wrote:tohellandback, why is minimum 10?? i'm not sure i follow...

i get the maximum part. it is the minimum what i struggle with.

thanks!!!

ok the explanation is really just reasoning..I hope someone come up with a simple algebraic solution

lets say houses are
1,2,3,4,....100
1-75 own DVD
now we want to minimize the number of people who have all the things in common
we can assume 1-80 own cell phone or the last 80 i.e 21-100 own a cell phone. there are other options too like households from 11-90 etc but
21-100 makes sure that number of people who hold both cell phone and dvd is minimum

now we have the MP3 player
how can we distribute it to the households in such a way that households with all three is minimum
now householdsfrom 1-20 only have dvd, lets give them 20 MP3. so they hold only two items and not three
households from 76-100 have only cell phones, lets give them 25 MP3, so those people hold cell phones and MP3 and not DVD
the rest 10 MP3 we must give to the households who already hold two of the other items
.so thats why the minimum is 10.

you get it??

I hope.