A group of burglars is trying to decide whether to target the houses

Total =100
Neither =0(zero)
x- television
y- artwork
z- car

p- households owning all three items
z < y< x

\(\frac{p}{100} < \frac{1}{2}\) ???
--> p < 50 ???

\(x+y+z = (a+b+c ) + 2(a_1+b_1+c_1) + 3p\)
\(x+y+z = 100+ (a_1+b_1+c_1) + 2p\)

--> \(p= \frac{((x+y+z) -100 -(a_1+b_1+c_1))}{2}\)

(Statement1): x+y=120
--> \(p= \frac{(120+z- 100-(a_1+b_1+c_1))}{2}= \frac{(z+20 -(a_1+b_1+c_1))}{2}\)

here, z could be maximum 58 --> \(p= \frac{(58+20)}{2} -\frac{(a_1+b_1+c_1)}{2}= 39-\frac{(a_1+b_1+c_1)}{2}\)
--> p is less than 50 and be never greater than 50.
(Always YES)-- Sufficient


(Statement2) z=55

\(p= \frac{(x+y+55- 100-(a_1+b_1+c_1))}{2}= \frac{(x+y- 45-(a_1+b_1+c_1))}{2}\)
here x and y are greater than 55. Also we don't know what \((a_1+b_1+c_1)\) is.

--> p could be greater than 50 or less than 50

insufficient

The answer is A

z<x<y

(1)x+y=120
-If x=100 and y=20 and z=19 and z are all inside y, then the burglars will not target the houses.
-If x=61 and y=59 and x+y intersect=20, then the burglars will not target the houses.
SUFFICIENT.

(2)z=55
-If z=55 and y=75 and x=80 and z-y-x intersect=55, then the burglars will target the houses.
-If z=55 and y=75 and x=80 and z-y-x intersect=10, then the burglars will not target the houses.
NOT SUFFICIENT

Answer is (A)
Check my other post for more explanation

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Camach700

Consider following sxenario: x=61, y=59, total =100, and entire z is within x and y domains. Thus, the x-y intersection is 61 + 59 - 100 = 20 and as a result, x-y-z intersection can only be up to 20.

Consider another scenario: x=61, y=59, z=58 (z<y<x) and entire y is inside x domain. Yes, the x-y intersection is 59 and x-y-z intersection, seemingly, can shoot beyond 50. But remember that there are (100-61)=39 left within the "empty" domain that must be covered by z solely. As a result, x-y-z intersection can only reach up to 19.

Running any other scenarios for x+y=120 (Statement 1), you will further validate that x-y-z intersection will never be > 20.

A group of burglars is trying to decide whether to target the houses in a particular neighborhood in which there are 100 households. x of the households own a high-definition television, y of the households own a valuable piece of artwork, z of the households own a luxury car, and all of the households own at least one of these items. The burglars will target the houses in the neighborhood unless the probability of any given household owning all three items exceeds 50%, as this is correlated with tighter security. If z < y < x, do the burglars target the houses in the neighborhood?
x = households with HDTV
y = households with Artwork
z = households with a luxury car

Let d = number of households with all three items
If d ≤ 50% then burglars will target, if d > 50% then no.

(1) x+y=120
Nothing about z is given so.

INSUFFICIENT.

(2) z=55
Nothing about x and y.

INSUFFICIENT.

Together 1 and 2

Refer snapshot.

X + Y + Z + a + b + c + d = 100 --> Eqn. 1
X + Y + a + b + c + d = 120 --> Eqn. 2
Z + b + d + c = 55 --> Eqn. 3

Adding Eqn. 2 and Eqn. 3
X + Y + a + b + c + d + Z + b + d + c = 120 + 55
(X + Y + Z + a + b + c + d) + b + d + c = 175 [from Eqn. 1]
100 + b + d + c = 175
b + d + c = 75
So,
d ≤ 50 OR d > 50

INSUFFICIENT.

Answer E.
_________________

z < y < x

if p(mid/100)>50%, mid>50 they will not target the neighborhood


Ans (A)

chondro48

Hi, could you pls explain in more details how x+y intersect = 20 when x = 61 and y = 59?

I guess when x= 61 and y= 59 the intersect is greater than 20, therefore we also have p >20:

--> Total = 100, so XUY has to be less than 100.

Then we have:

<100 = X + Y - intersection(xy) + neither

<100 = 120 - intersection

Intersection = 120 - <100 which leads to Intersection(XY) > 20. Meaning that if z was all included in the intersection it could be greater than 20.

How do you derive z < 50?

Bunuel do you mind providing a clear answer here? The responses listed are only slightly helpful and written unclearly. Thank you!

This is not entirely correct.

Rephrased the question is asking us whether the maximum number of people with all three appliances is less than 50 (to make things simple I'm assuming we have 100 people). To maximize the number of people with all three appliances, we need to have as many appliances as possible. Given that x+y = 120, then the maximim x + y + z is reached wehn x = 61, y = 59, z = 58 (x+y+z = 178). Also we know everyone has at least one appliance, so 100 = 178 - 2 * (people who have all 3). In this case, people who have all 3 = 39. This is smaller than 50, so 1 is sufficient.