Problem of the Day


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?

(1) x+y=120

(2) z=55

Reveal Answer

Answer

A. Statement 1 is sufficient alone, but statement 2 is not sufficient alone.

See the Solution

Solution

[latexpage]

Statement 1: If $x+y=120$, and $z<y<x$, then the biggest value $z$ could have is if $x=61, y=59, z=58$. In this case $x+y+z=178$. But there are only $100$ households, so there is an excess of 78 households that must have been counted multiple times. Every household that is in all three categories gets counted two extra times when you add $x,y$ and $z$. So there could be at most $39$ households that have all three items. If at most $39$ out of $100$ households have all three items, then the probability of any given household having all three items is no greater then $39$ percent. Thus, they will target the houses in this neighborhood. SUFFICIENT.

Statement 2: If $z$ is 55, then if $x=65,y=60$, then the maximum overlap would be $40$, in which case they would target the houses. If $x=90,y=85$, then the maximum overlap is $55$ in which case they would not target the houses. INSUFFICIENT.


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