How to find the number of elements which share all the characteristics when there are many sets? The problem is as follows:

In a toddler's room there are 120 toys, 95 of them uses batteries, 86
  have wheels, 94 are red color, 110 are made of plastic, 100 emit
  sound. How many of the toys share all the characteristics?

I'm stuck at this situation since I don't know how should I build up the sets in order they can be arranged in such a way that I can made an intersection of all thus finding the answer. Can somebody help me to be in the right direction?
 A: Hint: 
\begin{eqnarray*}
b_1+a=95 \\
w_1+a=86 \\
r_1+a=94 \\
p_1+a=110 \\
s_1+a =100 \\
b_1+w_1+r_1+p_1+s_1+a=120
\end{eqnarray*}
add $4a$ to both sides of the last equation & ... 
Edit :
To obtain the least value for the number of toys that has all $5$ popeerties ...
Let $a$ be the number of toys with all five properties. 
Let $b$ be the number of toys with all the properties apart from needing batteries. etc ... 
We have the following equations.
\begin{eqnarray*}
a &+& b &+&  w &+&  r &+&  p &+&  s &=& 120 \\
a & &  &+&  w &+&  r &+&  p &+&  s &=& 95 \\
a &+& b & &   &+&  r &+&  p &+&  s &=& 86 \\
a &+& b &+&  w & &   &+&  p &+&  s &=& 94 \\
a &+& b &+&  w &+&  r & &   &+&  s &=& 110 \\
a &+& b &+&  w &+&  r &+&  p & &   &=& 100 \\
\end{eqnarray*}
A: Using Venn diagram step by step:
$$\begin{align} &\underbrace{n(p/s)}_{20}+\underbrace{n(p\cap s)}_{90}+\underbrace{n(s)}_{10}= \\
&\underbrace{n(p\cap b/s)}_{20}+\underbrace{n(p\cap s\cap b)}_{65}+\underbrace{n(p\cap s/b)}_{25}+\underbrace{n(s\cap b/p)}_{10}= \\
&\underbrace{n(p\cap b\cap r)}_{20}+\underbrace{n(p\cap s\cap b\cap r)}_{39}+\underbrace{n(p\cap s\cap b/r)}_{26}+\underbrace{n(p\cap s\cap r/b)}_{25}+\underbrace{n(s\cap b\cap r/p)}_{10}= \\
&\underbrace{n(p\cap b\cap r\cap w/s)}_{20}+\color{red}{\underbrace{n(p\cap s\cap b\cap r\cap w)}_{5}}+\underbrace{n(p\cap s\cap b\cap r/w)}_{34}+\underbrace{n(p\cap s\cap b\cap w/r)}_{26}+\underbrace{n(p\cap s\cap r\cap w/b)}_{25}+\underbrace{n(s\cap b\cap r\cap w/p)}_{10}.\end{align}$$
Addendum: $A\B=A-B$ is a relative complement.
Imagine there are $120$ toys of only two kinds: $110$ plastic and $100$ sound. The minimum number of toys that share both characteristics is $90$ as indicated in the first Venn diagram (first expression). We move the numbers away from each other as much as we can. Then imagine $95$ toys with batteries are added. Again you indicate it in Venn diagram by the same logic. 
Yes, I agree, it could be not the best (easiest) way to solve the problem. Well, my teacher used to say: "You better solve one problem in $10$ different ways rather $10$ similar problems in one way".
A: The simple approach is that the minimum number that have all the characteristics comes when the ones that don't have all the characteristics are only missing one each.  There are $25$ without batteries, $34$ without wheels, $26$ not red, $10$ not plastic, and $20$ do not emit sound.  The minimum number that share all the characteristics is then $120-25-34-26-10-20=5$
