Why 2*10 for (A∩B∩C)

35 Don't use brain bros

Refer to the figure Below:

L refers to leftovers i.e. Not participating in any group

S = a + d + x + f
B = b + e + x + d
H = c + e + x + f

S + B + H = a + b + c + 2e + 2d + 3x


We know,
Total - L = a + b + c + d + e + f + x

Now, if we write Total - L = A + B + C - d - e - f = a + b + c + 2e + 2d + 3x - d - e - f = a + b + c + d + e + f + 3x
We need to subtract 2x from it in order to get the correct form.

Hence, we subrtract 2x.

Carcass, I have a doubt in this soln, when d+e+f = 10
So the given question states "40% of seniors, playing only two games, play soccer" means 4 out 10 play soccer. Am I right ??

So answer should be 35 - 10 + 4 ...am I correct here ??

TIA!

10 seniors are already playing all 3 games, how can only soccer be greater than 25 (35 - 10)???
What is it that I am missing?

AUBUC = A + B + C - (d + e + f) - 2(A∩B∩C)
100 = 35 + 45 + 50 - (d + e + f) - 2(10)
(d + e + f) = 110 - 100 = 10

Now, 40% of seniors, playing only two games, play soccer.
d + f = \(\frac{40}{100} (10) = 4\)

So, seniors playing only soccer = 35 - 4 -10 = 21[/quote]

- Total seniors = 100

- Soccer (S) $=35$

- Basketball $\((\mathrm{B})=45\)$

- Hockey $\((\mathrm{H})=50\)$

- All three games $\((\mathrm{S} \cap \mathrm{B} \cap \mathrm{H})=10\)$

- $40 %$ of seniors who play exactly two games include soccer (i.e., Soccer + Basketball or Soccer + Hockey). THIS IS WHAT YOU ASKED FOR


Let:
- $\(\mathbf{a}=\)$ Soccer \& Basketball only $\((S \cap B\)$, not H)
- $\(\mathbf{b}=\)$ Soccer \& Hockey only ( $\(\mathrm{S} \cap \mathrm{H}\)$, not B)
- $\(\mathbf{c}=\)$ Basketball \& Hockey only $\((B \cap H\)$, not $S)$
- All three = 10 (given)

Step 2: Total Playing Exactly Two Games
Total playing exactly two games $\(=\mathrm{a}+\mathrm{b}+\mathrm{c}\)$
Since 40\% of these include soccer, we have:

$$
\(a+b=0.4(a+b+c)\)
$$


Simplify:

$$
\(\begin{gathered}
a+b=0.4 a+0.4 b+0.4 c \\
0.6 a+0.6 b=0.4 c \\
6 a+6 b=4 c \\
3 a+3 b=2 c \\
c=\frac{3(a+b)}{2}
\end{gathered}\)
$$


Since $\(\mathbf{c}\)$ must be an integer, $\((\mathbf{a}+\mathbf{b})\)$ must be even.

1. Soccer (S) $=35$

(Only S) $+a+b+10=35$
Only $\(\mathrm{S}=25-a-b\)$

2. Basketball $(B)=45$

(Only B) $+a+c+10=45$
Only B $=35-a-c$

3. Hockey $(H)=50$

(Only H) $+b+c+10=50$
Only $\mathrm{H}=40-b-c$
Step 4: Total Seniors Equation
Only S + Only B + Only H + $a+b+c+10=100$
Substitute the expressions:

$$
\((25-a-b)+(35-a-c)+(40-b-c)+a+b+c+10=100\)
$$


Simplify:

$$
\(\begin{gathered}
25+35+40+10-a-b-a-c-b-c+a+b+c=100 \\
110-a-b-c=100 \\
a+b+c=10
\end{gathered}\)
$$

From earlier:

$$
\(c=\frac{3(a+b)}{2}\)
$$


And:

$$
\(a+b+c=10\)
$$


Substitute c:

$$
\(a+b+\frac{3(a+b)}{2}=10\)
$$


Multiply by 2:

$$
\(\begin{gathered}
2(a+b)+3(a+b)=20 \\
5(a+b)=20 \\
a+b=4
\end{gathered}\)
$$


Then:

$$
\(c=\frac{3(4)}{2}=6\)
$$


Step 6: Find "Only Soccer"

$$
\(\text { Only } \mathrm{S}=25-a-b=25-4=21\)
$$


Final Answer:
(E) 21 seniors play only soccer.

not a GRE question.

No that much because is tough but because you need too many steps to solve. My personal opinion

Thanks for the explanation