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+1 vote
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in Numerical Ability by Junior (965 points) | 89 views
0
Ans would be 7.
0
how please expain
0
$7$ ?

2 Answers

+3 votes

                

$\Rightarrow42=22+28+24-(6+x)-(4+x)-(8+x)+x$

$\Rightarrow2x=14 \Rightarrow x=7$

$\Rightarrow\#NY=\#NY(28)-(6+8+7)$

$\Rightarrow \#NY=\color{Red}{7}.$

by Boss (11.9k points)
+2 votes

Actually, this little bit tricky question ( I think can't apply directly inclusion exclusion principal).

Let's look on the question what is given:

S--> Singapore

N-->New York

L--> London

n(S) = 22n,

n(N) = 28,

n(L) = 24,

n(S AND N BUT NOT L) = 6

n(S AND L BUT NOT N) = 4

n(N AND L BUT NOT S) = 8

n( N U S U L) = 42

According to question formula would be like that

n(S U N U N) = n( alone S) + n( alone N) + n( alone L) + n(S AND N BUT NOT L) + n(S AND L BUT NOT N) + n(N AND L BUT NOT S) + n(S AND L AND N)

Let's assume n(S AND L AND N) = x 

42 = { 22 - ( 8 + 4 + x)} + { 28 - ( 8 + 6 + x)} + { 24 - ( 4 + 8 + x) } + { 8 + 6 + 4 } + x

2x = 56 - 42

X = 7

So we wanted to calculate

n(alone N) = 28 - { 8 + 6 + x} = 7

I hope it will help you to understand. 

by Active (2k points)

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