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16 Dec 2025, 11:58
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Did this by making the double set matrix table.
Firstly we have X: both X& Y as 3:2, consider it as 3k & 2k
then we have Y: both X&Y as 6:2, which will be 6k & 2k
lastly we have neither: both X& Y as 1.5:2, which will be 1.5k & 2k
Put this in matrix, we get only X as k & only Y as 4k, giving ratio as 1:4
Ans (A) 1:4
Firstly we have X: both X& Y as 3:2, consider it as 3k & 2k
then we have Y: both X&Y as 6:2, which will be 6k & 2k
lastly we have neither: both X& Y as 1.5:2, which will be 1.5k & 2k
Put this in matrix, we get only X as k & only Y as 4k, giving ratio as 1:4
| Y | Y not | Total | |
| X | 2k | k | 3k |
| X not | 4k | 1.5k | |
| Total | 6k |
Ans (A) 1:4
16 Dec 2025, 12:06
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Give:
X : Both (X&Y) = 3/2 [X= 3, Both = 2]
Y : Both (X&Y) = 2*3/2= 6/2 [Y= 6, Both = 2]
[Try to keep the value of "Both" same ]
As the question only asks about ONLY X : ONLY Y we can forget the details about Neither.
So, ONLY X= X - Both = 3 - 2 = 1
ONLY Y= Y - Both = 6 - 2 = 4
ONLY X : ONLY Y = 1:4
X : Both (X&Y) = 3/2 [X= 3, Both = 2]
Y : Both (X&Y) = 2*3/2= 6/2 [Y= 6, Both = 2]
[Try to keep the value of "Both" same ]
As the question only asks about ONLY X : ONLY Y we can forget the details about Neither.
So, ONLY X= X - Both = 3 - 2 = 1
ONLY Y= Y - Both = 6 - 2 = 4
ONLY X : ONLY Y = 1:4
16 Dec 2025, 12:10
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Let's denote:
Then, our ratios are:
\((x+b):b=3:2\)
\((y+b):b=6:2\)
\(n:b=3:4\)
To find \(x:y\), let's define them through b:
\(2(x+b)=3b\), so \(x=b/2\)
\(2(y+b)=6b\), so \(y=2b\)
Therefore, \(x:y=(\frac{b}{2}):2b=1:4\). The right answer is A.
- X - number of people only supporting proposition X
- Y - number of people only supporting proposition Y
- B - number of supporters of both, and
- N - number of supporters of neither.
Then, our ratios are:
\((x+b):b=3:2\)
\((y+b):b=6:2\)
\(n:b=3:4\)
To find \(x:y\), let's define them through b:
\(2(x+b)=3b\), so \(x=b/2\)
\(2(y+b)=6b\), so \(y=2b\)
Therefore, \(x:y=(\frac{b}{2}):2b=1:4\). The right answer is A.
