Only two words are common between I and II
From I: P> M, T
From II: J > W > P Combining, we get K > J > W > P > M, T
Hence, K is the heaviest and J lighter than only the heaviest.
The calendar method is helpful in either case.
From the question:
G > Sh: D > M
From I: Sa > Su
Not sufficient.
From II: Sa > Sh, M, D, But who is taller between Sa and G? Not sufficient.
I leads us nowhere. II will give us the present strength as we have the base and percentage increases.
From I: 22 students don't play any game. Which means 64 - 22 =42 play either chess or cricket.Now, either chess or cricket = only chess + only cricket + Chess and Cricket or 42 = only chess = 0
From II; We don't get information about boys.Hence II is also sufficient.
From I: 5$#3 = flowers are really good ..........(i)
From II: 7#35 = good flowers are available .......(ii)
From I and II: 5#3 = Flowers are good ...........(iii)
Putting (iii) in (i), we get $ = Really
From I : M (+) - P - T ( - )
Combining, we get M ( + ) - P - T ( - )
But II also say's J's husband has one son and two daughters.
Hence, P must be daughter of J.
From I: n Suresh = 12th from left
Mohan = 17th from right = (50 - 17 + 1 = ) 34th from left
No student between them = 34 - 12 - 1 = 21
From II: No data about Mohan.
The data in Statement I alone or in statement II alone are sufficient to answer the question.
The data in both the statement I and II together are necessary to answer the question
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