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.
Feb 29 can occur only in a leap year. 2008 happens to be the only leap years between 2005 and 2011. Hence Yesir was born on Feb 29, 2008. Using Calendar Method we can find out the day. Hence I is sufficient.
From II: Yasir was born on Feb 29, 2008 (2012 -4 =2008). hence II is also sufficient.
Only two words are common between I and II
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.
Comments
There are no comments.Copyright ©CuriousTab. All rights reserved.