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Six children P, Q, R, S, T and U, were all born on the same day of the year, but each was born in a different year during a single six-year period.
P is older than R.
Q is older than both S and T.
U is two years older than S.
P was born either in 2007 or in 2008.
The oldest children was born in 2005.
If P is older than U, then in how many different orders could the six children have been born ?
We know:
P>R
Q>S and T
U>S (2 years)
P- 2007/2008
Oldest in 2005
Now, we can see that P is not the oldest. Hence, the oldest shall be either U or Q.
The birth years shall be: 2005, 2006, 2007, 2008, 2009, 2010.
Case 1: When the oldest is U
Hence, U (2005), S (2007), Q (2006), P (2008), R/T (2009/2010)
Case 2: When Q is the oldest
Hence, Q (2005), U (2006), S (2008), P (2007), R/T (2009/2010)
It is case 2 where P is older than U. And there, two cases can be made depending on the year of R/T.
