from rpf

A humble try for Q1....

CASE I : When (x-y) is divisible by 5
The following pairs of integers from the set when subtracted give 5 :
(6, 1) , (1, 6)
(7, 2), (2, 7)
(8, 3), (3, 8)
(9, 4), (4, 9)
(10, 5), (5, 10)
(11, 6), (6, 11)
(12, 7), (7, 12)
(13, 8), (8, 13)
(14, 9), (9, 14)
(15, 10), (10, 15)
And for 10,
(15, 5), (5, 15)
Also there are cases where the end digit can be zero. There would be 14 such cases.
Probability : Number of Pairs/Subtractions Possible
= 35/2¹⁴ (There are 14 numbers in the set)

CASE II : When (x+y) is divisible by 5
The following pairs of integers when added yield 5 :
(1, 4), (4, 1)
(2, 3), (3, 2)
And there are no more.
But for further multiples of 5, such as 10,
(1, 9), (9, 1)
(2, 8), (8, 2)
(3, 7), (7, 3)
(4, 6), (6, 4)
And similarly for 15,
(1, 14), (14, 1)
.
.
.
.
.
.
(8, 7), (7, 8)

Probability : 28/2¹⁴

Needless to say either the first case or the second case must hold for overall divisibility by 5.
Adding the two,
Probability : 63/2¹⁴
Am not at all sure of this answer. Just a try.

2) My trial...
n⁴4n²-1 = 116x [16n⁴-14n²-1 + 14n²-1]

=116x [ (4n²+1) + 12x {12n-1-12n+1 }]

Now this can be done i suppose... :)

Not giving BITSAT, still trying my hand at it.

The no. of ways of choosing x and y is C^{15}_{2}

We can divide the given set into 5 groups
....5n-4
....5n-3
....5n-2
....5n-1
....5n

Also, a^{4}-b^{4}=(a-b)(a+b)(a^{2}+b^{2})

Hence, a^{4}-b^{4} will be divisible by 5 if both a and b belong to the last group or they belong to any of the remaining four groups. Thus no. of favourable ways is C_{2}+C^{4n}_{2}=C_{2}+C^{12}_{2}

Hence the required probability is
\frac{23}{35}