1@gallardo dese are not my doubts [1]
so y r u giving hint ?
some questions for bitsat practice
Q-1
\texttt{let x,y be selected from SET}\left\{1,2,3\cdots15 \right\}\\
what is the probaility that
x^2-y^2 is divisible by 5
Q-2
find
\sum_{1}^{16}{\frac{n^4}{4n^2-1}}
-
UP 0 DOWN 0 0 10
10 Answers
1@ur hint may be its suitable her for small range , but for bigger ranges ur calclations become cumbersome
39A 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.
132) 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... :)
1Not 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}
1@avik right [1]
for first answer is
\frac{5\binom{3}{2}+\left(\binom{3}{1} \right)^2}{\binom{15}{2}}
13RPF...Give sol.n fr the 1st one, since noone seems to be getting it rite.
1i think ray got it
bt sadly he has worked it out for x^4-y4
weras the expression is x^2-y^2