IIT JEE , AIEEE Forum - Mathematics , Algebra

bhanu

1) from the first 15 natural numbers,3 are chosen at random.the chance that they may be in A.P is

a)3/65 b)7/65 c)9/65 d)11/65

2)4 dice are thrown at a time.the chance that sum of digits is 8 is

a)5/1296 b)25/1296 c)35/1296 d)75/1296.

3)there are 10 points in a plane of which 4 are collinear.the number of different quadrilaterals that can be formed is

a)15 b)80 c)90 d)185.

ankurgupta91

1)the ans is 7/65 i.e (b)
2) the ans is 35/1296 i.e (c)
3)the ans is 185 i.e (d)
if u want the sol. tl me i will post it
thanks.......

ramyadiamond

The answer for the first question is b) 7/65

In this question,u wll have to consider different cases.

No. of sets of 3 no.s in AP that can be made which have the common diff. as 1

(1,2,3) , (2,3,4,), (3,4,5)..................(13,14,15) =13 (no.of sets)

No. of sets with common difference 2

(1,3,5), (2,4,6).............(11,13,15) =11 cases

Similarly, u can make sets with common difference 3,4, and so on upto 7, which forms the set as (1,8,15)

Hence the total no of cases in which an AP is formed are 13+11+9+.......+3+1=49

Hence Probability=49/15C3

=7/65

ramyadiamond

for teh second question,

there is a possibility of getting either 1 or 2 or 3 and so on upto6 for one die, and similarly for the other three dice. So for finding the sum as 8, u need to find the coefficient of x⁸ in (x¹+x²+x³+x⁴+x⁵+x⁶)⁴ which on solving will give the answer.

abhijeet_0201

hey , iam getting the 3rd one as (d) 185
quad is formed by 4 pts so nop of quad is
10C4=210
but these should not hav more than 2 collinear points (otherwise a quad wud not be formed)
therefore we have to substract quad which hav 3 or 4 of the collinear points .
with 3 of the collinear pt 4C3x6C1=24

with 4 collinear pts 4C4=1
total no=210-24-1=185 (i wonder y ankur is getting 15)
plz rate me if sol. is correct.plz correct me if it is not
thank u

amar.gupta

good work ramya and abhijeet

abhijeet_0201

thank u

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