Elitmus
Exam
Numerical Ability
Geometry
If a>b>c>d>e>0 and the avg of a & e is 7 and b & d is 4.5.How many possible combination will satisfy the above condition?
i think question was smthing like that came in 22 nov elitmus ph test....
Read Solution (Total 4)
-
- a+e=14 so (13,1) (12,2) (11,3) (10,4) (9,5) b/w a and e 3 digits should be there min and a>e so we cant take other otions simillerly
b+d=9 can be (7,2) (6,3) min 1 digit gap and b>d
but (11,4) (10,3) (9,4) do not valid for the above pair of b and d
(1 2 3 7 13)
(1 2 4 7 13)
(1 2 5 7 13)
(1 2 6 7 13)
(1 3 4 6 13)
(1 3 5 6 13)
(2 3 5 6 12)
(2 3 4 6 12)
ans 8
- 9 years agoHelpfull: Yes(25) No(1)
- (e,d,c,b,a) = (1 2 3 7 13) (1 2 4 7 13) (2 3 4 6 12) (1 2 5 7 13) (1 2 6 7 13) (1 3 4 6 13) (2 3 5 6 12) (1 3 5 6 13)
Therefore answer is 8. - 9 years agoHelpfull: Yes(2) No(0)
- a+e=14
∆ (13,1),(12,2),(11,3),(10,4),(9,5)
possible pairs since minimum difference between the digits must be 3 (b,c,d)
b+d= 9
∆ (7,2),(6,3),(8,1)
difference 1 (c)
8 pairs - 8 years agoHelpfull: Yes(1) No(0)
- Priya why u don't take combination with 11 in ( e,d,c,b,a)
- 8 years agoHelpfull: Yes(0) No(0)
Elitmus Other Question